Engineering Mechanics - Statics
Chapter 1
Problem 1-1
Represent each of the following combinations of units in the correct SI form using an
appropriate prefix:
(a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN
Units Used:
μN = 10
−6
μkm = 10
N
−6
km
9
Gs = 10 s
3
ks = 10 s
mN = 10
−3
−3
ms = 10
N
s
Solution:
( a)
m
3m
= 1 × 10
ms
s
m
km
=1
ms
s
( b)
μkm = 1 × 10
−3
m
μkm = 1 mm
( c)
ks
9 s
= 1 × 10
mg
kg
ks
Gs
=1
mg
kg
( d)
−3
km⋅ μN = 1 × 10
mN
km⋅ μN = 1 mm⋅ N
1
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Engineering Mechanics - Statics
Chapter 1
Problem 1-2
Wood has a density d. What is its density expressed in SI units?
Units Used:
Mg = 1000 kg
Given:
d = 4.70
slug
ft
3
Solution:
1slug = 14.594 kg
d = 2.42
Mg
m
3
Problem 1-3
Represent each of the following combinations of units in the correct SI form using an
appropriate prefix:
(a) Mg/mm (b) mN/μs (c) μm⋅ Mg
Solution:
3
( a)
6
Mg
10 kg
10 kg
Gg
=
=
=
−3
mm
m
m
10
m
Mg
Gg
=
mm
m
( b)
mN
μs
mN
μs
( c)
=
10
−3
10
=
N
−6
3
=
s
10 N
kN
=
s
s
kN
s
( − 6 m)(103 kg) = 10− 3 m⋅ kg
μm⋅ Mg = 10
μm⋅ Mg = mm⋅ kg
2
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Engineering Mechanics - Statics
Chapter 1
Problem 1-4
Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b)
N/mm, (c) mN/( kg⋅ μs).
Solution:
3
( a)
6
Mg
10 kg
10 kg
Gg
=
=
=
−3
ms
s
s
10
s
Mg
Gg
=
ms
s
N
=
mm
( b)
1N
10
−3
= 10
3N
m
m
=
kN
m
N
kN
=
mm
m
mN
( c)
kg⋅ μs
10
=
mN
kg⋅ μs
−3
−6
10
=
N
=
kg⋅ s
kN
kg⋅ s
kN
kg⋅ s
Problem 1-5
Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3.
Units Used:
kg = 1000 g
−3
ms = 10
s
3
kN = 10 N
Given:
S1 = 8653 ms
S2 = 8368 N
S3 = 0.893 kg
Solution:
( a)
S1 = 8.653 s
3
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Engineering Mechanics - Statics
( b)
S2 = 8.368 kN
( c)
S3 = 893 g
Chapter 1
Problem 1-6
Represent each of the following to three significant figures and express each answer in SI units
using an appropriate prefix: (a) x, (b) y, and (c) z.
Units Used:
6
MN = 10 N
μg = 1 × 10
−6
gm
3
kN = 10 N
Given:
x = 45320 kN
(
y = 568 × 10
5
) mm
z = 0.00563 mg
Solution:
( a)
x = 45.3 MN
( b)
y = 56.8 km
( c)
z = 5.63 μg
Problem 1-7
Evaluate ( a⋅ b)/c to three significant figures and express the answer in SI units using an
appropriate prefix.
Units Used:
μm = 10
−6
m
Given:
a = ( 204 mm)
4
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Engineering Mechanics - Statics
Chapter 1
b = ( 0.00457 kg)
c = ( 34.6 N)
Solution:
l =
ab
l = 26.945
c
μm⋅ kg
N
Problem 1-8
If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second.
Given:
v = 55
mi
hr
Solution:
v = 88.514
km
hr
m
v = 24.6
s
Problem 1-9
Convert: (a) S1 to N ⋅ m , (b) S2 to kN/m3, (c) S3 to mm/s. Express the result to three
significant figures. Use an appropriate prefix.
Units Used:
3
kN = 10 N
Given:
S1 = 200g lb⋅ ft
S2 = 350g
lb
ft
S3 = 8
3
ft
hr
5
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Engineering Mechanics - Statics
Chapter 1
Solution:
( a)
S1 = 271 N⋅ m
( b)
S2 = 55.0
kN
m
S3 = 0.677
( c)
3
mm
s
Problem 1-10
What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Express
the result to three significant figures. Use an appropriate prefix.
Units Used:
3
Mg = 10 kg
mN = 10
−3
N
3
kN = 10 N
Given:
m1 = 10 kg
m2 = 0.5 gm
m3 = 4.50 Mg
Solution:
( a)
W = m1 g
W = 98.1 N
( b)
W = m2 g
W = 4.90 mN
( c)
W = m3 g
W = 44.1 kN
6
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Engineering Mechanics - Statics
Chapter 1
Problem 1-11
If an object has mass m, determine its mass in kilograms.
Given:
m = 40 slug
Solution:
m = 584 kg
Problem 1-12
The specific weight (wt./vol.) of brass is ρ. Determine its density (mass/vol.) in SI units.
Use an appropriate prefix.
Units Used:
3
Mg = 10 kg
Given:
lb
ρ = 520
ft
3
Solution:
Mg
ρ = 8.33
m
3
Problem 1-13
A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ,
determine the weight of the column in pounds.
Units Used:
3
Mg = 10 kg
3
kip = 10 lb
Given:
d = 350 mm
L = 2m
7
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Engineering Mechanics - Statics
ρ = 2.45
Chapter 1
Mg
m
3
Solution:
2
d
V = π ⎛⎜ ⎟⎞ L
⎝ 2⎠
V = 192.423 L
W = ρ V
W = 1.04 kip
Problem 1-14
The density (mass/volume) of aluminum is ρ. Determine its density in SI units. Use an
appropriate prefix.
Units Used:
Mg = 1000 kg
Given:
ρ = 5.26
slug
ft
3
Solution:
ρ = 2.17
Mg
m
3
Problem 1-15
Determine your own mass in kilograms, your weight in newtons, and your height in meters.
Solution:
Example
W = 150 lb
m = W
m = 68.039 kg
W g = 667.233 N
h = 72 in
h = 1.829 m
8
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Engineering Mechanics - Statics
Chapter 1
Problem 1-16
Two particles have masses m1 and m2, respectively. If they are a distance d apart, determine
the force of gravity acting between them. Compare this result with the weight of each particle.
Units Used:
G = 66.73 × 10
− 12
m
3
2
kg⋅ s
nN = 10
−9
N
Given:
m1 = 8 kg
m2 = 12 kg
d = 800 mm
Solution:
F =
G m1 m2
d
2
F = 10.0 nN
W1 = m1 g
W1 = 78.5 N
W2 = m2 g
W2 = 118 N
W1
F
W2
F
= 7.85 × 10
9
= 1.18 × 10
10
Problem 1-17
Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionally
homogeneous equation which gives F in newtons. Compute the gravitational force acting
between two identical spheres that are touching each other. The mass of each sphere is m1, and
the radius is r.
Units Used:
μN = 10
−6
N
G = 66.73 10
− 12
m
3
2
kg⋅ s
9
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Engineering Mechanics - Statics
Chapter 1
Given:
m1 = 150 kg
r = 275 mm
Solution:
F =
G m1
( 2r)
2
2
F = 4.96 μN
Since the force F is measured in Newtons, then the equation is dimensionally homogeneous.
Problem 1-18
Evaluate each of the following to three significant figures and express each answer in SI units
using an appropriate prefix: (a) x, (b) y, (c) z.
Units Used:
6
MN = 10 N
3
kN = 10 N
μm = 10
−6
m
Given:
x = ( 200 kN)
2
y = ( 0.005 mm)
z = ( 400 m)
2
3
Solution:
2
( a)
x = 0.040 MN
( b)
y = 25.0 μm
( c)
z = 0.0640 km
2
3
10
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Engineering Mechanics - Statics
Chapter 1
Problem 1-19
Evaluate each of the following to three significant figures and express each answer in SI units
using an appropriate prefix: (a) a 1/b1, (b) a2b2/c2, (c) a3b3.
Units Used:
μm = 10
−6
6
m
Mm = 10 m
Mg = 10 gm
kg = 10 gm
6
−3
ms = 10
3
s
Given:
a1 = 684 μm
b1 = 43 ms
a2 = 28 ms
b2 = 0.0458 Mm
c2 = 348 mg
a3 = 2.68 mm
b3 = 426 Mg
Solution:
a1
( a)
b1
= 15.9
a2 b2
( b)
c2
mm
s
= 3.69 Mm
s
kg
a3 b3 = 1.14 km⋅ kg
( c)
Problem 1-20
Evaluate each of the following to three significant figures and express each answer in SI units
using an appropriate prefix: (a) a1/b12 (b) a22b23.
Units Used:
6
Mm = 10 m
11
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Engineering Mechanics - Statics
Chapter 1
Given:
a1 = 0.631 Mm
b1 = 8.60 kg
a2 = 35 mm
b2 = 48 kg
Solution:
( a)
a1
b1
( b)
2
2
= 8.532
km
kg
3
2
3
a2 b2 = 135.48 kg ⋅ m
2
12
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Engineering Mechanics - Statics
Chapter 2
Problem 2-1
Determine the magnitude of the resultant force FR = F1 + F 2 and its direction, measured
counterclockwise from the positive x axis.
Given:
F 1 = 600 N
F 2 = 800 N
F 3 = 450 N
α = 45 deg
β = 60 deg
γ = 75 deg
Solution:
ψ = 90 deg − β + α
FR =
F1 + F 2 − 2 F1 F 2 cos ( ψ)
2
2
F R = 867 N
FR
sin ( ψ)
=
F2
sin ( θ )
⎛
⎝
θ = asin ⎜F 2
θ = 63.05 deg
sin ( ψ) ⎞
⎟
FR ⎠
φ = θ+α
φ = 108 deg
Problem 2-2
Determine the magnitude of the resultant force and its direction measured counterclockwise
from the positive x axis.
Given:
F 1 = 80 lb
F 2 = 60 lb
13
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Engineering Mechanics - Statics
Chapter 2
θ = 120 deg
Solution:
FR =
F1 + F 2 − 2 F1 F 2 cos ( 180 deg − θ )
2
2
F R = 72.1 lb
⎛
⎝
β = asin ⎜ F1
sin ( 180 deg − θ ) ⎞
⎟
FR
⎠
β = 73.9 deg
Problem 2-3
Determine the magnitude of the resultant force F R = F1 + F 2 and its direction, measured
counterclockwise from the positive x axis.
Given:
F 1 = 250 lb
F 2 = 375 lb
θ = 30 deg
φ = 45 deg
Solution:
FR =
F1 + F 2 − 2 F1 F 2 cos ( 90 deg + θ − φ )
2
2
F R = 178 kg
FR
sin ( 90 deg + θ − φ )
⎛ F1
β = asin ⎜
⎝ FR
=
F1
sin ( β )
⎞
sin ( 90 deg + θ − φ )⎟
⎠
β = 37.89 deg
14
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Engineering Mechanics - Statics
Chapter 2
Angle measured ccw from x axis
360 deg − φ + β = 353 deg
Problem 2-4
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured
counterclockwise from the positive u axis.
Given:
F 1 = 300 N
F 2 = 500 N
α = 30 deg
β = 45 deg
γ = 70 deg
Solution:
FR =
F1 + F 2 − 2 F1 F 2 cos ( 180 deg − β − γ + α )
2
2
F R = 605 N
FR
sin ( 180 deg − β − γ + α )
⎛
⎝
θ = asin ⎜F 2
=
F2
sin ( θ )
sin ( 180 deg − β − γ + α ) ⎞
⎟
FR
⎠
θ = 55.40 deg
φ = θ+α
φ = 85.4 deg
Problem 2-5
Resolve the force F 1 into components acting along the u and v axes and determine the
magnitudes of the components.
Given:
F 1 = 300 N
α = 30 deg
15
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Engineering Mechanics - Statics
F 2 = 500 N
Chapter 2
β = 45 deg
γ = 70 deg
Solution:
F1u
=
sin ( γ − α )
F1
sin ( 180 deg − γ )
sin ( γ − α )
F 1u = F1
sin ( 180 deg − γ )
F 1u = 205 N
F 1v
sin ( α )
=
F 1v = F 1
F1
sin ( 180 deg − γ )
sin ( α )
sin ( 180 deg − γ )
F 1v = 160 N
Problem 2-6
Resolve the force F2 into components acting along the u and v axes and determine the
magnitudes of the components.
Given:
F 1 = 300 N
F 2 = 500 N
α = 30 deg
β = 45 deg
γ = 70 deg
Solution:
⎛ sin ( β ) ⎞
⎟
⎝ sin ( γ ) ⎠
F 2u = F2 ⎜
16
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Engineering Mechanics - Statics
Chapter 2
F 2u = 376.2 N
⎡ sin ⎡⎣180 deg − ( β + γ )⎤⎦⎤
⎥
sin ( γ )
⎣
⎦
F 2v = F 2 ⎢
F 2v = 482.2 N
Problem 2-7
Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction measured
counterclockwise from the positive u axis.
Given:
F 1 = 25 lb
F 2 = 50 lb
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 45 deg
Solution:
α = 180 deg − ( θ 3 + θ 1 )
FR =
F2 + F 1 − 2 F1 F 2 cos ( α )
2
2
F R = 61.4 lb
sin ( θ' )
sin ( α )
=
F2
FR
⎛
F2 ⎞
⎝
FR ⎠
θ' = asin ⎜ sin ( α )
⎟
θ' = 51.8 deg
θ = θ' − θ 3
θ = 6.8 deg
Problem 2-8
Resolve the force F 1 into components acting along the u and v axes and determine the components.
17
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 25 lb
F 2 = 50 lb
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 45 deg
Solution:
−F u
=
sin ( θ 3 − θ 2 )
Fu =
F1
sin ( θ 2 )
−F1 sin ( θ 3 − θ 2 )
sin ( θ 2 )
F u = −12.9 lb
Fv
sin ( 180 deg − θ 3 )
Fv =
=
F1
sin ( θ 2 )
F 1 sin ( 180 deg − θ 3 )
sin ( θ 2 )
F v = 35.4 lb
Problem 2-9
Resolve the force F2 into components acting along
the u and v axes and determine the components.
Given:
F 1 = 25 lb
F 2 = 50 lb
θ 1 = 30 deg
18
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Engineering Mechanics - Statics
Chapter 2
θ 2 = 30 deg
θ 3 = 45 deg
Solution:
Fu
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
=
F2
sin ( θ 2 )
F2 sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
Fu =
sin ( θ 2 )
F u = 86.6 lb
−Fv
=
sin ( θ 1 )
Fv =
F2
sin ( θ 2 )
−F 2 sin ( θ 1 )
sin ( θ 2 )
F v = −50 lb
Problem 2-10
Determine the components of the F force acting along the u and v axes.
Given:
θ 1 = 70 deg
θ 2 = 45 deg
θ 3 = 60 deg
F = 250 N
Solution:
Fu
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
Fu =
=
F
sin ( θ 2 )
F sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
sin ( θ 2 )
F u = 320 N
19
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Engineering Mechanics - Statics
Fv
sin ( θ 1 )
Fv =
=
Chapter 2
F
sin ( θ 2 )
F sin ( θ 1 )
F v = 332 N
sin ( θ 2 )
Problem 2-11
The force F acts on the gear tooth. Resolve this force into two components acting along the
lines aa and bb.
Given:
F = 20 lb
θ 1 = 80 deg
θ 2 = 60 deg
Solution:
F
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
Fa =
Fa
sin ( θ 1 )
F sin ( θ 1 )
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
F
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
Fb =
=
F sin ( θ 2 )
=
F a = 30.6 lb
Fb
sin ( θ 2 )
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
F b = 26.9 lb
Problem 2-12
The component of force F acting along line aa is required to be Fa. Determine the magnitude
of F and its component along line bb.
20
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Engineering Mechanics - Statics
Chapter 2
Given:
F a = 30 lb
θ 1 = 80 deg
θ 2 = 60 deg
Solution:
Fa
=
sin ( θ 1 )
F
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
⎛ sin ( 180 deg − θ 1 − θ 2) ⎞
⎟
sin ( θ 1 )
⎝
⎠
F = Fa ⎜
Fa
=
sin ( θ 1 )
Fb =
F = 19.6 lb
Fb
sin ( θ 2 )
Fa sin ( θ 2 )
sin ( θ 1 )
F b = 26.4 lb
Problem 2-13
A resultant force F is necessary to hold the ballon in place. Resolve this force into components
along the tether lines AB and AC, and compute the magnitude of each component.
Given:
F = 350 lb
θ 1 = 30 deg
θ 2 = 40 deg
Solution:
F AB
sin ( θ 1 )
=
F
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
21
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 2
sin ( θ 1 )
⎡
⎤
⎥
⎣ sin ⎡⎣180 deg − ( θ 1 + θ 2)⎤⎦⎦
F AB = F ⎢
F AB = 186 lb
F AC
sin ( θ 2 )
F
=
sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦
sin ( θ 2 )
⎡
⎤
⎥
⎣sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ ⎦
F AC = F ⎢
F AC = 239 lb
Problem 2-14
The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to force
F 1 and is directed at angle θ1 from the horizontal. If the resultant force acting on the post is to be
F R, vertically upward, determine the force T in rope B and the corresponding angle θ.
Given:
F R = 1200 lb
F 1 = 600 lb
θ 1 = 60 deg
Solution:
T =
F 1 + FR − 2 F1 F R cos ( 90 deg − θ 1 )
2
2
T = 744 lb
sin ( 90 − θ 1 )
sin ( θ )
=
FR
T
⎛
⎝
θ = asin ⎜ sin ( 90 deg − θ 1 )
F1 ⎞
⎟
T ⎠
θ = 23.8 deg
22
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Engineering Mechanics - Statics
Chapter 2
Problem 2-15
Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes
of the components.
Given:
F 1 = 250 N
F 2 = 150 N
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 105 deg
Solution:
F 1v
sin ( θ 1 )
=
F1
sin ( θ 3 )
⎛ sin ( θ 1 ) ⎞
⎟
⎝ sin ( θ 3 ) ⎠
F 1v = F 1 ⎜
F 1v = 129 N
F 1u
sin ( 180 deg − θ 1 − θ 3 )
=
F1
sin ( θ 3 )
⎛ sin ( 180 deg − θ 1 − θ 3 ) ⎞
⎟
sin ( θ 3 )
⎝
⎠
F 1u = F1 ⎜
F 1u = 183 N
Problem 2-16
Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes
of the components.
Given:
F 1 = 250 N
23
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Engineering Mechanics - Statics
Chapter 2
F 2 = 150 N
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 105 deg
Solution:
F 1v
sin ( θ 1 )
F2
=
sin ( 180 deg − θ 3 )
⎛
sin ( θ 1 )
⎞
⎟
⎝ sin ( 180 deg − θ 3 ) ⎠
F 1v = F 2 ⎜
F 1v = 77.6 N
F2u
sin ( 180 deg − θ 3 )
=
F2
sin ( 180 deg − θ 3 )
⎛ sin ( 180 deg − θ 3) ⎞
⎟
⎝ sin ( 180 deg − θ 3) ⎠
F 2u = F2 ⎜
F 2u = 150 N
Problem 2-17
Determine the magnitude and direction of the resultant force F R. Express the result in terms of
the magnitudes of the components F 1 and F 2 and the angle φ.
Solution:
F R = F 1 + F2 − 2F1 F 2 cos ( 180 deg − φ )
2
2
2
24
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Engineering Mechanics - Statics
Chapter 2
Since cos ( 180 deg − φ ) = −cos ( φ ),
F 1 + F2 + 2 F 1 F2 cos ( φ )
2
FR =
2
From the figure,
tan ( θ ) =
F1 sin ( φ )
F 2 + F 1 cos ( φ )
Problem 2-18
If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting
on the pulley. This angle defines the same angle θ of line AB on the tailboard block.
Given:
F 1 = 400 N
θ 1 = 30 deg
Solution:
FR =
F1 + F 1 − 2F 1 F 1 cos ( 90 deg − θ 1 )
2
2
F R = 400 N
sin ( θ 1 )
sin ( 90 deg − θ )
=
FR
F1
⎛ FR
θ = 90 deg − asin ⎜
⎝ F1
⎞
sin ( θ 1 )⎟
⎠
θ = 60 deg
25
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Engineering Mechanics - Statics
Chapter 2
Problem 2-19
The riveted bracket supports two forces. Determine the angle θ so that the resultant force
is directed along the negative x axis. What is the magnitude of this resultant force?
Given:
F 1 = 60 lb
F 2 = 70 lb
θ 1 = 30 deg
Solution:
sin ( θ 1 )
sin ( θ )
=
F1
F2
⎛
F1 ⎞
⎝
F2 ⎠
θ = asin ⎜ sin ( θ 1 )
⎟
θ = 25.4 deg
φ = 180 deg − θ − θ 1
φ = 124.6 deg
F1 + F 2 − 2F 1 F 2 cos ( φ )
2
R =
2
R = 115 lb
Problem 2-20
The plate is subjected to the forces acting on members A and B as shown. Determine the magnitude
of the resultant of these forces and its direction measured clockwise from the positive x axis.
Given:
F A = 400 lb
F B = 500 lb
26
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Engineering Mechanics - Statics
Chapter 2
θ 1 = 30 deg
θ = 60 deg
Solution:
Cosine law:
FR =
FB + FA − 2F B FA cos ( 90 deg − θ + θ 1 )
2
2
F R = 458 lb
Sine law:
sin ( 90 deg − θ + θ 1 )
FR
=
sin ( θ − α )
FA
⎛
FA ⎞
⎝
FR ⎠
α = θ − asin ⎜ sin ( 90 deg − θ + θ 1 )
⎟
α = 10.9 deg
Problem 2-21
Determine the angle θ for connecting member B to the plate so that the resultant of FA and FB
is directed along the positive x axis. What is the magnitude of the resultant force?
Given:
F A = 400 lb
F B = 500 lb
θ 1 = 30 deg
27
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Engineering Mechanics - Statics
Chapter 2
Solution:
Sine law:
sin ( 90 deg − θ 1 )
sin ( θ )
=
FA
FB
⎛
FA ⎞
⎝
FB ⎠
θ = asin ⎜ sin ( 90 deg − θ 1 )
⎟
θ = 43.9 deg
FR
sin ( 90 deg + θ 1 − θ )
FR = FA
=
FA
sin ( θ )
sin ( 90 deg − θ + θ 1 )
sin ( θ )
F R = 561 lb
Problem 2-22
Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces
by first finding the resultant F' = F1 + F 2 and then forming F R = F' + F 3.
Given:
F 1 = 30 N
F 2 = 20 N
F 3 = 50 N
θ = 20 deg
c = 3
d = 4
28
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Engineering Mechanics - Statics
Chapter 2
Solution:
c
α = atan ⎛⎜ ⎟⎞
⎝ d⎠
F' =
F1 + F 2 − 2F 1 F 2 cos ( 90 deg + θ − α )
2
2
F' = 30.9 N
F'
sin ( ( 90 deg − θ + α ) )
=
⎛
⎝
F1
sin ( 90 deg − θ − β )
β = 90 deg − θ − asin ⎜F 1
sin ( 90 deg − θ + α ) ⎞
⎟
F'
⎠
β = 1.5 deg
Now add in force F3.
FR =
F' + F3 − 2F' F3 cos ( β )
2
2
F R = 19.2 N
FR
sin ( β )
F'
=
sin ( φ )
⎛
⎝
φ = asin ⎜F'
sin ( β ) ⎞
⎟
FR ⎠
φ = 2.4 deg
Problem 2-23
Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by
first finding the resultant F' = F 2 + F 3 and then forming F R = F' + F 1.
Given:
F 1 = 30 N
F 2 = 20 N
29
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Engineering Mechanics - Statics
Chapter 2
F 3 = 50 N
θ = 20 deg
c = 3
d = 4
Solution:
F' =
F2 + F 3 − 2F 2 F 3 cos ( ( 90 deg − θ ) )
2
2
F' = 47.07 N
F2
sin ( β )
F'
=
sin ( 90 deg − θ )
⎛
⎝
β = asin ⎜ F2
sin ( 90 deg − θ ) ⎞
⎟
F'
⎠
β = 23.53 deg
c
α = atan ⎛⎜ ⎟⎞
⎝ d⎠
γ = α−β
FR =
F' + F1 − 2F' F 1 cos ( γ )
2
2
F R = 19.2 N
FR
sin ( γ )
=
F1
sin ( φ )
⎛
⎝
φ = asin ⎜F 1
sin ( γ ) ⎞
⎟
FR ⎠
φ = 21.16 deg
ψ = β−φ
ψ = 2.37 deg
30
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Engineering Mechanics - Statics
Chapter 2
Problem 2-24
Resolve the force F into components acting along (a) the x and y axes, and (b) the x and y'
axes.
Given:
F = 50 lb
α = 65 deg
β = 45 deg
γ = 30 deg
Solution:
(a)
F x = F cos ( β )
F x = 35.4 lb
F y = F sin ( β )
F y = 35.4 lb
(b)
Fx
sin ( 90 deg − β − γ )
=
F
sin ( 90 deg + γ )
sin ( 90 deg − β − γ )
Fx = F
sin ( 90 deg + γ )
F x = 14.9 lb
Fy'
sin ( β )
=
F y' = F
F
sin ( 90 deg + γ )
sin ( β )
sin ( 90 deg + γ )
31
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Engineering Mechanics - Statics
Chapter 2
F y' = 40.8 lb
Problem 2-25
The boat is to be pulled onto the shore using two ropes. Determine the magnitudes of forces
T and P acting in each rope in order to develop a resultant force F1, directed along the keel
axis aa as shown.
Given:
θ = 40 deg
θ 1 = 30 deg
F 1 = 80 lb
Solution:
F1
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
T = F1
=
T
sin ( θ )
sin ( θ )
sin ( 180 deg − θ − θ 1 )
T = 54.7 lb
F1
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
=
P = sin ( θ 1 )
F1
P
sin ( θ 1 )
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
P = 42.6 lb
32
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
Problem 2-26
The boat is to be pulled onto the shore using two ropes. If the resultant force is to be F1, directed
along the keel aa as shown, determine the magnitudes of forces T and P acting in each rope and
the angle θ of P so that the magnitude of P is a minimum. T acts at θ from the keel as shown.
Given:
θ 1 = 30 deg
F 1 = 80 lb
Solution:
From the figure, P is minimum when
θ + θ 1 = 90 deg
θ = 90 deg − θ 1
θ = 60 deg
P
sin ( θ 1 )
P =
F1
=
sin ( 90 deg)
F 1 sin ( θ 1 )
sin ( 90 deg)
P = 40 lb
T
sin ( θ )
T = F1
=
F1
sin ( 90 deg)
sin ( θ )
sin ( 90 deg)
T = 69.3 lb
33
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Engineering Mechanics - Statics
Chapter 2
Problem 2-27
The beam is to be hoisted using two chains. Determine the magnitudes of forces FA and FB
acting on each chain in order to develop a resultant force T directed along the positive y axis.
Given:
T = 600 N
θ 1 = 30 deg
θ = 45 deg
Solution:
FA
sin ( θ )
FA =
T
=
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
T sin ( θ )
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
F A = 439 N
FB
sin ( θ 1 )
FB = T
=
T
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
sin ( θ 1 )
sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦
F B = 311 N
34
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
Problem 2-28
The beam is to be hoisted using two chains. If the resultant force is to be F, directed along the
positive y axis, determine the magnitudes of forces FA and F B acting on each chain and the
orientation θ of F B so that the magnitude of FB is a minimum.
Given:
F = 600 N
θ 1 = 30 deg
Solution:
For minimum FB, require
θ = 90 deg − θ 1
θ = 60 deg
F A = F cos ( θ 1 )
F A = 520 N
F B = F sin ( θ 1 )
F B = 300 N
Problem 2-29
Three chains act on the bracket such that they create a resultant force having magnitude F R. If
two of the chains are subjected to known forces, as shown, determine the orientation θ of the
third chain,measured clockwise from the positive x axis, so that the magnitude of force F in this
chain is a minimum. All forces lie in the x-y plane.What is the magnitude of F ? Hint: First find
the resultant of the two known forces. Force F acts in this direction.
35
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Engineering Mechanics - Statics
Chapter 2
Given:
F R = 500 lb
F 1 = 200 lb
F 2 = 300 lb
φ = 30 deg
Solution:
Cosine Law:
F R1 =
F 1 + F2 − 2 F 1 F2 cos ( 90 deg − φ )
2
2
F R1 = 264.6 lb
Sine Law:
Make F parallel to FR1
sin ( φ + θ )
sin ( 90 deg − φ )
=
F1
F R1
⎛
F1 ⎞
⎝
FR1 ⎠
θ = −φ + asin ⎜ sin ( 90 deg − φ )
⎟
θ = 10.9 deg
When F is directed along F R1, F will be minimum to create the resultant forces.
F = F R − FR1
F = 235 lb
Problem 2-30
Three cables pull on the pipe such that they create a resultant force having magnitude F R. If two
of the cables are subjected to known forces, as shown in the figure, determine the direction θ of
the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the
36
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Engineering Mechanics - Statics
Chapter 2
g
x–y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces.
Given:
F R = 900 lb
F 1 = 600 lb
F 2 = 400 lb
α = 45 deg
β = 30 deg
Solution:
F' =
F1 + F 2 − 2F 1 F 2 cos ( 90 deg + α − β )
2
2
F' = 802.64 lb
F = F R − F'
F = 97.4 lb
sin ( φ )
F1
=
sin ( 90 deg + α − β )
F'
⎛
⎝
φ = asin ⎜ sin ( 90 deg + α − β )
F1 ⎞
⎟
F' ⎠
φ = 46.22 deg
θ = φ−β
θ = 16.2 deg
Problem 2-31
Determine the x and y components of the force F.
37
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 800 lb
α = 60 deg
β = 40 deg
Solution:
F x = F sin ( β )
F y = −F cos ( β )
F x = 514.2 lb
F y = −612.8 lb
Problem 2-32
Determine the magnitude of the resultant force and its direction, measured clockwise from
the positive x axis.
Given:
F 1 = 70 N
F 2 = 50 N
F 3 = 65 N
θ = 30 deg
φ = 45 deg
Solution:
+
→
F Rx = ΣFx;
F RX = F 1 + F 2 cos ( θ ) − F 3 cos ( φ )
38
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Engineering Mechanics - Statics
+
↑
F Ry = ΣF y;
FR =
FRX + FRY
2
Chapter 2
F RY = −F2 sin ( θ ) − F 3 sin ( φ )
2
⎛ FRY ⎞
⎟
⎝ FRX ⎠
θ = atan ⎜
F R = 97.8 N
θ = 46.5 deg
Problem 2-33
Determine the magnitude of the resultant force and its direction measured counterclockwise
from the positive x axis.
Given:
F 1 = 50 lb
F 2 = 35 lb
α = 120 deg
β = 25 deg
Solution:
+
F Rx = ΣF x;
→
F Rx = F 1 sin ( α ) − F2 sin ( β )
F Rx = 28.5 lb
+
↑ FRy = ΣFy;
F Ry = −F 1 cos ( α ) − F 2 cos ( β )
39
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
F Ry = −6.7 lb
FR =
2
FRx + FRy
2
F R = 29.3 lb
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ' = atan ⎜
θ' = 13.3 deg
θ = 360 deg − θ'
θ = 347 deg
Problem 2-34
Determine the magnitude of the resultant force and its direction , measured counterclockwise
from the positive x axis.
Given:
F 1 = 850 N
F 2 = 625 N
F 3 = 750 N
θ = 45 deg
φ = 30 deg
c = 3
d = 4
Solution:
+
F Rx = SF x;
→
F RX = F 1
d
2
c +d
2
− F 2 sin ( φ ) − F3 sin ( θ )
40
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Engineering Mechanics - Statics
+
↑
F Ry = SF y;
F RY = −F1
c
2
2
− F2 cos ( φ ) + F3 cos ( θ )
c +d
F RX = −162.8 N
FR =
Chapter 2
2
FRX + FRY
F RY = −520.9 N
2
F R = 546 N
⎛ FRY ⎞
⎟
⎝ FRX ⎠
α = atan ⎜
α = 72.64 deg
β = α + 180 deg
β = 252.6 deg
Problem 2-35
Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the
resultant force is directed along the positive x' axis and has a magnitude of FR.
Units Used:
3
kN = 10 N
Given:
F R = 1 kN
F 2 = 450 N
F 3 = 200 N
α = 45 deg
β = 30 deg
Solution:
+
→
+
F Rx = SF x;
↑ FRy = SFy;
F R cos ( β ) = F 3 + F 2 cos ( α ) + F 1 cos ( θ + β )
−F R sin ( β ) = F2 sin ( α ) − F 1 sin ( θ + β )
41
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
F 1 cos ( θ + β ) = FR cos ( β ) − F3 − F2 cos ( α )
F 1 sin ( θ + β ) = F 2 sin ( α ) + FR sin ( β )
⎛
F2 sin ( α ) + F R sin ( β )
⎞
⎟−β
⎝ FR cos ( β ) − F3 − F2 cos ( α ) ⎠
θ = atan ⎜
θ = 37 deg
(FR cos (β ) − F3 − F2 cos (α ))2 + (F2 sin (α ) + FR sin (β ))2
F1 =
F 1 = 889 N
Problem 2-36
Determine the magnitude and direction, measured counterclockwise from the x' axis, of the
resultant force of the three forces acting on the bracket.
Given:
F 1 = 300 N
F 2 = 450 N
F 3 = 200 N
α = 45 deg
β = 30 deg
θ = 20 deg
Solution:
F Rx = F 1 cos ( θ + β ) + F3 + F2 cos ( α )
F Rx = 711.03 N
F Ry = −F 1 sin ( θ + β ) + F 2 sin ( α )
F Ry = 88.38 N
42
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
FR =
2
FRx + FRy
Chapter 2
2
F R = 717 N
φ (angle from x axis)
⎛ FRy ⎞
⎟
⎝ FRx ⎠
φ = atan ⎜
φ = 7.1 deg
φ' (angle from x' axis)
φ' = β + φ
φ' = 37.1 deg
Problem 2-37
Determine the magnitude of the resultant force and its direction, measured counterclockwise
from the positive x axis.
Given:
F 1 = 800 N
F 2 = 600 N
θ = 40 deg
c = 12
d = 5
Solution:
+
F Rx = ΣF x;
→
F Rx = F 1 cos ( θ ) − F 2 ⎜
2
+
F Ry = F 1 sin ( θ ) + F2 ⎜
⎛
d
↑
F Ry = ΣF y;
⎛
c
⎞
2⎟
⎝ c +d ⎠
2
⎞
2⎟
⎝ c +d ⎠
FR =
2
FRx + FRy
2
F Rx = 59 N
F Ry = 745 N
F R = 747 N
43
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = atan ⎜
θ = 85.5 deg
Problem 2-38
Determine the magnitude of the resultant force and its direction, measured counterclockwise
from the positive x axis.
Units Used:
3
kN = 10 N
Given:
F 1 = 30 kN
F 2 = 26 kN
θ = 30 deg
c = 5
d = 12
Solution:
+ F Rx = ΣF x;
→
F Rx = −F 1 sin ( θ ) − ⎜
⎛
c
+
F Ry = −F 1 cos ( θ ) + ⎜
⎛
d
↑ FRy = ΣFy;
⎞F
2
2
2⎟
+
d
c
⎝
⎠
⎞F
2
⎟
2
2
+
d
c
⎝
⎠
FR =
2
FRx + FRy
2
F Rx = −25 kN
F Ry = −2 kN
F R = 25.1 kN
⎛ FRy ⎞
⎟
⎝ FRx ⎠
φ = atan ⎜
φ = 4.5 deg
β = 180 deg + φ
β = 184.5 deg
44
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
Problem 2-39
Determine the magnitude of the resultant force and its direction measured counterclockwise
from the positive x axis.
Given:
F 1 = 60 lb
F 2 = 70 lb
F 3 = 50 lb
θ 1 = 60 deg
θ 2 = 45 deg
c = 1
d = 1
Solution:
d
θ 3 = atan ⎛⎜ ⎟⎞
⎝c⎠
F Rx = −F 1 cos ( θ 3 ) − F2 sin ( θ 1 )
F Rx = −103 lb
F Ry = F 1 sin ( θ 3 ) − F 2 cos ( θ 1 ) − F3
F Ry = −42.6 lb
FR =
2
FRx + FRy
2
F R = 111.5 lb
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = 180 deg + atan ⎜
θ = 202 deg
Problem 2-40
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured
counterclockwise from the positive x axis by summing the rectangular or x, y components
of the forces to obtain the resultant force.
45
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 600 N
F 2 = 800 N
F 3 = 450 N
θ 1 = 60 deg
θ 2 = 45 deg
θ 3 = 75 deg
Solution:
+
F Rx = ΣF x;
→
F Rx = F 1 cos ( θ 2 ) − F2 sin ( θ 1 )
F Rx = −268.556 N
+
F Ry = F 1 sin ( θ 2 ) + F 2 cos ( θ 1 )
F Ry = 824.264 N
↑ FRy = ΣFy;
FR =
2
FRx + FRy
2
F R = 867 N
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = 180 deg − atan ⎜
θ = 108 deg
Problem 2-41
Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces
by summing the rectangular or x, y components of the forces to obtain the resultant force.
Given:
F 1 = 30 N
F 2 = 20 N
46
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Engineering Mechanics - Statics
Chapter 2
F 3 = 50 N
θ = 20 deg
c = 3
d = 4
Solution:
⎛
⎞ − F ( sin ( θ ) ) + F
d
F Rx = −F 1 ⎜
2
⎛
c
2⎟
2
⎝ c +d ⎠
F Ry = F 1 ⎜
2
⎞ − F cos ( θ )
2⎟
⎝ c +d ⎠
FR =
2
FRx + FRy
F Rx = 19.2 N
3
F Ry = −0.8 N
2
2
F R = 19.2 N
⎛ −FRy ⎞
⎟
⎝ FRx ⎠
θ = atan ⎜
θ = 2.4 deg
Problem 2-42
Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the
resultant force acting on the bracket.
Given:
F A = 700 N
47
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Engineering Mechanics - Statics
Chapter 2
F B = 600 N
θ = 20 deg
φ = 30 deg
Solution:
Scalar Notation: Suming the force components algebraically, we have
F Rx = ΣF x;
F Rx = F A sin ( φ ) − F B cos ( θ )
F Rx = −213.8 N
F Ry = ΣF y;
F Ry = F A cos ( φ ) + FB sin ( θ )
F Ry = 811.4 N
The magnitude of the resultant force F R is
2
FR =
FRx + FRy
2
F R = 839 N
The directional angle θ measured counterclockwise
from the positive x axis is
⎛ FRx ⎞
⎟
⎝ FRy ⎠
θ = atan ⎜
θ = 14.8 deg
Problem 2-43
Determine the magnitude and direction, measured counterclockwise from the positive x'
axis, of the resultant force of the three forces acting on the bracket.
Given:
F 1 = 300 N
48
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Engineering Mechanics - Statics
Chapter 2
F 2 = 200 N
F 3 = 180 N
θ = 10 deg
θ 1 = 60 deg
c = 5
d = 12
Solution:
+
F Rx = ΣF x;
→
F Rx = F 1 sin ( θ 1 + θ ) − ⎛
+
F Ry = F 1 cos ( θ 1 + θ ) + F2 + ⎛
↑
F Ry = ΣF y;
d
⎞
⎜ 2 2 ⎟ F3
⎝ c +d ⎠
c
⎞
⎜ 2 2 ⎟ F3
⎝ c +d ⎠
FR =
2
FRx + FRy
2
F Rx = 115.8 N
F Ry = 371.8 N
F R = 389 N
φ = atan ⎜
⎛ FRy ⎞
⎟
⎝ FRx ⎠
φ = 72.7 deg
ψ = φ − ( 90 deg − θ 1 )
ψ = 42.7 deg
Problem 2-44
Determine the x and y components of F 1 and F 2.
49
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 200 N
F 2 = 150 N
θ = 45 deg
φ = 30 deg
Solution:
F 1x = F 1 sin ( θ )
F 1x = 141.4 N
F 1y = F 1 cos ( θ )
F 1y = 141.4 N
F 2x = −F 2 cos ( φ )
F 2x = −129.9 N
F 2y = F 2 sin ( φ )
F 2y = 75 N
Problem 2-45
Determine the magnitude of the resultant force and its direction, measured counterclockwise
from the positive x axis.
Given:
F 1 = 200 N
F 2 = 150 N
θ = 45 deg
φ = 30 deg
50
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Engineering Mechanics - Statics
Chapter 2
Solution:
+
F Rx = ΣF x;
→
F Rx = F 1 sin ( θ ) − F2 cos ( φ )
F Rx = 11.5 N
+
F Ry = F 1 cos ( θ ) + F 2 sin ( φ )
F Ry = 216.4 N
↑ FRy = ΣFy;
F =
2
FRx + FRy
2
F = 217 N
⎛ FRy ⎞
⎟
⎝ FRx ⎠
β = atan ⎜
β = 87 deg
Problem 2-46
Determine the x and y components of each force acting on the gusset plate of the bridge truss.
Given:
F 1 = 200 lb
c = 3
F 2 = 400 lb
d = 4
F 3 = 300 lb
e = 3
F 4 = 300 lb
f = 4
Solution:
F 1x = −F 1
F 1x = −200 lb
F 1y = 0 lb
F 2x = F 2 ⎛
d
⎞
⎜ 2 2⎟
⎝ c +d ⎠
F 2x = 320 lb
F 2y = −F 2 ⎛
c
⎞
⎜ 2 2⎟
⎝ c +d ⎠
F 2y = −240 lb
F 3x = F 3 ⎛
e
⎞
⎜ 2 2⎟
⎝ e +f ⎠
51
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Engineering Mechanics - Statics
Chapter 2
F 3x = 180 lb
F 3y = F 3 ⎛
f
⎞
⎜ 2 2⎟
⎝ e +f ⎠
F 3y = 240 lb
F 4x = −F 4
F 4x = −300 lb
F 4y = 0 lb
Problem 2-47
Determine the magnitude of the resultant force and its direction measured clockwise from
the positive x axis.
Units Used:
3
kN = 10 N
Given:
F 1 = 20 kN
F 2 = 40 kN
F 3 = 50 kN
θ = 60 deg
c = 1
d = 1
e = 3
f = 4
Solution:
+
F Rx = ΣF x;
→
F Rx = F 3 ⎛
f
⎛ d ⎞
⎞
⎜ 2 2 ⎟ + F2 ⎜ 2 2 ⎟ − F1 cos ( θ )
⎝ e +f ⎠
⎝ c +d ⎠
F Rx = 58.28 kN
52
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Engineering Mechanics - Statics
+
Chapter 2
⎛
c
⎞−F ⎛
⎞
2
⎜ 2 2 ⎟ − F1 sin ( θ )
2
2⎟
⎝ e +f ⎠
⎝ c +d ⎠
↑ FRy = ΣFy;
F Ry = F 3 ⎜
e
F Ry = −15.6 kN
F =
2
FRx + FRy
2
F = 60.3 kN
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = atan ⎜
θ = 15 deg
Problem 2-48
Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the
resultant force is directed along the positive x' axis and has magnitude FR.
Given:
F 2 = 200 N
F 3 = 180 N
θ 1 = 60 deg
F R = 800 N
c = 5
d = 12
Solution:
Initial Guesses:
F 1 = 100 N
θ = 10 deg
Given
+
F Rx = ΣF x;
→
⎛
F R sin ( θ 1 ) = F1 sin ( θ 1 + θ ) − ⎜
d
2
⎞F
2⎟
⎝ c +d ⎠
3
53
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Engineering Mechanics - Statics
+
↑
F Ry = ΣF y;
Chapter 2
F R( cos ( θ 1 ) ) = F 1 cos ( θ 1 + θ ) + F2 + ⎛
c
⎞
⎜ 2 2 ⎟ F3
⎝ c +d ⎠
⎛ F1 ⎞
⎜ ⎟ = Find ( F1 , θ )
⎝θ ⎠
F 1 = 869 N
θ = 21.3 deg
Problem 2-49
Determine the magnitude and direction, measured counterclockwise from the positive x'
axis, of the resultant force acting on the bracket.
Given:
F 1 = 300 N
F 2 = 200 N
F 3 = 180 N
θ 1 = 60 deg
θ = 10 deg
c = 5
d = 12
Solution:
Guesses
F Rx = 100 N
F Ry = 100 N
Given
+
F Rx = ΣF x;
→
F Rx = F1 sin ( θ 1 + θ ) − ⎛
+
F Ry = F1 cos ( θ 1 + θ ) + F 2 + ⎛
↑
F Ry = ΣF y;
d
⎞
⎜ 2 2 ⎟ F3
⎝ c +d ⎠
c
⎞
⎜ 2 2 ⎟ ( F3 )
⎝ c +d ⎠
54
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Engineering Mechanics - Statics
Chapter 2
⎛ FRx ⎞
⎜
⎟ = Find ( FRx , FRy)
⎝ FRy ⎠
FR =
2
FRx + FRy
2
⎛ FRx ⎞ ⎛ 115.8 ⎞
⎜
⎟=⎜
⎟N
⎝ FRy ⎠ ⎝ 371.8 ⎠
F R = 389 N
φ = atan ⎜
⎛ FRy ⎞
⎟
⎝ FRx ⎠
φ = 72.7 deg
φ' = ⎡⎣φ − ( 90 deg − θ 1 )⎤⎦
φ' = 42.7 deg
Problem 2-50
Express each of the three forces acting on the column in Cartesian vector form and compute
the magnitude of the resultant force.
Given:
F 1 = 150 lb
θ = 60 deg
F 2 = 275 lb
c = 4
F 3 = 75 lb
d = 3
Solution:
Find the components of each force.
⎛
⎛
⎞
⎟
2
2
⎝ c +d ⎠
F 1y = F 1 ⎜
F 1v = ⎜
⎛ F1x ⎞
⎟
⎝ F1y ⎠
F 1v = ⎜
F 2x = 0 lb
F 2y = −F 2
F 1x = F 1 ⎜
d
⎛ F2x ⎞
⎟
⎝ F2y ⎠
F 2v = ⎜
⎞
⎟
2
2
⎝ c +d ⎠
−c
⎛ 90 ⎞
⎟ lb
⎝ −120 ⎠
⎛ 0 ⎞
⎟ lb
⎝ −275 ⎠
F 2v = ⎜
55
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Engineering Mechanics - Statics
F 3x = −F 3 cos ( θ )
Chapter 2
F 3y = −F 3 sin ( θ )
⎛ F3x ⎞
⎟
⎝ F3y ⎠
⎛ −37.5 ⎞
⎟ lb
⎝ −65 ⎠
F 3v = ⎜
F 3v = ⎜
Now find the magnitude of the resultant force.
F R = F 1v + F 2v + F 3v
F R = 462.9 lb
Problem 2-51
Determine the magnitude of force F so that the resultant F R of the three forces is as small
as possible. What is the minimum magnitude of FR?
Units Used:
kN = 1000 N
Given:
F 1 = 5 kN
F 2 = 4 kN
θ = 30 deg
Solution:
Scalar Notation: Suming the force components
algebrically, we have
+
→
+
↑
F Rx = ΣF x;
F Rx = F1 − F sin ( θ )
F Ry = ΣF y;
F Ry = F cos ( θ ) − F 2
The magnitude of the resultant force F R is
FR =
2
2
F Rx + F Ry =
(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2
56
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Engineering Mechanics - Statics
Chapter 2
F R = F 1 + F2 + F − 2FF1 sin ( θ ) − 2F2 Fcos( θ )
2
2FR
2
dF R
dF
2
= 2F − 2F1 sin ( θ ) − 2F2 cos ( θ )
If F is a minimum, then
FR =
2
⎞
⎛ dFR
= 0⎟
⎜
⎝ dF
⎠
F = F 1 sin ( θ ) + F2 cos ( θ )
(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2
F = 5.96 kN
F R = 2.3 kN
Problem 2-52
Express each of the three forces acting on the bracket in Cartesian vector form with respect
to the x and y axes. Determine the magnitude and direction θ of F1 so that the resultant force
is directed along the positive x' axis and has magnitude F R.
Units Used:
kN = 1000 N
Given:
F R = 600 N
F 2 = 350 N
F 3 = 100 N
φ = 30 deg
Solution:
⎛ F2 ⎞
⎟
⎝0 ⎠
F 3v = ⎜
⎛ F1 cos ( θ ) ⎞
⎟
⎝ F1 sin ( θ ) ⎠
F 2v = ⎜
F 2v = ⎜
F 1v = ⎜
⎛ 0 ⎞
⎟
⎝ −F 3 ⎠
⎛ 350 ⎞
⎟N
⎝ 0 ⎠
⎛ 0 ⎞
⎟N
⎝ −100 ⎠
F 3v = ⎜
57
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Engineering Mechanics - Statics
Chapter 2
F 1 = 20 N θ = 10 deg
The initial guesses:
Given
⎛ F1 cos ( θ ) ⎞
⎛ FR cos ( φ ) ⎞
+
F
+
F
=
⎜
⎟
⎟
2v
3v ⎜
⎝ F1 sin ( θ ) ⎠
⎝ FR sin ( φ ) ⎠
⎛ F1 ⎞
⎜ ⎟ = Find ( F1 , θ )
⎝θ ⎠
F 1 = 434.5 N
θ = 67 deg
Problem 2-53
The three concurrent forces acting on the post produce a resultant force F R = 0. If F 2 = (1/2)F 1,
and F 1 is to be 90° from F 2 as shown, determine the required magnitude F 3 expressed in terms
of F1 and the angle θ.
Solution:
Use the primed coordiates.
ΣFRx = 0
F 3 cos ( θ − 90 deg) − F1 = 0
ΣFRy = 0
−F 3 sin ( θ − 90 deg) + F 2 = 0
tan ( θ − 90 deg) =
F2
F1
=
1
2
1
θ = 90 deg + atan ⎛⎜ ⎟⎞
⎝ 2⎠
θ = 117 deg
k =
1
cos ( θ − 90 deg)
k = 1.1
F3 = k F1
58
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Engineering Mechanics - Statics
Chapter 2
Problem 2-54
Three forces act on the bracket. Determine the magnitude and orientation θ of F2 so that the
resultant force is directed along the positive u axis and has magnitude F R.
Given:
F R = 50 lb
F 1 = 80 lb
F 3 = 52 lb
φ = 25 deg
c = 12
d = 5
Solution:
Guesses
F 2 = 1 lb
θ = 120 deg
Given
⎛
⎞F
3
2
2⎟
⎝ c +d ⎠
F R cos ( φ ) = F1 + F2 cos ( φ + θ ) + ⎜
⎛
d
⎞F
3
2
2⎟
⎝ c +d ⎠
−F R sin ( φ ) = −F 2 sin ( φ + θ ) + ⎜
⎛ F2 ⎞
⎜ ⎟ = Find ( F2 , θ )
⎝θ ⎠
c
θ = 103.3 deg
F 2 = 88.1 lb
59
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
Problem 2-55
Determine the magnitude and orientation, measured clockwise from the positive x axis, of
the resultant force of the three forces acting on the bracket.
Given:
F 1 = 80 lb
F 2 = 150 lb
F 3 = 52lb
θ = 55 deg
φ = 25 deg
c = 12 m
d = 5m
Solution:
⎛
F Rx = F 1 + F 3 ⎜
d
2
⎞ + F cos ( θ + φ )
2⎟
⎝ c +d ⎠
⎛
F Rx = 126.05 lb
2
⎞ − F sin ( θ + φ )
2
⎟
2
2
⎝ c +d ⎠
c
F Ry = F 3 ⎜
FR =
2
FRx + FRy
F Ry = −99.7 lb
2
F R = 161 lb
⎛ FRy ⎞
⎟
⎝ FRx ⎠
β = atan ⎜
β = 38.3 deg
Problem 2-56
Three forces act on the ring. Determine the range of values for the magnitude of P so that the
magnitude of the resultant force does not exceed F. Force P is always directed to the right.
60
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Engineering Mechanics - Statics
Chapter 2
Units Used:
3
kN = 10 N
Given:
F = 2500 N
F 1 = 1500 N
F 2 = 600 N
θ 1 = 60 deg
θ 2 = 45 deg
Solution:
Initial Guesses:
F Rx = 100 N
F Ry = 100 N
P = 100 N
Given
+
F Rx = ΣF x;
→
F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )
+
F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )
↑ FRy = ΣFy;
F=
2
F Rx + F Ry
⎛ FRx ⎞
⎜
⎟
⎜ FRy ⎟ = Find ( FRx , FRy , P)
⎜P ⎟
⎝ max ⎠
Initial Guesses:
2
P max = 1.6 kN
F Rx = −100 N
F Ry = 100 N
P = −2000 N
Given
+
F Rx = ΣF x;
→
F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )
+
F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )
↑ FRy = ΣFy;
F=
⎛ FRx ⎞
⎜
⎟
⎜ FRy ⎟ = Find ( FRx , FRy , P)
⎜P ⎟
⎝ min ⎠
2
F Rx + F Ry
2
P min = −1.7 kN
61
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Engineering Mechanics - Statics
Since P > 0
we conclude that
Chapter 2
0 <= P <= P max = 1.6 kN
Problem 2-57
Determine the magnitude and coordinate direction angles of F1 and F2. Sketch each force
on an x, y, z reference.
Given:
⎛ 60 ⎞
F 1 = ⎜ −50 ⎟ N
⎜
⎟
⎝ 40 ⎠
⎛ −40 ⎞
F 2 = ⎜ −85 ⎟ N
⎜
⎟
⎝ 30 ⎠
Solution:
F 1 = 87.7 N
⎛ α1 ⎞
⎜ ⎟
⎛ F1 ⎞
⎜ β 1 ⎟ = acos ⎜
⎟
⎝ F1 ⎠
⎜γ ⎟
⎝ 1⎠
⎛ α 1 ⎞ ⎛ 46.9 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 1 ⎟ = ⎜ 124.7 ⎟ deg
⎜ γ ⎟ ⎝ 62.9 ⎠
⎝ 1⎠
F 2 = 98.6 N
62
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Engineering Mechanics - Statics
Chapter 2
⎛ α2 ⎞
⎜ ⎟
⎛ F2 ⎞
⎜ β 2 ⎟ = acos ⎜
⎟
⎝ F2 ⎠
⎜γ ⎟
⎝ 2⎠
⎛ α 2 ⎞ ⎛ 113.9 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 2 ⎟ = ⎜ 149.5 ⎟ deg
⎜ γ ⎟ ⎝ 72.3 ⎠
⎝ 2⎠
Problem 2-58
Express each force in Cartesian vector form.
Units Used:
3
kN = 10 N
Given:
F 1 = 5 kN
F 2 = 2 kN
θ 1 = 60 deg
θ 2 = 60 deg
θ 3 = 45 deg
Solution:
⎛ cos ( θ 2) ⎞
⎜
⎟
F 1v = F1 ⎜ cos ( θ 3 ) ⎟
⎜ cos θ ⎟
⎝ ( 1) ⎠
⎛ 2.5 ⎞
F 1v = ⎜ 3.5 ⎟ kN
⎜ ⎟
⎝ 2.5 ⎠
⎛0⎞
F 2v = F2 ⎜ −1 ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
F 2v = ⎜ −2 ⎟ kN
⎜ ⎟
⎝0⎠
63
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Engineering Mechanics - Statics
Chapter 2
Problem 2-59
Determine the magnitude and coordinate direction angles of the force F acting on the stake.
Given:
F h = 40 N
θ = 70 deg
c = 3
d = 4
Solution:
⎛ c2 + d2 ⎞
⎟
F = Fh ⎜
⎝ d ⎠
F = 50 N
F x = F h cos ( θ )
F y = F h sin ( θ )
Fz = ⎛
F x = 13.7 N
F y = 37.6 N
F z = 30 N
⎛ Fx ⎞
⎟
⎝F⎠
c
⎞
⎜ 2 2 ⎟F
⎝ c +d ⎠
⎛ Fy ⎞
⎟
⎝F⎠
⎛ Fz ⎞
⎟
⎝F⎠
α = acos ⎜
β = acos ⎜
γ = acos ⎜
α = 74.1 deg
β = 41.3 deg
γ = 53.1 deg
Problem 2-60
Express each force in Cartesian vector form.
64
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 400 lb
F 2 = 600 lb
θ 1 = 45 deg
θ 2 = 60 deg
θ 3 = 60 deg
θ 4 = 45 deg
θ 5 = 30 deg
Solution:
⎛ −cos ( θ 2) ⎞
⎜
⎟
F 1v = F1 ⎜ cos ( θ 3 ) ⎟
⎜ cos θ ⎟
( 1) ⎠
⎝
⎛ −200 ⎞
F 1v = ⎜ 200 ⎟ lb
⎜
⎟
⎝ 282.8 ⎠
⎛ cos ( θ 5) cos ( θ 4 ) ⎞
⎜
⎟
F 2v = F2 ⎜ cos ( θ 5 ) sin ( θ 4 ) ⎟
⎜ −sin θ
( 5) ⎟⎠
⎝
⎛ 367.4 ⎞
F 2v = ⎜ 367.4 ⎟ lb
⎜
⎟
⎝ −300 ⎠
Problem 2-61
The stock S mounted on the lathe is subjected to a
force F , which is caused by the die D. Determine the
coordinate direction angle β and express the force as
a Cartesian vector.
Given:
F = 60 N
α = 60 deg
γ = 30 deg
65
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Engineering Mechanics - Statics
Chapter 2
Solution:
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
(
2
β = acos 1 − cos ( α ) − cos ( γ )
2
⎛⎜ −cos ( α ) ⎞⎟
F v = F⎜ −cos ( β ) ⎟
⎜ −cos ( γ ) ⎟
⎝
⎠
2
)
β = 90 deg
⎛ −30 ⎞
Fv = ⎜ 0 ⎟ N
⎜
⎟
⎝ −52 ⎠
Problem 2-62
Determine the magnitude and coordinate direction angles of the resultant force.
Given:
F 1 = 80 lb
F 2 = 130 lb
θ = 40 deg
φ = 30 deg
Solution:
⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞
F 1v = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟
⎜
⎟
sin ( φ )
⎝
⎠
⎛ 53.1 ⎞
F 1v = ⎜ −44.5 ⎟ lb
⎜
⎟
⎝ 40 ⎠
⎛0⎞
F 2v = F2 ⎜ 0 ⎟
⎜ ⎟
⎝ −1 ⎠
⎛ 0 ⎞
F 2v = ⎜ 0 ⎟ lb
⎜
⎟
⎝ −130 ⎠
F R = F1v + F2v
F R = 113.6 lb
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 62.1 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 113.1 ⎟ deg
⎜ γ ⎟ ⎝ 142.4 ⎠
⎝ ⎠
66
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Engineering Mechanics - Statics
Chapter 2
Problem 2-63
Specify the coordinate direction angles of F 1 and F 2 and express each force as a cartesian
vector.
Given:
F 1 = 80 lb
F 2 = 130 lb
φ = 30 deg
θ = 40 deg
Solution:
⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞
F 1v = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟
⎜
⎟
sin ( φ )
⎝
⎠
⎛ 53.1 ⎞
⎜
⎟
F 1v = −44.5 lb
⎜
⎟
⎝ 40 ⎠
⎛ α1 ⎞
⎜ ⎟
⎛ F1v ⎞
⎜ β 1 ⎟ = acos ⎜
⎟
⎝ F1 ⎠
⎜γ ⎟
⎝ 1⎠
⎛ α 1 ⎞ ⎛ 48.4 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 1 ⎟ = ⎜ 123.8 ⎟ deg
⎜ γ ⎟ ⎝ 60 ⎠
⎝ 1⎠
⎛0⎞
⎜ ⎟
F 2v = F2 0
⎜ ⎟
⎝ −1 ⎠
⎛ 0 ⎞
⎜ 0 ⎟ lb
F 2v =
⎜
⎟
⎝ −130 ⎠
⎛ α2 ⎞
⎜ ⎟
⎛ F2v ⎞
⎜ β 2 ⎟ = acos ⎜
⎟
⎝ F2 ⎠
⎜γ ⎟
⎝ 2⎠
⎛ α 2 ⎞ ⎛ 90 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 2 ⎟ = ⎜ 90 ⎟ deg
⎜ γ ⎟ ⎝ 180 ⎠
⎝ 2⎠
Problem 2-64
The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1
of F1 so that the resultant force acting on the mast is FRi.
Given:
F R = 350 N
67
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Engineering Mechanics - Statics
Chapter 2
F 1 = 500 N
F 2 = 200 N
F 3 = 300 N
Solution:
Guesses
α = 20 deg
β = 20 deg
γ = 20 deg
Given
⎛⎜ cos ( α ) ⎟⎞
⎛0⎞
⎛0⎞
⎛1⎞
⎜
⎟
⎜
⎟
F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = FR⎜ 0 ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ cos ( γ ) ⎟
⎝ −1 ⎠
⎝0⎠
⎝0⎠
⎝
⎠
⎛⎜ α ⎞⎟
⎜ β ⎟ = Find ( α , β , γ )
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 45.6 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 53.1 ⎟ deg
⎜ γ ⎟ ⎝ 66.4 ⎠
⎝ ⎠
Problem 2-65
The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1
of F1 so that the resultant force acting on the mast is zero .
Given:
F 1 = 500 N
F 2 = 200 N
F 3 = 300 N
Solution:
Guesses
α = 20 deg
β = 20 deg
γ = 20 deg
68
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Engineering Mechanics - Statics
Chapter 2
⎛⎜ cos ( α ) ⎞⎟
⎛0⎞
⎛0⎞
⎜ ⎟
⎜ ⎟
F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = 0
⎜ ⎟
⎜ ⎟
⎜ cos ( γ ) ⎟
⎝ −1 ⎠
⎝0⎠
⎝
⎠
Given
⎛⎜ α ⎞⎟
⎜ β ⎟ = Find ( α , β , γ )
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 90 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 53.1 ⎟ deg
⎜ γ ⎟ ⎝ 66.4 ⎠
⎝ ⎠
Problem 2-66
The shaft S exerts three force
components on the die D. Find the
magnitude and direction of the resultant
force. Force F2 acts within the octant
shown.
Given:
F 1 = 400 N
F 2 = 300 N
F 3 = 200 N
α 2 = 60 deg
γ 2 = 60 deg
c = 3
d = 4
Solution:
cos ( α 2 ) + cos ( β 2 ) + cos ( γ 2 ) = 1
2
2
2
Solving for the positive root,
β 2 = acos ⎛⎝ 1 − cos ( α 2 ) − cos ( γ 2 ) ⎞⎠
2
⎛⎜ F1 ⎞⎟
F 1v = ⎜ 0 ⎟
⎜0 ⎟
⎝ ⎠
2
β 2 = 45 deg
⎛ cos ( α 2) ⎞
⎜
⎟
F 2v = F2 ⎜ cos ( β 2 ) ⎟
⎜ cos γ ⎟
⎝ ( 2) ⎠
F 3v =
⎛0⎞
⎜ −d ⎟
⎟
2
2⎜
c +d ⎝ c ⎠
F3
69
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Engineering Mechanics - Statics
Chapter 2
F R = F1v + F2v + F3v
F R = 615 N
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 26.6 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 85.1 ⎟ deg
⎜ γ ⎟ ⎝ 64.0 ⎠
⎝ ⎠
Problem 2-67
The beam is subjected to the two forces shown. Express each force in Cartesian vector form
and determine the magnitude and coordinate direction angles of the resultant force.
Given:
F 1 = 630 lb
α = 60 deg
F 2 = 250 lb
β = 135 deg
c = 24
γ = 60 deg
d = 7
Solution:
F 1v =
⎛0⎞
⎜d⎟
2
2⎜ ⎟
c + d ⎝ −c ⎠
F1
⎛ 0 ⎞
⎜
⎟
F 1v = 176.4 lb
⎜
⎟
⎝ −604.8 ⎠
⎛⎜ cos ( α ) ⎟⎞
F 2v = F2 ⎜ cos ( β ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
⎛ 125 ⎞
⎜
⎟
F 2v = −176.8 lb
⎜
⎟
⎝ 125 ⎠
F R = F1v + F2v
F R = 495.8 lb
70
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Engineering Mechanics - Statics
Chapter 2
⎛ αR ⎞
⎜ ⎟
⎛ FR ⎞
⎜ β R ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ R⎠
⎛ α R ⎞ ⎛ 75.4 ⎞
⎜ ⎟ ⎜
⎟
⎜ β R ⎟ = ⎜ 90 ⎟ deg
⎜ γ ⎟ ⎝ 165.4 ⎠
⎝ R⎠
Problem 2-68
Determine the magnitude and coordinate direction angles of the resultant force.
Given:
F 1 = 350 N
α = 60 deg
F 2 = 250N
β = 60 deg
c = 3
γ = 45 deg
d = 4
θ = 30 deg
Solution:
⎛⎜ cos ( α ) ⎟⎞
F 1v = F1 ⎜ cos ( β ) ⎟
⎜ −cos ( γ ) ⎟
⎝
⎠
⎛ 175 ⎞
F 1v = ⎜ 175 ⎟ N
⎜
⎟
⎝ −247.5 ⎠
F 2h = F2 ⎛
F 2y = F 2 ⎛
⎛ F2h cos ( θ ) ⎞
⎜
⎟
F 2v = ⎜ −F 2h sin ( θ ) ⎟
⎜ F
⎟
2y
⎝
⎠
⎛ 173.2 ⎞
F 2v = ⎜ −100 ⎟ N
⎜
⎟
⎝ 150 ⎠
d
⎞
⎜ 2 2⎟
⎝ c +d ⎠
c
⎞
⎜ 2 2⎟
⎝ c +d ⎠
F R = F1v + F2v
F R = 369.3 N
⎛ αR ⎞
⎜ ⎟
⎛ FR ⎞
⎜ β R ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ R⎠
⎛ α R ⎞ ⎛ 19.5 ⎞
⎜ ⎟ ⎜
⎟
⎜ β R ⎟ = ⎜ 78.3 ⎟ deg
⎜ γ ⎟ ⎝ 105.3 ⎠
⎝ R⎠
Problem 2-69
Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three
71
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Engineering Mechanics - Statics
Chapter 2
forces acts along the positive y axis and has magnitude F.
Given:
F = 600 lb
F 1 = 180 lb
F 2 = 300 lb
α 1 = 30 deg
α 2 = 40 deg
Solution:
Initial guesses:
α = 40 deg
γ = 50 deg
β = 50 deg
F 3 = 45 lb
Given
F Rx = ΣF x;
0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )
F Ry = ΣF y;
F = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )
F Rz = ΣF z;
0 = −F2 sin ( α 1 ) + F3 cos ( γ )
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
⎛ F3 ⎞
⎜ ⎟
⎜ α ⎟ = Find ( F , α , β , γ )
3
⎜β ⎟
⎜ ⎟
⎝γ ⎠
2
2
F 3 = 428 lb
⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 20.6 ⎟ deg
⎜ γ ⎟ ⎝ 69.5 ⎠
⎝ ⎠
Problem 2-70
Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three
forces is zero.
72
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 180 lb
α 1 = 30 deg
F 2 = 300 lb
α 2 = 40 deg
Solution:
Initial guesses:
α = 40 deg
γ = 50 deg
β = 50 deg
F 3 = 45 lb
Given
F Rx = ΣF x;
0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )
F Ry = ΣF y;
0 = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )
F Rz = ΣF z;
0 = −F2 sin ( α 1 ) + F3 cos ( γ )
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
⎛ F3 ⎞
⎜ ⎟
⎜ α ⎟ = Find ( F , α , β , γ )
3
⎜β ⎟
⎜ ⎟
⎝γ ⎠
2
2
F 3 = 250 lb
⎛⎜ α ⎞⎟ ⎛ 87.0 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 142.9 ⎟ deg
⎜ γ ⎟ ⎝ 53.1 ⎠
⎝ ⎠
Problem 2-71
Specify the magnitude F 3 and directions α3, β3, and
γ3 of F 3 so that the resultant force of the three forces
is FR.
Units Used:
3
kN = 10 N
Given:
F 1 = 12 kN
c = 5
F 2 = 10 kN
d = 12
θ = 30 deg
73
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Engineering Mechanics - Statics
Chapter 2
⎛0⎞
F R = ⎜ 9 ⎟ kN
⎜ ⎟
⎝0⎠
Solution:
F 3x = 1 kN
Initial Guesses:
F 3y = 1 kN
⎛ F3x ⎞
⎛ 0 ⎞
⎜
⎟
F R = ⎜ F3y ⎟ + F 1 ⎜ cos ( θ ) ⎟ +
⎜
⎟
⎜F ⎟
(
)
−
sin
θ
⎝
⎠
⎝ 3z ⎠
Given
⎛ F3x ⎞
⎜
⎟
⎜ F3y ⎟ = Find ( F3x , F3y , F3z)
⎜F ⎟
⎝ 3z ⎠
⎛ α3 ⎞
⎜ ⎟
⎛ F3 ⎞
⎜ β 3 ⎟ = acos ⎜
⎟
⎝ F3 ⎠
⎜γ ⎟
⎝ 3⎠
F 3z = 1 kN
⎛ −d ⎞
⎜0⎟
⎟
2
2⎜
c +d ⎝ c ⎠
F2
⎛ F3x ⎞
⎜
⎟
F 3 = ⎜ F 3y ⎟
⎜F ⎟
⎝ 3z ⎠
⎛ 9.2 ⎞
F 3 = ⎜ −1.4 ⎟ kN
⎜
⎟
⎝ 2.2 ⎠
F 3 = 9.6 kN
⎛ α 3 ⎞ ⎛ 15.5 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 3 ⎟ = ⎜ 98.4 ⎟ deg
⎜ γ ⎟ ⎝ 77.0 ⎠
⎝ 3⎠
Problem 2-72
The pole is subjected to the force F , which has
components acting along the x,y,z axes as
shown. Given β and γ, determine the magnitude
of the three components of F .
Units Used:
kN = 1000 N
Given:
F = 3 kN
β = 30 deg
γ = 75 deg
Solution:
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
(
2
)
74
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Engineering Mechanics - Statics
α = acos
Chapter 2
( −cos (β )2 − cos (γ )2 + 1)
α = 64.67 deg
F x = F cos ( α )
F y = F cos ( β )
F z = F cos ( γ )
F x = 1.28 kN
F y = 2.60 kN
F z = 0.8 kN
Problem 2-73
The pole is subjected to the force F which has components F x and Fz. Determine the
magnitudes of F and Fy.
Units Used:
kN = 1000 N
Given:
F x = 1.5 kN
F z = 1.25 kN
β = 75 deg
Solution:
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
2
2
2
⎛ Fx ⎞
2 ⎛ Fz ⎞
⎜ ⎟ + cos ( β ) + ⎜ ⎟ = 1
⎝F⎠
⎝F⎠
2
F =
2
Fx + Fz
2
1 − cos ( β )
F y = F cos ( β )
F = 2.02 kN
F y = 0.5 kN
Problem 2-74
The eye bolt is subjected to the cable force F which has a component F x along the x axis, a
component F z along the z axis, and a coordinate direction angle β. Determine the magnitude of F .
75
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 2
Given:
F x = 60 N
F z = −80 N
β = 80 deg
Solution:
F y = F cos ( β )
F x + F z + Fy cos ( β )
2
Fy =
2
2
Fy =
2
Fx + Fz
1 − cos ( β )
2
F =
2
2
2
cos ( β )
F y = 17.6 N
2
F = 102 N
Fx + Fy + Fz
Problem 2-75
Three forces act on the hook. If the resultant force FR has a magnitude and direction as
shown, determine the magnitude and the coordinate direction angles of force F3.
Given:
F R = 120 N
F 1 = 80 N
F 2 = 110 N
c = 3
d = 4
θ = 30 deg
φ = 45 deg
76
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Engineering Mechanics - Statics
Chapter 2
Solution:
⎛d⎞
⎛ F1 ⎞ ⎜ ⎟
F 1v = ⎜
⎟ 0
2
2 ⎜ ⎟
c
+
d
⎝
⎠⎝ c ⎠
⎛ 64 ⎞
F 1v = ⎜ 0 ⎟ N
⎜ ⎟
⎝ 48 ⎠
⎛0⎞
F 2v = F2 ⎜ 0 ⎟
⎜ ⎟
⎝ −1 ⎠
⎛ 0 ⎞
F 2v = ⎜ 0 ⎟ N
⎜
⎟
⎝ −110 ⎠
⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟
F Rv = F R⎜ cos ( φ ) cos ( θ ) ⎟
⎜ sin ( φ ) ⎟
⎝
⎠
⎛ 42.4 ⎞
F Rv = ⎜ 73.5 ⎟ N
⎜
⎟
⎝ 84.9 ⎠
F 3v = FRv − F 1v − F 2v
⎛ −21.6 ⎞
F 3v = ⎜ 73.5 ⎟ N
⎜
⎟
⎝ 146.9 ⎠
⎛⎜ α ⎞⎟
⎛ F3v ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ F3v ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 97.5 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 63.7 ⎟ deg
⎜ γ ⎟ ⎝ 27.5 ⎠
⎝ ⎠
F 3v = 165.6 N
Problem 2-76
Determine the coordinate direction angles of F 1 and FR.
Given:
F R = 120 N
F 1 = 80 N
F 2 = 110 N
c = 3
d = 4
θ = 30 deg
φ = 45 deg
77
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Engineering Mechanics - Statics
Chapter 2
Solution:
⎛d⎞
⎛ F1 ⎞ ⎜ ⎟
F 1v = ⎜
⎟ 0
2
2 ⎜ ⎟
c
+
d
⎝
⎠⎝ c ⎠
⎛ 64 ⎞
F 1v = ⎜ 0 ⎟ N
⎜ ⎟
⎝ 48 ⎠
⎛ α1 ⎞
⎜ ⎟
⎛ F1v ⎞
⎜ β 1 ⎟ = acos ⎜
⎟
⎝ F1v ⎠
⎜γ ⎟
⎝ 1⎠
⎛ α 1 ⎞ ⎛ 36.9 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 1 ⎟ = ⎜ 90 ⎟ deg
⎜ γ ⎟ ⎝ 53.1 ⎠
⎝ 1⎠
⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟
F Rv = F R⎜ cos ( φ ) cos ( θ ) ⎟
⎜ sin ( φ ) ⎟
⎝
⎠
⎛ 42.4 ⎞
F Rv = ⎜ 73.5 ⎟ N
⎜
⎟
⎝ 84.9 ⎠
⎛ αR ⎞
⎜ ⎟
⎛ FRv ⎞
⎜ β R ⎟ = acos ⎜
⎟
⎝ FRv ⎠
⎜γ ⎟
⎝ R⎠
⎛ α R ⎞ ⎛ 69.3 ⎞
⎜ ⎟ ⎜
⎟
⎜ β R ⎟ = ⎜ 52.2 ⎟ deg
⎜ γ ⎟ ⎝ 45 ⎠
⎝ R⎠
Problem 2-77
The pole is subjected to the force F , which has components acting along the x, y, z axes as
shown. Given the magnitude of F and the angles α and γ , determine the magnitudes of the
components of F .
Given:
F = 80 N α = 60 deg
γ = 45 deg
Solution:
(
β = acos − 1 − cos ( α ) − cos ( γ )
2
2
)
β = 120 deg
F x = F cos ( α )
F y = F cos ( β )
F z = F cos ( γ )
F x = 40 N
F y = 40 N
F z = 56.6 N
78
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Engineering Mechanics - Statics
Chapter 2
Problem 2-78
Two forces F1 and F 2 act on the bolt. If the resultant force F R has magnitude F R and
coordinate direction angles α and β, as shown, determine the magnitude of F2 and its
coordinate direction angles.
Given:
F 1 = 20 lb
F R = 50 lb
α = 110 deg
β = 80 deg
Solution:
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
(
2
γ = acos − 1 − cos ( α ) − cos ( β )
2
Initial Guesses
Given
F 2x = 1 lb
2
)
γ = 157.44 deg
F 2y = 1 lb
F 2z = 1 lb
⎛⎜ cos ( α ) ⎟⎞
⎛ 0 ⎞ ⎛⎜ F2x ⎟⎞
⎜ ⎟
F R⎜ cos ( β ) ⎟ = F 1 0 + ⎜ F2y ⎟
⎜ ⎟
⎜ cos ( γ ) ⎟
⎝ −1 ⎠ ⎜⎝ F2z ⎟⎠
⎝
⎠
⎛ F2x ⎞
⎜
⎟
F 2 = ⎜ F 2y ⎟
⎜F ⎟
⎝ 2z ⎠
⎛ F2x ⎞
⎜
⎟
⎜ F2y ⎟ = Find ( F2x , F2y , F2z)
⎜F ⎟
⎝ 2z ⎠
⎛ −17.1 ⎞
⎜ 8.7 ⎟ lb
F2 =
⎜
⎟
⎝ −26.2 ⎠
F 2 = 32.4 lb
⎛ α2 ⎞
⎜ ⎟
⎛ F2 ⎞
⎜ β 2 ⎟ = acos ⎜
⎟
⎝ F2 ⎠
⎜γ ⎟
⎝ 2⎠
⎛ α 2 ⎞ ⎛ 121.8 ⎞
⎜ ⎟ ⎜
⎟
⎜ β 2 ⎟ = ⎜ 74.5 ⎟ deg
⎜ γ ⎟ ⎝ 143.8 ⎠
⎝ 2⎠
Problem 2-79
Given r1, r2, and r3, determine the magnitude and direction of r = 2r1 − r2 + 3r3.
79
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Engineering Mechanics - Statics
Chapter 2
Given:
⎛3⎞
⎜ ⎟
r1 = −4 m
⎜ ⎟
⎝3⎠
⎛4⎞
⎜ ⎟
r2 = 0 m
⎜ ⎟
⎝ −5 ⎠
⎛3⎞
⎜ ⎟
r3 = −2 m
⎜ ⎟
⎝5⎠
Solution:
r = 2r1 − r2 + 3r3
⎛ 11 ⎞
⎜
⎟
r = −14 m
⎜
⎟
⎝ 26 ⎠
r = 31.5 m
⎛⎜ α ⎞⎟
⎛ r ⎞
⎜ β ⎟ = acos ⎜ ⎟
⎝ r ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 69.6 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 116.4 ⎟ deg
⎜ γ ⎟ ⎝ 34.4 ⎠
⎝ ⎠
Problem 2-80
Represent the position vector r acting from point A(a, b, c) to point B(d, e, f) in Cartesian vector
form. Determine its coordinate direction angles and find the distance between points A and B.
Given:
a = 3m
b = 5m
c = 6m
d = 5m
e = −2 m
f = 1m
Solution:
⎛d − a⎞
⎜
⎟
r = e−b
⎜
⎟
⎝ f− c⎠
⎛2⎞
⎜ ⎟
r = −7 m
⎜ ⎟
⎝ −5 ⎠
r = 8.8 m
80
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Engineering Mechanics - Statics
Chapter 2
α = acos ⎛⎜
d − a⎞
α = 76.9 deg
β = acos ⎛⎜
e − b⎞
β = 142 deg
γ = acos ⎛⎜
f − c⎞
γ = 124 deg
⎟
⎝ r ⎠
⎟
⎝ r ⎠
⎟
⎝ r ⎠
Problem 2-81
A position vector extends from the origin to point A(a, b, c). Determine the angles α, β, γ which
the tail of the vector makes with the x, y, z axes, respectively.
Given:
a = 2m b = 3m
c = 6m
Solution:
⎛a⎞
r = ⎜b⎟
⎜ ⎟
⎝c ⎠
⎛2⎞
r = ⎜3⎟ m
⎜ ⎟
⎝6⎠
⎛⎜ α ⎞⎟
⎛ r ⎞
⎜ β ⎟ = acos ⎜ ⎟
⎝ r ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 73.4 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 64.6 ⎟ deg
⎜ γ ⎟ ⎝ 31.0 ⎠
⎝ ⎠
Problem 2-82
Express the position vector r in Cartesian vector form; then determine its magnitude and
coordinate direction angles.
Given:
a = 4m
81
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Engineering Mechanics - Statics
Chapter 2
b = 8m
c = 3m
d = 4m
Solution:
⎛ −c ⎞
r = ⎜ −d − b ⎟
⎜
⎟
⎝ a ⎠
⎛⎜ α ⎞⎟
⎛ r ⎞
⎜ β ⎟ = acos ⎜ ⎟
⎝ r ⎠
⎜γ ⎟
⎝ ⎠
⎛ −3 ⎞
r = ⎜ −12 ⎟ m
⎜
⎟
⎝ 4 ⎠
r = 13 m
⎛⎜ α ⎞⎟ ⎛ 103.3 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 157.4 ⎟ deg
⎜ γ ⎟ ⎝ 72.1 ⎠
⎝ ⎠
Problem 2-83
Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate
direction angles.
Given:
a = 8 ft
b = 2 ft
c = 5 ft
θ = 30 deg
φ = 20 deg
Solution:
⎛⎜ −c cos ( φ ) sin ( θ ) ⎞⎟
r = ⎜ a − c cos ( φ ) cos ( θ ) ⎟
⎜ b + c sin ( φ ) ⎟
⎝
⎠
⎛ −2.35 ⎞
r = ⎜ 3.93 ⎟ ft
⎜
⎟
⎝ 3.71 ⎠
r = 5.89 ft
82
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Engineering Mechanics - Statics
Chapter 2
⎛⎜ α ⎞⎟
⎛ r ⎞
⎜ β ⎟ = acos ⎜ ⎟
⎝ r ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 113.5 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 48.2 ⎟ deg
⎜ γ ⎟ ⎝ 51 ⎠
⎝ ⎠
Problem 2-84
Determine the length of the connecting rod AB by first formulating a Cartesian position vector
from A to B and then determining its magnitude.
Given:
b = 16 in
a = 5 in
α = 30 deg
Solution:
⎛ a sin ( α ) + b ⎞
⎟
⎝ −a cos ( α ) ⎠
r = ⎜
⎛ 18.5 ⎞
⎟ in
⎝ −4.3 ⎠
r=⎜
r = 19 in
Problem 2-85
Determine the length of member
AB of the truss by first
establishing a Cartesian position
vector from A to B and then
determining its magnitude.
Given:
a = 1.2 m
b = 0.8 m
c = 0.3 m
d = 1.5 m
θ = 40 deg
83
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Engineering Mechanics - Statics
Chapter 2
Solution:
⎛ c + d cot ( θ ) ⎞
⎟
⎝ d−a ⎠
⎛ 2.09 ⎞
⎟m
⎝ 0.3 ⎠
r = ⎜
r=⎜
r = 2.11 m
Problem 2- 86
The positions of point A on the building and point B on the antenna have been measured relative
to the electronic distance meter (EDM) at O. Determine the distance between A and B. Hint:
Formulate a position vector directed from A to B; then determine its magnitude.
Given:
a = 460 m
b = 653 m
α = 60 deg
β = 55 deg
θ = 30 deg
φ = 40 deg
Solution:
⎛⎜ −a cos ( φ ) sin ( θ ) ⎟⎞
rOA = ⎜ a cos ( φ ) cos ( θ ) ⎟
⎜
⎟
a sin ( φ )
⎝
⎠
⎛⎜ −b cos ( β ) sin ( α ) ⎟⎞
rOB = ⎜ −b cos ( β ) cos ( α ) ⎟
⎜
⎟
b sin ( β )
⎝
⎠
rAB = rOB − rOA
⎛ −148.2 ⎞
⎜
⎟
rAB = −492.4 m
⎜
⎟
⎝ 239.2 ⎠
rAB = 567.2 m
Problem 2-87
Determine the lengths of cords ACB and CO. The knot at C is located midway between A and B.
84
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Engineering Mechanics - Statics
Chapter 2
Given:
a = 3 ft
b = 6 ft
c = 4 ft
Solution:
⎛c ⎞
⎜ ⎟
rAB = b
⎜ ⎟
⎝ −a ⎠
rAC =
⎛0⎞
⎜ ⎟
rOA = 0
⎜ ⎟
⎝a⎠
rAB
2
rAB = 7.8 ft
rOC = rOA + rAC
rOC = 3.91 ft
Problem 2-88
Determine the length of the crankshaft AB by first formulating a Cartesian position vector from
A to B and then determining its magnitude.
85
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Engineering Mechanics - Statics
Chapter 2
Given:
a = 400
b = 125
θ = 25 deg
Solution:
⎡a + b sin ( θ ) ⎤
⎢
⎥
rAB = −( b cos ( θ ) ) mm
⎢
⎥
0
⎣
⎦
rAB = 467 mm
Problem 2-89
Determine the length of wires AD, BD, and CD. The ring at D is midway between A and B.
Given:
a = 0.5 m
b = 1.5 m
c = 2m
d = 2m
e = 0.5 m
86
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Engineering Mechanics - Statics
Chapter 2
Solution:
⎛ −c ⎞
⎜ 2 ⎟
⎜
⎟
d ⎟
⎜
rAD =
⎜ 2 ⎟
⎜e b⎟
⎜ − ⎟
⎝2 2⎠
⎛ −1 ⎞
⎜ 1 ⎟m
rAD =
⎜
⎟
⎝ −0.5 ⎠
rAD = 1.5 m
rBD = −rAD
rBD = 1.5 m
c
⎞
⎛
⎜
⎟
2
⎜
⎟
d
⎜
⎟
rCD =
2
⎜
⎟
⎜
b e⎟
⎜a + − ⎟
2 2⎠
⎝
⎛1⎞
⎜ ⎟
rCD = 1 m
⎜ ⎟
⎝1⎠
rCD = 1.7 m
Problem 2-90
Express force F as a Cartesian vector; then determine its coordinate direction angles.
Given:
F = 600 lb
c = 3 ft
a = 1.5 ft
φ = 60 deg
b = 5 ft
Solution:
r = bi + ( a + c sin ( φ ) ) j + ( 0 − c cos ( φ ) ) k
r =
b + ( a + c sin ( φ ) ) + ( c cos ( φ ) )
2
2
2
r=2m
87
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Engineering Mechanics - Statics
Chapter 2
⎡ −( c cos ( φ ) )⎤
⎥
r
⎣
⎦
b
d = F
r
⎛ a + c sin ( φ ) ⎞
e = F⎜
⎟
r
⎝
⎠
f = F⎢
d = 452 lb
e = 370 lb
f = −136 lb
F = ( di + ej + fk) lb
d
α = acos ⎛⎜ ⎞⎟
α = 41.1 deg
e
β = acos ⎛⎜ ⎞⎟
β = 51.9 deg
f
γ = acos ⎛⎜ ⎞⎟
γ = 103 deg
⎝F⎠
⎝ F⎠
⎝ F⎠
Problem 2-91
Express force F as a Cartesian vector; then
determine its coordinate direction angles.
Given:
a = 1.5 ft
b = 5 ft
c = 3 ft
θ = 60 deg
F = 600 lb
Solution:
b
⎛
⎞
⎜
⎟
r = a + c sin ( θ )
⎜
⎟
⎝ −c cos ( θ ) ⎠
F = F
⎛ 452 ⎞
⎜
⎟
F = 370 lb
⎜
⎟
⎝ −136 ⎠
r
r
⎛⎜ α ⎞⎟
⎛ F ⎞
⎜ β ⎟ = acos ⎜ ⎟
⎝ F ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 41.1 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 51.9 ⎟ deg
⎜ γ ⎟ ⎝ 103.1 ⎠
⎝ ⎠
88
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Engineering Mechanics - Statics
Chapter 2
Problem 2-92
Determine the magnitude and coordinate direction angles of the resultant force acting at point A.
Given:
F 1 = 150 N
F 2 = 200 N
a = 1.5 m
b = 4m
c = 3m
d = 2m
e = 3m
θ = 60 deg
Solution:
Define the position vectors and then the forces
⎛ e cos ( θ ) ⎞
rAB = ⎜ a + e sin ( θ ) ⎟
⎜
⎟
−b
⎝
⎠
⎛ c ⎞
rAC = ⎜ a − d ⎟
⎜
⎟
⎝ −b ⎠
F 1v = F1
F 2v = F2
⎛ 38 ⎞
F 1v = ⎜ 103.8 ⎟ N
⎜
⎟
⎝ −101.4 ⎠
rAB
rAB
⎛ 119.4 ⎞
F 2v = ⎜ −19.9 ⎟ N
⎜
⎟
⎝ −159.2 ⎠
rAC
rAC
Add the forces and find the magnitude of the resultant
F R = F1v + F2v
⎛ 157.4 ⎞
F R = ⎜ 83.9 ⎟ N
⎜
⎟
⎝ −260.6 ⎠
F R = 316 N
Find the direction cosine angles
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 60.1 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 74.6 ⎟ deg
⎜ γ ⎟ ⎝ 145.6 ⎠
⎝ ⎠
89
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Engineering Mechanics - Statics
Chapter 2
Problem 2-93
The plate is suspended using the three
cables which exert the forces shown.
Express each force as a Cartesian vector.
Given:
F BA = 350 lb
F CA = 500 lb
F DA = 400 lb
a = 3 ft
b = 3 ft
c = 6 ft
d = 14 ft
e = 3 ft
f = 3 ft
g = 2 ft
Solution:
⎛ −e − g ⎞
⎜
⎟
rBA = a + b
⎜
⎟
⎝ d ⎠
⎛f⎞
⎜ ⎟
rCA = b
⎜ ⎟
⎝d⎠
⎛ −g ⎞
⎜ ⎟
rDA = −c
⎜ ⎟
⎝d⎠
F BAv = FBA
F CAv = FCA
F DAv = FDA
rBA
rBA
rCA
rCA
rDA
rDA
⎛ −109.2 ⎞
⎜ 131 ⎟ lb
F BAv =
⎜
⎟
⎝ 305.7 ⎠
⎛ 102.5 ⎞
⎜
⎟
F CAv = 102.5 lb
⎜
⎟
⎝ 478.5 ⎠
⎛ −52.1 ⎞
⎜
⎟
F DAv = −156.2 lb
⎜
⎟
⎝ 364.5 ⎠
90
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Engineering Mechanics - Statics
Chapter 2
Problem 2-94
The engine of the lightweight plane
is supported by struts that are
connected to the space truss that
makes up the structure of the
plane. The anticipated loading in
two of the struts is shown.
Express each of these forces as a
Cartesian vector.
Given:
F 1 = 400 lb
F 2 = 600 lb
a = 0.5 ft
b = 0.5 ft
c = 3.0 ft
d = 2.0 ft
e = 0.5 ft
f = 3.0 ft
Solution:
⎛c ⎞
⎜ ⎟
rCD = −b
⎜ ⎟
⎝a⎠
⎛ −c ⎞
⎜ ⎟
rAB = b
⎜ ⎟
⎝ −e ⎠
F 1v = F1
F 2v = F2
rCD
rCD
rAB
rAB
⎛ 389.3 ⎞
⎜
⎟
F 1v = −64.9 lb
⎜
⎟
⎝ 64.9 ⎠
⎛ −584.0 ⎞
⎜ 97.3 ⎟ lb
F 2v =
⎜
⎟
⎝ −97.3 ⎠
Problem 2-95
The window is held open by cable AB. Determine the length of the cable and express the force
F acting at A along the cable as a Cartesian vector.
91
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Engineering Mechanics - Statics
Chapter 2
Given:
a = 300 mm
b = 500 mm
c = 150 mm
d = 250 mm
θ = 30 deg
F = 30 N
Solution:
⎛ −a cos ( θ ) ⎞
⎜ c−b ⎟
rAB =
⎜
⎟
⎝ d + a sin ( θ ) ⎠
Fv = F
rAB
rAB
rAB = 591.6 mm
⎛ −13.2 ⎞
⎜
⎟
F v = −17.7 N
⎜
⎟
⎝ 20.3 ⎠
Problem 2-96
The force acting on the man, caused by his pulling on the anchor cord, is F. If the length of
the cord is L, determine the coordinates A(x, y, z) of the anchor.
92
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Engineering Mechanics - Statics
Chapter 2
Given:
⎛ 40 ⎞
F = ⎜ 20 ⎟ N
⎜
⎟
⎝ −50 ⎠
L = 25 m
Solution:
r = L
F
F
⎛ 14.9 ⎞
r = ⎜ 7.5 ⎟ m
⎜
⎟
⎝ −18.6 ⎠
Problem 2-97
Express each of the forces in Cartesian vector form and determine the magnitude and coordinate
direction angles of the resultant force.
Given:
F 1 = 80 lb
c = 4 ft
93
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Engineering Mechanics - Statics
Chapter 2
F 2 = 50 lb
d = 2.5 ft
a = 6 ft
e = 12
b = 2 ft
f = 5
Solution:
⎛⎜ −d ⎞⎟
−c ⎟
rAC = ⎜
⎜ d⎟
⎜e f ⎟
⎝
⎠
⎛b⎞
rAB = ⎜ −c ⎟
⎜ ⎟
⎝ −a ⎠
F R = F1v + F2v
F 1v = F1
F 2v = F2
⎛ −26.2 ⎞
F 1v = ⎜ −41.9 ⎟ lb
⎜
⎟
⎝ 62.9 ⎠
rAC
rAC
⎛ 6.1 ⎞
F 2v = ⎜ −12.1 ⎟ kg
⎜
⎟
⎝ −18.2 ⎠
rAB
rAB
⎛ −12.8 ⎞
F R = ⎜ −68.7 ⎟ lb
⎜
⎟
⎝ 22.8 ⎠
F R = 73.5 lb
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 1.7 ⎞
⎜ ⎟
⎜ β ⎟ = ⎜ 2.8 ⎟
⎜ γ ⎟ ⎝ 1.3 ⎠
⎝ ⎠
Problem 2-98
The cable attached to the tractor at B exerts force F on the framework. Express this force as a
Cartesian vector
94
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 350 lb
a = 35 ft
b = 50 ft
θ = 20 deg
Solution:
Find the position vector and then the force
⎛ b sin ( θ ) ⎞
⎜
⎟
rAB = b cos ( θ )
⎜
⎟
⎝ −a ⎠
Fv = F
⎛ 98.1 ⎞
⎜
⎟
F v = 269.4 lb
⎜
⎟
⎝ −200.7 ⎠
rAB
rAB
Problem 2-99
The cable OA exerts force F on point O. If the length of the cable is L, what are the coordinates
(x, y, z) of point A?
Given:
⎛ 40 ⎞
⎜ ⎟
F = 60 N
⎜ ⎟
⎝ 70 ⎠
L = 3m
Solution:
r = L
F
F
95
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Engineering Mechanics - Statics
Chapter 2
⎛ 1.2 ⎞
r = ⎜ 1.8 ⎟ m
⎜ ⎟
⎝ 2.1 ⎠
Problem 2-100
Determine the position (x, y, 0) for fixing cable BA so that the resultant of the forces exerted on the
pole is directed along its axis, from B toward O, and has magnitude F R. Also, what is the magnitude
of force F 3?
Given:
F 1 = 500 N
F 2 = 400 N
F R = 1000 N
a = 1m
b = 2m
c = 2m
d = 3m
Solution:
Initial Guesses
F3 = 1 N
x = 1m
y = 1m
Given
⎛0⎞
⎛ −c ⎞
⎛a⎞
⎛x ⎞
F3
F2
⎛ F1 ⎞ ⎜ ⎟ ⎛
⎞⎜ ⎟ ⎛
⎞⎜ ⎟
⎜
⎟
FR 0 = ⎜
⎟ 0 ⎟ + ⎜ 2 2 2 ⎟ ⎜ −b ⎟ + ⎜ 2 2 2 ⎟ ⎜ y ⎟
⎜ ⎟
2
2 ⎜
c
+
d
⎠ ⎝ −d ⎠ ⎝ a + b + d ⎠ ⎝ −d ⎠ ⎝ x + y + d ⎠ ⎝ −d ⎠
⎝ −1 ⎠ ⎝
⎛⎜ F3 ⎞⎟
⎜ x ⎟ = Find ( F3 , x , y)
⎜ y ⎟
⎝ ⎠
⎛ x ⎞ ⎛ 1.9 ⎞
⎜ ⎟=⎜ ⎟m
⎝ y ⎠ ⎝ 2.4 ⎠
F 3 = 380 N
96
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Engineering Mechanics - Statics
Chapter 2
Problem 2-101
The cord exerts a force F on the hook. If the cord is length L, determine the location x, y of the
point of attachment B, and the height z of the hook.
Given:
⎛ 12 ⎞
F = ⎜ 9 ⎟ lb
⎜ ⎟
⎝ −8 ⎠
L = 8 ft
a = 2 ft
Solution:
Initial guesses
Given
x = 1 ft
⎛x − a⎞
⎜ y ⎟=L F
⎜
⎟
F
⎝ −z ⎠
y = 1 ft
z = 1 ft
⎛x⎞
⎜ y ⎟ = Find ( x , y , z)
⎜ ⎟
⎝z⎠
⎛ x ⎞ ⎛ 7.65 ⎞
⎜ y ⎟ = ⎜ 4.24 ⎟ ft
⎜ ⎟ ⎜
⎟
⎝ z ⎠ ⎝ 3.76 ⎠
Problem 2-102
The cord exerts a force of magnitude F on the hook. If the cord length L, the distance z, and the
x component of the force, Fx, are given, determine the location x, y of the point of attachment B
of the cord to the ground.
Given:
F = 30 lb
L = 8 ft
z = 4 ft
F x = 25 lb
a = 2 ft
Solution:
Guesses
x = 1 ft
97
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Engineering Mechanics - Statics
Chapter 2
y = 1 ft
Given
F x = ⎛⎜
x − a⎞
2
2
2
2
⎟F
⎝ L ⎠
L = ( x − a) + y + z
⎛x⎞
⎜ ⎟ = Find ( x , y)
⎝ y⎠
⎛ x ⎞ ⎛ 8.67 ⎞
⎜ ⎟=⎜
⎟ ft
⎝ y ⎠ ⎝ 1.89 ⎠
Problem 2-103
Each of the four forces acting at E has magnitude F. Express each force as a Cartesian vector
and determine the resultant force.
Units used:
3
kN = 10 N
Given:
F = 28 kN
a = 4m
b = 6m
c = 12 m
Solution:
Find the position vectors and
then the forces
⎛b⎞
rEA = ⎜ −a ⎟
⎜ ⎟
⎝ −c ⎠
F EA = F
rEA
rEA
⎛ 12 ⎞
F EA = ⎜ −8 ⎟ kN
⎜
⎟
⎝ −24 ⎠
98
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Engineering Mechanics - Statics
⎛b⎞
rEB = ⎜ a ⎟
⎜ ⎟
⎝ −c ⎠
⎛ −b ⎞
rEC = ⎜ a ⎟
⎜ ⎟
⎝ −c ⎠
⎛ −b ⎞
rED = ⎜ −a ⎟
⎜ ⎟
⎝ −c ⎠
F EB = F
F EC = F
F ED = F
Chapter 2
rEB
rEB
rEC
rEC
rED
rED
Find the resultant sum
F R = FEA + F EB + F EC + FED
⎛ 12 ⎞
F EB = ⎜ 8 ⎟ kN
⎜
⎟
⎝ −24 ⎠
⎛ −12 ⎞
F EC = ⎜ 8 ⎟ kN
⎜
⎟
⎝ −24 ⎠
⎛ −12 ⎞
F ED = ⎜ −8 ⎟ kN
⎜
⎟
⎝ −24 ⎠
⎛ 0 ⎞
F R = ⎜ 0 ⎟ kN
⎜
⎟
⎝ −96 ⎠
Problem 2-104
The tower is held in place by three cables. If the force of each cable acting on the tower is
shown, determine the magnitude and coordinate direction angles α, β, γ of the resultant force.
Units Used:
3
kN = 10 N
Given:
x = 20 m
a = 16 m
y = 15 m
b = 18 m
F 1 = 600 N
c = 6m
F 2 = 400 N
d = 4m
F 3 = 800 N
e = 24 m
99
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Engineering Mechanics - Statics
Chapter 2
Solution:
Find the position vectors, then the force vectors
⎛a⎞
rDC = ⎜ −b ⎟
⎜ ⎟
⎝ −e ⎠
⎛x⎞
rDA = ⎜ y ⎟
⎜ ⎟
⎝ −e ⎠
⎛ −c ⎞
rDB = ⎜ d ⎟
⎜ ⎟
⎝ −e ⎠
F 1v = F1
F 2v = F2
F 3v = F3
⎛ 282.4 ⎞
F 1v = ⎜ −317.6 ⎟ N
⎜
⎟
⎝ −423.5 ⎠
rDC
rDC
⎛ 230.8 ⎞
F 2v = ⎜ 173.1 ⎟ N
⎜
⎟
⎝ −277 ⎠
rDA
rDA
⎛ −191.5 ⎞
F 3v = ⎜ 127.7 ⎟ N
⎜
⎟
⎝ −766.2 ⎠
rDB
rDB
Find the resultant, magnitude, and direction angles
F R = F1v + F2v + F3v
⎛ 0.322 ⎞
F R = ⎜ −0.017 ⎟ kN
⎜
⎟
⎝ −1.467 ⎠
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 77.6 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 90.6 ⎟ deg
⎜ γ ⎟ ⎝ 167.6 ⎠
⎝ ⎠
F R = 1.502 kN
Problem 2-105
The chandelier is supported by three chains which are concurrent at point O. If the force in each
chain has magnitude F, express each force as a Cartesian vector and determine the magnitude and
coordinate direction angles of the resultant force.
100
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 60 lb
a = 6 ft
b = 4 ft
θ 1 = 120 deg
θ 2 = 120 deg
Solution:
θ 3 = 360 deg − θ 1 − θ 2
⎛ b sin ( θ 1 ) ⎞
⎜
⎟
rOA = ⎜ b cos ( θ 1 ) ⎟
⎜
⎟
⎝ −a ⎠
⎛ b sin ( θ 1 + θ 2 ) ⎞
⎜
⎟
rOB = ⎜ b cos ( θ 1 + θ 2 ) ⎟
⎜
⎟
−a
⎝
⎠
⎛0⎞
⎜ ⎟
rOC = b
⎜ ⎟
⎝ −a ⎠
FA = F
FB = F
FC = F
rOA
rOA
rOB
rOB
rOC
rOC
⎛ 28.8 ⎞
⎜
⎟
F A = −16.6 lb
⎜
⎟
⎝ −49.9 ⎠
⎛ −28.8 ⎞
⎜
⎟
F B = −16.6 lb
⎜
⎟
⎝ −49.9 ⎠
⎛ 0 ⎞
⎜
⎟
F C = 33.3 lb
⎜
⎟
⎝ −49.9 ⎠
FR = FA + FB + FC
F R = 149.8 lb
⎛⎜ α ⎞⎟
⎛ FR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ FR ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 90 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 90 ⎟ deg
⎜ γ ⎟ ⎝ 180 ⎠
⎝ ⎠
101
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Engineering Mechanics - Statics
Chapter 2
Problem 2-106
The chandelier is supported by three chains which are concurrent at point O. If the resultant
force at O has magnitude F R and is directed along the negative z axis, determine the force in
each chain assuming F A = F B = F C = F.
Given:
a = 6 ft
b = 4 ft
F R = 130 lb
Solution:
2
2
a +b
FR
3a
F =
F = 52.1 lb
Problem 2-107
Given the three vectors A, B, and D, show that A⋅ ( B + D) = ( A⋅ B) + ( A⋅ D).
Solution:
Since the component of (B + D) is equal to the sum of the components of B and D, then
A⋅ ( B + D) = A⋅ B + A⋅ D
(QED)
Also,
A⋅ ( B + D) = ( A xi + A yj + Azk) ⎡⎣( B x + Dx) i + ( By + Dy) j + ( Bz + Dz) k⎤⎦
=
Ax ( B x + Dx) + Ay ( B y + Dy) + Az( B z + Dz)
102
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Engineering Mechanics - Statics
Chapter 2
=
( Ax Bx + Ay By + Az Bz) + ( Ax Dx + Ay Dy + Az Dz)
=
( A⋅ B) + ( A⋅ D)
(QED)
Problem 2-108
Cable BC exerts force F on the top of the flagpole. Determine the projection of this force along
the z axis of the pole.
Given:
F = 28 N
a = 12 m
b = 6m
c = 4m
Solution:
⎛b⎞
⎜ ⎟
rBC = −c
⎜ ⎟
⎝ −a ⎠
Fv = F
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
rBC
rBC
F z = −Fv k
F z = 24 N
Problem 2-109
Determine the angle θ between the tails of the two vectors.
103
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Engineering Mechanics - Statics
Chapter 2
Given:
r1 = 9 m
r2 = 6 m
α = 60 deg
β = 45 deg
γ = 120 deg
φ = 30 deg
ε = 40 deg
Solution:
Determine the two position vectors and use the dot
product to find the angle
⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞
r1v = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟
⎜ cos ( ε ) ⎟
⎝
⎠
⎛ r1v⋅ r2v ⎞
⎟
⎝ r1v r2v ⎠
θ = acos ⎜
⎛⎜ cos ( α ) ⎟⎞
r2v = r2 ⎜ cos ( β ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
θ = 109.4 deg
Problem 2-110
Determine the magnitude of the projected component of r1 along r2, and the projection of r2
along r1.
Given:
r1 = 9 m
r2 = 6 m
α = 60 deg
β = 45 deg
γ = 120 deg
φ = 30 deg
ε = 40 deg
104
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Engineering Mechanics - Statics
Chapter 2
Solution:
Write the vectors and unit vectors
⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞
r1v = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟
⎜ cos ( ε ) ⎟
⎝
⎠
⎛ 5.01 ⎞
r1v = ⎜ −2.89 ⎟ m
⎜
⎟
⎝ 6.89 ⎠
⎛⎜ cos ( α ) ⎟⎞
r2v = r2 ⎜ cos ( β ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
⎛ 3 ⎞
r2v = ⎜ 4.24 ⎟ m
⎜
⎟
⎝ −3 ⎠
u1 =
r1v
r1v
u2 =
r2v
r2v
⎛ 0.557 ⎞
u1 = ⎜ −0.321 ⎟
⎜
⎟
⎝ 0.766 ⎠
⎛ 0.5 ⎞
u2 = ⎜ 0.707 ⎟
⎜
⎟
⎝ −0.5 ⎠
The magnitude of the projection of r1 along r2.
r1v⋅ u2 = 2.99 m
The magnitude of the projection of r2 along r1.
r2v⋅ u1 = 1.99 m
Problem 2-111
Determine the angles θ and φ between the wire segments.
Given:
a = 0.6
b = 0.8
c = 0.5
d = 0.2
Solution:
e = a−d
105
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Engineering Mechanics - Statics
Chapter 2
⎛0⎞
rBA = ⎜ −e ⎟ m
⎜ ⎟
⎝ −c ⎠
⎛b⎞
rBC = ⎜ d ⎟ ft
⎜ ⎟
⎝ −c ⎠
θ = acos ⎜
⎛ −b ⎞
rCB = ⎜ −d ⎟ m
⎜ ⎟
⎝c ⎠
⎛ −b ⎞
rCD = ⎜ 0 ⎟ ft
⎜ ⎟
⎝0⎠
φ = acos ⎜
⎛ rBA⋅ rBC ⎞
⎟
⎝ rBA rBC ⎠
θ = 74.0 deg
⎛ rCB⋅ rCD ⎞
⎟
⎝ rCB rCD ⎠
φ = 33.9 deg
⎛ rAC⋅ rAB ⎞
⎟
⎝ rAC rAB ⎠
θ = 64.6 deg
Problem 2-112
Determine the angle θ between the two cords.
Given:
a = 3m
b = 2m
c = 6m
d = 3m
e = 4m
Solution:
⎛b⎞
rAC = ⎜ a ⎟ ft
⎜ ⎟
⎝c⎠
⎛0⎞
rAB = ⎜ −d ⎟ ft
⎜ ⎟
⎝e ⎠
θ = acos ⎜
Problem 2-113
Determine the angle θ between the two cables.
Given:
a = 8 ft
b = 10 ft
106
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Engineering Mechanics - Statics
Chapter 2
c = 8 ft
d = 10 ft
e = 4 ft
f = 6 ft
F AB = 12 lb
Solution:
⎛a − f⎞
rAC = ⎜ −c ⎟ ft
⎜
⎟
⎝ b ⎠
⎛ −f ⎞
rAB = ⎜ d − c ⎟ ft
⎜
⎟
⎝ e ⎠
⎛ rAC⋅ rAB ⎞
⎟
⎝ rAC rAB ⎠
θ = acos ⎜
θ = 82.9 deg
Problem 2-114
Determine the projected component of the force F acting in the direction of cable AC. Express
the result as a Cartesian vector.
Given:
F = 12 lb
107
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Engineering Mechanics - Statics
Chapter 2
a = 8 ft
b = 10 ft
c = 8 ft
d = 10 ft
e = 4 ft
f = 6 ft
Solution:
⎛a − f⎞
⎜ −c ⎟ m
rAC =
⎜
⎟
⎝ b ⎠
uAC =
rAC
rAC
⎛ −f ⎞
⎜
⎟
rAB = d − c
⎜
⎟
⎝ e ⎠
F AB = F
F AC = ( F AB⋅ uAC) uAC
⎛ 0.229 ⎞
⎜
⎟
F AC = −0.916 lb
⎜
⎟
⎝ 1.145 ⎠
rAB
rAB
⎛ 0.2 ⎞
⎜
⎟
uAC = −0.6
⎜
⎟
⎝ 0.8 ⎠
⎛ −9.6 ⎞
⎜
⎟
F AB = 3.2 lb
⎜
⎟
⎝ 6.4 ⎠
Problem 2-115
Determine the components of F that act along rod AC and perpendicular to it. Point B is located
at the midpoint of the rod.
108
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 600 N c = 4 m
a = 4m
d = 3m
b = 6m
e = 4m
Solution:
Find the force vector and the unit vector uAC.
⎛c + d ⎞
⎜
2⎟
⎜
⎟
e⎟
⎜
rBD = b −
⎜
2⎟
⎜ −a ⎟
⎜
⎟
⎝ 2 ⎠
⎛ 5.5 ⎞
⎜ ⎟m
rBD = 4
⎜ ⎟
⎝ −2 ⎠
⎛ −d ⎞
⎜ ⎟
rAC = e
⎜ ⎟
⎝ −a ⎠
⎛ −3 ⎞
⎜ ⎟
rAC = 4 m
⎜ ⎟
⎝ −4 ⎠
Fv = F
uAC =
rBD
rBD
rAC
rAC
⎛ 465.5 ⎞
⎜
⎟
F v = 338.6 N
⎜
⎟
⎝ −169.3 ⎠
⎛ −0.5 ⎞
⎜
⎟
uAC = 0.6
⎜
⎟
⎝ −0.6 ⎠
Now find the component parallel to AC.
F parallel = Fv ⋅ uAC
F parallel = 99.1 N
The perpendicular component is now found
F perpendicular =
2
Fv ⋅ F v − F parallel
F perpendicular = 591.8 N
Problem 2-116
Determine the components of F that act along rod AC and perpendicular to it. Point B is located
a distance f along the rod from end C.
109
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 600 N
c = 4m
a = 4m
d = 3m
b = 6m
e = 4m
f = 3m
Solution:
f
r =
2
2
d +e +a
2
Find the force vector and the unit vector uAC.
⎡c + d( 1 − r) ⎤
⎢
⎥
rBD = b − e( 1 − r)
⎢
⎥
⎣ −a r ⎦
Fv = F
⎛ 5.5944 ⎞
⎜
⎟
rBD = 3.8741 m
⎜
⎟
⎝ −1.8741 ⎠
⎛ 475.6 ⎞
⎜
⎟
F v = 329.3 N
⎜
⎟
⎝ −159.3 ⎠
rBD
rBD
⎛ −d ⎞
⎜ ⎟
rAC = e
⎜ ⎟
⎝ −a ⎠
⎛ −3 ⎞
⎜ ⎟
rAC = 4 m
⎜ ⎟
⎝ −4 ⎠
uAC =
rAC
rAC
⎛ −0.5 ⎞
⎜
⎟
uAC = 0.6
⎜
⎟
⎝ −0.6 ⎠
Now find the component parallel to AC.
F parallel = Fv ⋅ uAC
F parallel = 82.4 N
The perpendicular component is now found
F perpendicular =
2
Fv ⋅ F v − F parallel
F perpendicular = 594.3 N
Problem 2-117
Determine the magnitude of the projected component of the length of cord OA along the Oa axis.
110
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Engineering Mechanics - Statics
Chapter 2
Given:
a = 10 ft
b = 5 ft
c = 15 ft
d = 5 ft
θ 1 = 45 deg
θ 2 = 60 deg
Solution:
⎛ cos ( θ 1) cos ( θ 2 ) ⎞
⎜
⎟
rOA = d ⎜ cos ( θ 1 ) sin ( θ 2 ) ⎟
⎜
⎟
sin ( θ 1 )
⎝
⎠
rOa = rOA ⋅ uOa
⎛c ⎞
rOa = ⎜ a ⎟
⎜ ⎟
⎝ −b ⎠
uOa =
rOa
rOa
rOa = 2.1 ft
111
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Engineering Mechanics - Statics
Chapter 2
Problem 2-118
Force F acts at the end of the pipe.
Determine the magnitudes of the
components F1 and F 2 which are directed
along the pipe's axis and perpendicular to
it.
Given:
a = 5 ft
⎛ 0 ⎞
⎜ 0 ⎟ lb
F =
⎜
⎟
⎝ −40 ⎠
b = 3 ft
c = 3 ft
Solution:
⎛b⎞
⎜ ⎟
r = a
⎜ ⎟
⎝ −c ⎠
u =
F1 = F⋅ u
F2 =
F⋅ F − F1
r
r
F 1 = 18.3 lb
2
F 2 = 35.6 lb
Problem 2-119
Determine the projected component of the force F acting along the axis AB of the pipe.
112
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 80 N
a = 4m
b = 3m
c = 12 m
d = 2m
e = 6m
Solution:
Find the force and the unit vector
⎛ −e ⎞
⎜
⎟
rA = −a − b
⎜
⎟
⎝ d−c ⎠
⎛ −6 ⎞
⎜
⎟
rA = −7 m
⎜
⎟
⎝ −10 ⎠
⎛ −e ⎞
⎜ ⎟
rAB = −b
⎜ ⎟
⎝d⎠
⎛ −6 ⎞
⎜ ⎟
rAB = −3 m
⎜ ⎟
⎝2⎠
Fv = F
uAB =
rA
rA
rAB
rAB
⎛ −35.3 ⎞
⎜
⎟
F v = −41.2 N
⎜
⎟
⎝ −58.8 ⎠
⎛ −0.9 ⎞
⎜
⎟
uAB = −0.4
⎜
⎟
⎝ 0.3 ⎠
Now find the projection using the Dot product.
F AB = F v ⋅ uAB
F AB = 31.1 N
Problem 2-120
Determine the angles θ and φ between the axis OA of the pole and each cable, AB and AC.
Given:
F 1 = 50 N
113
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Engineering Mechanics - Statics
Chapter 2
F 2 = 35 N
a = 1m
b = 3m
c = 2m
d = 5m
e = 4m
f = 6m
g = 4m
Solution:
⎛0⎞
⎜ ⎟
rAO = −g
⎜ ⎟
⎝−f ⎠
⎛e⎞
⎜ ⎟
rAB = a
⎜ ⎟
⎝−f⎠
⎛ −c ⎞
⎜
⎟
rAC = a + b
⎜
⎟
⎝ −f ⎠
⎛ rAO⋅ rAB ⎞
⎟
⎝ rAO rAB ⎠
θ = 52.4 deg
⎛ rAO⋅ rAC ⎞
⎟
⎝ rAO rAC ⎠
φ = 68.2 deg
θ = acos ⎜
φ = acos ⎜
Problem 2-121
The two cables exert the forces shown on
the pole. Determine the magnitude of the
projected component of each force acting
along the axis OA of the pole.
Given:
F 1 = 50 N
F 2 = 35 N
a = 1m
b = 3m
c = 2m
114
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Engineering Mechanics - Statics
Chapter 2
d = 5m
e = 4m
f = 6m
g = 4m
Solution:
⎛e⎞
⎜ ⎟
rAB = a
⎜ ⎟
⎝−f⎠
F 1v = F1
F 2v = F2
⎛ −c ⎞
⎜
⎟
rAC = a + b
⎜
⎟
⎝ −f ⎠
rAC
rAC
rAB
rAB
⎛0⎞
⎜ ⎟
rAO = −g
⎜ ⎟
⎝−f ⎠
F 1AO = F 1v⋅ uAO
F 1AO = 18.5 N
F 2AO = F 2v⋅ uAO
F 2AO = 21.3 N
uAO =
rAO
rAO
Problem 2-122
Force F is applied to the handle of the wrench. Determine the angle θ between the tail of the force
and the handle AB.
115
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Engineering Mechanics - Statics
Chapter 2
Given:
a = 300 mm
b = 500 mm
F = 80 N
θ 1 = 30 deg
θ 2 = 45 deg
Solution:
⎛ −cos ( θ 1) sin ( θ 2) ⎞
⎜
⎟
F v = F ⎜ cos ( θ 1 ) cos ( θ 2 ) ⎟
⎜
⎟
sin ( θ 1 )
⎝
⎠
⎛0⎞
⎜ ⎟
uab = −1
⎜ ⎟
⎝0⎠
⎛ Fv⋅ uab ⎞
⎟
⎝ F ⎠
θ = 127.8 deg
θ = acos ⎜
Problem 2-123
Two cables exert forces on the pipe. Determine the magnitude of the projected component of
F 1 along the line of action of F2.
Given:
F 1 = 30 lb
β = 30 deg
F 2 = 25 lb
γ = 60 deg
α = 30 deg
ε = 60 deg
Solution:
We first need to find the third
angle ( > 90 deg) that locates
force F 2.
Initial Guess:
φ = 120 deg
116
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Engineering Mechanics - Statics
Chapter 2
Given
cos ( ε ) + cos ( γ ) + cos ( φ ) = 1
2
2
φ = Find ( φ )
2
φ = 135 deg
Find the force F 1v and the unit vector u2.
⎛⎜ cos ( α ) sin ( β ) ⎞⎟
F 1v = F1 ⎜ cos ( α ) cos ( β ) ⎟
⎜ −sin ( α ) ⎟
⎝
⎠
⎛ 13 ⎞
F 1v = ⎜ 22.5 ⎟ lb
⎜
⎟
⎝ −15 ⎠
⎛⎜ cos ( φ ) ⎞⎟
u2 = ⎜ cos ( ε ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
⎛ −0.7 ⎞
u2 = ⎜ 0.5 ⎟
⎜
⎟
⎝ 0.5 ⎠
Now find the projection
F 12 = F1v⋅ u2
F 12 = 5.4 lb
Problem 2-124
Determine the angle θ between the two cables attached to the pipe.
Given:
F 1 = 30 lb
β = 30 deg
F 2 = 25 lb
γ = 60 deg
α = 30 deg
ε = 60 deg
Solution:
We first need to find the third
angle ( > 90 deg) that locates
force F 2.
Initial Guesses:
φ = 120 deg
117
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Engineering Mechanics - Statics
Chapter 2
Given
cos ( ε ) + cos ( γ ) + cos ( φ ) = 1
2
2
2
φ = Find ( φ )
φ = 135 deg
Find the unit vectors u1 and u2.
⎛⎜ cos ( α ) sin ( β ) ⎞⎟
u1 = ⎜ cos ( α ) cos ( β ) ⎟
⎜ −sin ( α ) ⎟
⎝
⎠
⎛ 0.4 ⎞
u1 = ⎜ 0.8 ⎟
⎜
⎟
⎝ −0.5 ⎠
⎛⎜ cos ( φ ) ⎞⎟
u2 = ⎜ cos ( ε ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
⎛ −0.7 ⎞
u2 = ⎜ 0.5 ⎟
⎜
⎟
⎝ 0.5 ⎠
Find the angle using the dot product
θ = acos ( u1 ⋅ u2 )
θ = 100.4 deg
Problem 2-125
Determine the angle θ between the two cables.
Given:
a = 7.5 ft
b = 2 ft
c = 3 ft
d = 2 ft
e = 3 ft
f = 3 ft
F 1 = 60 lb
F 2 = 30 lb
118
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Engineering Mechanics - Statics
Chapter 2
Solution:
⎛ −d − b ⎞
rAC = ⎜ −c − f ⎟
⎜
⎟
⎝ a−e ⎠
⎛ −d ⎞
rAB = ⎜ −c − f ⎟
⎜
⎟
⎝ −e ⎠
⎛ rAC⋅ rAB ⎞
⎟
⎝ rAC rAB ⎠
θ = acos ⎜
θ = 59.2 deg
Problem 2-126
Determine the projection of the
force F 1 along cable AB. Determine
the projection of the force F 2 along
cable AC.
Given:
a = 7.5 ft
b = 2 ft
c = 3 ft
d = 2 ft
e = 3 ft
f = 3 ft
F 1 = 60 lb
F 2 = 30 lb
Solution:
⎛ −d − b ⎞
rAC = ⎜ −c − f ⎟
⎜
⎟
⎝ a−e ⎠
⎛ −d ⎞
rAB = ⎜ −c − f ⎟
⎜
⎟
⎝ −e ⎠
F 1v = F1 uAC
F 1AB = F1v⋅ uAB
F 1AB = 30.8 lb
F 2v = F2 uAB
F 2AC = F2v⋅ uAC
F 2AC = 15.4 lb
uAC =
rAC
rAC
uAB =
rAB
rAB
119
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Engineering Mechanics - Statics
Chapter 2
Problem 2-127
Determine the angle θ between the edges of
the sheet-metal bracket.
Given:
a = 50 mm
b = 300 mm
c = 250 mm
d = 400 mm
Solution:
Find the unit vectors and use the dot
product
⎛d⎞
⎜ ⎟
r1 = 0
⎜ ⎟
⎝c⎠
u1 =
⎛a⎞
⎜ ⎟
r2 = b
⎜ ⎟
⎝0⎠
u2 =
r1
r1
r2
r2
⎛ 0.848 ⎞
⎛ 0.164 ⎞
⎜
⎟
⎜
⎟
u1 = 0.000
u2 = 0.986
⎜
⎟
⎜
⎟
⎝ 0.530 ⎠
⎝ 0.000 ⎠
θ = acos ( u1 ⋅ u2 )
θ = 82 deg
Problem 2-128
Determine the magnitude of the projected component of the force F acting along the axis BC
of the pipe.
120
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 100 lb
a = 2 ft
b = 8 ft
c = 6 ft
d = 4 ft
e = 2 ft
Solution:
⎛ −c ⎞
⎜ ⎟
rCD = b
⎜ ⎟
⎝e⎠
uCD =
F BC = ( FuCD) ⋅ uCB
⎛ −c ⎞
⎜ ⎟
rCB = −d
⎜ ⎟
⎝e ⎠
rCD
rCD
uCB =
rCB
rCB
F BC = 10.5 lb
Problem 2-129
Determine the angle θ between pipe segments BA and BC.
Given:
F = 100 lb
a = 3 ft
b = 8 ft
c = 6 ft
d = 4 ft
e = 2 ft
Solution:
⎛c⎞
⎜ ⎟
rBC = d
⎜ ⎟
⎝ −e ⎠
⎛ −a ⎞
⎜ ⎟
rBA = 0
⎜ ⎟
⎝0⎠
⎛ rBC⋅ rBA ⎞
⎟
⎝ rBC rBA ⎠
θ = acos ⎜
θ = 143.3 deg
121
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Engineering Mechanics - Statics
Chapter 2
Problem 2-130
Determine the angles θ and φ made between the axes OA of the flag pole and AB and AC,
respectively, of each cable.
Given:
F B = 55 N
c = 2m
F c = 40 N
d = 4m
a = 6m
e = 4m
b = 1.5 m
f = 3m
Solution:
⎛0⎞
⎜ ⎟
rAO = −e
⎜ ⎟
⎝−f⎠
⎛ b ⎞
⎜ −e ⎟
rAB =
⎜
⎟
⎝a − f⎠
⎛ −c ⎞
⎜ −e ⎟
rAC =
⎜
⎟
⎝d − f⎠
⎛ rAB⋅ rAO ⎞
⎟
⎝ rAB rAO ⎠
θ = 74.4 deg
⎛ rAC⋅ rAO ⎞
⎟
⎝ rAC rAO ⎠
φ = 55.4 deg
θ = acos ⎜
φ = acos ⎜
Problem 2-131
Determine the magnitude and coordinate direction angles of F3 so that resultant of the three
forces acts along the positive y axis and has magnitude FR.
122
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Engineering Mechanics - Statics
Chapter 2
Given:
F R = 600 lb
F 1 = 180 lb
F 2 = 300 lb
φ = 40 deg
θ = 30 deg
Solution:
The initial guesses:
F 3 = 100 lb
β = 30 deg
α = 10 deg
γ = 60 deg
Given
F Rx = ΣF x;
−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0
F Ry = ΣF y;
F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = FR
F Rz = ΣF z;
−F 2 sin ( θ ) + F3 cos ( γ ) = 0
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
2
Solving:
⎛ F3 ⎞
⎜ ⎟
⎜ α ⎟ = Find ( F , α , β , γ )
3
⎜β ⎟
⎜ ⎟
⎝γ ⎠
⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 20.6 ⎟ deg
⎜ γ ⎟ ⎝ 69.5 ⎠
⎝ ⎠
F 3 = 428.3 lb
123
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Engineering Mechanics - Statics
Chapter 2
Problem 2-132
Determine the magnitude and coordinate direction angles of F3 so that resultant of the three
forces is zero.
Given:
F 1 = 180 lb
F 2 = 300 lb
φ = 40 deg
θ = 30 deg
Solution:
The initial guesses:
α = 10 deg
β = 30 deg
γ = 60 deg
F 3 = 100 lb
Given
F Rx = ΣF x;
−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0
F Ry = ΣF y;
F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0
F Rz = ΣF z;
−F 2 sin ( θ ) + F3 cos ( γ ) = 0
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
2
Solving:
⎛ F3 ⎞
⎜ ⎟
⎜ α ⎟ = Find ( F , α , β , γ )
3
⎜β ⎟
⎜ ⎟
⎝γ ⎠
⎛⎜ α ⎞⎟ ⎛ 87 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 142.9 ⎟ deg
⎜ γ ⎟ ⎝ 53.1 ⎠
⎝ ⎠
F 3 = 249.6 lb
124
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Engineering Mechanics - Statics
Chapter 2
Problem 2-133
Resolve the force F into two components, one acting parallel and the other acting perpendicular
to the u axis.
Given:
F = 600 lb
θ 1 = 60 deg
θ 2 = 20 deg
Solution:
F perpendicular = F cos ( θ 1 − θ 2 )
F perpendicular = 460 lb
F parallel = F sin ( θ 1 − θ 2 )
F parallel = 386 lb
Problem 2-134
The force F has a magnitude F and acts at the midpoint C of the thin rod. Express the force as
a Cartesian vector.
125
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Engineering Mechanics - Statics
Chapter 2
Given:
F = 80 lb
a = 2 ft
b = 3 ft
c = 6 ft
Solution:
⎛ −b ⎞
⎜ 2 ⎟
⎜ ⎟
a ⎟
rCO = ⎜
⎜ 2 ⎟
⎜ −c ⎟
⎜ ⎟
⎝ 2 ⎠
Fv = F
rCO
rCO
⎛ −34.3 ⎞
⎜
⎟
F v = 22.9 lb
⎜
⎟
⎝ −68.6 ⎠
Problem 2-135
Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by
first finding the resultant F' = F1 + F 3 and then forming FR = F ' + F 2. Specify its direction
measured counterclockwise from the positive x axis.
126
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Engineering Mechanics - Statics
Chapter 2
Given:
F 1 = 80 N
F 2 = 75 N
F 3 = 50 N
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 45 deg
Solution:
⎛ −sin ( θ 1 ) ⎞
⎟
⎝ cos ( θ 1) ⎠
F 1v = F1 ⎜
⎛ cos ( θ 2 + θ 3 ) ⎞
⎟
⎝ sin ( θ 2 + θ 3 ) ⎠
⎛ cos ( θ 3) ⎞
⎟
⎝ sin ( θ 3) ⎠
F 2v = F2 ⎜
F 3v = F3 ⎜
⎛ −4.6 ⎞
⎟N
⎝ 104.6 ⎠
i = ⎜
⎛ 14.8 ⎞
⎟N
⎝ 177.1 ⎠
F R = 177.7 N
F' = F1v + F3v
F' = ⎜
F R = F' + F2v
FR = ⎜
⎛1⎞
⎟
⎝0⎠
⎛ FR j ⎞
⎟
⎝ FR i ⎠
θ = atan ⎜
⎛0⎞
⎟
⎝1⎠
j = ⎜
θ = 85.2 deg
Problem 2-136
The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If the force
exerted on this muscle by the pelvis is F , in the direction shown, determine the stabilizing force
component acting along the positive y axis and the supporting force component acting along the
negative x axis.
Given:
F = 85 N
127
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Engineering Mechanics - Statics
Chapter 2
θ 1 = 55 deg
θ 2 = 45 deg
Solution:
F x = F cos ( θ 1 − θ 2 )
F x = 83.7 N
F y = F sin ( θ 1 − θ 2 )
F y = 14.8 N
Problem 2-137
Determine the magnitudes of the projected components of the force F in the direction of the
cables AB and AC .
Given:
⎛ 60 ⎞
⎜
⎟
F = 12 N
⎜
⎟
⎝ −40 ⎠
a = 3m
b = 1.5 m
c = 1m
d = 0.75 m
e = 1m
128
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Engineering Mechanics - Statics
Chapter 2
Solution:
Find the unit vectors, then use the dot product
⎛ −a ⎞
⎜ ⎟
rAB = −d
⎜ ⎟
⎝e ⎠
⎛ −a ⎞
⎜ ⎟
rAC = c
⎜ ⎟
⎝b⎠
⎛ −3 ⎞
⎜
⎟
rAB = −0.8 m
⎜
⎟
⎝ 1 ⎠
⎛ −3 ⎞
⎜ ⎟
rAC = 1 m
⎜ ⎟
⎝ 1.5 ⎠
⎛ −78.5 ⎞
⎜
⎟
F AB = −19.6 N
⎜
⎟
⎝ 26.2 ⎠
F AB = FuAB
uAB =
uAC =
rAB
rAB
rAC
rAC
F AC = FuAC
⎛ −0.9 ⎞
⎜
⎟
uAB = −0.2
⎜
⎟
⎝ 0.3 ⎠
⎛ −0.9 ⎞
⎜
⎟
uAC = 0.3
⎜
⎟
⎝ 0.4 ⎠
⎛ −72.9 ⎞
⎜
⎟
F AC = 24.3 N
⎜
⎟
⎝ 36.4 ⎠
Problem 2-138
Determine the magnitude and
coordinate direction angles of F3 so
that resultant of the three forces is
zero.
Given:
F 1 = 180 lb
φ = 40 deg
F 2 = 300 lb
θ = 30 deg
Solution:
The initial guesses:
α = 10 deg
β = 30 deg
γ = 60 deg
F 3 = 100 lb
Given
F Rx = ΣFx;
−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0
F Ry = ΣF y;
F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0
129
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Engineering Mechanics - Statics
Chapter 2
−F 2 sin ( θ ) + F3 cos ( γ ) = 0
F Rz = ΣF z;
cos ( α ) + cos ( β ) + cos ( γ ) = 1
2
2
2
Solving:
⎛ F3 ⎞
⎜ ⎟
⎜ α ⎟ = Find ( F , α , β , γ )
3
⎜β ⎟
⎜ ⎟
⎝γ ⎠
⎛⎜ α ⎞⎟ ⎛ 87 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 142.9 ⎟ deg
⎜ γ ⎟ ⎝ 53.1 ⎠
⎝ ⎠
F 3 = 249.6 lb
Problem 2-139
Determine the angles θ and φ so that the resultant force is directed along the positive x axis
and has magnitude FR. .
Given:
F 1 = 30 lb
F 2 = 30 lb
F R = 20 lb
Solution:
Initial Guesses:
θ = 20 deg
φ = 20 deg
Given
F1
sin ( φ )
=
F2
sin ( θ )
F R = F 1 + F2 − 2 F 1 F2 cos ( 180 deg − θ − φ )
2
2
2
130
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Engineering Mechanics - Statics
Chapter 2
⎛θ⎞
⎜ ⎟ = Find ( θ , φ )
⎝φ⎠
θ = 70.5 deg
φ = 70.5 deg
Problem 2-140
Determine the magnitude of the resultant force and its direction measured counterclockwise
from the x axis.
Given:
F 1 = 300 lb
F 2 = 200 lb
θ 1 = 40 deg
θ 2 = 100 deg
Solution:
F Rx = F 1 cos ( 180 deg − θ 2 ) + F2 cos ( θ 1 )
F Rx = 205.3 lb
F Ry = F 1 sin ( 180 deg − θ 2 ) − F 2 sin ( θ 1 )
F Ry = 166.9 lb
FR =
2
FRx + FRy
2
F R = 265 lb
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = atan ⎜
θ = 39.1 deg
131
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Engineering Mechanics - Statics
Chapter 3
Problem 3-1
Determine the magnitudes of F1 and F2 so that the particle is in equilibrium.
Given:
F = 500 N
θ 1 = 45 deg
θ 2 = 30deg
Solution:
Initial Guesses
F 1 = 1N
F 2 = 1N
Given
+
ΣF x = 0;
→
F 1 cos ( θ 1 ) + F2 cos ( θ 2 ) − F = 0
+
F 1 sin ( θ 1 ) − F 2 sin ( θ 2 ) = 0
↑ ΣFy = 0;
⎛ F1 ⎞
⎜ ⎟ = Find ( F1 , F2)
⎝ F2 ⎠
⎛ F1 ⎞ ⎛ 259 ⎞
⎜ ⎟=⎜
⎟N
⎝ F2 ⎠ ⎝ 366 ⎠
Problem 3-2
Determine the magnitude and direction θ of F so that
the particle is in equilibrium.
Units Used:
3
kN = 10 N
Given:
F 1 = 7 kN
F 2 = 3 kN
c = 4
d = 3
132
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Engineering Mechanics - Statics
Chapter 3
Solution:
F = 1kN
The initial guesses:
θ = 30deg
Given
Equations of equilibrium:
+
Σ F x = 0;
→
+
↑
Σ F y = 0;
⎛ −d ⎞ F + F cos ( θ ) = 0
⎜ 2 2⎟ 1
⎝ c +d ⎠
c
⎛
⎞
⎜ 2 2 ⎟ F1 − F2 − F sin ( θ ) = 0
⎝ c +d ⎠
⎛F⎞
⎜ ⎟ = Find ( F , θ )
⎝θ ⎠
F = 4.94 kN
θ = 31.8 deg
Problem 3-3
Determine the magnitude of F and the orientation θ of the force F 3 so that the particle is in equilibrium.
Given:
F 1 = 700 N
F 2 = 450 N
F 3 = 750 N
θ 1 = 15 deg
θ 2 = 30 deg
Solution:
Initial Guesses:
F = 1N
θ = 10deg
Given
+
ΣF x = 0;
→
F 1 cos ( θ 1 ) − F2 sin ( θ 2 ) − F3 cos ( θ ) = 0
133
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Engineering Mechanics - Statics
+
↑ ΣFy = 0;
Chapter 3
F + F2 cos ( θ 2 ) + F 1 sin ( θ 1 ) − F 3 sin ( θ ) = 0
⎛F⎞
⎜ ⎟ = Find ( F , θ )
⎝θ ⎠
F = 28.25 N
θ = 53.02 deg
Problem 3-4
Determine the magnitude and angle θ of F so that the particle is in equilibrium.
Units Used:
3
kN = 10 N
Given:
F 1 = 4.5 kN
F 2 = 7.5 kN
F 3 = 2.25 kN
α = 60 deg
φ = 30 deg
Solution:
Guesses:
F = 1 kN
θ = 1
Given
Equations of Equilibrium:
+
→
Σ F x = 0;
F cos ( θ ) − F2 sin ( φ ) − F 1 + F 3 cos ( α ) = 0
+
F sin ( θ ) − F 2 cos ( φ ) − F 3 sin ( α ) = 0
↑ Σ F y = 0;
⎛F⎞
⎜ ⎟ = Find ( F , θ )
⎝θ ⎠
F = 11.05 kN
θ = 49.84 deg
134
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Engineering Mechanics - Statics
Chapter 3
Problem 3-5
The members of a truss are connected to the gusset plate. If the forces are concurrent at point O,
determine the magnitudes of F and T for equilibrium.
Units Used:
3
kN = 10 N
Given:
F 1 = 8 kN
F 2 = 5 kN
θ 1 = 45 deg
θ = 30 deg
Solution:
+
ΣF x = 0;
→
−T cos ( θ ) + F 1 + F 2 sin ( θ 1 ) = 0
T =
F1 + F2 sin ( θ 1 )
cos ( θ )
T = 13.3 kN
+
↑ ΣFy = 0;
F − T sin ( θ ) − F 2 cos ( θ 1 ) = 0
F = T sin ( θ ) + F2 cos ( θ 1 )
F = 10.2 kN
Problem 3-6
The gusset plate is subjected to the forces of four members. Determine the force in member B and its
proper orientation θ for equilibrium. The forces are concurrent at point O.
Units Used:
3
kN = 10 N
135
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Engineering Mechanics - Statics
Chapter 3
Given:
F = 12 kN
F 1 = 8 kN
F 2 = 5 kN
θ 1 = 45 deg
Solution:
Initial Guesses
T = 1kN
θ = 10deg
Given
+
ΣF x = 0;
→
F 1 − T cos ( θ ) + F 2 sin ( θ 1 ) = 0
+
−T sin ( θ ) − F2 cos ( θ 1 ) + F = 0
↑ ΣFy = 0;
⎛T⎞
⎜ ⎟ = Find ( T , θ )
⎝θ⎠
T = 14.31 kN
θ = 36.27 deg
Problem 3-7
Determine the maximum weight of the engine that
can be supported without exceeding a tension of T1
in chain AB and T2 in chain AC.
Given:
θ = 30 deg
T1 = 450 lb
T2 = 480 lb
Solution:
Initial Guesses
F AB = T1
F AC = T2
W = 1lb
136
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Engineering Mechanics - Statics
Chapter 3
Given
Assuming cable AB reaches the maximum
tension F AB = T1.
+
ΣF x = 0;
→
F AC cos ( θ ) − F AB = 0
+
F AC sin ( θ ) − W = 0
↑ ΣFy = 0;
⎛ FAC1 ⎞
⎜
⎟ = Find ( FAC , W)
⎝ W1 ⎠
Given
W1 = 259.81 lb
Assuming cable AC reaches the maximum tension FAC = T2.
+
ΣF x = 0;
→
F AC cos ( θ ) − F AB = 0
+
F AC sin ( θ ) − W = 0
↑ ΣFy = 0;
⎛ FAB2 ⎞
⎜
⎟ = Find ( FAB , W)
⎝ W2 ⎠
W = min ( W1 , W2 )
W2 = 240.00 lb
W = 240.00 lb
Problem 3-8
The engine of mass M is suspended from a vertical chain at A. A second
chain is wrapped around the engine and held in position by the spreader
bar BC. Determine the compressive force acting along the axis of the bar
and the tension forces in segments BA and CA of the chain. Hint:
Analyze equilibrium first at A, then at B.
Units Used:
3
kN = 10 N
Given:
M = 200 kg
θ 1 = 55 deg
g = 9.81
m
2
s
137
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Engineering Mechanics - Statics
Chapter 3
Solution:
Initial guesses:
Given
F BA = 1 kN
F CA = 2 kN
Point A
+
ΣF x = 0;
→
F CA cos ( θ 1 ) − FBA cos ( θ 1 ) = 0
+
M( g) − FCA sin ( θ 1 ) − FBA sin ( θ 1 ) = 0
↑ ΣFy = 0;
⎛ FBA ⎞
⎜
⎟ = Find ( FBA , FCA)
⎝ FCA ⎠
⎛ FBA ⎞ ⎛ 1.20 ⎞
⎜
⎟=⎜
⎟ kN
⎝ FCA ⎠ ⎝ 1.20 ⎠
At point B:
+
ΣF x = 0;
→
F BA cos ( θ 1 ) − F BC = 0
F BC = FBA cos ( θ 1 )
F BC = 687 N
Problem 3-9
Cords AB and AC can each sustain a maximum tension T. If the drum has weight W,
determine the smallest angle θ at which they can be attached to the drum.
Given:
T = 800 lb
W = 900 lb
Solution:
+
↑ Σ Fy = 0;
W − 2T sin ( θ ) = 0
W⎞
⎟
⎝ 2T ⎠
θ = asin ⎛⎜
θ = 34.2 deg
138
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Engineering Mechanics - Statics
Chapter 3
Problem 3-10
The crate of weight W is hoisted using the ropes AB and AC. Each rope can withstand a
maximum tension T before it breaks. If AB always remains horizontal, determine the smallest
angle θ to which the crate can be hoisted.
Given:
W = 500 lb
T = 2500 lb
Solution:
TAB = T
Case 1: Assume
The initial guess
θ = 30 deg
TAC = 2000 lb
Given
+
Σ F x = 0;
→
TAB − TAC cos ( θ ) = 0
+
TAC sin ( θ ) − W = 0
↑ Σ F y = 0;
⎛ TAC1 ⎞
⎜
⎟ = Find ( TAC , θ )
⎝ θ1 ⎠
θ 1 = 11.31 deg
TAC1 = 2550 lb
TAC = T
Case 1: Assume
The initial guess
θ = 30 deg
TAB = 2000 lb
Given
+
Σ F x = 0;
→
TAB − TAC cos ( θ ) = 0
+
TAC sin ( θ ) − W = 0
↑ Σ F y = 0;
⎛ TAB2 ⎞
⎜
⎟ = Find ( TAB , θ )
⎝ θ2 ⎠
θ = max ( θ 1 , θ 2 )
θ 2 = 11.54 deg
TAB2 = 2449 lb
θ = 11.54 deg
139
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Engineering Mechanics - Statics
Chapter 3
Problem 3-11
Two electrically charged pith balls, each having mass M, are suspended from light threads of equal
length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured
distance between them is r.
Given:
M = 0.2 gm
r = 200 mm
l = 150 mm
d = 50 mm
Solution:
The initial guesses :
T = 200 N
F = 200 N
Given
+
↑
Σ F x = 0;
⎛ r − d⎞ = 0
⎟
⎝ 2l ⎠
F − T⎜
2
⎡⎢ 2
r − d ⎞ ⎥⎤
⎛
⎢ l − ⎜⎝ 2 ⎟⎠ ⎥
+
→ Σ Fy = 0; T⎢⎣
⎥ − Mg = 0
l
⎦
⎛T ⎞
⎜ ⎟ = Find ( T , F)
⎝F⎠
F = 1.13 × 10
−3
T = 0.00 N
N
Problem 3-12
The towing pendant AB is subjected to the force F which is developed from a tugboat. Determine
the force that is in each of the bridles, BC and BD, if the ship is moving forward with constant
velocity.
Units Used:
3
kN = 10 N
140
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Engineering Mechanics - Statics
Chapter 3
Given:
F = 50 kN
θ 1 = 20 deg
θ 2 = 30 deg
Solution:
Initial guesses:
TBC = 1 kN
TBD = 2 kN
Given
+
ΣF x = 0;
→
TBC sin ( θ 2 ) − TBD sin ( θ 1 ) = 0
+
TBC cos ( θ 2 ) + TBD cos ( θ 1 ) − F = 0
↑ ΣFy = 0;
⎛ TBC ⎞
⎜
⎟ = Find ( TBC , TBD)
⎝ TBD ⎠
⎛ TBC ⎞ ⎛ 22.32 ⎞
⎜
⎟=⎜
⎟ kN
⎝ TBD ⎠ ⎝ 32.64 ⎠
Problem 3-13
Determine the stretch in each spring for
equilibrium of the block of mass M. The
springs are shown in the equilibrium position.
141
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Engineering Mechanics - Statics
Chapter 3
Given:
M = 2 kg
a = 3m
b = 3m
c = 4m
kAB = 30
N
kAC = 20
N
kAD = 40
N
g = 9.81
m
m
m
m
2
s
Solution:
The initial guesses:
F AB = 1 N F AC = 1 N
Given
+
→
Σ F x = 0;
+
Σ F y = 0;
↑
F AB⎛
c
b
⎛
⎞
⎞
⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0
⎝ a +c ⎠
⎝ a +b ⎠
F AC⎛
a
⎛ a ⎞
⎞
⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − M g = 0
⎝ a +b ⎠
⎝ a +c ⎠
⎛ FAC ⎞
⎜
⎟ = Find ( FAC , FAB)
⎝ FAB ⎠
xAC =
xAB =
F AC
kAC
FAB
kAB
⎛ FAC ⎞ ⎛ 15.86 ⎞
⎜
⎟=⎜
⎟N
⎝ FAB ⎠ ⎝ 14.01 ⎠
xAC = 0.79 m
xAB = 0.47 m
142
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Engineering Mechanics - Statics
Chapter 3
Problem 3-14
The unstretched length of spring AB is δ. If the block is held in the equilibrium position shown,
determine the mass of the block at D.
Given:
δ = 2m
a = 3m
b = 3m
c = 4m
kAB = 30
N
kAC = 20
N
kAD = 40
N
g = 9.81
m
m
m
m
2
s
Solution:
F AB = kAB
( a2 + c2 − δ )
The initial guesses:
mD = 1 kg
F AC = 1 N
Given
+
Σ F x = 0;
→
+
↑
Σ F y = 0;
F AB⎛
c
b
⎛
⎞
⎞
⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0
⎝ a +c ⎠
⎝ a +b ⎠
F AC⎛
⎛ FAC ⎞
⎜
⎟ = Find ( FAC , mD)
⎝ mD ⎠
a
⎛ a ⎞
⎞
⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − mD g = 0
⎝ a +b ⎠
⎝ a +c ⎠
F AC = 101.8 N
mD = 12.8 kg
143
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Engineering Mechanics - Statics
Chapter 3
Problem 3-15
The springs AB and BC have stiffness k and unstretched
lengths l/2. Determine the horizontal force F applied to
the cord which is attached to the small pulley B so that
the displacement of the pulley from the wall is d.
Given:
l = 6m
N
k = 500
m
d = 1.5 m
Solution:
⎡
T = k⎢ ⎛⎜
l⎞
2
⎟
⎣ ⎝ 2⎠
2
+d −
+
Σ F x = 0;
→
l⎤
⎥
d
d
( 2T) − F = 0
d + ⎛⎜
2
F =
T = 177.05 N
2⎦
( 2T)
l⎞
2
⎟
⎝ 2⎠
F = 158.36 N
2
⎛ l ⎞ + d2
⎜ ⎟
⎝ 2⎠
Problem 3-16
The springs AB and BC have stiffness k and an unstretched length of l. Determine the
displacement d of the cord from the wall when a force F is applied to the cord.
144
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Engineering Mechanics - Statics
Chapter 3
Given:
l = 6m
k = 500
N
m
F = 175 N
Solution:
The initial guesses:
d = 1m
T = 1N
Given
+
Σ F x = 0;
→
d
−F + ( 2T)
2
=0
⎛l⎞
⎟
⎝ 2⎠
2
d +⎜
⎡
Spring
⎛T⎞
⎜ ⎟ = Find ( T , d)
⎝d ⎠
2
⎤
⎛ l ⎞ − l⎥
⎟
2⎦
⎝ 2⎠
T = k⎢ d + ⎜
⎣
2
T = 189.96 N
d = 1.56 m
Problem 3-17
Determine the force in each cable and the force F needed to hold the lamp of mass M in the position
shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the
equilibrium at C.
Given:
M = 4 kg
θ 1 = 30 deg
θ 2 = 60 deg
θ 3 = 30 deg
Solution:
Initial guesses:
145
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Engineering Mechanics - Statics
TBC = 1 N
Chapter 3
TBA = 2 N
Given
At B:
+
ΣF x = 0;
→
TBC cos ( θ 1 ) − TBA cos ( θ 2 ) = 0
+
TBA sin ( θ 2 ) − TBC sin ( θ 1 ) − M g = 0
↑ ΣFy = 0;
⎛ TBC ⎞
⎜
⎟ = Find ( TBC , TBA)
⎝ TBA ⎠
At C:
TCD = 1 N
⎛ TBC ⎞ ⎛ 39.24 ⎞
⎜
⎟=⎜
⎟N
⎝ TBA ⎠ ⎝ 67.97 ⎠
F = 2N
Given
+
ΣF x = 0;
→
−TBC cos ( θ 1 ) + TCD cos ( θ 3 ) = 0
+
TBC sin ( θ 1 ) + TCD sin ( θ 3 ) − F = 0
↑ ΣFy = 0;
⎛ TCD ⎞
⎜
⎟ = Find ( TCD , F)
⎝ F ⎠
⎛ TCD ⎞ ⎛ 39.24 ⎞
⎜
⎟=⎜
⎟N
⎝ F ⎠ ⎝ 39.24 ⎠
Problem 3-18
The motor at B winds up the cord attached to the crate of weight W with a constant speed.
Determine the force in cord CD supporting the pulley and the angle θ for equilibrium. Neglect the
146
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Engineering Mechanics - Statics
Chapter 3
size of the pulley at C.
Given:
W = 65 lb
c = 12
d = 5
Solution:
The initial guesses: θ = 100 deg
F CD = 200 lb
Given
Equations of Equilibrium:
+
Σ F x = 0;
→
⎛
⎞=0
2
2⎟
⎝ c +d ⎠
F CD cos ( θ ) − W⎜
⎛
d
⎞
↑Σ Fy = 0; FCD sin(θ ) − W⎜ c2 + d2 ⎟ − W = 0
+
⎝
c
⎠
⎛ θ ⎞
⎜
⎟ = Find ( θ , FCD)
⎝ FCD ⎠
θ = 78.69 deg
F CD = 127.5 lb
Problem 3-19
The cords BCA and CD can each support a maximum load T. Determine the maximum weight
of the crate that can be hoisted at constant velocity, and the angle θ for equilibrium.
147
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Engineering Mechanics - Statics
Chapter 3
Given:
T = 100 lb
c = 12
d = 5
The maximum will occur in CD rather than in BCA.
Solution :
The initial guesses: θ = 100 deg
W = 200 lb
Given
Equations of Equilibrium:
+
Σ F x = 0;
→
+
↑
⎛
T cos ( θ ) − W⎜
d
2
⎞=0
2⎟
⎝ c +d ⎠
⎛
Σ F y = 0; T sin ( θ ) − W⎜
c
2
⎞−W=0
2⎟
⎝ c +d ⎠
⎛θ⎞
⎜ ⎟ = Find ( θ , W)
⎝W⎠
θ = 78.69 deg
W = 51.0 lb
148
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Engineering Mechanics - Statics
Chapter 3
Problem 3-20
The sack has weight W and is supported by
the six cords tied together as shown.
Determine the tension in each cord and the
angle θ for equilibrium. Cord BC is horizontal.
Given:
W = 15 lb
θ 1 = 30 deg
θ 2 = 45 deg
θ 3 = 60 deg
Solution:
Guesses
TBE = 1 lb
TAB = 1 lb
θ = 20deg
TBC = 1 lb
TAC = 1 lb
TCD = 1 lb
TAH = 1 lb
Given
At H:
+
↑ ΣFy = 0;
TAH − W = 0
At A:
+
ΣF x = 0;
→
−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0
+
TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0
↑ ΣFy = 0;
At B:
+
ΣF x = 0;
→
TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0
+
TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0
↑ ΣFy = 0;
At C:
+
ΣF x = 0;
→
TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0
+
TCD sin ( θ ) − TAC sin ( θ 3 ) = 0
↑ ΣFy = 0;
149
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Engineering Mechanics - Statics
Chapter 3
⎛ TBE ⎞
⎜
⎟
⎜ TAB ⎟
⎜ TBC ⎟
⎜
⎟
⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ )
⎜T ⎟
⎜ CD ⎟
⎜ TAH ⎟
⎜
⎟
⎝ θ ⎠
⎛⎜ TBE ⎞⎟
⎛⎜ 10.98 ⎞⎟
T
⎜ AB ⎟
7.76 ⎟
⎜
⎟ ⎜
⎜ TBC ⎟ = ⎜ 4.02 ⎟ lb
⎜ TAC ⎟ ⎜ 10.98 ⎟
⎟
⎜
⎟ ⎜
13.45
⎜
⎟
⎜ TCD ⎟
⎜
⎜ TAH ⎟ ⎝ 15.00 ⎟⎠
⎝
⎠
θ = 45.00 deg
Problem 3-21
Each cord can sustain a maximum tension T. Determine the largest weight of the sack that can be
supported. Also, determine θ of cord DC for equilibrium.
Given:
T = 200 lb
θ 1 = 30 deg
θ 2 = 45 deg
θ 3 = 60 deg
Solution:
Solve for W = 1 and then scale the answer at the end.
Guesses
TBE = 1
TAB = 1
TBC = 1
TAC = 1
TCD = 1
TAH = 1
150
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Engineering Mechanics - Statics
Chapter 3
θ = 20 deg
Given
At H:
+
↑ ΣFy = 0;
TAH − W = 0
At A:
+
ΣF x = 0;
→
−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0
+
TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0
↑ ΣFy = 0;
At B:
+
ΣF x = 0;
→
TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0
+
TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0
↑ ΣFy = 0;
At C:
+
ΣF x = 0;
→
TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0
+
TCD sin ( θ ) − TAC sin ( θ 3 ) = 0
↑ ΣFy = 0;
⎛ TBE ⎞
⎜
⎟
⎜ TAB ⎟
⎜ TBC ⎟
⎜
⎟
⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ )
⎜T ⎟
⎜ CD ⎟
⎜ TAH ⎟
⎜
⎟
⎝ θ ⎠
W =
T
⎛⎜ TBE ⎞⎟
⎛⎜ 0.73 ⎞⎟
T
⎜ AB ⎟
0.52 ⎟
⎜
⎟ ⎜
TBC
⎜
⎟ = ⎜ 0.27 ⎟
⎜ TAC ⎟ ⎜ 0.73 ⎟
⎟
⎜
⎟ ⎜
0.90
⎜
⎟
⎜ TCD ⎟
⎜
⎜ TAH ⎟ ⎝ 1.00 ⎟⎠
⎝
⎠
W = 200.00 lb
max ( TBE , TAB , TBC , TAC , TCD , TAH )
θ = 45.00 deg
151
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Engineering Mechanics - Statics
Chapter 3
Problem 3-22
The block has weight W and is being hoisted at uniform
velocity. Determine the angle θ for equilibrium and the
required force in each cord.
Given:
W = 20 lb
φ = 30 deg
Solution:
The initial guesses:
θ = 10 deg
TAB = 50 lb
Given
TAB sin ( φ ) − W sin ( θ ) = 0
TAB cos ( φ ) − W − W cos ( θ ) = 0
⎛ θ ⎞
⎜
⎟ = Find ( θ , TAB)
⎝ TAB ⎠
θ = 60.00 deg
TAB = 34.6 lb
Problem 3-23
Determine the maximum weight W of the block that can be suspended in the position shown if
each cord can support a maximum tension T. Also, what is the angle θ for equilibrium?
152
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Engineering Mechanics - Statics
Chapter 3
Given:
T = 80 lb
φ = 30 deg
The maximum load will occur in cord AB.
Solution:
TAB = T
The initial guesses:
θ = 100deg W = 200lb
Given
+
↑Σ Fy = 0;
TAB cos ( φ ) − W − W cos ( θ ) = 0
+
Σ F x = 0;
→
TAB sin ( φ ) − W sin ( θ ) = 0
⎛θ⎞
⎜ ⎟ = Find ( θ , W)
⎝W⎠
W = 46.19 lb
θ = 60.00 deg
Problem 3-24
Two spheres A and B have an equal mass M and are electrostatically charged such that the
repulsive force acting between them has magnitude F and is directed along line AB. Determine the
angle θ, the tension in cords AC and BC, and the mass M of each sphere.
153
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Engineering Mechanics - Statics
Chapter 3
Unit used:
mN = 10
−3
N
Given:
F = 20 mN
g = 9.81
m
2
s
θ 1 = 30 deg
θ 2 = 30 deg
Solution:
Guesses
TB = 1 mN
M = 1 gm
TA = 1 mN
θ = 30 deg
Given
For B:
+
ΣF x = 0; F cos ( θ 2 ) − TB sin ( θ 1 ) = 0
→
↑ ΣFy = 0; F sin(θ2) + TB cos (θ 1) − M g = 0
+
For A:
+
ΣF x = 0; TA sin ( θ ) − F cos ( θ 2 ) = 0
→
↑ ΣFy = 0; TA cos (θ ) − F sin(θ2) − M g = 0
+
⎛ TA ⎞
⎜ ⎟
⎜ TB ⎟ = Find ( T , T , θ , M)
A B
⎜θ ⎟
⎜ ⎟
⎝M⎠
⎛ TA ⎞ ⎛ 52.92 ⎞
⎜ ⎟=⎜
⎟ mN
⎝ TB ⎠ ⎝ 34.64 ⎠
θ = 19.11 deg
M = 4.08 gm
Problem 3-25
Blocks D and F weigh W1 each and block E weighs W2. Determine the sag s for equilibrium.
Neglect the size of the pulleys.
154
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Engineering Mechanics - Statics
Chapter 3
Given:
W1 = 5 lb
W2 = 8 lb
a = 4 ft
Solution:
Sum forces in the y direction
Guess
s = 1 ft
Given
2⎜
⎛
⎞W − W = 0
1
2
⎝ s +a ⎠
s
2⎟
2
s = Find ( s)
s = 5.33 ft
Problem 3-26
If blocks D and F each have weight W1, determine the weight of block E if the sag is s. Neglect
the size of the pulleys.
Given:
W1 = 5 lb
s = 3 ft
a = 4 ft
Solution:
Sum forces in the y direction
⎛
2⎜
s
2
⎞W − W = 0
2⎟
⎝ s +a ⎠
1
155
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Engineering Mechanics - Statics
⎛
Chapter 3
⎞W
1
⎟
2
2
+
a
s
⎝
⎠
s
W = 2⎜
W = 6.00 lb
Problem 3-27
The block of mass M is supported by two springs having the stiffness shown. Determine the
unstretched length of each spring.
Units Used:
3
kN = 10 N
Given:
M = 30 kg
l1 = 0.6 m
l2 = 0.4 m
l3 = 0.5 m
kAC = 1.5
kN
m
kAB = 1.2
kN
m
g = 9.81
m
2
s
Solution:
Initial guesses:
F AC = 20 N
F AB = 30 N
Given
+
ΣF x = 0;
→
+
↑
ΣF y = 0;
l2 FAB
2
l2 + l3
−
2
l3 FAB
2
l2 + l3
2
l1 + l3
+
2
l1 F AC
=0
2
l3 F AC
2
l1 + l3
−Mg=0
2
156
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Engineering Mechanics - Statics
Chapter 3
⎛ FAC ⎞
⎜
⎟ = Find ( FAC , FAB)
⎝ FAB ⎠
LAB = 0.1 m
Then guess
Given
⎛ FAC ⎞ ⎛ 183.88 ⎞
⎜
⎟=⎜
⎟N
⎝ FAB ⎠ ⎝ 226.13 ⎠
LAC = 0.1 m
F AC = kAC ⎛⎝ l1 + l3 − LAC⎞⎠
2
2
F AB = kAB⎛⎝ l2 + l3 − LAB⎞⎠
2
2
⎛ LAB ⎞
⎜
⎟ = Find ( LAB , LAC)
L
AC
⎝
⎠
⎛ LAB ⎞ ⎛ 0.452 ⎞
⎜
⎟=⎜
⎟m
0.658
L
⎝
⎠
AC
⎝
⎠
Problem 3-28
Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC = W,
WB = kW, determine the angle θ for equilibrium.
Given:
k = 0.25
Solution:
+
ΣF x = 0;
→
W cos ( φ ) − k W cos ( θ ) = 0
+
W sin ( φ ) + k W sin ( θ ) − W = 0
↑ ΣFy = 0;
cos ( φ ) = k cos ( θ )
sin ( φ ) = 1 − k sin ( θ )
157
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Engineering Mechanics - Statics
Chapter 3
1 = k cos ( θ ) + ( 1 − k sin ( θ ) ) = 1 + k − 2k sin ( θ )
2
2
k
θ = asin ⎛⎜ ⎟⎞
⎝ 2⎠
2
2
θ = 7.18 deg
Problem 3-29
A continuous cable of total length l is wrapped around the small pulleys at A, B, C, and D. If each
spring is stretched a distance b, determine the mass M of each block. Neglect the weight of the pulleys
and cords. The springs are unstretched when d = l/2.
Given:
l = 4m
k = 500
N
m
b = 300 mm
g = 9.81
m
2
s
Solution:
Fs = k b
F s = 150.00 N
Guesses
T = 1N
θ = 10 deg
M = 1 kg
Given
2T sin ( θ ) − F s = 0
−2T cos ( θ ) + M g = 0
b+
l
4
sin ( θ ) =
l
4
⎛T⎞
⎜ θ ⎟ = Find ( T , θ , M)
⎜ ⎟
⎝M⎠
158
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Engineering Mechanics - Statics
Chapter 3
θ = 44.43 deg
T = 107.14 N
M = 15.60 kg
Problem 3-30
Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each is
proportional to the sine of the angle of the other two; that is, P/sin α = Q/sin β = R/sin γ.
Solution:
Sine law:
R
sin ( 180deg − γ )
=
However, in general
R
sin ( γ )
=
Q
sin ( β )
=
Q
sin ( 180deg − β )
=
P
sin ( 180deg − α )
sin ( 180deg − φ ) = sin ( φ ) , hence
P
sin ( α )
Q.E.D.
Problem 3-31
A vertical force P is applied to the ends
of cord AB of length a and spring AC.
If the spring has an unstretched length
δ, determine the angle θ for equilibrium.
Given:
P = 10 lb
δ = 2 ft
k = 15
lb
ft
a = 2 ft
159
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Engineering Mechanics - Statics
Chapter 3
b = 2 ft
Guesses
θ = 10 deg
φ = 10 deg
T = 1 lb
F = 1 lb
x = 1ft
Given
−T cos ( θ ) + F cos ( φ ) = 0
T sin ( θ ) + F sin ( φ ) − P = 0
F = k( x − δ )
a sin ( θ ) = x sin ( φ )
a cos ( θ ) + x cos ( φ ) = a + b
⎛θ ⎞
⎜ ⎟
⎜φ ⎟
⎜ T ⎟ = Find ( θ , φ , T , F , x)
⎜F⎟
⎜ ⎟
⎝x⎠
⎛ T ⎞ ⎛ 10.30 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ F ⎠ ⎝ 9.38 ⎠
θ = 35 deg
Problem 3-32
Determine the unstretched length δ of
spring AC if a force P causes the
angle θ for equilibrium. Cord AB has
length a.
Given:
P = 80 lb
θ = 60 deg
k = 50
lb
ft
a = 2 ft
160
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Engineering Mechanics - Statics
Chapter 3
b = 2 ft
Guesses
δ = 1ft
φ = 10 deg
T = 1 lb
F = 1 lb
x = 1ft
Given
−T cos ( θ ) + F cos ( φ ) = 0
T sin ( θ ) + F sin ( φ ) − P = 0
F = k( x − δ )
a sin ( θ ) = x sin ( φ )
a cos ( θ ) + x cos ( φ ) = a + b
⎛δ ⎞
⎜ ⎟
⎜φ ⎟
⎜ T ⎟ = Find ( δ , φ , T , F , x)
⎜F⎟
⎜ ⎟
⎝x⎠
⎛ T ⎞ ⎛ 69.28 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ F ⎠ ⎝ 40.00 ⎠
δ = 2.66 ft
Problem 3-33
The flowerpot of mass M is
suspended from three wires and
supported by the hooks at B and C.
Determine the tension in AB and AC
for equilibrium.
Given:
M = 20 kg
l1 = 3.5 m
l2 = 2 m
l3 = 4 m
161
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Engineering Mechanics - Statics
Chapter 3
l4 = 0.5 m
g = 9.81
m
2
s
Solution:
Initial guesses:
TAB = 1 N
TAC = 1 N
θ = 10 deg
φ = 10 deg
Given
−TAC cos ( φ ) + TAB cos ( θ ) = 0
TAC sin ( φ ) + TAB sin ( θ ) − M g = 0
l1 cos ( φ ) + l2 cos ( θ ) = l3
l1 sin ( φ ) = l2 sin ( θ ) + l4
⎛ TAB ⎞
⎜
⎟
⎜ TAC ⎟ = Find ( T , T , θ , φ )
AB AC
⎜ θ ⎟
⎜
⎟
⎝ φ ⎠
⎛ θ ⎞ ⎛ 53.13 ⎞
⎜ ⎟=⎜
⎟ deg
⎝ φ ⎠ ⎝ 36.87 ⎠
⎛ TAB ⎞ ⎛ 156.96 ⎞
⎜
⎟=⎜
⎟N
⎝ TAC ⎠ ⎝ 117.72 ⎠
Problem 3-34
A car is to be towed using the rope arrangement shown. The towing force required is P. Determine
the minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T.
Hint: Use the equilibrium condition at point A to determine the required angle θ for attachment, then
determine l using trigonometry applied to triangle ABC.
Given:
P = 600 lb
T = 750 lb
φ = 30 deg
162
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Engineering Mechanics - Statics
Chapter 3
d = 4 ft
Solution:
The initial guesses
TAB = T
TAC = T
θ = 30 deg
l = 2 ft
Case 1:
Assume TAC = T
Given
+
Σ F x = 0;
→
TAC cos ( φ ) − TAB cos ( θ ) = 0
+
P − TAC sin ( φ ) − TAB sin ( θ ) = 0
↑ Σ Fy = 0;
l
sin ( φ )
=
⎛ TAB ⎞
⎜
⎟
⎜ θ ⎟ = Find ( TAB , θ , l)
⎜ l ⎟
⎝ 1 ⎠
Case 2:
d
sin ( 180deg − θ − φ )
TAB = 687.39 lb
θ = 19.11 deg
l1 = 2.65 ft
θ = 13.85 deg
l2 = 2.89 ft
Assume TAB = T
Given
+
Σ F x = 0;
→
TAC cos ( φ ) − TAB cos ( θ ) = 0
+
P − TAC sin ( φ ) − TAB sin ( θ ) = 0
↑ Σ Fy = 0;
l
sin ( φ )
⎛ TAC ⎞
⎜
⎟
⎜ θ ⎟ = Find ( TAC , θ , l)
⎜ l ⎟
⎝ 2 ⎠
l = min ( l1 , l2 )
=
d
sin ( 180deg − θ − φ )
TAC = 840.83 lb
l = 2.65 ft
163
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Engineering Mechanics - Statics
Chapter 3
Problem 3-35
Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended
from the rings at A and B. Note that s = 0 when the cylinders are removed.
Given:
d = 0.5 m
l1 = 1.5 m
l2 = 2 m
l3 = 1 m
k = 100
N
g = 9.81
m
m
2
s
Solution:
TAC = k⎡⎣ ( l1 + d) + l2 −
2
2
l1 + l2 ⎤⎦
2
2
TAC = 32.84 N
⎛ l1 + d ⎞
⎟
⎝ l2 ⎠
θ = atan ⎜
θ = 45 deg
M =
TAC sin ( θ )
g
M = 2.37 kg
164
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Engineering Mechanics - Statics
Chapter 3
Problem 3-36
The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling
and plot its value T (ordinate) as a function of its orientation θ , where 0 ≤ θ ≤ 90° .
Solution:
W − 2T cos ( θ ) = 0
T=
W ⎞
⎜
⎟
2 ⎝ cos ( θ ) ⎠
1 ⎛
Problem 3-37
The lamp fixture has weight W and is suspended from two springs, each having unstretched length L
and stiffness k. Determine the angle θ for equilibrium.
Units Used:
3
kN = 10 N
Given:
W = 10 lb
L = 4 ft
k = 5
lb
ft
a = 4 ft
165
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
Solution:
The initial guesses: T = 200 lb
θ = 10 deg
Given
Spring
+
↑Σ Fy = 0;
⎛T⎞
⎜ ⎟ = Find ( T , θ )
⎝θ⎠
⎛ a − L⎞
⎟
⎝ cos ( θ )
⎠
T = k⎜
2T sin ( θ ) − W = 0
T = 7.34 lb
θ = 42.97 deg
Problem 3-38
The uniform tank of weight W is suspended by means of a cable, of length l, which is attached
to the sides of the tank and passes over the small pulley located at O. If the cable can be
attached at either points A and B, or C and D, determine which attachment produces the least
amount of tension in the cable.What is this tension?
Given:
W = 200 lb
l = 6 ft
a = 1 ft
b = 2 ft
c = b
d = 2a
Solution:
Free Body Diagram: By
observation, the force F has to
support the entire weight of the
tank. Thus, F = W. The tension in
166
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Engineering Mechanics - Statics
Chapter 3
cable is the same throughout the
cable.
Equations of Equilibrium
W − 2T sin ( θ ) = 0
ΣF y = 0;
Attached to CD
θ 1 = acos ⎛⎜
2 a⎞
Attached to AB
θ 2 = acos ⎛⎜
2 b⎞
⎟
⎝ l ⎠
⎟
⎝ l ⎠
θ 1 = 70.53 deg
θ 2 = 48.19 deg
We choose the largest angle (which will produce the smallest force)
θ = max ( θ 1 , θ 2 )
T =
1⎛ W
⎜
⎞
⎟
2 ⎝ sin ( θ ) ⎠
θ = 70.53 deg
T = 106 lb
Problem 3-39
A sphere of mass ms rests on the smooth parabolic surface. Determine the normal force it exerts
on the surface and the mass mB of block B needed to hold it in the equilibrium position shown.
Given:
ms = 4 kg
a = 0.4 m
b = 0.4 m
θ = 60 deg
g = 9.81
m
2
s
Solution:
k =
a
2
b
Geometry: The angle θ1 which the surface make with the horizontal is to be determined first.
167
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Engineering Mechanics - Statics
tan ( θ 1 ) =
dy
dx
=2kx
Chapter 3
evaluated at x = a
θ 1 = atan ( 2 k a)
θ 1 = 63.43 deg
Free Body Diagram : The tension in the cord is the same throughout the cord and is equal
to the weight of block B, mBg.
The initial guesses: mB = 200 kg
F N = 200 N
Given
+
Σ F x = 0; mB g cos ( θ ) − FN sin ( θ 1 ) = 0
→
+
↑Σ Fy = 0;
mB g sin ( θ ) + F N cos ( θ 1 ) − ms g = 0
⎛ mB ⎞
⎜ ⎟ = Find ( mB , FN)
⎝ FN ⎠
F N = 19.66 N
mB = 3.58 kg
Problem 3-40
The pipe of mass M is supported at A by a system of five cords. Determine the force in each
cord for equilibrium.
Given:
M = 30 kg
c = 3
m
d = 4
g = 9.81
2
s
θ = 60 deg
Solution:
Initial guesses:
TAB = 1 N
TAE = 1 N
TBC = 1 N
TBD = 1 N
Given
TAB sin ( θ ) − M g = 0
TAE − TAB cos ( θ ) = 0
168
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Engineering Mechanics - Statics
Chapter 3
TBD⎛
c
⎞
⎜ 2 2 ⎟ − TAB sin ( θ ) = 0
⎝ c +d ⎠
TBD⎛
d
⎞
⎜ 2 2 ⎟ + TAB cos ( θ ) − TBC = 0
⎝ c +d ⎠
⎛ TAB ⎞
⎜
⎟
⎜ TAE ⎟ = Find ( T , T , T , T )
AB AE BC BD
⎜ TBC ⎟
⎜
⎟
⎝ TBD ⎠
⎛ TAB ⎞ ⎛ 339.8 ⎞
⎜
⎟ ⎜
⎟
⎜ TAE ⎟ = ⎜ 169.9 ⎟ N
⎜ TBC ⎟ ⎜ 562.3 ⎟
⎜
⎟ ⎜
⎟
⎝ TBD ⎠ ⎝ 490.5 ⎠
Problem 3-41
The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OB lies
in the y-z plane. Determine the forces acting in each of the three struts required for equilibrium.
Units Used:
3
kN = 10 N
Given:
F = 2 kN
θ 1 = 45 deg
θ 2 = 40 deg
Solution:
ΣF x = 0;
−R sin ( θ 1 ) = 0
R = 0
169
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Engineering Mechanics - Statics
Chapter 3
P sin ( θ 2 ) − F = 0
ΣF z = 0;
P =
F
sin ( θ 2 )
P = 3.11 kN
Q − P cos ( θ 2 ) = 0
ΣF y = 0;
Q = P cos ( θ 2 )
Q = 2.38 kN
Problem 3-42
Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle.
Units Used:
3
kN = 10 N
Given:
F 4 = 800 N
α = 60 deg
β = 30 deg
γ = 30 deg
c = 3
d = 4
Solution:
The initial guesses: F 1 = 100 N
F 2 = 100 N
F 3 = 100 N
Given
⎛ cos ( α ) ⎞
⎜ 0 ⎟+
F1
⎜
⎟
⎝ sin ( α ) ⎠
⎛ −cos ( γ ) ⎞
⎛ 0 ⎞
⎛c ⎞
⎜ −d ⎟ + F ⎜ −sin ( γ ) ⎟ + F ⎜ sin ( β ) ⎟ = 0
3
4
⎟
⎜
⎟
⎜
⎟
2
2⎜
c +d ⎝ 0 ⎠
⎝ 0 ⎠
⎝ −cos ( β ) ⎠
F2
170
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Engineering Mechanics - Statics
⎛ F1 ⎞
⎜ ⎟
⎜ F2 ⎟ = Find ( F1 , F2 , F3)
⎜F ⎟
⎝ 3⎠
Chapter 3
⎛ F1 ⎞ ⎛ 800 ⎞
⎜ ⎟ ⎜
⎟
⎜ F2 ⎟ = ⎜ 147 ⎟ N
⎜ F ⎟ ⎝ 564 ⎠
⎝ 3⎠
Problem 3-43
Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle.
Units Used:
kN = 1000 N
Given:
F 4 = 8.5 kN
F 5 = 2.8 kN
α = 15 deg
β = 30 deg
c = 7
d = 24
Solution:
Initial Guesses:
F 1 = 1 kN
F 2 = 1 kN
F 3 = 1 kN
Given
⎛ −cos ( β ) ⎞
⎜ 0 ⎟+
F1
⎜
⎟
⎝ sin ( β ) ⎠
⎛ −sin ( α ) ⎞
⎛ −c ⎞
⎛1⎞
⎛0⎞
⎜ −d ⎟ + F ⎜ 0 ⎟ + F ⎜ cos ( α ) ⎟ + F ⎜ 0 ⎟ = 0
3
4
5
⎟
⎜ ⎟
⎜
⎟
⎜ ⎟
2
2⎜
c +d ⎝ 0 ⎠
⎝0⎠
⎝ −1 ⎠
⎝ 0 ⎠
F2
171
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Engineering Mechanics - Statics
⎛ F1 ⎞
⎜ ⎟
⎜ F2 ⎟ = Find ( F1 , F2 , F3)
⎜F ⎟
⎝ 3⎠
Chapter 3
⎛ F1 ⎞ ⎛ 5.60 ⎞
⎜ ⎟ ⎜
⎟
⎜ F2 ⎟ = ⎜ 8.55 ⎟ kN
⎜ F ⎟ ⎝ 9.44 ⎠
⎝ 3⎠
Problem 3-44
Determine the magnitudes of F1, F 2 and F3 for equilibrium of the particle F = {- 9i - 8j - 5k}.
Units Used:
3
kN = 10 N
Given:
⎛ −9 ⎞
F = ⎜ −8 ⎟ kN
⎜ ⎟
⎝ −5 ⎠
a = 4m
b = 2m
c = 4m
θ 1 = 30 deg
θ 2 = 60 deg
θ 3 = 135 deg
θ 4 = 60 deg
θ 5 = 60 deg
Solution:
Initial guesses:
F 1 = 8 kN F 2 = 3 kN F 3 = 12 kN
172
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Engineering Mechanics - Statics
Chapter 3
Given
⎛ cos ( θ 2 ) cos ( θ 1) ⎞
⎛ cos ( θ 3 ) ⎞
⎜
⎟
⎜
⎟
F 1 ⎜ −cos ( θ 2 ) sin ( θ 1 ) ⎟ + F 2 ⎜ cos ( θ 5 ) ⎟ +
⎜
⎟
⎜ cos θ ⎟
sin ( θ 2 )
⎝
⎠
⎝ ( 4) ⎠
⎛ F1 ⎞
⎜ ⎟
⎜ F2 ⎟ = Find ( F1 , F2 , F3)
⎜F ⎟
⎝ 3⎠
⎛a⎞
⎜ c ⎟+F=0
⎟
2
2
2⎜
a + b + c ⎝ −b ⎠
F3
⎛ F1 ⎞ ⎛ 8.26 ⎞
⎜ ⎟ ⎜
⎟
⎜ F2 ⎟ = ⎜ 3.84 ⎟ kN
⎜ F ⎟ ⎝ 12.21 ⎠
⎝ 3⎠
Problem 3-45
The three cables are used to support the lamp of weight W. Determine the force developed in
each cable for equilibrium.
Units Used:
3
kN = 10 N
Given:
W = 800 N
b = 4m
a = 4m
c = 2m
Solution:
Initial Guesses:
F AB = 1 N
F AC = 1 N
F AD = 1 N
173
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Engineering Mechanics - Statics
Chapter 3
Given
⎛0⎞
⎛1⎞
⎜
⎟
F AB 1 + FAC ⎜ 0 ⎟ +
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
⎛ −c ⎞
⎛0⎞
⎜ −b ⎟ + W⎜ 0 ⎟ = 0
⎟
⎜ ⎟
2
2
2⎜
a +b +c ⎝ a ⎠
⎝ −1 ⎠
F AD
⎛ FAB ⎞
⎜
⎟
F
⎜ AC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
⎛ FAB ⎞ ⎛ 800 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AC ⎟ = ⎜ 400 ⎟ N
⎜ F ⎟ ⎝ 1200 ⎠
⎝ AD ⎠
Problem 3-46
Determine the force in each cable needed to support the load W.
Given:
a = 8 ft
b = 6 ft
c = 2 ft
d = 2 ft
e = 6 ft
W = 500 lb
Solution:
Initial guesses:
F CD = 600 lb
F CA = 195 lb
F CB = 195 lb
174
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Engineering Mechanics - Statics
Chapter 3
Given
⎛c⎞
⎜ −e ⎟ +
2
2⎜ ⎟
c +e ⎝0 ⎠
FCA
⎛ −d ⎞
⎜ −e ⎟ +
⎟
2
2⎜
d +e ⎝ 0 ⎠
F CB
⎛0⎞
⎛0⎞
⎜ b ⎟ + W⎜ 0 ⎟ = 0
⎜ ⎟
2
2⎜ ⎟
a + b ⎝a⎠
⎝ −1 ⎠
F CD
⎛ FCD ⎞
⎜
⎟
F
⎜ CA ⎟ = Find ( FCD , FCA , FCB)
⎜F ⎟
⎝ CB ⎠
⎛ FCD ⎞ ⎛ 625 ⎞
⎜
⎟ ⎜
⎟
F
⎜ CA ⎟ = ⎜ 198 ⎟ lb
⎜ F ⎟ ⎝ 198 ⎠
⎝ CB ⎠
Problem 3-47
Determine the stretch in each of the two
springs required to hold the crate of mass
mc in the equilibrium position shown. Each
spring has an unstretched length δ and a
stiffness k.
Given:
mc = 20 kg
δ = 2m
k = 300
N
m
a = 4m
b = 6m
c = 12 m
Solution:
Initial Guesses
F OA = 1 N
F OB = 1 N
F OC = 1 N
Given
⎛0⎞
⎛ −1 ⎞
⎜
⎟
F OA −1 + F OB⎜ 0 ⎟ +
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
⎛b⎞
⎛0⎞
⎜ a ⎟ + m g⎜ 0 ⎟ = 0
c
⎜ ⎟
2
2
2⎜ ⎟
a + b + c ⎝c⎠
⎝ −1 ⎠
F OC
175
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Engineering Mechanics - Statics
Chapter 3
⎛ FOA ⎞
⎜
⎟
⎜ FOB ⎟ = Find ( FOA , FOB , FOC)
⎜F ⎟
⎝ OC ⎠
δ OA =
δ OB =
F OA
k
F OB
k
⎛ FOA ⎞ ⎛ 65.40 ⎞
⎜
⎟ ⎜
⎟
⎜ FOB ⎟ = ⎜ 98.10 ⎟ N
⎜ F ⎟ ⎝ 228.90 ⎠
⎝ OC ⎠
δ OA = 218 mm
δ OB = 327 mm
Problem 3-48
If the bucket and its contents have total weight W, determine the force in the supporting cables
DA, DB, and DC.
Given:
W = 20 lb
a = 3 ft
b = 4.5 ft
c = 2.5 ft
d = 3 ft
e = 1.5 ft
f = 1.5 ft
176
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Engineering Mechanics - Statics
Chapter 3
Solution:
The initial guesses:
F DA = 40 lb
F DB = 20 lb
F DC = 30 lb
Given
Σ F x = 0;
b−e
e
⎡
⎤F − ⎡
⎤F = 0
DA ⎢
DC
⎢
⎥
2
2
2
2
2
2⎥
⎣ ( b − e) + f + a ⎦
⎣ e + d + ( c − f) ⎦
Σ F y = 0;
−f
c− f
⎡
⎤F + ⎡
⎤F − F = 0
DA ⎢
DC
DB
⎢
⎥
2
2
2
2
2
2⎥
+
f
+
a
+
d
+
(
c
−
f
)
(
b
−
e
)
e
⎣
⎦
⎣
⎦
Σ F z = 0;
a
d
⎡
⎤F + ⎡
⎤F − W = 0
DA ⎢
DC
⎢
⎥
2
2
2
2
2
2⎥
⎣ ( b − e) + f + a ⎦
⎣ e + d + ( c − f) ⎦
⎛ FDA ⎞
⎜
⎟
⎜ FDB ⎟ = Find ( FDA , FDB , FDC)
⎜F ⎟
⎝ DC ⎠
⎛ FDA ⎞ ⎛ 10.00 ⎞
⎜
⎟ ⎜
⎟
⎜ FDB ⎟ = ⎜ 1.11 ⎟ lb
⎜ F ⎟ ⎝ 15.56 ⎠
⎝ DC ⎠
Problem 3-49
The crate which of weight F is to be hoisted with constant velocity from the hold of a ship
using the cable arrangement shown. Determine the tension in each of the three cables for
equilibrium.
177
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Engineering Mechanics - Statics
Chapter 3
Units Used:
3
kN = 10 N
Given:
F = 2.5 kN
a = 3m
b = 1m
c = 0.75 m
d = 1m
e = 1.5 m
f = 3m
Solution:
The initial guesses
F AD = 3 kN
F AC = 3 kN
−c
Given
2
2
c +b +a
2
b
2
2
c +b +a
2
F AD +
2
c +b +a
d
2
2
d +e +a
F AD +
2
F AD +
⎛ FAD ⎞
⎜
⎟
F
⎜ AC ⎟ = Find ( FAD , FAC , FAB)
⎜F ⎟
⎝ AB ⎠
2
e
2
2
d +e +a
−a
2
F AB = 3 kN
2
F AC +
2
d +e +a
2
2
d + f +a
F AC +
2
F AC +
2
−f
2
2
d + f +a
−a
2
d
2
F AB = 0
F AB = 0
−a
2
2
d + f +a
2
F AB + F = 0
⎛ FAD ⎞ ⎛ 1.55 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AC ⎟ = ⎜ 0.46 ⎟ kN
⎜ F ⎟ ⎝ 0.98 ⎠
⎝ AB ⎠
178
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Engineering Mechanics - Statics
Chapter 3
Problem 3-50
The lamp has mass ml and is supported by pole AO and cables AB and AC. If the force in the
pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium.
Given:
ml = 15 kg
d = 1.5 m
a = 6m
e = 4m
b = 1.5 m
f = 1.5 m
c = 2m
g = 9.81
m
2
s
Solution:
The initial guesses:
F AO = 100 N
F AB = 200 N
F AC = 300 N
Given
Equilibrium equations:
c
2
2
c +b +a
2
F AO −
2
2
c +b +a
a
2
2
c +b +a
2
2
2
( c + e) + ( b + f) + a
b
−
c+e
2
2
2
( c + e) + ( b + f) + a
a
2
2
( c + e) + ( b + f) + a
⎛ FAO ⎞
⎜
⎟
⎜ FAB ⎟ = Find ( FAO , FAB , FAC)
⎜F ⎟
⎝ AC ⎠
2
c
2
2
c + ( b + d) + a
b+ f
F AO +
F AO −
2
F AB −
2
F AC = 0
b+d
F AB +
F AB −
2
2
2
c + ( b + d) + a
a
2
2
c + ( b + d) + a
2
2
F AC = 0
F AC − ml g = 0
⎛ FAO ⎞ ⎛ 318.82 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB ⎟ = ⎜ 110.36 ⎟ N
⎜ F ⎟ ⎝ 85.84 ⎠
⎝ AC ⎠
179
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Engineering Mechanics - Statics
Chapter 3
Problem 3-51
Cables AB and AC can sustain a maximum tension Tmax, and the pole can support a maximum
compression Pmax. Determine the maximum weight of the lamp that can be supported in the
position shown. The force in the pole acts along the axis of the pole.
Given:
Tmax = 500 N
c = 2m
Pmax = 300 N
d = 1.5 m
a = 6m
e = 4m
b = 1.5 m
f = 1.5 m
Solution:
Lengths
2
2
2
AO =
a +b +c
AB =
a + ( c + e) + ( b + d)
AC =
a + c + ( b + d)
2
2
2
2
2
2
The initial guesses:
F AO = Pmax
F AB = Tmax
F AC = Tmax
W = 300N
Case 1 Assume the pole reaches maximum compression
Given
⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞
⎛0⎞
AB ⎜
AC ⎜
⎟
⎟
⎟
−b +
b+ f +
b + d + W⎜ 0 ⎟ = 0
⎜
⎟
⎜
⎟
⎜
⎟
⎜ ⎟
AO
AB
AC
⎝a⎠
⎝ −a ⎠
⎝ −a ⎠
⎝ −1 ⎠
F AO ⎜
⎛ W1 ⎞
⎜
⎟
⎜ FAB1 ⎟ = Find ( W , FAB , FAC)
⎜F
⎟
⎝ AC1 ⎠
⎛ W1 ⎞ ⎛ 138.46 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB1 ⎟ = ⎜ 103.85 ⎟ N
⎜F
⎟
⎝ AC1 ⎠ ⎝ 80.77 ⎠
180
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Engineering Mechanics - Statics
Chapter 3
Case 2 Assume that cable AB reaches maximum tension
Given
⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞
⎛0⎞
AB ⎜
AC ⎜
⎟
⎟
⎟
−b +
b+ f +
b + d + W⎜ 0 ⎟ = 0
⎟ AC ⎜
⎟
⎜ ⎟
AO ⎜ ⎟
AB ⎜
⎝a⎠
⎝ −a ⎠
⎝ −a ⎠
⎝ −1 ⎠
F AO ⎜
⎛ W2 ⎞
⎜
⎟
⎜ FAO2 ⎟ = Find ( W , FAO , FAC)
⎜F
⎟
⎝ AC2 ⎠
⎛ W2 ⎞ ⎛ 666.67 ⎞
⎜
⎟ ⎜
⎟
⎜ FAO2 ⎟ = ⎜ 1444.44 ⎟ N
⎜F
⎟
⎝ AC2 ⎠ ⎝ 388.89 ⎠
Case 3 Assume that cable AC reaches maximum tension
Given
⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞
⎛0⎞
AB ⎜
AC ⎜
⎟
⎟
⎟
−b +
b+ f +
b + d + W⎜ 0 ⎟ = 0
⎟ AC ⎜
⎟
⎜ ⎟
AO ⎜ ⎟
AB ⎜
⎝a⎠
⎝ −a ⎠
⎝ −a ⎠
⎝ −1 ⎠
F AO ⎜
⎛ W3 ⎞
⎜
⎟
⎜ FAO3 ⎟ = Find ( W , FAO , FAB)
⎜F
⎟
⎝ AB3 ⎠
Final Answer
⎛ W3 ⎞ ⎛ 857.14 ⎞
⎜
⎟ ⎜
⎟
⎜ FAO3 ⎟ = ⎜ 1857.14 ⎟ N
⎜F
⎟
⎝ AB3 ⎠ ⎝ 642.86 ⎠
W = min ( W1 , W2 , W3 )
W = 138.46 N
Problem 3-52
Determine the tension in cables AB, AC, and AD, required to hold the crate of weight W in
equilibrium.
Given:
W = 60 lb
a = 6 ft
b = 12 ft
c = 8 ft
181
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Engineering Mechanics - Statics
Chapter 3
d = 9 ft
e = 4 ft
f = 6 ft
Solution:
The initial guesses:
TB = 100 lb
TC = 100 lb
TD = 100 lb
Given
Σ F x = 0;
b
TB −
2
2
b +c +d
Σ F y = 0;
d
2
2
b +c +d
Σ F z = 0;
2
2
TC −
2
2
2
2
2
2
b +e + f
e
2
b +e + f
c
−W +
b
TC −
b +c +d
2
TC +
2
TD = 0
TD = 0
f
2
2
b +e + f
2
TD = 0
Solving
⎛ TB ⎞
⎜ ⎟
⎜ TC ⎟ = Find ( TB , TC , TD)
⎜T ⎟
⎝ D⎠
⎛ TB ⎞ ⎛ 108.84 ⎞
⎜ ⎟ ⎜
⎟
⎜ TC ⎟ = ⎜ 47.44 ⎟ lb
⎜ T ⎟ ⎝ 87.91 ⎠
⎝ D⎠
Problem 3-53
The bucket has weight W. Determine the tension developed in each cord for equilibrium.
Given:
W = 20 lb
182
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Engineering Mechanics - Statics
Chapter 3
a = 2 ft
b = 2 ft
c = 8 ft
d = 7 ft
e = 3 ft
f = a
Solution:
F DA = 20 lb
Initial Guesses:
F DB = 10 lb
F DC = 15 lb
Given
c−e
ΣF x = 0;
2
2
( c − e) + b + a
2
−b
ΣF x = 0;
2
2
( c − e) + b + a
2
a
ΣF y = 0;
2
2
( c − e) + b + a
2
F DA +
−e
2
2
e + ( d − b) + f
F DA +
d−b
2
2
e + ( d − b) + f
F DA +
2
2
f
2
2
e + ( d − b) + f
2
F DC +
−e
2
2
e +b + f
F DC +
−b
2
2
2
2
2
e +b + f
F DC +
2
f
2
e +b + f
F DB = 0
F DB = 0
F DB − W = 0
⎛ FDA ⎞
⎜
⎟
⎜ FDB ⎟ = Find ( FDA , FDB , FDC)
⎜F ⎟
⎝ DC ⎠
⎛ FDA ⎞ ⎛ 21.54 ⎞
⎜
⎟ ⎜
⎟
⎜ FDB ⎟ = ⎜ 13.99 ⎟ lb
⎜ F ⎟ ⎝ 17.61 ⎠
⎝ DC ⎠
Problem 3-54
The mast OA is supported by three cables. If cable AB is subjected to tension T, determine the tension
in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A.
Given:
T = 500 N
183
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Engineering Mechanics - Statics
Chapter 3
a = 6m
b = 3m
c = 6m
d = 3m
e = 2m
f = 1.5 m
g = 2m
Solution:
Initial Guesses:
F AC = 90 N
F AD = 350 N
F = 750 N
Given
Σ F x = 0;
e
2
2
e +d +a
Σ F y = 0;
2
2
−a
2
2
2
e +d +a
2
f +g +a
2
2
f +g +a
⎛ FAC ⎞
⎜
⎟
F
⎜ AD ⎟ = Find ( FAC , FAD , F)
⎜ F ⎟
⎝
⎠
2
a
T−
2
2
g
T+
e +d +a
Σ F z = 0;
2
d
2
f
T−
2
2
f +g +a
2
F AC −
b
2
2
2
2
2
2
2
b +c +a
F AC −
c
2
b +c +a
F AC −
a
2
b +c +a
F AD = 0
F AD = 0
F AD + F = 0
⎛ FAC ⎞ ⎛ 92.9 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AD ⎟ = ⎜ 364.3 ⎟ N
⎜ F ⎟ ⎝ 757.1 ⎠
⎝
⎠
Problem 3-55
The ends of the three cables are attached to a ring at A and to the edge of the uniform plate of mass
M. Determine the tension in each of the cables for equilibrium.
Given:
M = 150 kg
e = 4m
184
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Engineering Mechanics - Statics
Chapter 3
a = 2m
f = 6m
b = 10 m
g = 6m
c = 12 m
h = 6m
d = 2m
i = 2m
gravity = 9.81
m
2
s
Solution:
The initial guesses:
F B = 15 N
F C = 16 N
F D = 16 N
Given
Σ F x = 0;
F B( f − i )
2
2
+
2
2
( f − i) + h + c
Σ F y = 0;
F B ( − h)
2
2
F B( −c)
2
2
( f − i) + h + c
⎛ FB ⎞
⎜ ⎟
⎜ FC ⎟ = Find ( FB , FC , FD)
⎜F ⎟
⎝ D⎠
2
2
+
2
2
FC( −c)
2
2
( d + e) + ( h − a) + c
2
=0
2
FD( g)
2
2
=0
2
e +g +c
+
2
FD( −e)
e +g +c
( d + e) + ( h − a) + c
+
2
2
F C [ − ( h − a) ]
2
( f − i) + h + c
Σ F z = 0;
+
( d + e) + ( h − a) + c
+
2
F C( −d − e)
FD( −c)
2
2
+ M gravity = 0
2
e +g +c
⎛ FB ⎞ ⎛ 858 ⎞
⎜ ⎟ ⎜
⎟
⎜ FC ⎟ = ⎜ 0 ⎟ N
⎜ F ⎟ ⎝ 858 ⎠
⎝ D⎠
Problem 3-56
The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine
the largest mass the plate can have if each cable can support a maximum tension of T.
185
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Engineering Mechanics - Statics
Chapter 3
3
kN = 10 N
Given:
T = 15 kN
e = 4m
a = 2m
f = 6m
b = 10 m
g = 6m
c = 12 m
h = 6m
d = 2m
i = 2m
gravity = 9.81
m
2
s
Solution:
The initial guesses:
FB = T
FC = T
FD = T
M = 1 kg
Case 1: Assume that cable B reaches maximum tension
Given
Σ F x = 0;
F B( f − i )
2
2
+
2
2
( f − i) + h + c
Σ F y = 0;
F B ( − h)
2
2
F B( −c)
2
2
( f − i) + h + c
⎛ M1 ⎞
⎜
⎟
⎜ FC1 ⎟ = Find ( M , FC , FD)
⎜F ⎟
⎝ D1 ⎠
2
2
FC( −c)
2
2
( d + e) + ( h − a) + c
⎛ FC1 ⎞ ⎛ −0.00 ⎞
⎜
⎟=⎜
⎟ kN
15.00
F
⎝
⎠
D1
⎝
⎠
=0
2
FD( g)
2
2
=0
2
e +g +c
+
2
2
e +g +c
+
2
FD( −e)
2
( d + e) + ( h − a) + c
+
2
2
F C [ − ( h − a) ]
2
( f − i) + h + c
Σ F z = 0;
+
( d + e) + ( h − a) + c
+
2
F C( −d − e)
FD( −c)
2
2
+ M gravity = 0
2
e +g +c
M1 = 2621.23 kg
186
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Engineering Mechanics - Statics
Chapter 3
Case 2: Assume that cable D reaches maximum tension
Given
F B( f − i )
Σ F x = 0;
2
2
2
2
( f − i) + h + c
F B ( − h)
Σ F y = 0;
2
2
F B( −c)
2
2
2
( f − i) + h + c
⎛ M2 ⎞
⎜
⎟
⎜ FB2 ⎟ = Find ( M , FB , FC)
⎜F ⎟
⎝ C2 ⎠
2
2
FC( −c)
2
2
( d + e) + ( h − a) + c
⎛ FB2 ⎞ ⎛ 15.00 ⎞
⎜
⎟=⎜
⎟ kN
0.00
F
⎝
⎠
C2
⎝
⎠
=0
2
FD( g)
2
2
=0
2
e +g +c
+
2
2
e +g +c
+
2
FD( −e)
2
( d + e) + ( h − a) + c
+
2
2
F C [ − ( h − a) ]
+
2
+
( d + e) + ( h − a) + c
( f − i) + h + c
Σ F z = 0;
F C( −d − e)
+
FD( −c)
2
2
+ M gravity = 0
2
e +g +c
M2 = 2621.23 kg
For this set of number FC = 0 for any mass that is applied. For a different set of numbers it would be
necessary to also check case 3: Assume that the cable C reaches a maximum.
M = min ( M1 , M2 )
M = 2621.23 kg
Problem 3-57
The crate of weight W is suspended from the cable system shown. Determine the force in each
segment of the cable, i.e., AB, AC, CD, CE, and CF. Hint: First analyze the equilibrium of point A,
then using the result for AC, analyze the equilibrium of point C.
Units Used:
kip = 1000 lb
Given:
W = 500 lb
a = 10 ft
b = 24 ft
c = 24 ft
187
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Engineering Mechanics - Statics
Chapter 3
d = 7 ft
e = 7 ft
θ 1 = 20 deg
θ 2 = 35 deg
Solution:
At A:
Initial guesses:
F AC = 570 lb
F AB = 500 lb
Given
+
ΣF x = 0;
→
F AB cos ( θ 1 ) − F AC cos ( θ 2 ) = 0
+
F AB sin ( θ 1 ) + FAC sin ( θ 2 ) − W = 0
↑ ΣFy = 0;
⎛ FAC ⎞
⎜
⎟ = Find ( FAC , FAB)
⎝ FAB ⎠
At C:
⎛ FAC ⎞ ⎛ 574 ⎞
⎜
⎟=⎜
⎟ lb
⎝ FAB ⎠ ⎝ 500 ⎠
Initial Guesses
F CD = 1 lb
F CE = 1 lb
F CF = 1 lb
Given
⎛ cos ( θ 2 ) ⎞
⎜
⎟
F AC⎜
0
⎟+
⎜ −sin ( θ ) ⎟
2 ⎠
⎝
⎛ −a ⎞
⎜ 0⎟+
⎟
2
2⎜
a +b ⎝ b ⎠
FCD
⎛0⎞
⎜d⎟+
2
2⎜ ⎟
c + d ⎝ −c ⎠
F CE
⎛0⎞
⎜ −e ⎟ = 0
2
2⎜ ⎟
c + e ⎝ −c ⎠
F CF
188
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Engineering Mechanics - Statics
Chapter 3
⎛ FCD ⎞
⎜
⎟
F
⎜ CE ⎟ = Find ( FCD , FCE , FCF )
⎜F ⎟
⎝ CF ⎠
F CD = 1.22 kip
⎛ FCE ⎞ ⎛ 416 ⎞
⎜
⎟=⎜
⎟ lb
⎝ FCF ⎠ ⎝ 416 ⎠
Problem 3-58
The chandelier of weight W is supported by three wires as shown. Determine the force in each
wire for equilibrium.
Given:
W = 80 lb
r = 1 ft
h = 2.4 ft
Solution:
The initial guesses:
F AB = 40 lb
F AC = 30 lb
F AD = 30 lb
Given
r
Σ F x = 0;
2
2
FAC −
r +h
Σ F y = 0;
−r
2
2
r +h
FAD +
r cos ( 45 deg)
2
r +h
2
r cos ( 45 deg)
2
r +h
2
FAB = 0
FAB = 0
189
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Engineering Mechanics - Statics
Chapter 3
h
Σ F z = 0;
2
2
FAC +
r +h
h
2
r +h
⎛ FAB ⎞
⎜
⎟
⎜ FAC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
2
F AD +
h
2
2
FAB − W = 0
r +h
⎛ FAB ⎞ ⎛ 35.9 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC ⎟ = ⎜ 25.4 ⎟ lb
⎜ F ⎟ ⎝ 25.4 ⎠
⎝ AD ⎠
Problem 3-59
If each wire can sustain a maximum tension Tmax before it fails, determine the greatest weight
of the chandelier the wires will support in the position shown.
Given:
Tmax = 120 lb
r = 1 ft
h = 2.4 ft
Solution:
The initial guesses:
F AB = Tmax
F AC = Tmax
F AD = Tmax
W = Tmax
Case 1 Assume that cable AB has maximum tension
Given
r
Σ F x = 0;
2
2
FAC −
r +h
Σ F y = 0;
−r
2
2
r +h
FAD +
r cos ( 45deg)
2
r +h
2
r cos ( 45deg)
2
r +h
2
F AB = 0
F AB = 0
190
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Engineering Mechanics - Statics
Chapter 3
h
Σ Fz = 0;
2
2
FAC +
r +h
h
2
r +h
⎛ W1 ⎞
⎜
⎟
⎜ FAC1 ⎟ = Find ( W , FAC , FAD)
⎜F
⎟
⎝ AD1 ⎠
2
F AD +
h
2
2
FAB − W = 0
r +h
⎛ W1 ⎞ ⎛ 267.4 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC1 ⎟ = ⎜ 84.9 ⎟ lb
⎜F
⎟
⎝ AD1 ⎠ ⎝ 84.9 ⎠
Case 2 Assume that cable AC has maximum tension
Given
r
Σ F x = 0;
2
2
FAC −
r cos ( 45 deg)
2
r +h
Σ F y = 0;
−r
2
r +h
FAD +
2
r cos ( 45 deg)
r +h
h
Σ F z = 0;
2
2
2
FAC +
r +h
FAB = 0
FAB = 0
2
r +h
2
h
F AD +
2
r +h
⎛ W2 ⎞
⎜
⎟
⎜ FAB2 ⎟ = Find ( W , FAB , FAD)
⎜F
⎟
⎝ AD2 ⎠
2
h
2
2
FAB − W = 0
r +h
⎛ W2 ⎞ ⎛ 378.2 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB2 ⎟ = ⎜ 169.7 ⎟ lb
⎜F
⎟
⎝ AD2 ⎠ ⎝ 120 ⎠
Case 3 Assume that cable AD has maximum tension
Given
r
Σ F x = 0;
2
2
FAC −
r +h
Σ F y = 0;
−r
2
2
2
r +h
h
2
r +h
2
r cos ( 45deg)
r +h
Σ F z = 0;
2
r +h
FAD +
2
r cos ( 45deg)
FAC +
2
h
2
r +h
2
F AB = 0
F AB = 0
F AD +
h
2
2
FAB − W = 0
r +h
191
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Engineering Mechanics - Statics
Chapter 3
⎛ W3 ⎞
⎜
⎟
⎜ FAB3 ⎟ = Find ( W , FAB , FAC)
⎜F
⎟
⎝ AC3 ⎠
W = min ( W1 , W2 , W3 )
⎛ W3 ⎞ ⎛ 378.2 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB3 ⎟ = ⎜ 169.7 ⎟ lb
⎜F
⎟
⎝ AC3 ⎠ ⎝ 120 ⎠
W = 267.42 lb
Problem 3-60
Determine the force in each cable used to lift the surge arrester of mass M at constant velocity.
Units Used:
3
kN = 10 N
3
Mg = 10 kg
Given:
M = 9.50 Mg
a = 2m
b = 0.5 m
θ = 45 deg
Solution:
Initial guesses:
F B = 50 kN
F C = 30 kN
Given
b
ΣF x = 0;
FB
ΣF y = 0;
FC
ΣF z = 0;
M g − FB
2
b +a
− FC
2
2
=0
2
b +a
−b sin ( θ )
2
b cos ( θ )
F D = 10 kN
+ FD
2
b +a
b
2
=0
b +a
a
2
2
b +a
− FC
2
a
2
b +a
2
− FD
a
2
=0
2
b +a
192
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Engineering Mechanics - Statics
Chapter 3
⎛ FB ⎞
⎜ ⎟
⎜ FC ⎟ = Find ( FB , FC , FD)
⎜F ⎟
⎝ D⎠
⎛ FB ⎞ ⎛ 28.13 ⎞
⎜ ⎟ ⎜
⎟
⎜ FC ⎟ = ⎜ 39.78 ⎟ kN
⎜ F ⎟ ⎝ 28.13 ⎠
⎝ D⎠
Problem 3-61
The cylinder of weight W is supported by three chains as shown. Determine the force in each
chain for equilibrium.
Given:
W = 800 lb
r = 1 ft
d = 1 ft
Solution:
The initial guesses:
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
Given
Σ F x = 0;
r
2
2
FAC −
r cos ( 45 deg)
2
r +d
Σ F y = 0;
−r
2
2
r +d
FAD +
r +d
d
Σ F z = 0;
2
2
2
r sin ( 45 deg)
FAB = 0
F AB = 0
2
2
d
F AC +
r +d
FAD +
r +d
⎛ FAB ⎞
⎜
⎟
⎜ FAC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
2
r +d
2
d
2
2
FAB − W = 0
r +d
⎛ FAB ⎞ ⎛ 469 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC ⎟ ⎜ 331 ⎟
⎜ F ⎟ = ⎝ 331 ⎠ lb
⎝ AD ⎠
193
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Engineering Mechanics - Statics
Chapter 3
Problem 3-62
The triangular frame ABC can be adjusted vertically between the three equal-length cords. If it
remains in a horizontal plane, determine the required distance s so that the tension in each of the
cords, OA, OB, and OC, equals F . The lamp has a mass M.
Given:
F = 20 N
M = 5 kg
m
g = 9.81
2
s
d = 0.5 m
Solution:
3F cos ( γ ) = M g
ΣF z = 0;
γ = acos ⎛⎜
M g⎞
⎟
⎝ 3F ⎠
2d cos ( 30 deg)
Geometry
3
s =
γ = 35.16 deg
= s tan ( γ )
2d cos ( 30 deg)
3 tan ( γ )
s = 410 mm
Problem 3-63
Determine the force in each cable needed to support the platform of weight W.
Units Used:
3
kip = 10 lb
194
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Engineering Mechanics - Statics
Chapter 3
Given:
W = 3500 lb
d = 4 ft
a = 2 ft
e = 3 ft
f = 3 ft
b = 4 ft
c = 4 ft
g = 10 ft
Solution:
The initial guesses:
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
Given
−b
2
2
2
c−d
FAD +
2
g + ( e − a) + b
e−a
2
2
2
2
2
2
g + ( e − a) + b
2
( c − d) + e + g
e
FAD +
2
g + ( e − a) + b
−g
2
2
2
g
2
2
2
e + ( c − d) + g
⎛ FAB ⎞
⎜
⎟
⎜ FAC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
2
2
2
2
2
c + f +g
f
FAC −
e + ( c − d) + g
FAD −
c
FAC +
FAC −
2
c + f +g
g
2
2
2
F AB = 0
F AB = 0
F AB + W = 0
g + f +c
⎛ FAB ⎞ ⎛ 1.467 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC ⎟ = ⎜ 0.914 ⎟ kip
⎜ F ⎟ ⎝ 1.42 ⎠
⎝ AD ⎠
195
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Engineering Mechanics - Statics
Chapter 3
Problem 3-64
A flowerpot of mass M is supported at A by the three cords. Determine the force acting in each
cord for equilibrium.
Given:
M = 25 kg
g = 9.81
m
2
s
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 60 deg
θ 4 = 45 deg
Solution:
Initial guesses:
F AB = 1 N
F AD = 1 N
F AC = 1 N
Given
ΣF x = 0;
F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0
ΣF y = 0;
−F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0
ΣF z = 0;
F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − M g = 0
⎛ FAB ⎞
⎜
⎟
⎜ FAC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
⎛ FAB ⎞ ⎛ 219.89 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC ⎟ = ⎜ 103.65 ⎟ N
⎜ F ⎟ ⎝ 103.65 ⎠
⎝ AD ⎠
Problem 3-65
If each cord can sustain a maximum tension of T before it fails, determine the greatest weight of the
flowerpot the cords can support.
196
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Engineering Mechanics - Statics
Chapter 3
Given:
T = 50 N
θ 1 = 30 deg
θ 2 = 30 deg
θ 3 = 60 deg
θ 4 = 45 deg
Solution:
Initial guesses:
F AB = T
F AD = T
F AC = T
W = T
Case 1
Assume that AB reaches maximum tension
Given
ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0
ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0
ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0
⎛ W1 ⎞
⎜
⎟
⎜ FAC1 ⎟ = Find ( W , FAC , FAD)
⎜F
⎟
⎝ AD1 ⎠
Case 2
⎛ W1 ⎞ ⎛ 55.77 ⎞
⎜
⎟ ⎜
⎟
⎜ FAC1 ⎟ = ⎜ 23.57 ⎟ N
⎜F
⎟
⎝ AD1 ⎠ ⎝ 23.57 ⎠
Assume that AC reaches maximum tension
197
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Engineering Mechanics - Statics
Chapter 3
Given
ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0
ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0
ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0
⎛ W2 ⎞
⎜
⎟
⎜ FAB2 ⎟ = Find ( W , FAB , FAD)
⎜F
⎟
⎝ AD2 ⎠
Case 3
⎛ W2 ⎞ ⎛ 118.30 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB2 ⎟ = ⎜ 106.07 ⎟ N
⎜F
⎟
⎝ AD2 ⎠ ⎝ 50.00 ⎠
Assume that AD reaches maximum tension
Given
ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0
ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0
ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0
⎛ W3 ⎞
⎜
⎟
⎜ FAB3 ⎟ = Find ( W , FAB , FAC)
⎜F
⎟
⎝ AC3 ⎠
W = min ( W1 , W2 , W3 )
⎛ W3 ⎞ ⎛ 118.30 ⎞
⎜
⎟ ⎜
⎟
⎜ FAB3 ⎟ = ⎜ 106.07 ⎟ N
⎜F
⎟
⎝ AC3 ⎠ ⎝ 50.00 ⎠
W = 55.77 N
198
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Engineering Mechanics - Statics
Chapter 3
Problem 3-66
The pipe is held in place by the vice. If the bolt
exerts force P on the pipe in the direction
shown, determine the forces F A and F B that the
smooth contacts at A and B exerton the pipe.
Given:
P = 50 lb
θ = 30 deg
c = 3
d = 4
Solution:
Initial Guesses
F A = 1 lb
F B = 1 lb
Given
+
Σ F x = 0;
→
F B − FA sin ( θ ) − P⎜
+
−F A cos ( θ ) + P⎜
↑Σ Fy = 0;
⎛ FA ⎞
⎜ ⎟ = Find ( FA , FB)
⎝ FB ⎠
⎛
⎞=0
⎟
2
2
⎝ c +d ⎠
⎛
d
⎞=0
⎟
2
2
⎝ c +d ⎠
c
⎛ FA ⎞ ⎛ 34.6 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ FB ⎠ ⎝ 57.3 ⎠
Problem 3-67
When y is zero, the springs sustain force F0. Determine the magnitude of the applied vertical
forces F and -F required to pull point A away from point B a distance y1. The ends of cords
CAD and CBD are attached to rings at C and D.
199
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Engineering Mechanics - Statics
Chapter 3
Given:
F 0 = 60 lb
k = 40
lb
ft
d = 2 ft
y1 = 2 ft
Solution:
Initial spring stretch:
s1 =
F0
s1 = 1.50 ft
k
Initial guesses:
F s = 1 lb
T = 1 lb
F = 1 lb
Given
2
⎛ y1 ⎞
d −⎜ ⎟
⎝ 2⎠ T−F =0
2
F−
y1
2d
T=0
⎛⎜ Fs ⎞⎟
⎜ T ⎟ = Find ( Fs , T , F)
⎜F⎟
⎝ ⎠
d
s
⎛ Fs ⎞ ⎛ 70.72 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ T ⎠ ⎝ 81.66 ⎠
2
⎡
⎤
2 ⎛ y1 ⎞
⎢
F s = k d − d − ⎜ ⎟ + s1⎥
⎣
⎝2⎠
⎦
F = 40.83 lb
Problem 3-68
When y is zero, the springs are each stretched a distance δ. Determine the distance y if a force
F is applied to points A and B as shown. The ends of cords CAD and CBD are attached to
200
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Engineering Mechanics - Statics
Chapter 3
pp
p
rings at C and D.
Given:
δ = 1.5 ft
k = 40
lb
ft
d = 2 ft
F = 60 lb
Solution:
Initial guesses:
F s = 1 lb
T = 1 lb
y = 1 ft
Given
F−
y
T=0
2d
2
y
d − ⎛⎜ ⎟⎞
⎝ 2⎠ T − F = 0
2
d
s
⎛⎜ Fs ⎞⎟
⎜ T ⎟ = Find ( Fs , T , y)
⎜y⎟
⎝ ⎠
⎡
F s = k⎢d −
⎣
2
2
⎤
⎛ y ⎞ + δ⎥
⎟
⎝ 2⎠
⎦
d −⎜
⎛ Fs ⎞ ⎛ 76.92 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ T ⎠ ⎝ 97.55 ⎠
y = 2.46 ft
Problem 3-69
Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other
end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a
weight W is suspended from B, determine the angle θ of cord AB for equilibrium.
201
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Engineering Mechanics - Statics
Chapter 3
Given:
a = 5 ft
b = 5 ft
k = 10
lb
ft
W = 10 lb
Solution:
Initial Guesses
F BA = 1 lb
F sp = 1 lb
θ = 30 deg
Given
F sp − FBA cos ( θ ) = 0
F BA sin ( θ ) − W = 0
F sp = k( a − a cos ( θ ) )
⎛ Fsp ⎞
⎜
⎟
⎜ FBA ⎟ = Find ( Fsp , FBA , θ )
⎜
⎟
⎝ θ ⎠
⎛ Fsp ⎞ ⎛ 11.82 ⎞
⎜
⎟=⎜
⎟ lb
⎝ FBA ⎠ ⎝ 15.49 ⎠
θ = 40.22 deg
Problem 3-70
The uniform crate of mass M is suspended by using a cord of length l that is attached to the sides of
the crate and passes over the small pulley at O. If the cord can be attached at either points A and B, or
C and D, determine which attachment produces the least amount of tension in the cord and specify
the cord tension in this case.
202
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
Given:
M = 50 kg
g = 9.81
m
2
s
a = 0.6 m
b = 1.5 m
l = 2m
c =
a
d =
b
2
2
Solution:
Case 1 Attached at A and B
Given
⎡⎢ l 2 2 ⎥⎤
⎛ ⎞ −d
⎢ ⎜⎝ 2 ⎟⎠
⎥
Mg − ⎢
⎥ 2T = 0
l
⎢
⎥
2
⎣
⎦
Case 2 Attached at C and D
Given
Guess
Guess
⎡⎢ l 2 2 ⎥⎤
⎛ ⎞ −c
⎢ ⎜⎝ 2 ⎟⎠
⎥
Mg − ⎢
⎥ 2T = 0
l
⎢
⎥
2
⎣
⎦
Choose the arrangement that
gives the smallest tension.
T = 1N
T1 = Find ( T)
T1 = 370.78 N
T = 1N
T2 = Find ( T)
T = min ( T1 , T2 )
T2 = 257.09 N
T = 257.09 N
Problem 3-71
The man attempts to pull the log at C by using the three ropes. Determines the direction
θ in which he should pull on his rope with a force P, so that he exerts a maximum
203
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Engineering Mechanics - Statics
Chapter 3
p
p
force on the log. What is the force on the log for this case ? Also, determine the direction
in which he should pull in order to maximize the force in the rope attached to B.What is this
maximum force?
Given:
P = 80 lb
φ = 150 deg
Solution:
+
Σ F x = 0;
→
F AB + P cos ( θ ) − F AC sin ( φ − 90 deg) = 0
+
P sin ( θ ) − F AC cos ( φ − 90 deg) = 0
↑Σ Fy = 0;
F AC =
P sin ( θ )
In order to maximize FAC we choose
cos ( φ − 90deg)
Thus
θ = 90 deg
F AC =
P sin ( θ )
cos ( φ − 90deg)
sin ( θ ) = 1.
F AC = 160.00 lb
Now let's find the force in the rope AB.
F AB = −P cos ( θ ) + F AC sin ( φ − 90 deg)
F AB = −P cos ( θ ) +
F AB = P
P sin ( θ ) sin ( φ − 90 deg)
cos ( φ − 90 deg)
sin ( θ ) sin ( φ − 90 deg) − cos ( θ ) cos ( φ − 90 deg)
cos ( φ − 90 deg)
= −P
cos ( θ + φ − 90deg)
cos ( φ − 90 deg)
204
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Engineering Mechanics - Statics
Chapter 3
cos ( θ + φ − 90 deg) = −1
In order to maximize the force we set
θ + φ − 90 deg = 180 deg
θ = 270 deg − φ
F AB = −P
θ = 120.00 deg
cos ( θ + φ − 90 deg)
F AB = 160.00 lb
cos ( φ − 90 deg)
Problem 3-72
The "scale" consists of a known weight W which is suspended at A from a cord of total length L.
Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of
the pulleys.
Solution:
+
↑ ΣFy = 0;
2W sin ( θ ) − w = 0
Geometry
h=
2
2
⎛ L − y ⎞ − ⎛ d ⎞ = 1 ( L − y) 2 − d2
⎜
⎟ ⎜ ⎟
2
⎝ 2 ⎠ ⎝ 2⎠
205
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
⎡ 1 ( L − y) 2 + d2 ⎤
⎢2
⎥
w = 2W⎢
⎥
L− y
⎢
⎥
2
⎣
⎦
w=
2W
L−y
2
2
( L − y) − d
Problem 3-73
Determine the maximum weight W that can be supported in the
position shown if each cable AC and AB can support a maximum
tension of F before it fails.
Given:
θ = 30 deg
F = 600 lb
c = 12
d = 5
Solution:
Initial Guesses
F AB = F
F AC = F
W = F
Case 1 Assume that cable AC reaches maximum tension
Given
F AC sin ( θ ) −
d
2
FAB = 0
2
c +d
F AC cos ( θ ) +
c
2
2
FAB − W = 0
c +d
⎛ W1 ⎞
⎜
⎟ = Find ( W , FAB)
⎝ FAB1 ⎠
⎛ W1 ⎞ ⎛ 1239.62 ⎞
⎜
⎟=⎜
⎟ lb
⎝ FAB1 ⎠ ⎝ 780.00 ⎠
Case 2 Assume that cable AB reaches maximum tension
206
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Engineering Mechanics - Statics
F AC sin ( θ ) −
Given
Chapter 3
d
2
FAB = 0
2
c +d
F AC cos ( θ ) +
c
2
2
FAB − W = 0
c +d
⎛ W2 ⎞
⎜
⎟ = Find ( W , FAC)
⎝ FAC2 ⎠
W = min ( W1 , W2 )
⎛ W2 ⎞ ⎛ 953.55 ⎞
⎜
⎟=⎜
⎟ lb
⎝ FAC2 ⎠ ⎝ 461.54 ⎠
W = 953.6 lb
Problem 3-74
If the spring on rope OB has been stretched a distance δ. and fixed in place as shown, determine the
tension developed in each of the other three ropes in order to hold the weight W in equilibrium. Rope
OD lies in the x-y plane.
Given:
a = 2 ft
b = 4 ft
c = 3 ft
d = 4 ft
e = 4 ft
f = 4 ft
xB = −2 ft
yB = −3 ft
zB = 3 ft
θ = 30 deg
k = 20
lb
in
δ = 2 in
W = 225 lb
207
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Engineering Mechanics - Statics
Chapter 3
Solution:
F OA = 10 lb
Initial Guesses
F OC = 10 lb
F OD = 10 lb
Given
c
2
2
2
a +b +c
−b
2
2
2
2
2
2
a +b +c
2
2
2
yB
F OA + kδ
2
2
2
⎛ FOA ⎞
⎜
⎟
⎜ FOC ⎟ = Find ( FOA , FOC , FOD)
⎜F ⎟
⎝ OD ⎠
2
2
2
2
2
f
2
d +e + f
e
+
xB + yB + zB
2
d +e + f
2
zB
2
2
+
xB + yB + zB
F OA + kδ
−d
+
xB + yB + zB
a +b +c
a
xB
F OA + kδ
2
2
d +e + f
FOC + F OD sin ( θ ) = 0
FOC + F OD cos ( θ ) = 0
FOC − W = 0
⎛ FOA ⎞ ⎛ 201.6 ⎞
⎜
⎟ ⎜
⎟
⎜ FOC ⎟ = ⎜ 215.7 ⎟ lb
⎜ F ⎟ ⎝ 58.6 ⎠
⎝ OD ⎠
Problem 3-75
The joint of a space frame is subjected to four
member forces. Member OA lies in the x - y plane
and member OB lies in the y - z plane. Determine the
forces acting in each of the members required for
equilibrium of the joint.
Given:
F 4 = 200 lb
θ = 40 deg
φ = 45 deg
Solution:
The initial guesses : F 1 = 200 lb
F 2 = 200 lb
F 3 = 200 lb
208
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Engineering Mechanics - Statics
Chapter 3
Given
Σ F y = 0;
F 3 + F 1 cos ( φ ) − F 2 cos ( θ ) = 0
Σ F x = 0;
−F 1 sin ( φ ) = 0
Σ F z = 0;
F 2 sin ( θ ) − F4 = 0
⎛ F1 ⎞
⎜ ⎟
⎜ F2 ⎟ = Find ( F1 , F2 , F3)
⎜F ⎟
⎝ 3⎠
⎛ F1 ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ F2 ⎟ = ⎜ 311.1 ⎟ lb
⎜ F ⎟ ⎝ 238.4 ⎠
⎝ 3⎠
209
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Engineering Mechanics - Statics
Chapter 4
Problem 4-1
If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e.,
A × ( B + D) = ( A × B) + ( A × D).
Solution:
Consider the three vectors; with A vertical.
Note triangle obd is perpendicular to A.
od = A × ( B + D) = A ( B + D ) sin ( θ 3 )
ob = A × B = A B sin ( θ 1 )
bd = A × D = A B sin ( θ 2 )
Also, these three cross products all lie in the plane
obd since they are all perpendicular to A. As noted
the magnitude of each cross product is
proportional to the length of each side of the
triangle.
The three vector cross - products also form a
closed triangle o'b'd' which is similar to triangle obd.
Thus from the figure,
A × ( B + D) = A × B + A × D
(QED)
Note also,
A = Axi + Ayj + A zk
B = Bxi + B yj + B zK
D = Dxi + Dyj + Dzk
j
k
⎞
⎛ i
⎜
⎟
Ay
Az ⎟
A × ( B + D) = ⎜ Ax
⎜B + D B + D B + D ⎟
x
y
y
z
z⎠
⎝ x
=
=
⎡⎣Ay( Bz + Dz) − Az( By + Dy)⎤⎦ i − ⎡⎣Ax( Bz + Dz) − Az( Bx + Dx)⎤⎦ j + ⎡⎣Ax( By + Dy) − Ay( Bx − Dx)⎤⎦ k
⎡⎣( Ay Bz − Az By) i − ( Ax Bz − Az Bx) j + ( Ax By − Ay Bx) k⎤⎦ ...
+ ⎡⎣( Ay Dz − A z Dy) i − ( Ax Dz − A z Dx) j + ( A x Dy − A y Dx) k⎤⎦
210
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Engineering Mechanics - Statics
Chapter 4
⎛ i j k⎞
⎛ i j k⎞
⎜
⎟
⎜
⎟
= ⎜ Ax Ay Az ⎟ + ⎜ Ax Ay Az ⎟
⎜B B B ⎟
⎜D D D ⎟
⎝ x y z⎠
⎝ x y z⎠
( A × B) + ( A × D)
=
(QED)
Problem 4-2
Prove the triple scalar product identity A⋅ ( B × C) = ( A × B) ⋅ C.
Solution:
As shown in the figure
Area = B ( C sin ( θ ) ) = B × C
Thus,
Volume of parallelopiped is B × C h
But,
h = A⋅ u
B× C
⎛ B× C ⎞
⎟
⎝ B× C ⎠
= A⋅ ⎜
Thus,
Volume = A⋅ ( B × C)
Since A × B⋅ C represents this same volume then
A⋅ ( B × C) = ( A × B) ⋅ C
(QED)
Also,
LHS = A⋅ ( B × C)
⎛ i j k⎞
⎜
⎟
= ( A xi + A yj + Azk) ⎜ Bx By Bz ⎟
⎜C C C ⎟
⎝ x y z⎠
= Ax( By Cz − B z Cy) − Ay( Bx Cz − B z Cx) + Az( B x Cy − B y Cx)
= Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx
RHS = ( A × B) ⋅ C
211
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Engineering Mechanics - Statics
Chapter 4
⎛ i j k⎞
⎜
⎟
A
A
A
x
y
z
⎜
⎟ ( Cxi + Cyj + Czk)
⎜B B B ⎟
⎝ x y z⎠
=
= Cx( Ay B z − Az By) − Cy( A x Bz − A z B x) + Cz( A x By − Ay B x)
=
Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx
Thus,
LHS = RHS
A⋅ B × C = A × B⋅ C
(QED)
Problem 4-3
Given the three nonzero vectors A, B, and C, show that if A⋅ ( B × C) = 0, the three vectors
must lie in the same plane.
Solution:
Consider,
A⋅ ( B × C) = A B × C cos ( θ )
=
( A cos (θ ) ) B × C
=
h
B× C
= BC h sin ( φ )
= volume of parallelepiped.
If A⋅ ( B × C) = 0, then the volume equals zero, so that A, B, and C are coplanar.
Problem 4-4
Determine the magnitude and directional sense of the resultant moment of the forces at A and B
about point O.
Given:
F 1 = 40 lb
F 2 = 60 lb
212
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Engineering Mechanics - Statics
Chapter 4
θ 1 = 30 deg
θ 2 = 45 deg
a = 5 in
b = 13 in
c = 3 in
d = 6 in
e = 3 in
f = 6 in
Solution:
MRO =ΣMO; MRO = F 1 cos ( θ 2 ) e − F1 sin ( θ 2 ) f − F2 cos ( θ 1 ) b − a − F2 sin ( θ 1 ) a
2
MRO = −858 lb⋅ in
2
MRO = 858 lb⋅ in
Problem 4-5
Determine the magnitude and directional sense of the
resultant moment of the forces at A and B about
point P.
Units Used:
kip = 1000 lb
Given:
F 1 = 40 lb
b = 13 in
F 2 = 60 lb
c = 3 in
θ 1 = 30 deg
d = 6 in
θ 2 = 45 deg
e = 3 in
a = 5 in
f = 6 in
Solution:
MRP = ΣMP; MRP = F1 cos ( θ 2 ) ( e + c) − F 1 sin ( θ 2 ) ( d + f) − F2 cos ( θ 1 )
+ −F2 sin ( θ 1 ) ( a − c)
( b2 − a2 + d) ...
213
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Engineering Mechanics - Statics
Chapter 4
MRP = −1165 lb⋅ in
MRP = 1.17 kip⋅ in
Problem 4-6
Determine the magnitude of the force F that
should be applied at the end of the lever such
that this force creates a clockwise moment M
about point O.
Given:
M = 15 N m
φ = 60 deg
θ = 30 deg
a = 50 mm
b = 300 mm
Solution:
M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )
F =
M
cos ( θ ) ( a + b sin ( φ ) ) − sin ( θ ) ( b cos ( φ ) )
F = 77.6 N
Problem 4-7
Determine the angle θ (0 <= θ <= 90 deg) so that the force F develops a clockwise moment M
about point O.
Given:
F = 100 N
φ = 60 deg
214
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Engineering Mechanics - Statics
Chapter 4
M = 20 N⋅ m a = 50 mm
θ = 30 deg
b = 300 mm
Solution:
θ = 30 deg
Initial Guess
Given
M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )
θ = Find ( θ )
θ = 28.6 deg
Problem 4-8
Determine the magnitude and directional sense of the moment of the forces about point O.
Units Used:
3
kN = 10 N
Given:
F B = 260 N
e = 2m
a = 4m
f = 12
b = 3m
g = 5
c = 5m
θ = 30 deg
d = 2m
F A = 400 N
Solution:
Mo = F A sin ( θ ) d + FA cos ( θ ) c + F B
f
2
( a + e)
2
f +g
Mo = 3.57 kN⋅ m
(positive means counterclockwise)
215
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Engineering Mechanics - Statics
Chapter 4
Problem 4-9
Determine the magnitude and directional sense of the moment of the forces about point P.
Units Used:
3
kN = 10 N
Given:
F B = 260 N
e = 2m
a = 4m
f = 12
b = 3m
g = 5
c = 5m
θ = 30 deg
d = 2m
F A = 400 N
Solution:
Mp = F B
g
2
f +g
2
b + FB
Mp = 3.15 kN⋅ m
f
2
2
e − FA sin ( θ ) ( a − d) + FA cos ( θ ) ( b + c)
f +g
(positive means counterclockwise)
Problem 4-10
A force F is applied to the wrench. Determine the moment of this force about point O. Solve
the problem using both a scalar analysis and a vector analysis.
216
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 40 N
θ = 20 deg
a = 30 mm
b = 200 mm
Scalar Solution
MO = −F cos ( θ ) b + F sin ( θ ) a
MO = −7.11 N⋅ m
MO = 7.11 N⋅ m
Vector Solution
⎛ b ⎞ ⎛ −F sin ( θ ) ⎞
⎜ ⎟ ⎜
⎟
MO = a × −F cos ( θ )
⎜ ⎟ ⎜
⎟
0
⎝0⎠ ⎝
⎠
⎛ 0 ⎞
⎜ 0 ⎟ N⋅ m
MO =
⎜
⎟
⎝ −7.11 ⎠
MO = 7.107 N⋅ m
Problem 4-11
Determine the magnitude and directional sense of the resultant moment of the forces about
point O.
Units Used:
3
kip = 10 lb
Given:
F 1 = 300 lb
e = 10 ft
F 2 = 250 lb
f = 4
a = 6 ft
g = 3
b = 3 ft
θ = 30 deg
c = 4 ft
φ = 30 deg
d = 4 ft
217
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Engineering Mechanics - Statics
Chapter 4
Solution:
Mo = F 2
f
2
f +g
Mo = 2.42 kip⋅ ft
2
e sin ( φ ) + F2
g
2
2
e cos ( φ ) + F1 sin ( θ ) a − F1 cos ( θ ) b
f +g
positive means clockwise
Problem 4-12
To correct a birth defect, the tibia of the leg is straightened using three wires that are attached
through holes made in the bone and then to an external brace that is worn by the patient.
Determine the moment of each wire force about joint A.
Given:
F1 = 4 N
d = 0.15 m
F2 = 8 N
e = 20 mm
F3 = 6 N
f = 35 mm
a = 0.2 m
g = 15 mm
b = 0.35 m
θ 1 = 30 deg
c = 0.25 m
θ 2 = 15 deg
Solution:
Positive means counterclockwise
MA1 = F 1 cos ( θ 2 ) d + F1 sin ( θ 2 ) e
MA1 = 0.6 N⋅ m
MA2 = F 2 ( c + d)
MA2 = 3.2 N⋅ m
MA3 = F 3 cos ( θ 1 ) ( b + c + d) − F 3 sin ( θ 1 ) g MA3 = 3.852 N⋅ m
Problem 4-13
To correct a birth defect, the tibia of the leg is straightened using three wires that are attached
through holes made in the bone and then to an external brace that is worn by the patient.
Determine the moment of each wire force about joint B.
Given:
F1 = 4 N
d = 0.15 m
218
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Engineering Mechanics - Statics
F2 = 8 N
e = 20 mm
F3 = 6 N
f = 35 mm
a = 0.2 m
g = 15 mm
b = 0.35 m
θ 1 = 30 deg
c = 0.25 m
θ 2 = 15 deg
Chapter 4
Solution:
Positive means clockwise
MB1 = F 1 cos ( θ 2 ) ( a + b + c) − F 1 sin ( θ 2 ) e MB1 = 3.07 N⋅ m
MB2 = F 2 ( a + b)
MB2 = 4.4 N⋅ m
MB3 = F 3 cos ( θ 1 ) a + F 3 sin ( θ 1 ) g
MB3 = 1.084 N⋅ m
Problem 4-14
Determine the moment of each force about the bolt located at A.
Given:
F B = 40 lb
a = 2.5 ft
α = 20 deg
F C = 50 lb
b = 0.75 ft
β = 25 deg
γ = 30 deg
219
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Engineering Mechanics - Statics
Chapter 4
Solution:
MB = FB cos ( β ) a
MB = 90.6 lb⋅ ft
MC = F C cos ( γ ) ( a + b)
MC = 141 lb⋅ ft
Problem 4-15
Determine the resultant moment about the bolt located at A.
Given:
F B = 30 lb
F C = 45 lb
a = 2.5 ft
b = 0.75 ft
α = 20 deg
β = 25 deg
γ = 30 deg
Solution:
MA = FB cos ( β ) a + FC cos ( γ ) ( a + b)
MA = 195 lb⋅ ft
Problem 4-16
The elbow joint is flexed using the biceps brachii muscle, which remains essentially vertical as
the arm moves in the vertical plane. If this muscle is located a distance a from the pivot point
A on the humerus, determine the variation of the moment capacity about A if the constant
force developed by the muscle is F. Plot these results of M vs. θ for −60 ≤ θ ≤ 80.
220
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
a = 16 mm
F = 2.30 kN
θ = ( −60 .. 80)
Solution:
MA ( θ ) = F ( a) cos ( θ deg)
N.m
50
MA( θ )
0
50
0
50
100
θ
Problem 4-17
The Snorkel Co.produces the articulating boom platform that can support weight W. If the
boom is in the position shown, determine the moment of this force about points A, B, and C.
Units Used:
3
kip = 10 lb
221
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Engineering Mechanics - Statics
Chapter 4
Given:
a = 3 ft
b = 16 ft
c = 15 ft
θ 1 = 30 deg
θ 2 = 70 deg
W = 550 lb
Solution:
MA = W a
MA = 1.65 kip⋅ ft
MB = W( a + b cos ( θ 1 ) )
MB = 9.27 kip⋅ ft
MC = W( a + b cos ( θ 1 ) − c cos ( θ 2 ) )
MC = 6.45 kip⋅ ft
Problem 4-18
Determine the direction θ ( 0° ≤ θ ≤ 180°) of the force F so that it produces (a) the maximum
moment about point A and (b) the minimum moment about point A. Compute the moment in
each case.
Given:
F = 40 lb
a = 8 ft
b = 2 ft
222
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Engineering Mechanics - Statics
Chapter 4
Solution: The maximum occurs when the force is
perpendicular to the line between A and the point of
application of the force. The minimum occurs when the
force is parallel to this line.
( a)
MAmax = F
2
a +b
2
b
φ a = atan ⎛⎜ ⎟⎞
φ a = 14.04 deg
θ a = 90 deg − φ a
θ a = 76.0 deg
MAmin = 0 lb⋅ ft
MAmin = 0 lb⋅ ft
b
φ b = atan ⎛⎜ ⎟⎞
φ b = 14.04 deg
θ b = 180 deg − φ b
θ b = 166 deg
⎝ a⎠
( b)
MAmax = 329.848 lb⋅ ft
⎝ a⎠
Problem 4-19
The rod on the power control
mechanism for a business jet is
subjected to force F. Determine the
moment of this force about the
bearing at A.
Given:
F = 80 N
θ 1 = 20 deg
a = 150 mm θ 2 = 60 deg
Solution:
MA = F cos ( θ 1 ) ( a) sin ( θ 2 ) − F sin ( θ 1 ) ( a) cos ( θ 2 )
MA = 7.71 N⋅ m
Problem 4-20
The boom has length L, weight Wb, and mass center at G. If the maximum moment that can be
developed by the motor at A is M, determine the maximum load W, having a mass center at G',
that can be lifted.
223
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Engineering Mechanics - Statics
Chapter 4
Given:
L = 30 ft
Wb = 800 lb
a = 14 ft
b = 2 ft
θ = 30 deg
3
M = 20 × 10 lb⋅ ft
Solution:
M = Wb ( L − a) cos ( θ ) + W ( L cos ( θ ) + b)
W =
M − Wb ( L − a) cos ( θ )
W = 319 lb
L cos ( θ ) + b
Problem 4-21
The tool at A is used to hold a power lawnmower blade
stationary while the nut is being loosened with the
wrench. If a force P is applied to the wrench at B in the
direction shown, determine the moment it creates about
the nut at C. What is the magnitude of force F at A so
that it creates the opposite moment about C ?
Given:
P = 50 N
θ = 60 deg
a = 400 mm
b = 300 mm
c = 5
d = 12
Solution:
(a)
MA = P sin ( θ ) b
MA = 13.0 N⋅ m
(b)
MA − F
d
2
c +d
a=0
2
224
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Engineering Mechanics - Statics
Chapter 4
⎛ c2 + d2 ⎞
⎟
F = MA ⎜
⎝ da ⎠
F = 35.2 N
Problem 4-22
Determine the clockwise direction θ ( 0 deg ≤ θ ≤ 180 deg) of the force F so that it produces
(a) the maximum moment about point A and (b) no moment about point A. Compute the
moment in each case.
Given:
F = 80 lb
a = 4 ft
b = 1 ft
Solution:
(a)
2
MAmax = 330 lb⋅ ft
b
φ = atan ⎛⎜ ⎟⎞
φ = 14.0 deg
θ a = 90 deg + φ
θ a = 104 deg
⎝ a⎠
( b)
2
MAmax = F a + b
MAmin = 0
b
θ b = atan ⎛⎜ ⎟⎞
⎝ a⎠
θ b = 14.04 deg
Problem 4-23
The Y-type structure is used to support the high voltage transmission cables. If the supporting
cables each exert a force F on the structure at B, determine the moment of each force about
point A. Also, by the principle of transmissibility, locate the forces at points C and D and
determine the moments.
225
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Engineering Mechanics - Statics
Chapter 4
Units Used:
kip = 1000 lb
Given:
F = 275 lb
a = 85 ft
θ = 30 deg
Solution:
MA1 = F sin ( θ ) a
MA1 = 11.7 kip⋅ ft
MA2 = F sin ( θ ) a
MA2 = 11.7 kip⋅ ft
Also
b = ( a)tan ( θ )
MA1 = F cos ( θ ) b
MA1 = 11.7 kip⋅ ft
MA2 = F cos ( θ ) b
MA2 = 11.7 kip⋅ ft
Problem 4-24
The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point
A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the
same moment.
Given:
F = 70 N
a = 0.9 m
b = 0.3 m
c = 0.7 m
θ = 60 deg
Solution:
(a)
MA = F sin ( θ ) c + F cos ( θ ) a
MA = 73.9 N⋅ m
226
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Engineering Mechanics - Statics
Chapter 4
F C ( a) = MA
(b)
FC =
MA
F C = 82.2 N
a
Problem 4-25
The force F acts on the end of the pipe at B. Determine the angles θ ( 0° ≤ θ ≤ 180° ) of
the force that will produce maximum and minimum moments about point A. What are the
magnitudes of these moments?
Given:
F = 70 N
a = 0.9 m
b = 0.3 m
c = 0.7 m
Solution:
MA = F sin ( θ ) c + F cos ( θ ) a
For maximum moment
d
dθ
MA = c F cos ( θ ) − a F sin ( θ ) = 0
c
θ max = atan ⎛⎜ ⎟⎞
θ max = 37.9 deg
MAmax = F sin ( θ max) c + F cos ( θ max) a
MAmax = 79.812 N⋅ m
⎝ a⎠
For minimum moment
MA = F sin ( θ ) c + F cos ( θ ) a = 0
−a ⎞
⎟
⎝c⎠
θ min = 180 deg + atan ⎛⎜
θ min = 128 deg
MAmin = F c sin ( θ min ) + F ( a) cos ( θ min )
MAmin = 0 N⋅ m
Problem 4-26
The towline exerts force P at the end of the crane boom of length L. Determine the placement
227
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Engineering Mechanics - Statics
Chapter 4
g
p
x of the hook at A so that this force creates a maximum moment about point O. What is this
moment?
Unit Used:
3
kN = 10 N
Given:
P = 4 kN
L = 20 m
θ = 30 deg
a = 1.5 m
Solution:
Maximum moment, OB ⊥ BA
Guesses
x = 1m
d = 1 m (Length of the cable from B to A)
Given
L cos ( θ ) + d sin ( θ ) = x
a + L sin ( θ ) = d cos ( θ )
⎛x⎞
⎜ ⎟ = Find ( x , d)
⎝d⎠
x = 23.96 m
Mmax = P L
Mmax = 80 kN⋅ m
Problem 4-27
The towline exerts force P at the end of the crane boom of length L. Determine the position θ of
the boom so that this force creates a maximum moment about point O. What is this moment?
Units Used:
3
kN = 10 N
228
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Engineering Mechanics - Statics
Chapter 4
Given:
P = 4 kN
x = 25 m
L = 20 m
a = 1.5 m
Solution:
Maximum moment, OB ⊥ BA
Guesses
θ = 30 deg
Given
L cos ( θ ) + d sin ( θ ) = x
d = 1m
(length of cable from B to A)
a + L sin ( θ ) = d cos ( θ )
⎛θ⎞
⎜ ⎟ = Find ( θ , d)
⎝d ⎠
θ = 33.573 deg
Mmax = P L
Mmax = 80 kN⋅ m
Problem 4-28
Determine the resultant moment of the forces about point A. Solve the problem first by
considering each force as a whole, and then by using the principle of moments.
229
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
F 1 = 250 N
a = 2m
F 2 = 300 N
b = 3m
F 3 = 500 N
c = 4m
θ 1 = 60 deg
d = 3
θ 2 = 30 deg
e = 4
Solution Using Whole Forces:
Geometry
d
α = atan ⎛⎜ ⎟⎞
⎝e⎠
d ⎞
c⎟
e ⎠
L = ⎛⎜ a + b −
⎝
e
2
e +d
MA = −F1 ⎡⎣( a)cos ( θ 2 )⎤⎦ − F2 ( a + b) sin ( θ 1 ) − F3 L
2
MA = −2.532 kN⋅ m
Solution Using Principle of Moments:
MA = −F1 cos ( θ 2 ) a − F2 sin ( θ 1 ) ( a + b) + F3
d
2
2
d +e
c − F3
e
2
( a + b)
2
d +e
3
MA = −2.532 × 10 N⋅ m
Problem 4-29
If the resultant moment about point A is M clockwise, determine the magnitude of F 3.
230
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
M = 4.8 kN⋅ m a = 2 m
F 1 = 300 N
b = 3m
F 2 = 400 N
c = 4m
θ 1 = 60 deg
d = 3
θ 2 = 30 deg
e = 4
Solution:
Initial Guess
F3 = 1 N
Given
⎛
e
⎞c − F ⎛
⎞
3
⎜ 2 2 ⎟ ( a + b)
2
2⎟
⎝ d +e ⎠
⎝ d +e ⎠
−M = −F 1 cos ( θ 2 ) a − F 2 sin ( θ 1 ) ( a + b) + F 3 ⎜
F 3 = Find ( F3 )
d
F 3 = 1.593 kN
Problem 4-30
The flat-belt tensioner is manufactured by the Daton Co. and is used with V-belt drives on
poultry and livestock fans. If the tension in the belt is F, when the pulley is not turning,
determine the moment of each of these forces about the pin at A.
231
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 52 lb
a = 8 in
b = 5 in
c = 6 in
θ 1 = 30 deg
θ 2 = 20 deg
Solution:
MA1 = F cos ( θ 1 ) ( a + c cos ( θ 1 ) ) − F sin ( θ 1 ) ( b − c sin ( θ 1 ) )
MA1 = 542 lb⋅ in
MA2 = F cos ( θ 2 ) ( a − c cos ( θ 2 ) ) − F ( sin ( θ 2 ) ) ( b + c sin ( θ 2 ) )
MA2 = −10.01 lb⋅ in
Problem 4-31
The worker is using the bar to pull two pipes together in order to complete the connection. If he
applies a horizantal force F to the handle of the lever, determine the moment of this force about
the end A. What would be the tension T in the cable needed to cause the opposite moment about
point A.
Given:
F = 80 lb
θ 1 = 40 deg
θ 2 = 20 deg
a = 0.5 ft
b = 4.5 ft
232
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Engineering Mechanics - Statics
Chapter 4
Solution:
MA = F( a + b) cos ( θ 1 )
MA = 306 lb⋅ ft
Require
MA = T cos ( θ 2 ) ( a) cos ( θ 1 ) + T sin ( θ 2 ) ( a) sin ( θ 1 )
T =
MA
( a) ( cos ( θ 2 ) cos ( θ 1 ) + sin ( θ 2 ) sin ( θ 1 ) )
T = 652 lb
Problem 4-32
If it takes a force F to pull the nail out, determine the smallest vertical force P that must be
applied to the handle of the crowbar. Hint: This requires the moment of F about point A to be
equal to the moment of P about A. Why?
Given:
F = 125 lb
a = 14 in
b = 3 in
c = 1.5 in
θ 1 = 20 deg
θ 2 = 60 deg
233
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Engineering Mechanics - Statics
Chapter 4
Solution:
MF = F sin ( θ 2 ) ( b)
MF = 325 lb⋅ in
P ⎡⎣( a)cos ( θ 1 ) + ( c)sin ( θ 1 )⎤⎦ = MF
P =
MF
( a) cos ( θ 1 ) + ( c) sin ( θ 1 )
P = 23.8 lb
Problem 4-33
The pipe wrench is activated by pulling on the cable segment with a horizantal force F .
Determine the moment MA produced by the wrench on the pipe at θ. Neglect the size of the
pulley.
Given:
F = 500 N
a = 0.2 m
b = 0.5 m
c = 0.4 m
θ = 20 deg
Solution:
Initial Guesses
φ = 20 deg
MA = 1 N⋅ m
Given
b − c sin ( θ )
c cos ( θ ) − a
= tan ( φ − θ )
MA = F c sin ( φ )
⎛ φ ⎞
⎜ ⎟ = Find ( φ , MA)
⎝ MA ⎠
φ = 84.161 deg
MA = 199 N⋅ m
234
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Engineering Mechanics - Statics
Chapter 4
Problem 4-34
Determine the moment of the force at A about point O. Express the result as a Cartesian vector.
Given:
⎛ 60 ⎞
⎜
⎟
F = −30 N
⎜
⎟
⎝ −20 ⎠
a = 4m
d = 4m
b = 7m
e = 6m
c = 3m
f = 2m
Solution:
⎛ −c ⎞
⎜ ⎟
rOA = −b
⎜ ⎟
⎝a⎠
MO = rOA × F
⎛ 260 ⎞
⎜
⎟
MO = 180 N⋅ m
⎜
⎟
⎝ 510 ⎠
Problem 4-35
Determine the moment of the force at A about point P. Express the result as a Cartesian vector.
Given:
a = 4m
b = 7m
c = 3m
d = 4m
e = 6m
f = 2m
⎛ 60 ⎞
⎜
⎟
F = −30 N
⎜
⎟
⎝ −20 ⎠
Solution:
⎛ −c − d ⎞
⎜
⎟
rPA = −b − e
⎜
⎟
⎝ a+ f ⎠
MP = rPA × F
⎛ 440 ⎞
⎜
⎟
MP = 220 N⋅ m
⎜
⎟
⎝ 990 ⎠
235
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Engineering Mechanics - Statics
Chapter 4
Problem 4-36
Determine the moment of the force F at A about point O. Express the result as a cartesian
vector.
Units Used:
3
kN = 10 N
Given:
F = 13 kN
a = 6m
b = 2.5 m
c = 3m
d = 3m
e = 8m
f = 6m
g = 4m
h = 8m
Solution:
⎛b − g⎞
⎜
⎟
rAB = c + d
⎜
⎟
⎝h − a⎠
⎛ −b ⎞
⎜ ⎟
rOA = −c
⎜ ⎟
⎝a⎠
MO = rOA × F1
⎛ −84 ⎞
⎜
⎟
MO = −8 kN⋅ m
⎜
⎟
⎝ −39 ⎠
F1 = F
rAB
rAB
Problem 4-37
Determine the moment of the force F at A about point P. Express the result as a Cartesian
vector.
Units Used:
3
kN = 10 N
236
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 13 kN
a = 6m
b = 2.5 m
c = 3m
d = 3m
e = 8m
f = 6m
g = 4m
h = 8m
Solution:
⎛b − g⎞
⎜
⎟
rAB = c + d
⎜
⎟
⎝h − a⎠
⎛ −b − f ⎞
⎜
⎟
rPA = −c − e
⎜
⎟
⎝ a ⎠
MO = rPA × F 1
⎛ −116 ⎞
⎜ 16 ⎟ kN⋅ m
MO =
⎜
⎟
⎝ −135 ⎠
F1 = F
rAB
rAB
Problem 4-38
The curved rod lies in the x-y plane and has radius r. If a force F acts at its end as shown,
determine the moment of this force about point O.
Given:
r = 3m
a = 1m
θ = 45 deg
F = 80 N b = 2 m
237
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛a⎞
⎜ ⎟
rAC = −r
⎜ ⎟
⎝ −b ⎠
Fv = F
⎛1⎞
⎜ ⎟
rAC = −3 m
⎜ ⎟
⎝ −2 ⎠
⎛ 21.381 ⎞
⎜
⎟
F v = −64.143 N
⎜
⎟
⎝ −42.762 ⎠
rAC
rAC
⎛r⎞
⎜ ⎟
rOA = r
⎜ ⎟
⎝0⎠
⎛3⎞
⎜ ⎟
rOA = 3 m
⎜ ⎟
⎝0⎠
MO = rOA × Fv
⎛ −128.285 ⎞
⎜
⎟
MO = 128.285 N⋅ m
⎜
⎟
⎝ −256.571 ⎠
Problem 4-39
The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown,
determine the moment of this force about point B.
Given:
F = 80 N
c = 3m
a = 1m
r = 3m
238
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Engineering Mechanics - Statics
Chapter 4
θ = 45 deg
b = 2m
Solution:
⎛a⎞
⎜ ⎟
rAC = −c
⎜ ⎟
⎝ −b ⎠
Fv = F
⎛ r cos ( θ ) ⎞
⎜
⎟
rBA = r − r sin ( θ )
⎜
⎟
0
⎠
⎝
rAC
rAC
MB = rBA × F v
⎛ −37.6 ⎞
⎜ 90.7 ⎟ N⋅ m
MB =
⎜
⎟
⎝ −154.9 ⎠
Problem 4-40
The force F acts at the end of the beam.
Determine the moment of the force about
point A.
Given:
⎛ 600 ⎞
⎜
⎟
F = 300 N
⎜
⎟
⎝ −600 ⎠
a = 1.2 m
b = 0.2 m
c = 0.4 m
Solution:
⎛b⎞
⎜ ⎟
rAB = a
⎜ ⎟
⎝0⎠
MA = rAB × F
⎛ −720 ⎞
⎜
⎟
MA = 120 N⋅ m
⎜
⎟
⎝ −660 ⎠
239
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Engineering Mechanics - Statics
Chapter 4
Problem 4-41
The pole supports a traffic light of weight W. Using Cartesian vectors, determine the moment
of the weight of the traffic light about the base of the pole at A.
Given:
W = 22 lb
a = 12 ft
θ = 30 deg
Solution:
⎡ ( a)sin ( θ ) ⎤
⎢
⎥
r = ( a)cos ( θ )
⎢
⎥
⎣ 0 ⎦
⎛ 0 ⎞
⎜ 0 ⎟
F =
⎜ ⎟
⎝ −W ⎠
MA = r × F
⎛ −229 ⎞
⎜
⎟
MA = 132 lb⋅ ft
⎜
⎟
⎝ 0 ⎠
Problem 4-42
The man pulls on the rope with a force F. Determine the moment that this force exerts about the
base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A,
then O to B.
Given:
F = 20 N
a = 3m
240
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Engineering Mechanics - Statics
Chapter 4
b = 4m
c = 1.5 m
d = 10.5 m
Solution:
⎛ b ⎞
⎜ −a ⎟
rAB =
⎜
⎟
⎝c − d⎠
⎛b⎞
⎜ ⎟
rOB = −a
⎜ ⎟
⎝c ⎠
⎛0⎞
⎜ ⎟
rOA = 0
⎜ ⎟
⎝d⎠
Fv = F
rAB
rAB
MO1 = rOA × F v
⎛ 61.2 ⎞
⎜
⎟
MO1 = 81.6 N⋅ m
⎜
⎟
⎝ 0 ⎠
MO2 = rOB × Fv
⎛ 61.2 ⎞
⎜
⎟
MO2 = 81.6 N⋅ m
⎜
⎟
⎝ −0 ⎠
Problem 4-43
Determine the smallest force F that must be applied along the rope in order to cause the curved
rod, which has radius r, to fail at the support C. This requires a moment to be developed at C of
magnitude M.
241
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Engineering Mechanics - Statics
Chapter 4
Given:
r = 5 ft
M = 80 lb⋅ ft
θ = 60 deg
a = 7 ft
b = 6 ft
Solution:
b
⎞
⎛
⎜
rAB = a − r sin ( θ ) ⎟
⎜
⎟
⎝ −r cos ( θ ) ⎠
uAB =
Guess
F = 1 lb
Given
rCB × ( F uAB) = M
⎛b⎞
rCB = ⎜ a ⎟
⎜ ⎟
⎝ −r ⎠
rAB
rAB
F = Find ( F)
F = 18.6 lb
Problem 4-44
The pipe assembly is subjected to the force F . Determine the moment of this force about
point A.
242
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 80 N
a = 400 mm
b = 300 mm
c = 200 mm
d = 250 mm
θ = 40 deg
φ = 30 deg
Solution:
⎛b + d⎞
⎜ a ⎟
rAC =
⎜
⎟
⎝ −c ⎠
⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟
F v = F ⎜ cos ( φ ) cos ( θ ) ⎟
⎜ −sin ( φ ) ⎟
⎝
⎠
MA = rAC × Fv
⎛ −5.385 ⎞
⎜
⎟
MA = 13.093 N⋅ m
⎜
⎟
⎝ 11.377 ⎠
Problem 4-45
The pipe assembly is subjected to the force F . Determine the moment of this force about
point B.
243
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 80 N
a = 400 mm
b = 300 mm
c = 200 mm
d = 250 mm
θ = 40 deg
φ = 30 deg
Solution:
⎛b + d⎞
rBC = ⎜ 0 ⎟
⎜
⎟
⎝ −c ⎠
⎛ 550 ⎞
rBC = ⎜ 0 ⎟ mm
⎜
⎟
⎝ −200 ⎠
⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟
F v = F ⎜ cos ( φ ) cos ( θ ) ⎟
⎜ −sin ( φ ) ⎟
⎝
⎠
MB = rBC × F v
⎛ 44.534 ⎞
F v = ⎜ 53.073 ⎟ N
⎜
⎟
⎝ −40 ⎠
⎛ 10.615 ⎞
MB = ⎜ 13.093 ⎟ N⋅ m
⎜
⎟
⎝ 29.19 ⎠
Problem 4-46
The x-ray machine is used for medical diagnosis. If the camera and housing at C have mass M
and a mass center at G, determine the moment of its weight about point O when it is in the
position shown.
Units Used:
3
kN = 10 N
244
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Engineering Mechanics - Statics
Chapter 4
Given:
M = 150 kg
a = 1.2 m
b = 1.5 m
θ = 60 deg
g = 9.81
m
2
s
Solution:
⎡ −( b) cos ( θ ) ⎤ ⎛ 0 ⎞
⎢
⎥×⎜ 0 ⎟
MO =
a
⎢
⎥ ⎜
⎟
⎣ ( b)sin ( θ ) ⎦ ⎝ −M g ⎠
⎛ −1.77 ⎞
⎜
⎟
MO = −1.1 kN⋅ m
⎜
⎟
⎝ 0 ⎠
Problem 4-47
Using Cartesian vector analysis, determine the resultant moment of the three forces about the base
of the column at A.
Units Used :
3
kN = 10 N
Given:
⎛ 400 ⎞
⎜
⎟
F 1 = 300 N
⎜
⎟
⎝ 120 ⎠
⎛ 100 ⎞
⎜
⎟
F 2 = −100 N
⎜
⎟
⎝ −60 ⎠
⎛ 0 ⎞
⎜ 0 ⎟N
F3 =
⎜
⎟
⎝ −500 ⎠
a = 4m
b = 8m
245
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Engineering Mechanics - Statics
Chapter 4
c = 1m
Solution:
⎛ 0 ⎞
⎜ 0 ⎟
rAB =
⎜
⎟
⎝a + b⎠
⎛0⎞
⎜ ⎟
rA3 = −c
⎜ ⎟
⎝b⎠
The individual moments
MA1 = rAB × F1
MA2 = rAB × F2
MA3 = rA3 × F3
⎛ −3.6 ⎞
⎜
⎟
MA1 = 4.8 kN⋅ m
⎜
⎟
⎝ 0 ⎠
⎛ 1.2 ⎞
⎜ ⎟
MA2 = 1.2 kN⋅ m
⎜ ⎟
⎝ 0 ⎠
⎛ 0.5 ⎞
⎜ ⎟
MA3 = 0 kN⋅ m
⎜ ⎟
⎝ 0 ⎠
The total moment
MA = MA1 + MA2 + MA3
⎛ −1.9 ⎞
⎜ 6 ⎟ kN⋅ m
MA =
⎜
⎟
⎝ 0 ⎠
Problem 4-48
A force F produces a moment MO about the origin of coordinates, point O. If the force acts at a
point having the given x coordinate, determine the y and z coordinates.
Units Used :
3
kN = 10 N
Given:
⎛6⎞
⎜ ⎟
F = −2 kN
⎜ ⎟
⎝1⎠
⎛ 4 ⎞
⎜ 5 ⎟ kN⋅ m
MO =
⎜
⎟
⎝ −14 ⎠
x = 1m
Solution:
The initial guesses:
y = 1m
z = 1m
246
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Engineering Mechanics - Statics
Chapter 4
Given
⎛x⎞
⎜ y⎟ × F = M
O
⎜ ⎟
⎝z⎠
⎛ y⎞
⎜ ⎟ = Find ( y , z)
⎝z⎠
⎛ y⎞ ⎛2⎞
⎜ ⎟=⎜ ⎟m
⎝ z⎠ ⎝1⎠
Problem 4-49
The force F creates a moment about point O of MO. If the force passes through a point
having the given x coordinate, determine the y and z coordinates of the point. Also, realizing
that MO = Fd, determine the perpendicular distance d from point O to the line of action of F .
Given:
⎛6⎞
F = ⎜8 ⎟ N
⎜ ⎟
⎝ 10 ⎠
⎛ −14 ⎞
MO = ⎜ 8 ⎟ N⋅ m
⎜
⎟
⎝ 2 ⎠
x = 1m
Solution:
The initial guesses:
y = 1m
z = 1m
Given
⎛x⎞
⎜ y⎟ × F = M
O
⎜ ⎟
⎝z⎠
d =
MO
F
⎛ y⎞
⎜ ⎟ = Find ( y , z)
⎝z⎠
⎛ y⎞ ⎛1⎞
⎜ ⎟=⎜ ⎟m
⎝ z⎠ ⎝3⎠
d = 1.149 m
Problem 4-50
The force F produces a moment MO about the origin of coordinates, point O. If the force acts
at a point having the given x-coordinate, determine the y and z coordinates.
247
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
x = 1m
⎛6⎞
⎜ ⎟
F = −2 kN
⎜ ⎟
⎝1⎠
⎛ 4 ⎞
⎜ 5 ⎟ kN⋅ m
MO =
⎜
⎟
⎝ −14 ⎠
Solution:
y = 1m
Initial Guesses:
Given
⎛x ⎞
⎜ ⎟
MO = y × F
⎜ ⎟
⎝z⎠
z = 1m
⎛ y⎞
⎜ ⎟ = Find ( y , z)
⎝z⎠
⎛ y⎞ ⎛2⎞
⎜ ⎟=⎜ ⎟m
⎝ z⎠ ⎝1⎠
Problem 4-51
Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector.
Given:
⎛ 50 ⎞
⎜
⎟
F = −20 N
⎜
⎟
⎝ 20 ⎠
a = 6m
b = 2m
c = 1m
d = 3m
e = 4m
248
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛c ⎞
rOF = ⎜ −b ⎟
⎜ ⎟
⎝a⎠
⎛0⎞
rOa = ⎜ e ⎟
⎜ ⎟
⎝d⎠
MOa = ⎡⎣( rOF × F ) ⋅ uOa⎤⎦ uOa
uOa =
rOa
rOa
⎛ 0 ⎞
MOa = ⎜ 217.6 ⎟ N⋅ m
⎜
⎟
⎝ 163.2 ⎠
Problem 4-52
Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector.
Given:
F = 600 lb
a = 6 ft
b = 3 ft
c = 2 ft
d = 4 ft
e = 4 ft
f = 2 ft
Solution:
Fv =
⎛ −d ⎞
⎜e ⎟
⎟
2
2
2⎜
c +d +e ⎝ c ⎠
F
Maa = ⎡⎣( r × F v ) ⋅ uaa⎤⎦ uaa
⎛d⎞
r = ⎜0 ⎟
⎜ ⎟
⎝ −c ⎠
uaa =
⎛ −b ⎞
⎜−f ⎟
⎟
2
2
2⎜
a +b + f ⎝ a ⎠
1
⎛ −441 ⎞
Maa = ⎜ −294 ⎟ lb⋅ ft
⎜
⎟
⎝ 882 ⎠
249
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Engineering Mechanics - Statics
Chapter 4
Problem 4-53
Determine the resultant moment of the two forces about the Oa axis. Express the result as a
Cartesian vector.
Given:
F 1 = 80 lb
F 2 = 50 lb
α = 120 deg
β = 60 deg
γ = 45 deg
a = 5 ft
b = 4 ft
c = 6 ft
θ = 30 deg
φ = 30 deg
Solution:
⎛⎜ cos ( α ) ⎟⎞
F 1v = F1 ⎜ cos ( β ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
⎛⎜ 0 ⎞⎟
F 2v = ⎜ 0 ⎟
⎜ F2 ⎟
⎝ ⎠
⎡ ( b)sin ( θ ) ⎤
⎢
⎥
r1 = ( b)cos ( θ )
⎢
⎥
⎣ c
⎦
0
⎤
⎡
⎢
r2 = −( a) sin ( φ ) ⎥
⎢
⎥
0
⎣
⎦
⎛ cos ( φ ) ⎞
⎜
⎟
uaa = −sin ( φ )
⎜
⎟
⎝ 0 ⎠
Maa = ⎡⎣( r1 × F1v + r2 × F2v) uaa⎤⎦ uaa
⎛ 26.132 ⎞
⎜
⎟
Maa = −15.087 lb⋅ ft
⎜
⎟
⎝ 0 ⎠
250
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Engineering Mechanics - Statics
Chapter 4
Problem 4-54
The force F is applied to the handle of the box wrench. Determine the component of the
moment of this force about the z axis which is effective in loosening the bolt.
Given:
a = 3 in
b = 8 in
c = 2 in
⎛8⎞
F = ⎜ −1 ⎟ lb
⎜ ⎟
⎝1⎠
Solution:
⎛0⎞
k = ⎜0⎟
⎜ ⎟
⎝1⎠
⎛c ⎞
r = ⎜ −b ⎟
⎜ ⎟
⎝a⎠
Mz = ( r × F ) ⋅ k
Mz = 62 lb⋅ in
Problem 4-55
The force F acts on the gear in the direction shown. Determine the moment of this force about
the y axis.
Given:
F = 50 lb
a = 3 in
θ 1 = 60 deg
θ 2 = 45 deg
θ 3 = 120 deg
251
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛0⎞
j = ⎜1⎟
⎜ ⎟
⎝0⎠
⎛0⎞
r = ⎜0⎟
⎜ ⎟
⎝a⎠
⎛ −cos ( θ 3) ⎞
⎜
⎟
F v = F ⎜ −cos ( θ 2 ) ⎟ My = ( r × Fv ) ⋅ j
⎜ −cos θ ⎟
( 1) ⎠
⎝
My = 75 lb⋅ in
Problem 4-56
The RollerBall skate is an in-line tandem skate that uses two large spherical wheels on each
skate, rather than traditional wafer-shape wheels. During skating the two forces acting on
the wheel of one skate consist of a normal force F 2 and a friction force F1. Determine the
moment of both of these forces about the axle AB of the wheel.
Given:
θ = 30 deg
F 1 = 13 lb
F 2 = 78 lb
a = 1.25 in
Solution:
⎛ F1 ⎞
⎛0⎞
⎜ ⎟
F = ⎜ F2 ⎟ r = ⎜ −a ⎟
⎜ ⎟
⎜0 ⎟
⎝0⎠
⎝ ⎠
⎛ cos ( θ ) ⎞
ab = ⎜ −sin ( θ ) ⎟
⎜
⎟
⎝ 0 ⎠
Mab = ( r × F ) ⋅ ab
Mab = 0 lb⋅ in
Problem 4-57
The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the
moment of this force about the y axis of the shaft.
252
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
⎛6⎞
⎜ ⎟
F = −4 kN
⎜ ⎟
⎝ −7 ⎠
a = 30 mm
θ = 40 deg
Solution:
⎛ cos ( θ ) ⎞
⎜ 0 ⎟
r = a
⎜
⎟
⎝ sin ( θ ) ⎠
⎛0⎞
⎜ ⎟
j = 1
⎜ ⎟
⎝0⎠
My = ( r × F) ⋅ j
My = 0.277 kN⋅ m
Problem 4-58
The hood of the automobile is supported by the strut AB, which exerts a force F on the hood.
Determine the moment of this force about the hinged axis y.
Given:
F = 24 lb
a = 2 ft
b = 4 ft
c = 2 ft
d = 4 ft
253
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛b⎞
⎜ ⎟
rA = 0
⎜ ⎟
⎝0⎠
⎛ −b + c ⎞
⎜ a ⎟
rAB =
⎜
⎟
⎝ d ⎠
⎛0⎞
⎜ ⎟
j = 1
⎜ ⎟
⎝0⎠
My = ( rA × F v ) ⋅ j
Fv = F
rAB
rAB
⎛ −9.798 ⎞
⎜
⎟
F v = 9.798 lb
⎜
⎟
⎝ 19.596 ⎠
My = −78.384 lb⋅ ft
Problem 4-59
The lug nut on the wheel of the automobile is to be removed using the wrench and applying the
vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is
initially required to turn the nut. If the force F can be applied at A in any other direction, will it
be possible to turn the nut?
254
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 30 N
M = 14 N⋅ m
a = 0.25 m
b = 0.3 m
c = 0.5 m
d = 0.1 m
Solution:
2
Mx = F c − b
2
Mx = 12 N⋅ m
Mx < M
No
For Mxmax, apply force perpendicular to the handle and the x-axis.
Mxmax = F c
Mxmax = 15 N⋅ m
Mxmax > M
Yes
Problem 4-60
The lug nut on the wheel of the automobile is to be removed using the wrench and applying the
vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and
the F force can be applied at any point and in any direction on the assembly. Determine if this
force is adequate, provided a torque M about the x axis is initially required to turn the nut.
Given:
F 1 = 30 N
M = 14 N⋅ m
a = 0.25 m
b = 0.3 m
c = 0.5 m
d = 0.1 m
255
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Engineering Mechanics - Statics
Chapter 4
Solution:
Mx = F1
a+c
c
2
2
c −b
Mx = 18 N⋅ m
Mx > M
Yes
Mxmax occurs when force is applied perpendicular to both the handle and the x-axis.
Mxmax = F1 ( a + c)
Mxmax = 22.5 N⋅ m
Mxmax > M
Yes
Problem 4-61
The bevel gear is subjected to the force F
which is caused from contact with another
gear. Determine the moment of this force
about the y axis of the gear shaft.
Given:
a = 30 mm
b = 40 mm
⎛ 20 ⎞
⎜ 8 ⎟N
F =
⎜
⎟
⎝ −15 ⎠
256
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛ −b ⎞
⎜ ⎟
r = 0
⎜ ⎟
⎝a⎠
⎛0⎞
⎜ ⎟
j = 1
⎜ ⎟
⎝0⎠
My = ( r × F) ⋅ j
My = 0 N⋅ m
Problem 4-62
The wooden shaft is held in a lathe.
The cutting tool exerts force F on the
shaft in the direction shown.
Determine the moment of this force
about the x axis of the shaft. Express
the result as a Cartesian vector. The
distance OA is a.
Given:
a = 25 mm
θ = 30 deg
⎛ −5 ⎞
⎜ ⎟
F = −3 N
⎜ ⎟
⎝8⎠
Solution:
⎡ 0 ⎤
⎢
⎥
r = ( a)cos ( θ )
⎢
⎥
⎣ ( a)sin ( θ ) ⎦
⎛1⎞
⎜ ⎟
i = 0
⎜ ⎟
⎝0⎠
Mx = ⎡⎣( r × F ) ⋅ i⎤⎦ i
⎛ 0.211 ⎞
⎜ 0 ⎟ N⋅ m
Mx =
⎜
⎟
⎝ 0 ⎠
Problem 4-63
Determine the magnitude of the moment of the force F about the base line CA of the tripod.
Given:
⎛ 50 ⎞
⎜
⎟
F = −20 N
⎜
⎟
⎝ −80 ⎠
257
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Engineering Mechanics - Statics
Chapter 4
a = 4m
b = 2.5 m
c = 1m
d = 0.5 m
e = 2m
f = 1.5 m
g = 2m
Solution:
⎛ −g ⎞
rCA = ⎜ e ⎟
⎜ ⎟
⎝0⎠
uCA =
rCA
rCA
⎛b − g⎞
rCD = ⎜ e ⎟
⎜
⎟
⎝ a ⎠
MCA = ( rCD × F) ⋅ uCA
MCA = 226 N⋅ m
Problem 4-64
The flex-headed ratchet wrench is
subjected to force P, applied
perpendicular to the handle as shown.
Determine the moment or torque this
imparts along the vertical axis of the
bolt at A.
Given:
P = 16 lb
a = 10 in
θ = 60 deg b = 0.75 in
Solution:
M = P ⎡⎣b + ( a)sin ( θ )⎤⎦
M = 150.564 lb⋅ in
258
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Engineering Mechanics - Statics
Chapter 4
Problem 4-65
If a torque or moment M is
required to loosen the bolt at A,
determine the force P that must be
applied perpendicular to the handle
of the flex-headed ratchet wrench.
Given:
M = 80 lb⋅ in
θ = 60 deg
a = 10 in
b = 0.75 in
Solution:
M = P ⎡⎣b + ( a)sin ( θ )⎤⎦
P =
M
b + ( a)sin ( θ )
P = 8.50 lb
Problem 4-66
The A-frame is being hoisted into
an upright position by the vertical
force F . Determine the moment of
this force about the y axis when
the frame is in the position shown.
Given:
F = 80 lb
a = 6 ft
b = 6 ft
θ = 30 deg
φ = 15 deg
Solution:
Using the primed coordinates we have
259
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Engineering Mechanics - Statics
⎛ −sin ( θ ) ⎞
⎜
⎟
j = cos ( θ )
⎜
⎟
⎝ 0 ⎠
Chapter 4
⎛0⎞
⎜ ⎟
Fv = F 0
⎜ ⎟
⎝1⎠
My = ( rAC × F v ) ⋅ j
⎛ −b cos ( φ ) ⎞
⎜
⎟
a
⎟
rAC = ⎜
2
⎜
⎟
⎜ b sin ( φ ) ⎟
⎝
⎠
My = 281.528 lb⋅ ft
Problem 4-67
Determine the moment of each force acting on the handle of the wrench about the a axis.
Given:
⎛ −2 ⎞
⎜ ⎟
F 1 = 4 lb
⎜ ⎟
⎝ −8 ⎠
⎛3⎞
⎜ ⎟
F 2 = 2 lb
⎜ ⎟
⎝ −6 ⎠
b = 6 in
c = 4 in
d = 3.5 in
θ = 45 deg
Solution:
⎛ cos ( θ ) ⎞
⎜ 0 ⎟
ua =
⎜
⎟
⎝ sin ( θ ) ⎠
⎛ cos ( θ ) ⎞
⎛ sin ( θ ) ⎞
⎜
⎟
⎜ 0 ⎟
r1 = b
0
+ ( c + d)
⎜
⎟
⎜
⎟
⎝ sin ( θ ) ⎠
⎝ −cos ( θ ) ⎠
M1a = ( r1 × F1 ) ⋅ ua
⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞
⎜ 0 ⎟ + c⎜ 0 ⎟
r2 = b
⎜
⎟ ⎜
⎟
⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠
M1a = 30 lb⋅ in
260
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Engineering Mechanics - Statics
M2a = ( r2 × F2 ) ⋅ ua
Chapter 4
M2a = 8 lb⋅ in
Problem 4-68
Determine the moment of each force acting on the handle of the wrench about the z axis.
Given:
⎛ −2 ⎞
⎜ ⎟
F 1 = 4 lb
⎜ ⎟
⎝ −8 ⎠
⎛3⎞
⎜ ⎟
F 2 = 2 lb
⎜ ⎟
⎝ −6 ⎠
b = 6 in
c = 4 in
d = 3.5 in
θ = 45 deg
Solution:
⎛ cos ( θ ) ⎞
⎛ sin ( θ ) ⎞
⎜
⎟
⎜ 0 ⎟
r1 = b
0
+ ( c + d)
⎜
⎟
⎜
⎟
⎝ sin ( θ ) ⎠
⎝ −cos ( θ ) ⎠
⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞
⎜ 0 ⎟ + c⎜ 0 ⎟
r2 = b
⎜
⎟ ⎜
⎟
⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠
M1z = ( r1 × F 1 ) ⋅ k
M1z = 38.2 lb⋅ in
M2z = ( r2 × F 2 ) ⋅ k
M2z = 14.1 lb⋅ in
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
Problem 4-69
Determine the magnitude and sense of the couple moment.
Units Used:
3
kN = 10 N
261
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 5 kN
θ = 30 deg
a = 0.5 m
b = 4m
c = 2m
d = 1m
Solution:
MC = F cos ( θ ) ( a + c) + F sin ( θ ) ( b − d)
MC = 18.325 kN⋅ m
Problem 4-70
Determine the magnitude and sense of the couple moment. Each force has a magnitude F.
Given:
F = 65 lb
a = 2 ft
b = 1.5 ft
c = 4 ft
d = 6 ft
e = 3 ft
Solution:
Mc = ΣMB;
⎡ ⎛
⎞ ( d + a)⎤ + ⎡F⎛ e ⎞ ( c + a)⎤
⎥ ⎢ ⎜ 2 2⎟
⎥
2
2⎟
⎣ ⎝ c +e ⎠
⎦ ⎣ ⎝ c +e ⎠
⎦
MC = ⎢F⎜
c
MC = 650 lb⋅ ft
(Counterclockwise)
Problem 4-71
Determine the magnitude and sense of the couple moment.
262
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kip = 10 lb
Given:
F = 150 lb
a = 8 ft
b = 6 ft
c = 8 ft
d = 6 ft
e = 6 ft
f = 8 ft
Solution:
MC = ΣMA;
MC = F
d
2
d + f
MC = 3120 lb⋅ ft
f
( a + f) + F
2
2
d + f
( c + d)
2
MC = 3.120 kip⋅ ft
Problem 4-72
If the couple moment has magnitude
M, determine the magnitude F of the
couple forces.
Given:
M = 300 lb⋅ ft
a = 6 ft
b = 12 ft
c = 1 ft
d = 2 ft
e = 12 ft
f = 7 ft
263
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Engineering Mechanics - Statics
Chapter 4
Solution:
e( f + a)
M=F⎡
−
⎢
2
2
+
e
(
f
−
d
)
⎣
( f − d) ( b + e) ⎤
2⎥
2
( f − d) + e ⎦
M
F =
e( f + a)
2
−
F = 108 lb
( f − d) ( b + e)
2
2
( f − d) + e
2
( f − d) + e
Problem 4-73
A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of
the reactive forces −R and R which act at supports A and B so that the resultant of the two
couples is zero.
Given:
a = 150 mm
θ = 60 deg
M = 5 N⋅ m
Solution:
MC = −M +
2R a
tan ( θ )
=0
R =
M tan ( θ )
2
a
R = 28.9 N
Problem 4-74
The resultant couple moment created by the two couples acting on the disk is MR. Determine
the magnitude of force T.
264
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kip = 10 lb
Given:
⎛0⎞
⎜ ⎟
MR = 0 kip⋅ in
⎜ ⎟
⎝ 10 ⎠
a = 4 in
b = 2 in
c = 3 in
Solution:
Initial Guess
Given
T = 1 kip
⎛ a ⎞ ⎛ 0 ⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛ −b − c ⎞ ⎛ 0 ⎞
⎜ 0 ⎟ × ⎜ T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ = M
R
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
⎝0⎠ ⎝0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠
T = Find ( T)
T = 0.909 kip
Problem 4-75
Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the
bend angle θ so that the resultant couple moment is zero.
Given:
θ 1 = 45 deg
M1 = 900 N⋅ m
M2 = 500 N⋅ m
265
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Solution:
Initial guesses:
θ = 10 deg
M3 = 10 N⋅ m
Given
+
ΣMx = 0;
→
M1 − M3 cos ( θ ) − M2 cos ( θ 1 ) = 0
+
M3 sin ( θ ) − M2 sin ( θ 1 ) = 0
↑ ΣMy = 0;
⎛ θ ⎞
⎜ ⎟ = Find ( θ , M3)
⎝ M3 ⎠
θ = 32.9 deg
M3 = 651 N⋅ m
Problem 4-76
The floor causes couple moments
MA and MB on the brushes of the
polishing machine. Determine the
magnitude of the couple forces
that must be developed by the
operator on the handles so that the
resultant couple moment on the
polisher is zero. What is the
magnitude of these forces if the
brush at B suddenly stops so that
MB = 0?
Given:
a = 0.3 m
MA = 40 N⋅ m
MB = 30 N⋅ m
Solution:
MA − MB − F 1 a = 0
F1 =
MA − F 2 a = 0
F2 =
MA − MB
a
MA
F 1 = 33.3 N
F 2 = 133 N
a
266
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Engineering Mechanics - Statics
Chapter 4
Problem 4-77
The ends of the triangular plate are
subjected to three couples. Determine
the magnitude of the force F so that
the resultant couple moment is M
clockwise.
Given:
F 1 = 600 N
F 2 = 250 N
a = 1m
θ = 40 deg
M = 400 N⋅ m
Solution:
Initial Guess
Given
F = 1N
⎛ a ⎞ − F a − F ⎛ a ⎞ = −M
⎟
2
⎜
⎟
⎝ 2 cos ( θ ) ⎠
⎝ 2 cos ( θ ) ⎠
F1 ⎜
F = Find ( F)
F = 830 N
Problem 4-78
Two couples act on the beam. Determine the magnitude of F so that the resultant couple
moment is M counterclockwise. Where on the beam does the resultant couple moment act?
Given:
M = 450 lb⋅ ft
P = 200 lb
a = 1.5 ft
b = 1.25 ft
c = 2 ft
θ = 30 deg
267
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Engineering Mechanics - Statics
Chapter 4
Solution:
MR = Σ M
M = F b cos ( θ ) + P a
F =
M − Pa
b cos ( θ )
F = 139 lb
The resultant couple moment is a free vector. It can act at any point on the beam.
Problem 4-79
Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve
the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O.
Given:
⎛0⎞
F = ⎜0 ⎟ N
⎜ ⎟
⎝ 25 ⎠
a = 300 mm
b = 150 mm
c = 400 mm
d = 200 mm
e = 200 mm
Solution:
( a)
⎛ −e − b ⎞
rAB = ⎜ −c + d ⎟
⎜
⎟
⎝ 0 ⎠
( b)
⎛a⎞
rOB = ⎜ d ⎟
⎜ ⎟
⎝0⎠
⎛ −5 ⎞
M = ⎜ 8.75 ⎟ N⋅ m
⎜
⎟
⎝ 0 ⎠
M = rAB × F
⎛a + b + e⎞
⎟
c
rOA = ⎜
⎜
⎟
⎝ 0 ⎠
⎛ −5 ⎞
M = ⎜ 8.75 ⎟ N⋅ m
⎜
⎟
⎝ 0 ⎠
M = rOB × F + rOA × ( −F)
268
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Engineering Mechanics - Statics
Chapter 4
Problem 4-80
If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the
vertical force applied to each wrench.
Given:
M = 400 N⋅ m
a = 300 mm
b = 150 mm
c = 400 mm
d = 200 mm
e = 200 mm
Solution:
⎛0⎞
k = ⎜0⎟
⎜ ⎟
⎝1⎠
⎛ −e − b ⎞
rAB = ⎜ −c + d ⎟
⎜
⎟
⎝ 0 ⎠
Guesss
F = 1N
Given
rAB × ( Fk) = M
F = Find ( F)
F = 992.278 N
Problem 4-81
Determine the resultant couple moment acting on the beam. Solve the problem two ways:
(a) sum moments about point O; and (b) sum moments about point A.
Units Used:
3
kN = 10 N
269
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 2 kN
θ 1 = 30 deg
F 2 = 8 kN
θ 2 = 45 deg
a = 0.3 m
b = 1.5 m
c = 1.8 m
Solution:
MR = ΣMO;
( a)
MRa = ( F1 cos ( θ 1 ) + F 2 cos ( θ 2 ) ) c + ( F2 cos ( θ 2 ) − F 1 sin ( θ 1 ) ) a ...
+ −( F2 cos ( θ 2 ) + F 1 cos ( θ 1 ) ) ( b + c)
MRa = −9.69 kN⋅ m
MR = ΣMA;
( b)
MRb = ( F2 sin ( θ 2 ) − F1 sin ( θ 1 ) ) a − ( F 2 cos ( θ 2 ) + F1 cos ( θ 1 ) ) b
MRb = −9.69 kN⋅ m
Problem 4-82
Two couples act on the beam as shown. Determine the magnitude of F so that the resultant
couple moment is M counterclockwise. Where on the beam does the resultant couple act?
Given:
M = 300 lb⋅ ft
a = 4 ft
b = 1.5 ft
P = 200 lb
c = 3
d = 4
270
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Engineering Mechanics - Statics
Chapter 4
Solution:
c
M=
2
Fa +
c +d
F =
2
d
2
c +d
Fb − Pb
2
2 M + Pb⎞
c + d ⎛⎜
2
F = 167 lb
⎟
⎝ ca + d b⎠
Resultant couple can act anywhere.
Problem 4-83
Two couples act on the frame. If the resultant couple moment is to be zero, determine the
distance d between the couple forces F 1.
Given:
F 1 = 80 lb
F 2 = 50 lb
a = 1 ft
b = 3 ft
c = 2 ft
e = 3 ft
f = 3
g = 4
θ = 30 deg
Solution:
g
⎡−F cos ( θ ) e + ⎛
⎞ ⎤
⎢ 2
⎜ 2 2 ⎟ F1 d⎥ = 0
⎣
⎝ g +f ⎠ ⎦
d =
F2
F1
⎛ g2 + f 2 ⎞
⎟
g
⎝
⎠
cos ( θ ) e ⎜
d = 2.03 ft
271
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Engineering Mechanics - Statics
Chapter 4
Problem 4-84
Two couples act on the frame. Determine the resultant
couple moment. Compute the result by resolving each force
into x and y components and (a) finding the moment of each
couple (Eq. 4-13) and (b) summing the moments of all the
force components about point A.
Given:
F 1 = 80 lb
c = 2 ft
g = 4
F 2 = 50 lb
d = 4 ft
θ = 30 deg
a = 1 ft
e = 3 ft
b = 3 ft
f = 3
Solution:
( a)
M = Σ ( r × F)
⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F
⎛ −g ⎞⎤
1
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ − f ⎟⎥
M = 0 × F 2 −cos ( θ ) + d ×
⎜ ⎟ ⎢ ⎜
⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦
( b)
⎛ 0 ⎞
⎜ 0 ⎟ lb⋅ ft
M=
⎜
⎟
⎝ 126.096 ⎠
Summing the moments of all force components aboout point A.
⎛
g
⎞F b + ⎛
⎞
1
⎜ 2 2 ⎟ F 1 ( b + d)
2
2⎟
⎝ f +g ⎠
⎝ f +g ⎠
M1 = ⎜
−g
M2 = F 2 cos ( θ ) c − F 2 sin ( θ ) ( a + b + d) − F 2 cos ( θ ) ( c + e) + F 2 sin ( θ ) ( a + b + d)
M = M1 + M2
M = 126.096 lb⋅ ft
Problem 4-85
Two couples act on the frame. Determine the resultant couple moment. Compute the result by
resolving each force into x and y components and (a) finding the moment of each couple
(Eq. 4 -13) and (b) summing the moments of all the force components about point B.
272
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 80 lb
d = 4ft
F 2 = 50 lb
e = 3 ft
a = 1 ft
f = 3 ft
b = 3 ft
g = 4 ft
c = 2 ft
θ = 30 deg
Solution:
( a)
M = Σ ( r × F)
⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F
⎛ −g ⎞⎤
1
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ − f ⎟⎥
M = 0 × F 2 −cos ( θ ) + d ×
⎜ ⎟ ⎢ ⎜
⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦
( b)
⎛ 0 ⎞
⎟ lb⋅ ft
M=⎜
0
⎜
⎟
⎝ 126.096 ⎠
Summing the moments of all force components about point B.
M1 = ⎛
g
g
⎛
⎞
⎞
⎜ 2 2 ⎟ F1( a + d) − ⎜ 2 2 ⎟ F1 a
⎝ f +g ⎠
⎝ f +g ⎠
M2 = F 2 cos ( θ ) c − F 2 cos ( θ ) ( c + e)
M = M1 + M2
M = 126.096 lb⋅ ft
Problem 4-86
Determine the couple moment. Express the result as a Cartesian vector.
273
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Engineering Mechanics - Statics
Chapter 4
Given:
⎛8⎞
F = ⎜ −4 ⎟ N
⎜ ⎟
⎝ 10 ⎠
a = 5m
b = 3m
c = 4m
d = 2m
e = 3m
Solution:
⎛ −b − e ⎞
r = ⎜ c+d ⎟
⎜
⎟
⎝ −a ⎠
M = r× F
⎛ 40 ⎞
M = ⎜ 20 ⎟ N⋅ m
⎜
⎟
⎝ −24 ⎠
Problem 4-87
Determine the couple moment. Express the result as a Cartesian vector.
Given:
F = 80 N
a = 6m
b = 10 m
c = 10 m
d = 5m
e = 4m
f = 4m
274
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Engineering Mechanics - Statics
Chapter 4
Solution:
u =
⎛ b ⎞
⎜c + d⎟
⎟
2
2
2⎜
a + b + ( c + d) ⎝ − a ⎠
1
F v = Fu
⎛−f − b⎞
r = ⎜ −d − c ⎟
⎜
⎟
⎝ e+a ⎠
⎛ −252.6 ⎞
M = ⎜ 67.4 ⎟ N⋅ m
⎜
⎟
⎝ −252.6 ⎠
M = r × Fv
Problem 4-88
If the resultant couple of the two
couples acting on the fire hydrant is
MR = { −15i + 30j} N ⋅ m, determine
the force magnitude P.
Given:
a = 0.2 m
b = 0.150 m
⎛ −15 ⎞
M = ⎜ 30 ⎟ N⋅ m
⎜
⎟
⎝ 0 ⎠
F = 75 N
Solution:
Initial guess
P = 1N
Given
⎛ −F a ⎞
M = ⎜ Pb ⎟
⎜
⎟
⎝ 0 ⎠
P = Find ( P)
P = 200 N
Problem 4-89
If the resultant couple of the three couples acting on the triangular block is to be zero, determine
the magnitude of forces F and P .
275
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 150 N
a = 300 mm
b = 400 mm
d = 600 mm
Solution:
Initial guesses:
Given
F = 1N
P = 1N
⎛ d ⎞ ⎛ 0 ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ −F1 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F1 ⎞⎟
⎜ 0 ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ −P ⎟ + ⎜ b ⎟ ×
⎜ ⎟
+ 0 ×
=0
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜0 ⎟
⎝ a ⎠ ⎝ F ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ a ⎠ ⎜⎝ 0 ⎟⎠
⎛F⎞
⎜ ⎟ = Find ( F , P)
⎝P⎠
⎛ F ⎞ ⎛ 75 ⎞
⎜ ⎟=⎜
⎟N
⎝ P ⎠ ⎝ 100 ⎠
Problem 4-90
Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector.
Member BA lies in the x-y plane.
276
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Given:
⎛ 0 ⎞
⎜ 0 ⎟N
F =
⎜
⎟
⎝ 100 ⎠
a = 300 mm
b = 150 mm
c = 200 mm
d = 200 mm
θ = 60 deg
Solution:
⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤
⎢
⎥
r = −b sin ( θ ) + ( c + d)cos ( θ )
⎢
⎥
0
⎦
⎣
M = r× F
⎛ 7.01 ⎞
⎜
⎟
M = 42.14 N⋅ m
⎜
⎟
⎝ 0.00 ⎠
Problem 4-91
If the magnitude of the resultant
couple moment is M, determine
the magnitude F of the forces
applied to the wrenches.
Given:
M = 15 N⋅ m
c = 200 mm
a = 300 mm
d = 200 mm
b = 150 mm
θ = 60 deg
277
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Solution:
⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤
⎢
⎥
r = −b sin ( θ ) + ( c + d)cos ( θ )
⎢
⎥
0
⎦
⎣
Guess
Given
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
F = 1N
r × ( Fk) = M
F = Find ( F)
F = 35.112 N
Problem 4-92
The gears are subjected to the couple moments shown. Determine the magnitude and coordinate
direction angles of the resultant couple moment.
Given:
M1 = 40 lb⋅ ft
M2 = 30 lb⋅ ft
θ 1 = 20 deg
θ 2 = 15 deg
θ 3 = 30 deg
Solution:
⎛ M1 cos ( θ 1) sin ( θ 2) ⎞
⎜
⎟
M1 = ⎜ M1 cos ( θ 1 ) cos ( θ 2 ) ⎟
⎜ −M sin θ
( 1) ⎟⎠
1
⎝
MR = M1 + M2
⎛⎜ α ⎞⎟
⎛ MR ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ MR ⎠
⎜γ ⎟
⎝ ⎠
⎛ −M2 sin ( θ 3) ⎞
⎜
⎟
M2 = ⎜ M2 cos ( θ 3 ) ⎟
⎜
⎟
0
⎝
⎠
MR = 64.0 lb⋅ ft
⎛⎜ α ⎞⎟ ⎛ 94.7 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 13.2 ⎟ deg
⎜ γ ⎟ ⎝ 102.3 ⎠
⎝ ⎠
278
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Engineering Mechanics - Statics
Chapter 4
Problem 4-93
Express the moment of the couple acting on the rod in Cartesian vector form. What is the
magnitude of the couple moment?
Given:
⎛ 14 ⎞
⎜ ⎟
F = −8 N
⎜ ⎟
⎝ −6 ⎠
a = 1.5 m
b = 0.5 m
c = 0.5 m
d = 0.8 m
Solution:
⎛d⎞
⎛0⎞
⎜ ⎟
⎜ ⎟
M = a × F + 0 × ( −F)
⎜ ⎟
⎜ ⎟
⎝ −c ⎠
⎝b⎠
⎛ −17 ⎞
⎜
⎟
M = −9.2 N⋅ m
⎜
⎟
⎝ −27.4 ⎠
M = 33.532 N⋅ m
Problem 4-94
Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve
the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O.
279
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Engineering Mechanics - Statics
Chapter 4
Given:
a = 0.3 m
b = 0.4 m
c = 0.6 m
⎛ −6 ⎞
F = ⎜ 2 ⎟ N
⎜ ⎟
⎝3⎠
Solution:
(a)
⎛0⎞
M = ⎜ c ⎟×F
⎜ ⎟
⎝ −b ⎠
⎛ 2.6 ⎞
M = ⎜ 2.4 ⎟ N⋅ m
⎜ ⎟
⎝ 3.6 ⎠
(b)
⎛0⎞
⎛ 0 ⎞
⎜
⎟
M = 0 × ( −F ) + ⎜ c ⎟ × F
⎜ ⎟
⎜
⎟
⎝ −a ⎠
⎝ −a − b ⎠
⎛ 2.6 ⎞
M = ⎜ 2.4 ⎟ N⋅ m
⎜ ⎟
⎝ 3.6 ⎠
Problem 4-95
A couple acts on each of the handles
of the minidual valve. Determine the
magnitude and coordinate direction
angles of the resultant couple
moment.
Given:
F 1 = 35 N θ = 60 deg
F 2 = 25 N
r1 = 175 mm
r2 = 175 mm
Solution:
⎛⎜ −F1 2 r1 ⎞⎟ ⎛⎜ −F2 2r2 cos ( θ ) ⎞⎟
M = ⎜
0
⎟ + ⎜ −F2 2 r2 sin ( θ ) ⎟
⎜ 0 ⎟ ⎜
⎟
⎝
⎠ ⎝
0
⎠
⎛ −16.63 ⎞
M = ⎜ −7.58 ⎟ N⋅ m
⎜
⎟
⎝ 0 ⎠
M = 18.3 N⋅ m
280
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
⎛⎜ α ⎞⎟
⎛ M ⎞
⎜ β ⎟ = acos ⎜
⎟
⎝ M ⎠
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 155.496 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 114.504 ⎟ deg
⎜ γ ⎟ ⎝ 90 ⎠
⎝ ⎠
Problem 4-96
Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude
of the couple moment?
Given:
F = 125 N
a = 150 mm
b = 150 mm
c = 200 mm
d = 600 mm
Solution:
⎛ c ⎞ ⎛0⎞
M = ⎜a + b⎟ × ⎜ 0 ⎟
⎜
⎟ ⎜ ⎟
⎝ 0 ⎠ ⎝F⎠
⎛ 37.5 ⎞
M = ⎜ −25 ⎟ N⋅ m
⎜
⎟
⎝ 0 ⎠
M = 45.1 N⋅ m
Problem 4-97
If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the
forces applied to the wrenches.
281
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Engineering Mechanics - Statics
Chapter 4
Given:
M = 300 N⋅ m
a = 150 mm
b = 150 mm
c = 200 mm
d = 600 mm
Solution:
F = 1N
Initial guess:
Given
⎛ c ⎞ ⎛0⎞
⎜a + b⎟ × ⎜ 0 ⎟ = M
⎜
⎟ ⎜ ⎟
⎝ 0 ⎠ ⎝F⎠
F = Find ( F)
F = 832.1 N
Problem 4-98
Replace the force at A by an equivalent force and couple moment at point O.
Given:
F = 375 N
a = 2m
b = 4m
c = 2m
d = 1m
θ = 30 deg
282
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Solution:
⎛ sin ( θ ) ⎞
⎜
⎟
F v = F −cos ( θ )
⎜
⎟
⎝ 0 ⎠
⎛ 187.5 ⎞
⎜
⎟
F v = −324.76 N
⎜
⎟
⎝ 0 ⎠
⎛ −a ⎞
⎜ ⎟
MO = b × Fv
⎜ ⎟
⎝0⎠
⎛ 0
⎞
⎜
⎟ N⋅ m
0
MO =
⎜
⎟
⎝ −100.481 ⎠
Problem 4-99
Replace the force at A by an equivalent force and couple moment at point P .
Given:
F = 375 N
a = 2m
b = 4m
c = 2m
d = 1m
θ = 30 deg
Solution:
⎛ sin ( θ ) ⎞
⎜
⎟
F v = F −cos ( θ )
⎜
⎟
⎝ 0 ⎠
⎛ 187.5 ⎞
⎜
⎟
F v = −324.76 N
⎜
⎟
⎝ 0 ⎠
⎛ −a − c ⎞
⎜
⎟
MP = b − d × Fv
⎜
⎟
⎝ 0 ⎠
⎛ 0 ⎞
⎜ 0 ⎟ N⋅ m
MP =
⎜
⎟
⎝ 736.538 ⎠
Problem 4-100
Replace the force system by an equivalent resultant force and couple moment at point O.
283
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 60 lb
a = 2 ft
F 2 = 85 lb
b = 3 ft
F 3 = 25 lb
c = 6 ft
θ = 45 deg
d = 4 ft
e = 3
f = 4
Solution:
⎛⎜ 0 ⎟⎞
F = ⎜ −F1 ⎟ +
⎜ 0 ⎟
⎝
⎠
⎛ cos ( θ ) ⎞
⎛ −e ⎞
⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟
3
⎜
⎟
2
2⎜ ⎟
e + f ⎝0 ⎠
⎝ 0 ⎠
F2
⎛ −33.322 ⎞
F = ⎜ −110.322 ⎟ lb
⎜
⎟
0
⎝
⎠
F = 115.245 lb
⎛ −c ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎡ F
⎛ −e ⎞⎤ ⎛ d ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤
2
⎜
⎟
⎜
⎟
⎢
⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥
MO = 0 × ⎜ −F1 ⎟ + a ×
⎜ ⎟
⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3⎜
⎟⎥
⎜
⎟
⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦
⎛ 0 ⎞
MO = ⎜ 0 ⎟ lb⋅ ft
⎜
⎟
⎝ 480 ⎠
MO = 480 lb⋅ ft
Problem 4-101
Replace the force system by an equivalent resultant force and couple moment at point P.
284
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 60 lb
a = 2 ft
F 2 = 85 lb
b = 3 ft
F 3 = 25 lb
c = 6 ft
θ = 45 deg
d = 4 ft
e = 3
f = 4
Solution:
⎛⎜ 0 ⎟⎞
F = ⎜ −F1 ⎟ +
⎜ 0 ⎟
⎝
⎠
⎛ cos ( θ ) ⎞
⎛ −e ⎞
⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟
3
⎜
⎟
2
2⎜ ⎟
e + f ⎝0 ⎠
⎝ 0 ⎠
F2
⎛ −33.322 ⎞
⎜
⎟
F = −110.322 lb
⎜
⎟
0
⎝
⎠
F = 115.245 lb
⎛ −c − d ⎞ ⎛⎜ 0 ⎟⎞ ⎛ −d ⎞ ⎡ F
⎛ −e ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤
2
⎜
⎟
⎜
⎟
⎢
⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥
0
MP =
× ⎜ −F 1 ⎟ + a ×
⎜
⎟
⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3 ⎜
⎟⎥
⎜
⎟
⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎜ 0 ⎟ lb⋅ ft
MP =
⎜
⎟
⎝ 921 ⎠
MP = 921 lb⋅ ft
Problem 4-102
Replace the force system by an equivalent
force and couple moment at point O.
Units Used:
3
kip = 10 lb
Given:
F 1 = 430 lb
F 2 = 260 lb
a = 2 ft
e = 5 ft
285
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Engineering Mechanics - Statics
Chapter 4
b = 8 ft
f = 12
c = 3 ft
g = 5
d = a
θ = 60 deg
Solution:
⎛ −sin ( θ ) ⎞
⎜
⎟
F R = F1 −cos ( θ ) +
⎜
⎟
⎝ 0 ⎠
⎛g⎞
⎜f⎟
2
2⎜ ⎟
g + f ⎝0⎠
⎛ −272 ⎞
⎜ 25 ⎟ lb
FR =
⎜
⎟
⎝ 0 ⎠
F2
⎛ −d ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ e ⎞ ⎡ F
⎛ g ⎞⎤
2
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ f ⎟⎥
MO = b × F 1 −cos ( θ ) + 0 ×
⎜ ⎟ ⎢ ⎜
⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦
F R = 274 lb
⎛ 0 ⎞
⎜ 0 ⎟ kip⋅ ft
MO =
⎜
⎟
⎝ 4.609 ⎠
Problem 4-103
Replace the force system by an equivalent force and couple moment at point P.
Units Used:
3
kip = 10 lb
Given:
F 1 = 430 lb
F 2 = 260 lb
a = 2 ft
e = 5 ft
b = 8 ft
f = 12
c = 3 ft
g = 5
d = a
θ = 60 deg
Solution:
⎛ −sin ( θ ) ⎞
⎜
⎟
F R = F1 −cos ( θ ) +
⎜
⎟
⎝ 0 ⎠
⎛g⎞
⎜f⎟
2
2⎜ ⎟
g + f ⎝0⎠
⎛ −272 ⎞
⎜ 25 ⎟ lb
FR =
⎜
⎟
⎝ 0 ⎠
F2
F R = 274 lb
286
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Engineering Mechanics - Statics
Chapter 4
⎛ 0 ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ d + e ⎞ ⎡ F
⎛ g ⎞⎤
2
⎜
⎟ ⎢ ⎜
⎟⎥ ⎜ c ⎟ × ⎢
⎜ f ⎟⎥
MP = b + c × F 1 −cos ( θ ) +
⎜
⎟ ⎢ ⎜
⎟⎥ ⎜
⎟ ⎢ 2 2 ⎜ ⎟⎥
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎜ 0 ⎟ kip⋅ ft
MP =
⎜
⎟
⎝ 5.476 ⎠
Problem 4-104
Replace the loading system acting on the post by an equivalent resultant force and couple
moment at point O.
Given:
F 1 = 30 lb
a = 1 ft
d = 3
F 2 = 40 lb
b = 3 ft
e = 4
F 3 = 60 lb
c = 2 ft
Solution:
⎛0⎞
⎛1⎞
⎜ ⎟
⎜ ⎟
F R = F1 −1 + F2 0 +
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
⎛ 4 ⎞
⎜
⎟
F R = −78 lb
⎜
⎟
⎝ 0 ⎠
⎛ −d ⎞
⎜ −e ⎟
⎟
2
2⎜
d +e ⎝ 0 ⎠
F3
F R = 78.1 lb
⎛ 0 ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F
⎛ −d ⎞⎤
3
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ −e ⎟⎥
MO = a + b + c × F1 −1 + c × F 2 0 + b + c ×
⎜
⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜
⎟ ⎢ 2 2 ⎜ ⎟⎥
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎜ 0 ⎟ lb⋅ ft
MO =
⎜
⎟
⎝ 100 ⎠
Problem 4-105
Replace the loading system acting on the post by an equivalent resultant force and couple
moment at point P.
287
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 30 lb
F 2 = 40 lb
F 3 = 60 lb
a = 1 ft
b = 3 ft
c = 2 ft
d = 3
e = 4
Solution:
⎛0⎞
⎛1⎞
⎜
⎟
⎜ ⎟
F R = F1 −1 + F2 0 +
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
⎛ 4 ⎞
⎜
⎟
F R = −78 lb
⎜
⎟
⎝ 0 ⎠
⎛ −d ⎞
⎜ −e ⎟
⎟
2
2⎜
d +e ⎝ 0 ⎠
F3
F R = 78.1 lb
⎡⎛ 0 ⎞ ⎤ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F
⎛ −d ⎞⎤
3
⎢
⎜
⎟
⎥
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ −e ⎟⎥
MP = 0 ft × F1 −1 + −a − b × F2 0 + −a ×
⎢⎜ ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎜
⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥
⎣⎝ 0 ⎠ ⎦ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎜ 0 ⎟ lb⋅ ft
MP =
⎜
⎟
⎝ 124 ⎠
Problem 4-106
Replace the force and couple system by an equivalent force and couple moment at point O.
Units Used:
3
kN = 10 N
288
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Engineering Mechanics - Statics
Chapter 4
Given:
M = 8 kN m
θ = 60 deg
a = 3m
f = 12
b = 3m
g = 5
c = 4m
F 1 = 6 kN
d = 4m
F 2 = 4 kN
e = 5m
Solution:
FR =
⎛ −cos ( θ ) ⎞
⎛g⎞
⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟
2
⎜
⎟
2
2⎜ ⎟
f + g ⎝0⎠
⎝ 0 ⎠
F1
⎛ 0.308 ⎞
⎜
⎟
F R = 2.074 kN
⎜
⎟
⎝ 0 ⎠
F R = 2.097 kN
⎛ 0 ⎞ ⎛ −c ⎞ ⎡ F
⎛ g ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤
1
⎜
⎟
⎜
⎟
⎢
⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥
MO = 0 + −e ×
⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜
⎟⎥
⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎜ 0 ⎟ kN⋅ m
MO =
⎜
⎟
⎝ −10.615 ⎠
Problem 4-107
Replace the force and couple system by an equivalent force and couple moment at point P.
Units Used:
3
kN = 10 N
289
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Given:
M = 8 kN⋅ m
θ = 60 deg
a = 3m
f = 12
b = 3m
g = 5
c = 4m
F 1 = 6 kN
d = 4m
F 2 = 4 kN
e = 5m
Solution:
FR =
⎛ −cos ( θ ) ⎞
⎛g⎞
⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟
2
⎜
⎟
2
2⎜ ⎟
f + g ⎝0⎠
⎝ 0 ⎠
F1
⎛ 0.308 ⎞
F R = ⎜ 2.074 ⎟ kN
⎜
⎟
⎝ 0 ⎠
F R = 2.097 kN
⎛ 0 ⎞ ⎛ −c − b ⎞ ⎡ F
⎛ g ⎞⎤ ⎛ −b ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤
1
⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥
MP = ⎜ 0 ⎟ + ⎜ −e ⎟ × ⎢
⎜ ⎟ ⎜
⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜
⎟⎥
⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦
⎛ 0 ⎞
⎟ kN⋅ m
0
MP = ⎜
⎜
⎟
⎝ −16.838 ⎠
Problem 4-108
Replace the force system by a single force resultant and specify its point of application, measured
along the x axis from point O.
Given:
F 1 = 125 lb
F 2 = 350 lb
F 3 = 850 lb
290
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
a = 2 ft
b = 6 ft
c = 3 ft
d = 4 ft
Solution:
F Ry = F 3 − F 2 − F 1
F Ry = 375 lb
F Ry x = F3 ( b + c) − F 2 ( b) + F1 ( a)
x =
F3 ( b + c) − F 2 ( b) + F1 a
FRy
x = 15.5 ft
Problem 4-109
Replace the force system by a single force resultant and specify its point of application, measured
along the x axis from point P.
Given:
F 1 = 125 lb
a = 2 ft
F 2 = 350 lb
b = 6 ft
F 3 = 850 lb
c = 3 ft
d = 4 ft
Solution:
F Ry = F 3 − F 2 − F 1
F Ry = 375 lb
F Ry x = F2 ( d + c) − F 3 ( d) + F1 ( a + b + c + d)
x =
F2 d + F2 c − F3 d + F1 a + F1 b + F1 c + F1 d
x = 2.47 ft
F Ry
(to the right of P)
291
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Engineering Mechanics - Statics
Chapter 4
Problem 4-110
The forces and couple moments which are exerted on the toe and heel plates of a snow ski are
F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment
acting at point O. Express the results in Cartesian vector form.
Given:
a = 120 mm
b = 800 mm
Solution:
⎛ −50 ⎞
F t = ⎜ 80 ⎟ N
⎜
⎟
⎝ −158 ⎠
⎛ −20 ⎞
F h = ⎜ 60 ⎟ N
⎜
⎟
⎝ −250 ⎠
⎛ −6 ⎞
Mt = ⎜ 4 ⎟ N⋅ m
⎜ ⎟
⎝2⎠
⎛ −20 ⎞
Mh = ⎜ 8 ⎟ N⋅ m
⎜
⎟
⎝ 3 ⎠
FR = Ft + Fh
⎛ −70 ⎞
F R = ⎜ 140 ⎟ N
⎜
⎟
⎝ −408 ⎠
⎛a⎞
r0Ft = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
MRP = ( r0Ft × F t) + Mt + Mh
⎛ −26 ⎞
MRP = ⎜ 31 ⎟ N⋅ m
⎜
⎟
⎝ 14.6 ⎠
Problem 4-111
The forces and couple moments which are exerted on the toe and heel plates of a snow ski are
F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment
292
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Engineering Mechanics - Statics
Chapter 4
acting at point P. Express the results in Cartesian vector form.
Given:
a = 120 mm
b = 800 mm
⎛ −50 ⎞
⎜ 80 ⎟ N
Ft =
⎜
⎟
⎝ −158 ⎠
⎛ −6 ⎞
⎜ ⎟
Mt = 4 N⋅ m
⎜ ⎟
⎝2⎠
⎛ −20 ⎞
⎜ 60 ⎟ N
Fh =
⎜
⎟
⎝ −250 ⎠
⎛ −20 ⎞
⎜ 8 ⎟ N⋅ m
Mh =
⎜
⎟
⎝ 3 ⎠
Solution:
FR = Ft + Fh
⎛ −70 ⎞
⎜
⎟
F R = 140 N
⎜
⎟
⎝ −408 ⎠
⎛b⎞
⎛a + b⎞
⎜
⎟
⎜ 0 ⎟×F
MP = Mt + Mh + 0 × F h +
⎜ ⎟
⎜
⎟ t
⎝0⎠
⎝ 0 ⎠
⎛ −26 ⎞
⎜
⎟
MP = 357.4 N⋅ m
⎜
⎟
⎝ 126.6 ⎠
Problem 4-112
Replace the three forces acting on the shaft by a single resultant force. Specify where the force
acts, measured from end B.
293
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 500 lb
F 2 = 200 lb
F 3 = 260 lb
a = 5 ft
e = 3
b = 3 ft
f = 4
c = 2 ft
g = 12
d = 4 ft
h = 5
Solution:
FR =
⎛−f⎞
⎛0⎞
⎜ −e ⎟ + F ⎜ −1 ⎟ +
2
⎜ ⎟
2
2⎜ ⎟
e + f ⎝0 ⎠
⎝0⎠
F1
Initial guess:
⎛h⎞
⎜ −g ⎟
⎟
2
2⎜
g +h ⎝ 0 ⎠
F3
⎛ −300 ⎞
F R = ⎜ −740 ⎟ lb
⎜
⎟
⎝ 0 ⎠
F R = 798 lb
x = 1 ft
Given
⎛a⎞ ⎡ F
⎛ − f ⎞⎤ ⎛ a + b ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ a + b + c ⎞ ⎡ F
⎛ h ⎞⎤ ⎛ −x ⎞
1
3
⎜0⎟ × ⎢
⎜ −e ⎟⎥ + ⎜ 0 ⎟ × ⎢F ⎜ −1 ⎟⎥ + ⎜ 0 ⎟ × ⎢
⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F
⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜
⎟ ⎢ 2⎜ ⎟⎥ ⎜
⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ R
⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠
x = Find ( x)
x = −7.432 ft
Problem 4-113
Replace the three forces acting on the shaft by a single resultant force. Specify where the force
acts, measured from end B.
Given:
F 1 = 500 lb
F 2 = 200 lb
F 3 = 260 lb
a = 5 ft
e = 3
294
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Engineering Mechanics - Statics
b = 3 ft
f = 4
c = 2 ft
g = 12
d = 4 ft
h = 5
Chapter 4
Solution:
FR =
⎛−f⎞
⎛0⎞
⎜ −e ⎟ + F ⎜ −1 ⎟ +
2
⎜ ⎟
2
2⎜ ⎟
e + f ⎝0 ⎠
⎝0⎠
F1
Initial guess:
⎛h⎞
⎜ −g ⎟
⎟
2
2⎜
g +h ⎝ 0 ⎠
F3
⎛ −300 ⎞
F R = ⎜ −740 ⎟ lb
⎜
⎟
⎝ 0 ⎠
F R = 798 lb
x = 1ft
Given
⎛ −b − c − d ⎞ ⎡ F
⎛ − f ⎞⎤ ⎛ −c − d ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ −d ⎞ ⎡ F
⎛ h ⎞⎤ ⎛ −x ⎞
1
3
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F
0
−e +
×
0
× F 2 −1 + 0 ×
⎜
⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜
⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟
0
⎝
⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠
x = Find ( x)
x = 6.568 ft
measured to the left of B
Problem 4-114
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member AB, measured from A.
Given:
F 1 = 300 lb
F 2 = 200 lb
F 3 = 400 lb
F 4 = 200 lb
M = 600 lb⋅ ft
a = 3 ft
b = 4 ft
c = 2 ft
d = 7 ft
Solution:
F Rx = −F 4
F Rx = −200 lb
F Ry = −F 1 − F 2 − F 3
F Ry = −900 lb
295
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Engineering Mechanics - Statics
2
F =
FRx + FRy
Chapter 4
2
F = 922 lb
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ = atan ⎜
θ = 77.5 deg
F Ry x = −F2 a − F 3 ( a + b) − F 4 c + M
x = −
F2 ( a) + F 3 ( a + b) + F 4 c − M
x = 3.556 ft
F Ry
Problem 4-115
Replace the loading on the frame by a single resultant force. Specify where the force acts ,
measured from end A.
Given:
F 1 = 450 N
a = 2m
F 2 = 300 N
b = 4m
F 3 = 700 N
c = 3m
θ = 60 deg
M = 1500 N⋅ m
φ = 30 deg
Solution:
F Rx = F 1 cos ( θ ) − F 3 sin ( φ )
F Rx = −125 N
F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2
F Ry = −1.296 × 10 N
F =
2
FRx + FRy
3
3
2
F = 1.302 × 10 N
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
θ 1 = 84.5 deg
F Ry( x) = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M
x =
−F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M
F Ry
x = 7.36 m
Problem 4-116
Replace the loading on the frame by a single resultant force. Specify where the force acts ,
measured from end B.
296
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 450 N
a = 2m
F 2 = 300 N
b = 4m
F 3 = 700 N
c = 3m
θ = 60 deg
M = 1500 N⋅ m
φ = 30deg
Solution:
F Rx = F 1 cos ( θ ) − F 3 sin ( φ )
F Rx = −125 N
F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2
F Ry = −1.296 × 10 N
F =
2
FRx + FRy
3
3
2
F = 1.302 × 10 N
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
θ 1 = 84.5 deg
F Ry x = F1 sin ( θ ) b − F3 cos ( φ ) c − M
x =
F1 sin ( θ ) b − F3 cos ( φ ) c − M
x = 1.36 m
F Ry
(to the right)
Problem 4-117
Replace the loading system acting on
the beam by an equivalent resultant
force and couple moment at point O.
Given:
F 1 = 200 N
F 2 = 450 N
M = 200 N⋅ m
a = 0.2 m
b = 1.5 m
c = 2m
d = 1.5 m
297
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Engineering Mechanics - Statics
Chapter 4
θ = 30 deg
Solution:
⎛ −sin ( θ ) ⎞
⎛0⎞
⎜
⎟
⎜
⎟
F R = F1 1 + F2 −cos ( θ )
⎜ ⎟
⎜
⎟
⎝0⎠
⎝ 0 ⎠
⎛ −225 ⎞
⎜
⎟
F R = −190 N
⎜
⎟
⎝ 0 ⎠
F R = 294 N
⎛ b + c ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ b ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤
⎛0⎞
⎜
⎟
⎢
⎜
⎟
⎥
⎜
⎟
⎢
⎜
⎟
⎥
⎜ ⎟
a
MO =
× F 1 + a × F −cos ( θ ) + M 0
⎜
⎟ ⎢ 1⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜
⎟⎥
⎜ ⎟
⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦
⎝ −1 ⎠
⎛ 0 ⎞
⎜ 0 ⎟ N⋅ m
MO =
⎜
⎟
⎝ −39.6 ⎠
Problem 4-118
Determine the magnitude and direction θ
of force F and its placement d on the
beam so that the loading system is
equivalent to a resultant force FR acting
vertically downward at point A and a
clockwise couple moment M.
Units Used:
3
kN = 10 N
Given:
F 1 = 5 kN
a = 3m
F 2 = 3 kN
b = 4m
F R = 12 kN
c = 6m
M = 50 kN⋅ m
e = 7
f = 24
F = 1 kN
θ = 30 deg
Solution:
Initial guesses:
d = 2m
298
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Engineering Mechanics - Statics
Given
Chapter 4
⎛ −e ⎞ F + F cos ( θ ) = 0
⎜ 2 2⎟ 1
⎝ e +f ⎠
⎛ − f ⎞ F − F sin ( θ ) − F = −F
2
R
⎜ 2 2⎟ 1
⎝ e +f ⎠
f
⎛
⎞
⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M
⎝ e +f ⎠
⎛F⎞
⎜ θ ⎟ = Find ( F , θ , d)
⎜ ⎟
⎝d⎠
θ = 71.565 deg
F = 4.427 kN
d = 3.524 m
Problem 4-119
Determine the magnitude and direction θ of force F and its placement d on the beam so that the
loading system is equivalent to a resultant force F R acting vertically downward at point A and a
clockwise couple moment M.
Units Used:
3
kN = 10 N
Given:
F 1 = 5 kN
a = 3m
F 2 = 3 kN
b = 4m
F R = 10 kN
c = 6m
M = 45 kN⋅ m e = 7
f = 24
Solution:
Initial guesses:
Given
F = 1 kN
θ = 30 deg
d = 1m
⎛ −e ⎞ F + F cos ( θ ) = 0
⎜ 2 2⎟ 1
⎝ e +f ⎠
⎛ − f ⎞ F − F sin ( θ ) − F = −F
2
R
⎜ 2 2⎟ 1
⎝ e +f ⎠
299
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Engineering Mechanics - Statics
Chapter 4
f
⎛
⎞
⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M
⎝ e +f ⎠
⎛F⎞
⎜ θ ⎟ = Find ( F , θ , d)
⎜ ⎟
⎝d⎠
θ = 57.529 deg
F = 2.608 kN
d = 2.636 m
Problem 4-120
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member AB, measured from A.
Given:
F 1 = 500 N
a = 3m
b = 2m
F 2 = 300 N
c = 1m
F 3 = 250 N
d = 2m
M = 400 N⋅ m
e = 3m
θ = 60 deg
f = 3
g = 4
Solution:
⎛
⎞ − F ( cos ( θ ) )
1
⎟
2
2
⎝ g +f ⎠
g
F Rx = −F 3 ⎜
F Rx = −450 N
⎛
f
F Ry = −883.0127 N
FRx + FRy
2
F R = 991.066 N
⎞ − F sin ( θ )
1
⎟
2
2
⎝ f +g ⎠
F Ry = −F 2 − F 3 ⎜
FR =
2
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
θ 1 = 62.996 deg
300
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Engineering Mechanics - Statics
Chapter 4
−F Rx( y) = M + F 1 cos ( θ ) a + F3
g
2
g + f
M + F 1 cos ( θ ) a + F3
g
2
g + f
y =
2
2
( b + a) − F2 ( d) − F 3 ⎛
f
⎞
⎜ 2 2 ⎟ ( d + e)
⎝ g +f ⎠
( b + a) − F2 ( d) − F 3 ⎛
f
⎞
⎜ 2 2 ⎟ ( d + e)
⎝ g +f ⎠
−FRx
y = 1.78 m
Problem 4-121
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member CD, measured from end C.
Given:
F 1 = 500 N
a = 3m
b = 2m
F 2 = 300 N
c = 1m
F 3 = 250 N
d = 2m
M = 400 N⋅ m
e = 3m
θ = 60 deg
f = 3
g = 4
Solution:
F Rx = −F 3 ⎛
g
⎞
⎜ 2 2 ⎟ − F1( cos ( θ ) )
⎝ g +f ⎠
F Rx = −450 N
F Ry = −F 2 − F 3 ⎛
f
⎞
⎜ 2 2 ⎟ − F1 sin ( θ )
⎝ f +g ⎠
FR =
2
FRx + FRy
F Ry = −883.0127 N
2
F R = 991.066 N
301
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Engineering Mechanics - Statics
Chapter 4
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
θ 1 = 62.996 deg
⎛
F Ry( x) = M − F2 ( d + c) − F 3 ⎜
f
2
⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ )
2⎟
1
⎝ g +f ⎠
⎛
M − F2 ( d + c) − F 3 ⎜
f
2
⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ )
2⎟
1
⎝ g +f ⎠
x =
1
1
F Ry
x = 2.64 m
Problem 4-122
Replace the force system acting on the frame
by an equivalent resultant force and specify
where the resultant's line of action intersects
member AB, measured from point A.
Given:
F 1 = 35 lb a = 2 ft
F 2 = 20 lb b = 4 ft
F 3 = 25 lb c = 3 ft
θ = 30 deg d = 2 ft
Solution:
F Rx = F 1 sin ( θ ) + F3
F Rx = 42.5 lb
F Ry = −F 1 cos ( θ ) − F 2
F Ry = −50.311 lb
FR =
2
FRx + FRy
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
2
F R = 65.9 lb
θ 1 = −49.8 deg
F Ry x = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)
302
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Engineering Mechanics - Statics
x =
Chapter 4
−F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)
x = 2.099 ft
FRy
Problem 4-123
Replace the force system acting on the frame by an equivalent resultant force and specify where the
resultant's line of action intersects member BC, measured from point B.
Given:
F 1 = 35 lb
F 2 = 20 lb
F 3 = 25 lb
θ = 30 deg
a = 2 ft
b = 4 ft
c = 3 ft
d = 2 ft
Solution:
F Rx = F 1 sin ( θ ) + F3
F Rx = 42.5 lb
F Ry = −F 1 cos ( θ ) − F 2
F Ry = −50.311 lb
FR =
2
FRx + FRy
2
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
F R = 65.9 lb
θ 1 = −49.8 deg
F Rx y = F 1 cos ( θ ) b + F3 ( c)
303
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Engineering Mechanics - Statics
y =
F 1 cos ( θ ) b + F3 ( c)
FRx
Chapter 4
y = 4.617 ft
(Below point B)
Problem 4-124
Replace the force system acting on the frame by an equivalent resultant force and couple moment
acting at point A.
Given:
F 1 = 35 lb
a = 2 ft
F 2 = 20 lb
b = 4 ft
F 3 = 25 lb
c = 3 ft
θ = 30 deg
d = 2 ft
Solution:
F Rx = F 1 sin ( θ ) + F3
F Rx = 42.5 lb
F Ry = F 1 cos ( θ ) + F 2
F Ry = 50.311 lb
FR =
2
FRx + FRy
⎛ FRy ⎞
⎟
⎝ FRx ⎠
θ 1 = atan ⎜
2
F R = 65.9 lb
θ 1 = 49.8 deg
MRA = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)
MRA = −106 lb⋅ ft
Problem 4-125
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point O. Express the results in Cartesian vector form.
304
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
⎛8⎞
⎜ ⎟
F = 6 kN
⎜ ⎟
⎝8⎠
a = 3m
b = 3m
⎛ −20 ⎞
⎜
⎟
M = −70 kN⋅ m
⎜
⎟
⎝ 20 ⎠
c = 4m
e = 5m
f = 6m
g = 5m
d = 6m
Solution:
⎛−f⎞
⎜ ⎟
MR = M + e × F
⎜ ⎟
⎝g⎠
FR = F
⎛8⎞
⎜ ⎟
F R = 6 kN
⎜ ⎟
⎝8⎠
⎛ −10 ⎞
⎜
⎟
MR = 18 kN⋅ m
⎜
⎟
⎝ −56 ⎠
Problem 4-126
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point P. Express the results in Cartesian vector form.
Units Used:
3
kN = 10 N
Given:
⎛8⎞
⎜ ⎟
F = 6 kN
⎜ ⎟
⎝8⎠
⎛ −20 ⎞
⎜
⎟
M = −70 kN⋅ m
⎜
⎟
⎝ 20 ⎠
a = 3m
b = 3m
e = 5m
c = 4m
f = 6m
305
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Engineering Mechanics - Statics
d = 6m
Chapter 4
g = 5m
Solution:
⎛ −f ⎞
⎜ e ⎟×F
MR = M +
⎜
⎟
⎝d + g⎠
FR = F
⎛8⎞
⎜ ⎟
F R = 6 kN
⎜ ⎟
⎝8⎠
⎛ −46 ⎞
⎜
⎟
MR = 66 kN⋅ m
⎜
⎟
⎝ −56 ⎠
Problem 4-127
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point Q. Express the results in Cartesian vector form.
Units Used:
3
kN = 10 N
Given:
⎛8⎞
⎜ ⎟
F = 6 kN
⎜ ⎟
⎝8⎠
⎛ −20 ⎞
⎜
⎟
M = −70 kN⋅ m
⎜
⎟
⎝ 20 ⎠
a = 3m
b = 3m
e = 5m
c = 4m
f = 6m
d = 6m
g = 5m
Solution:
FR = F
⎛0⎞
⎜ ⎟
MR = M + e × F
⎜ ⎟
⎝g⎠
⎛8⎞
⎜ ⎟
F R = 6 kN
⎜ ⎟
⎝8⎠
⎛ −10 ⎞
⎜
⎟
MR = −30 kN⋅ m
⎜
⎟
⎝ −20 ⎠
Problem 4-128
The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction.
Replace these forces by an equivalent force and couple moment at point A. Express the result in
306
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Engineering Mechanics - Statics
Chapter 4
p
y
Cartesian vector form.
q
p
p
p
Given:
F 1 = 40 N
r = 80 mm
F 2 = 40 N
a = 300 mm
θ = 0 deg
Solution:
⎛0⎞
F 1v = F1 ⎜ 0 ⎟
⎜ ⎟
⎝ −1 ⎠
⎛ 0 ⎞
F 2v = F2 ⎜ −cos ( θ ) ⎟
⎜
⎟
⎝ −sin ( θ ) ⎠
⎛ −a ⎞
r2 = ⎜ −r sin ( θ ) ⎟
⎜
⎟
⎝ r cos ( θ ) ⎠
⎛ −a ⎞
r1 = ⎜ r ⎟
⎜ ⎟
⎝0⎠
F R = F1v + F2v
MA = r1 × F1v + r2 × F2v
⎛ 0 ⎞
F R = ⎜ −40 ⎟ N
⎜
⎟
⎝ −40 ⎠
⎛ 0 ⎞
MA = ⎜ −12 ⎟ N⋅ m
⎜
⎟
⎝ 12 ⎠
Problem 4-129
The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction.
Replace these forces by an equivalent force and couple moment at point A. Express the result in
Cartesian vector form.
307
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 40 N
F 2 = 40 N
θ = 0 deg
r = 80 mm
a = 300 mm
θ = 45 deg
Solution:
⎛0⎞
F 1v = F1 ⎜ 0 ⎟
⎜ ⎟
⎝ −1 ⎠
⎛ 0 ⎞
F 2v = F2 ⎜ −cos ( θ ) ⎟
⎜
⎟
⎝ −sin ( θ ) ⎠
F R = F1v + F2v
MA = r1 × F1v + r2 × F2v
⎛ 0 ⎞
F R = ⎜ −28.28 ⎟ N
⎜
⎟
⎝ −68.28 ⎠
⎛ 0 ⎞
MA = ⎜ −20.49 ⎟ N⋅ m
⎜
⎟
⎝ 8.49 ⎠
⎛ −a ⎞
r1 = ⎜ r ⎟
⎜ ⎟
⎝0⎠
⎛ −a ⎞
r2 = ⎜ −r sin ( θ ) ⎟
⎜
⎟
⎝ r cos ( θ ) ⎠
Problem 4-130
Replace this system by an equivalent resultant force and couple moment acting at O. Express the
results in Cartesian vector form.
Given:
F 1 = 50 N
F 2 = 80 N
308
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Engineering Mechanics - Statics
Chapter 4
F 3 = 180 N
a = 1.25 m
b = 0.5 m
c = 0.75 m
Solution:
⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎟⎞
FR = ⎜ 0 ⎟ + ⎜ 0 ⎟ + ⎜ 0 ⎟
⎜ F 1 ⎟ ⎜ −F2 ⎟ ⎜ −F 3 ⎟
⎝ ⎠ ⎝
⎠ ⎝
⎠
⎛ 0 ⎞
⎜ 0 ⎟N
FR =
⎜
⎟
⎝ −210 ⎠
⎛ a + c ⎞ ⎛⎜ 0 ⎞⎟ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞
⎜ b ⎟ × 0 + ⎜b⎟ × 0 + ⎜0⎟ × 0
MO =
⎟ ⎜ ⎟ ⎜
⎟
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎜
⎟
⎜
⎟
⎜
⎝ 0 ⎠ ⎝ F1 ⎠ ⎝ 0 ⎠ ⎝ −F2 ⎠ ⎝ 0 ⎠ ⎝ −F3 ⎟⎠
⎛ −15 ⎞
⎜
⎟
MO = 225 N⋅ m
⎜
⎟
⎝ 0 ⎠
Problem 4-131
Handle forces F 1 and F 2 are applied to the electric drill. Replace this system by an equivalent
resultant force and couple moment acting at point O. Express the results in Cartesian vector
form.
Given:
a = 0.15 m
b = 0.25 m
c = 0.3 m
⎛ 6 ⎞
⎜
⎟
F 1 = −3 N
⎜
⎟
⎝ −10 ⎠
⎛0⎞
⎜ ⎟
F2 = 2 N
⎜ ⎟
⎝ −4 ⎠
309
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Engineering Mechanics - Statics
Chapter 4
Solution:
FR = F1 + F2
⎛ 6 ⎞
⎜
⎟
F R = −1 N
⎜
⎟
⎝ −14 ⎠
⎛a⎞
⎛0⎞
⎜
⎟
⎜ ⎟
MO = 0 × F1 + −b × F 2
⎜ ⎟
⎜ ⎟
⎝c⎠
⎝c ⎠
⎛ 1.3 ⎞
⎜ 3.3 ⎟ N⋅ m
MO =
⎜
⎟
⎝ −0.45 ⎠
Problem 4-132
A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in
the four muscle groups consist of F R for the rectus, FO for the oblique, FL for the lumbar
latissimus dorsi, and F E for the erector spinae. These loadings are symmetric with respect to the
y - z plane. Replace this system of parallel forces by an equivalent force and couple moment
acting at the spine, point O. Express the results in Cartesian vector form.
Given:
F R = 35 N
a = 75 mm
F O = 45 N
b = 45 mm
F L = 23 N
c = 15 mm
F E = 32 N
d = 50 mm
e = 40 mm
f = 30 mm
Solution:
F Res = ΣFi;
F Res = 2( FR + F O + F L + F E)
F Res = 270 N
MROx = ΣMOx;
MRO = −2FR a + 2F E c + 2F L b
MRO = −2.22 N⋅ m
Problem 4-133
The building slab is subjected to four parallel column loadings.Determine the equivalent resultant
force and specify its location (x, y) on the slab.
310
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
F 1 = 30 kN
a = 3m
F 2 = 40 kN
b = 8m
F 3 = 20 kN
c = 2m
F 4 = 50 kN
d = 6m
e = 4m
Solution:
+
↑ FR = ΣFx;
FR = F1 + F2 + F3 + F4
F R = 140 kN
MRx = ΣMx;
−F R( y) = −( F4 ) ( a) − ⎡⎣( F1 ) ( a + b)⎤⎦ − ⎡⎣( F2 ) ( a + b + c)⎤⎦
y =
F4 a + F1 a + F1 b + F2 a + F2 b + F2 c
FR
y = 7.14 m
MRy = ΣMy;
(FR)x = (F4)( e) + (F3)( d + e) + (F2)( b + c)
x =
F4 e + F3 d + F3 e + F2 b + F2 c
FR
x = 5.71 m
Problem 4-134
The building slab is subjected to four parallel column loadings. Determine the equivalent resultant
force and specify its location (x, y) on the slab.
311
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Engineering Mechanics - Statics
Chapter 4
Units Used:
3
kN = 10 N
Given:
F 1 = 20 kN
a = 3m
F 2 = 50 kN
b = 8m
F 3 = 20 kN
c = 2m
F 4 = 50 kN
d = 6m
e = 4m
Solution:
FR = F1 + F2 + F3 + F4
F R = 140 kN
F R x = F2 e + F1 ( d + e) + F 2 ( d + e)
x =
2 F2 e + F1 d + F1 e + F2 d
x = 6.43 m
FR
−F R y = −F 2 a − F3 ( a + b) − F2 ( a + b + c)
y =
2 F2 a + F3 a + F3 b + F2 b + F2 c
FR
y = 7.29 m
Problem 4-135
The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the
magnitude F of the couple forces so that the system can be simplified to a wrench acting at
point C.
Given:
a = 0.6 m
312
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Engineering Mechanics - Statics
Chapter 4
b = 0.8 m
c = 0.25 m
d = 0.7 m
e = 0.3 m
f = 0.3 m
g = 0.5 m
h = 0.25 m
P = 60 N
Q = 40 N
Solution:
Initial Guess
F = 1N
MC = 1 N⋅ m
Given
⎛⎜ −MC ⎞⎟ ⎡ −P( c + h) ⎤ ⎡ 0 ⎤ ⎛ a ⎞ ⎛ −Q ⎞
⎢
⎥ + ⎢ 0 ⎥ + ⎜b⎟ × ⎜ 0 ⎟
⎜ 0 ⎟=⎢ 0
⎥ ⎢
⎥ ⎜ ⎟ ⎜ ⎟
⎜ 0 ⎟ ⎣ 0
⎦ ⎣ − F ( e + f) ⎦ ⎝ 0 ⎠ ⎝ 0 ⎠
⎝
⎠
⎛ F ⎞
⎜
⎟ = Find ( F , MC)
⎝ MC ⎠
MC = 30 N⋅ m
F = 53.3 N
Problem 4-136
The three forces acting on the block each have a magnitude F 1 = F2 = F3. Replace this system
by a wrench and specify the point where the wrench intersects the z axis, measured from
point O.
Given:
F 1 = 10 lb
a = 6 ft
F2 = F1
b = 6 ft
F3 = F1
c = 2 ft
Solution:
The vectors
313
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Engineering Mechanics - Statics
F 1v =
⎛b⎞
⎜ −a ⎟
⎟
2
2⎜
b +a ⎝ 0 ⎠
F1
Chapter 4
⎛⎜ 0 ⎞⎟
F 2v = ⎜ −F 2 ⎟
⎜ 0 ⎟
⎝
⎠
F 3v =
⎛ −b ⎞
⎜a⎟
⎟
2
2⎜
b +a ⎝ 0 ⎠
F3
Place the wrench in the x - z plane.
x = 1ft z = 1ft
Guesses
M = 1 lb⋅ ft
R x = 1 lb
R y = 1 lb
R z = 1 lb
⎛ Rx ⎞
⎜ ⎟
Given
⎜ Ry ⎟ = F1v + F2v + F3v
⎜R ⎟
⎝ z⎠
⎛ Rx ⎞ ⎛ 0 ⎞
⎛ x ⎞ ⎛⎜ Rx ⎟⎞
⎛0⎞
⎛b⎞
⎜
⎟
M
⎜0⎟ × R +
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ Ry ⎟ = ⎜ a ⎟ × F2v + ⎜ a ⎟ × F1v + ⎜ 0 ⎟ × F3v
⎜ ⎟ ⎜ y⎟
2
2
2
Rx + Ry + Rz ⎜ R ⎟ ⎝ c ⎠
⎝ z ⎠ ⎜⎝ Rz ⎟⎠
⎝0⎠
⎝c ⎠
⎝ z⎠
⎛x ⎞
⎜ ⎟
⎜ z ⎟
⎛ Rx ⎞
⎜ ⎟
⎜M ⎟
M
Mv =
⎜ R ⎟ = Find ( x , z , M , Rx , Ry , Rz)
⎜ Ry ⎟
2
2
2
⎜ x⎟
Rx + Ry + Rz ⎜ R ⎟
⎝ z⎠
R
⎜ y⎟
⎜ ⎟
⎝ Rz ⎠
⎛ Rx ⎞ ⎛ 0 ⎞
⎛ 0 ⎞
⎜ ⎟ ⎜
⎛x⎞ ⎛ 0 ⎞
⎜
⎟
⎟
Mv = −14.142 lb⋅ ft
⎜ ⎟=⎜
⎟ ft
⎜ Ry ⎟ = ⎜ −10 ⎟ lb
⎜
⎟
⎝ z ⎠ ⎝ 0.586 ⎠
⎜R ⎟ ⎝ 0 ⎠
⎝ 0 ⎠
⎝ z⎠
Problem 4-137
Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force
and couple moment for the wrench and the point P(x, y) where its line of action intersects the
plate.
Units Used:
3
kN = 10 N
314
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Engineering Mechanics - Statics
Chapter 4
Given:
F A = 500 N
F B = 800 N
F C = 300 N
a = 4m
b = 6m
Solution:
⎛ FA ⎞
⎜ ⎟
FR = ⎜ FC ⎟
⎜F ⎟
⎝ B⎠
Guesses
F R = 0.9899 kN
x = 1m
Given
y = 1m
M = 100 N⋅ m
⎛x⎞
⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ 0 ⎞⎟
⎜
⎟
M
+ y × FR = ⎜ a ⎟ × ⎜ FC ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
FR
⎝0⎠
⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ FB ⎟⎠
FR
⎛M⎞
⎜ x ⎟ = Find ( M , x , y)
⎜ ⎟
⎝y⎠
⎛ x ⎞ ⎛ 1.163 ⎞
⎜ ⎟=⎜
⎟m
⎝ y ⎠ ⎝ 2.061 ⎠
M = 3.07 kN⋅ m
Problem 4-138
Replace the three forces acting on the
plate by a wrench. Specify the magnitude
of the force and couple moment for the
wrench and the point P(y, z) where its
line of action intersects the plate.
Given:
F A = 80 lb
a = 12 ft
F B = 60 lb
b = 12 ft
F C = 40 lb
315
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Engineering Mechanics - Statics
Chapter 4
Solution:
⎛ −F C ⎞
⎜
⎟
F R = ⎜ −F B ⎟
⎜ −F ⎟
⎝ A⎠
F R = 108 lb
Guesses
y = 1 ft
z = 1 ft
M = 1 lb⋅ ft
Given
⎛0⎞
⎛ 0 ⎞ ⎛⎜ −FC ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞
⎜
⎟
M
+ y × FR = ⎜ a ⎟ × ⎜ 0 ⎟ + ⎜ a ⎟ × ⎜ −FB ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
FR
⎝z⎠
⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠
FR
⎛M⎞
⎜ y ⎟ = Find ( M , y , z)
⎜ ⎟
⎝z⎠
⎛ y ⎞ ⎛ 0.414 ⎞
⎜ ⎟=⎜
⎟ ft
⎝ z ⎠ ⎝ 8.69 ⎠
M = −624 lb⋅ ft
Problem 4-139
The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent
resultant location, measured from point O.
Given:
w1 = 2
lb
ft
w2 = 3.5
lb
ft
a = 2.75 ft
b = 4 ft
c = 1.5 ft
R = 1 lb
Solution: Guesses
Given
d = 1ft
w1 b + w2 c = R
w1 b⎛⎜ a −
⎝
⎛R⎞
⎜ ⎟ = Find ( R , d)
⎝d⎠
b⎞
⎞
⎛c
⎟ − w2 c⎜ + b − a⎟ = −d R
2⎠
⎝2
⎠
R = 13.25 lb
d = 0.34 ft
316
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Engineering Mechanics - Statics
Chapter 4
Problem 4-140
Replace the loading by an equivalent resultant force and couple moment acting at point A.
Units Used:
3
kN = 10 N
Given:
w1 = 600
N
w2 = 600
N
m
m
a = 2.5 m
b = 2.5 m
Solution:
F R = w1 a − w2 b
MRA = w1 a⎛⎜
a + b⎞
⎟
⎝ 2 ⎠
FR = 0 N
MRA = 3.75 kN⋅ m
Problem 4-141
Replace the loading by an equivalent force and couple moment acting at point O.
Units Used:
3
kN = 10 N
Given:
w = 6
kN
m
F = 15 kN
M = 500 kN⋅ m
a = 7.5 m
b = 4.5 m
317
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Engineering Mechanics - Statics
Chapter 4
Solution:
FR =
1
2
w( a + b) + F
F R = 51.0 kN
2
1
1
b
MR = −M − ⎛⎜ w a⎟⎞ ⎛⎜ a⎞⎟ − ⎛⎜ w b⎟⎞ ⎛⎜ a + ⎞⎟ − F( a + b)
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠
MR = −914 kN⋅ m
Problem 4-142
Replace the loading by a single resultant force, and specify the location of the force on the beam
measured from point O.
Units Used:
3
kN = 10 N
Given:
w = 6
kN
m
F = 15 kN
M = 500 kN⋅ m
a = 7.5 m
b = 4.5 m
Solution:
F R = 1 kN
Initial Guesses:
d = 1m
Given
FR =
1
2
w( a + b) + F
2
1
b
w a⎞⎟ ⎛⎜ a⎞⎟ − ⎛⎜ w b⎟⎞ ⎛⎜ a + ⎞⎟ − F ( a + b)
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠
−F R d = −M − ⎛⎜
1
⎛ FR ⎞
⎜ ⎟ = Find ( FR , d)
⎝ d ⎠
F R = 51 kN
d = 17.922 m
Problem 4-143
The column is used to support the floor which exerts a force P on the top of the column.
The effect of soil pressure along its side is distributed as shown. Replace this loading by an
318
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Engineering Mechanics - Statics
Chapter 4
p
g
p
g y
equivalent resultant force and specify where it acts along the column, measured from its
base A.
3
kip = 10 lb
Units Used:
Given:
P = 3000 lb
w1 = 80
lb
ft
w2 = 200
lb
ft
h = 9 ft
Solution:
F Rx = w1 h +
1
(w2 − w1)h
2
F Rx = 1260 lb
F Ry = P
2
FR =
FRx + P
2
θ = atan ⎛⎜
P ⎞
⎟
⎝ FRx ⎠
θ = 67.2 deg
1
h
h
w2 − w1 ) h + w1 h
(
2
3
2
F Rx y =
y =
F R = 3.25 kip
1 2 w2 + 2 w1
h
FRx
6
y = 3.86 ft
Problem 4-144
Replace the loading by an equivalent force and couple moment at point O.
Units Used:
3
kN = 10 N
Given:
kN
m
w1 = 15
w2 = 5
kN
m
319
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Engineering Mechanics - Statics
Chapter 4
d = 9m
Solution:
FR =
1
(w1 + w2)d
2
MRO = w2 d
F R = 90 kN
d 1
d
+ ( w1 − w2 ) d
2 2
3
MRO = 338 kN⋅ m
Problem 4-145
Replace the distributed loading by an equivalent resultant force, and specify its location on the
beam, measured from the pin at C.
Units Used:
3
kip = 10 lb
Given:
w = 800
lb
ft
a = 15 ft
b = 15 ft
θ = 30 deg
Solution:
FR = w a +
FR x = w a
wa
x =
a
2
wb
2
F R = 18 kip
a wb⎛
b⎞
+
⎜a + ⎟
2 ⎝
2
3⎠
+
b⎞
⎜a + ⎟
2 ⎝
3⎠
wb⎛
FR
x = 11.7 ft
Problem 4-146
The beam supports the distributed load caused by the sandbags. Determine the resultant force
on the beam and specify its location measured from point A.
320
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Engineering Mechanics - Statics
Units Used:
Chapter 4
3
kN = 10 N
Given:
w1 = 1.5
w2 = 1
kN
a = 3m
m
kN
b = 3m
m
w3 = 2.5
kN
c = 1.5 m
m
Solution:
F R = w1 a + w2 b + w3 c
MA = w1 a
a
2
+ w2 b⎛⎜ a +
⎝
F R = 11.25 kN
b⎞
c⎞
⎛
⎟ + w3 c⎜ a + b + ⎟
2⎠
2⎠
⎝
MA = 45.563 kN⋅ m
d =
MA
d = 4.05 m
FR
Problem 4-147
Determine the length b of the triangular load and its position a on the beam such that the
equivalent resultant force is zero and the resultant couple moment is M clockwise.
Units Used:
3
kN = 10 N
Given:
w1 = 4
kN
w2 = 2.5
m
M = 8 kN⋅ m
kN
m
c = 9m
Solution:
Initial Guesses:
Given
−1
2
a = 1m
w1 b +
1
2
b = 1m
w2 c = 0
321
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Engineering Mechanics - Statics
1
2
w1 b⎛⎜ a +
⎝
⎛a⎞
⎜ ⎟ = Find ( a , b)
⎝b⎠
Chapter 4
2b ⎞
1
2c
⎟ − w2 c = −M
3⎠ 2
3
⎛ a ⎞ ⎛ 1.539 ⎞
⎜ ⎟=⎜
⎟m
⎝ b ⎠ ⎝ 5.625 ⎠
Problem 4-148
Replace the distributed loading by an equivalent resultant force and specify its location,
measured from point A.
Units Used:
3
kN = 10 N
Given:
w1 = 800
N
w2 = 200
N
m
m
a = 2m
b = 3m
Solution:
F R = w2 b + w1 a +
x F R = w1 a
w1 a
x =
a
2
+
1
2
(w1 − w2)b
F R = 3.10 kN
(w1 − w2)b⎛⎜a + 3 ⎞⎟ + w2 b⎛⎜a + 2 ⎞⎟
2
1
b
⎝
b
⎠
⎝
a 1
⎛ b⎞
⎛ b⎞
+ ( w1 − w2 ) b⎜ a + ⎟ + w2 b⎜ a + ⎟
2 2
⎝ 3⎠
⎝ 2⎠
FR
⎠
x = 2.06 m
Problem 4-149
The distribution of soil loading on the bottom of a building slab is shown. Replace this loading
by an equivalent resultant force and specify its location, measured from point O.
Units Used:
322
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
3
kip = 10 lb
Given:
w1 = 50
lb
ft
w2 = 300
lb
w3 = 100
lb
ft
ft
a = 12 ft
b = 9 ft
Solution :
(w2 − w1)a + 2 (w2 − w3)b + w3 b
2
F R = w1 a +
1
F R d = w1 a
a
2
1
+
(w2 − w1)a 3 + 2 (w2 − w3)b⎛⎜a + 3 ⎞⎟ + w3 b⎛⎜a + 2 ⎞⎟
2
1
2a
1
b
⎝
2
d =
F R = 3.9 kip
2
2
b
⎠
3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b
6FR
⎝
⎠
2
d = 11.3 ft
Problem 4-150
The beam is subjected to the distributed
loading. Determine the length b of the
uniform load and its position a on the beam
such that the resultant force and couple
moment acting on the beam are zero.
Given:
w1 = 40
lb
w2 = 60
lb
ft
c = 10ft
d = 6 ft
ft
Solution:
Initial Guesses:
a = 1 ft
b = 1ft
323
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Engineering Mechanics - Statics
Chapter 4
Given
1
2
1
w2 d − w1 b = 0
⎛a⎞
⎜ ⎟ = Find ( a , b)
⎝b⎠
2
w2 d⎛⎜ c +
⎝
d⎞
⎛ b⎞
⎟ − w1 b⎜ a + ⎟ = 0
3⎠
⎝ 2⎠
⎛ a ⎞ ⎛ 9.75 ⎞
⎜ ⎟=⎜
⎟ ft
⎝ b ⎠ ⎝ 4.5 ⎠
Problem 4-151
Replace the loading by an equivalent resultant force and specify its location on the beam,
measured from point B.
Units Used:
3
kip = 10 lb
Given:
w1 = 800
lb
w2 = 500
lb
ft
ft
a = 12 ft
b = 9 ft
Solution:
FR =
1
2
a w1 +
1
2
(w1 − w2)b + w2 b
F R = 10.65 kip
a 1
1
b
b
F R x = − a w1 + ( w1 − w2 ) b + w2 b
3 2
2
3
2
x =
a 1
b
b
1
− a w1 + ( w1 − w2 ) b + w2 b
3 2
3
2
2
FR
x = 0.479 ft
( to the right of B )
Problem 4-152
Replace the distributed loading by an equivalent resultant force and specify where its line of action
intersects member AB, measured from A.
324
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Engineering Mechanics - Statics
Chapter 4
Given:
w1 = 200
N
m
w2 = 100
N
m
w3 = 200
N
m
a = 5m
b = 6m
Solution:
F Rx = −w3 a
F Ry =
F Rx = −1000 N
−1
(w1 + w2)b
2
− y FRx = w3 a
w3 a
y =
a
2
F Ry = −900 N
a
b 1
b
− w2 b − ( w1 − w2 ) b
2
2 2
3
− w2 b
b
−
(w1 − w2)b 3
2
1
b
2
−FRx
y = 0.1 m
Problem 4-153
Replace the distributed loading by an equivalent resultant force and specify where its line of
action intersects member BC, measured from C.
Units Used:
3
kN = 10 N
325
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 4
Given:
w1 = 200
N
w2 = 100
N
w3 = 200
N
m
m
m
a = 5m
b = 6m
Solution :
F Rx = −w3 a
F Ry =
−1
2
(w1 + w2)b
−x F Ry = −w3 a
−w3 a
x =
F Rx = −1000 N
a
2
+ w2 b
F Ry = −900 N
b
2
+
(w1 − w2)b 3
2
1
2b
a
b 1
2b
+ w2 b + ( w1 − w2 ) b
2
2 2
3
x = 0.556 m
−FRy
⎛ FRx ⎞
⎜
⎟ = 1.345 kN
⎝ FRy ⎠
Problem 4-154
Replace the loading by an equivalent resultant force and couple moment acting at point O.
Units Used:
3
kN = 10 N
Given:
w1 = 7.5
kN
m
326
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Engineering Mechanics - Statics
w2 = 20
Chapter 4
kN
m
a = 3m
b = 3m
c = 4.5 m
Solution:
(w2 − w1)c + w1 c + w1 b + 2 w1 a
2
1
FR =
1
F R = 95.6 kN
MRo = −
w2 − w1 ) c − w1 c − w1 b⎛⎜ c + ⎞⎟ − w1 a⎛⎜ b + c + ⎟⎞
(
2
3
2
2
2
3
1
c
c
b
⎝
1
⎠
a
⎝
⎠
MRo = −349 kN⋅ m
Problem 4-155
Determine the equivalent resultant force
and couple moment at point O.
Units Used:
3
kN = 10 N
Given:
a = 3m
wO = 3
kN
m
x
w ( x) = wO ⎛⎜ ⎟⎞
⎝ a⎠
2
Solution:
⌠a
F R = ⎮ w ( x) dx
⌡0
F R = 3 kN
⌠a
MO = ⎮ w ( x) ( a − x) dx
⌡0
MO = 2.25 kN⋅ m
327
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Engineering Mechanics - Statics
Chapter 4
Problem 4-156
Wind has blown sand over a platform such that the intensity of the load can be approximated by
3
⎛ x ⎞ . Simplify this distributed loading to an equivalent resultant force and
⎟
⎝d⎠
the function w = w0 ⎜
specify the magnitude and location of the force, measured from A.
Units Used:
3
kN = 10 N
Given:
w0 = 500
N
m
d = 10 m
⎛x⎞
⎟
⎝ d⎠
3
w ( x) = w0 ⎜
Solution:
d
⌠
F R = ⎮ w ( x) dx
⌡0
d =
⌠d
⎮ x w ( x) dx
⌡0
FR
F R = 1.25 kN
d=8m
Problem 4-157
Determine the equivalent resultant force and its location, measured from point O.
328
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Engineering Mechanics - Statics
Chapter 4
Solution:
L
⌠
2w0 L
⎛ πx⎞
F R = ⎮ w0 sin ⎜ ⎟ dx =
⎮
π
⎝L⎠
⌡
0
L
d=
⌠
⎮ x w sin ⎛ π x ⎞ dx
⎜ ⎟
0
⎮
⎝L⎠
⌡
0
FR
=
L
2
Problem 4-158
Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and
specify where it acts, measured from point A.
Given:
a = 3 ft
k = 86
lb
ft
3
2
w ( x) = k x
Solution:
a
⌠
F R = ⎮ w ( x) dx
⌡0
x =
⌠a
⎮ x w ( x) dx
⌡0
FR
F R = 774 lb
x = 2.25 ft
Problem 4-159
Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To
329
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Engineering Mechanics - Statics
Chapter 4
y g y
p
j
y
alleviate this problem, an automobile seat restraint has been developed that provides additional
pressure contact with the cranium. During dynamic tests the distribution of load on the cranium
has been plotted and shown to be parabolic. Determine the equivalent resultant force and its
location, measured from point A.
Given:
a = 0.5 ft
w0 = 12
lb
ft
lb
k = 24
ft
3
2
w ( x) = w0 + kx
Solution:
⌠a
F R = ⎮ w ( x) dx
⌡0
F R = 7 lb
a
x =
⌠
⎮ x w ( x) dx
⌡0
FR
x = 0.268 ft
Problem 4-160
Determine the equivalent resultant force of the distributed loading and its location, measured from
point A. Evaluate the integrals using Simpson's rule.
Units Used:
3
kN = 10 N
Given:
c1 = 5
c2 = 16
a = 3
330
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Engineering Mechanics - Statics
Chapter 4
b = 1
Solution:
a+ b
⌠
FR = ⎮
⌡0
2
c1 x +
c2 + x dx
x c1 x +
c2 + x dx
F R = 14.9
a+ b
d =
⌠
⎮
⌡0
2
d = 2.27
FR
Problem 4-161
Determine the coordinate direction angles of F, which is applied to the end A of the pipe
assembly, so that the moment of F about O is zero.
Given:
F = 20 lb
a = 8 in
b = 6 in
c = 6 in
d = 10 in
Solution:
Require Mo = 0. This happens when force F is directed either towards or away from point O.
⎛ c ⎞
r = ⎜a + b⎟
⎜
⎟
⎝ d ⎠
u =
r
r
⎛ 0.329 ⎞
u = ⎜ 0.768 ⎟
⎜
⎟
⎝ 0.549 ⎠
If the force points away from O, then
⎛⎜ α ⎞⎟
⎜ β ⎟ = acos ( u)
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 70.774 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 39.794 ⎟ deg
⎜ γ ⎟ ⎝ 56.714 ⎠
⎝ ⎠
If the force points towards O, then
331
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Engineering Mechanics - Statics
⎛⎜ α ⎞⎟
⎜ β ⎟ = acos ( −u)
⎜γ ⎟
⎝ ⎠
Chapter 4
⎛⎜ α ⎞⎟ ⎛ 109.226 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 140.206 ⎟ deg
⎜ γ ⎟ ⎝ 123.286 ⎠
⎝ ⎠
Problem 4-162
Determine the moment of the force F about
point O. The force has coordinate direction
angles α, β, γ. Express the result as a Cartesian
vector.
Given:
F = 20 lb
a = 8 in
α = 60 deg
b = 6 in
β = 120 deg
c = 6 in
γ = 45 deg
d = 10 in
Solution:
⎛ c ⎞
r = ⎜a + b⎟
⎜
⎟
⎝ d ⎠
⎛⎜ cos ( α ) ⎟⎞
F v = F⎜ cos ( β ) ⎟
⎜ cos ( γ ) ⎟
⎝
⎠
M = r × Fv
⎛ 297.99 ⎞
M = ⎜ 15.147 ⎟ lb⋅ in
⎜
⎟
⎝ −200 ⎠
Problem 4-163
Replace the force at A by an equivalent resultant force and couple moment at point P. Express
the results in Cartesian vector form.
Units Used :
332
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Engineering Mechanics - Statics
Chapter 4
3
kN = 10 N
Given:
a = 4m
b = 6m
c = 8m
d = 4m
⎛ −300 ⎞
⎜
⎟
F = 200 N
⎜
⎟
⎝ −500 ⎠
Solution :
FR = F
⎛ −300 ⎞
⎜
⎟
F R = 200 N
⎜
⎟
⎝ −500 ⎠
⎛ −a − c ⎞
⎜ b ⎟×F
MP =
⎜
⎟
⎝ d ⎠
⎛ −3.8 ⎞
⎜
⎟
MP = −7.2 kN⋅ m
⎜
⎟
⎝ −0.6 ⎠
Problem 4-164
Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian
vector.
333
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 250 N
b = 1m
c = 2.5 m
d = 1.5 m
e = 0.5 m
θ = 30 deg
Solution:
⎛ c−e ⎞
⎜
⎟
rCB = b + d cos ( θ )
⎜
⎟
⎝ −d sin ( θ ) ⎠
MA = rAB × F v
⎛0⎞
⎜ ⎟
rAB = b
⎜ ⎟
⎝0⎠
Fv = F
rCB
rCB
⎛ −59.7 ⎞
⎜ 0.0 ⎟ N⋅ m
MA =
⎜
⎟
⎝ −159.3 ⎠
Problem 4-165
Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.
334
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Engineering Mechanics - Statics
Chapter 4
Given:
F = 250 N
b = 1m
c = 2.5 m
d = 1.5 m
e = 0.5 m
θ = 30 deg
Solution:
⎛ c−e ⎞
⎜
⎟
rCB = b + d cos ( θ )
⎜
⎟
⎝ −d sin ( θ ) ⎠
Maa = ( rAB × Fv) ⋅ ua
⎛0⎞
⎜ ⎟
rAB = b
⎜ ⎟
⎝0⎠
Fv = F
rCB
rCB
⎛1⎞
⎜ ⎟
ua = 0
⎜ ⎟
⎝0⎠
Maa = −59.7 N⋅ m
Problem 4-166
A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force
about the bolt axis (z axis), which is directed at angle θ from the vertical.
335
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 80 N
a = 100 mm
b = 300 mm
c = 200 mm
θ = 15 deg
Solution:
⎛ −b ⎞
⎜
⎟
r = a+c
⎜
⎟
⎝ 0 ⎠
⎛ sin ( θ ) ⎞
⎜ 0 ⎟
F = F1
⎜
⎟
⎝ −cos ( θ ) ⎠
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
Mz = ( r × F ) k
Mz = −6.212 N⋅ m
Problem 4-167
Replace the force F having acting at point A by an equivalent force and couple moment at
point C.
Units Used:
3
kip = 10 lb
Given:
F = 50 lb
a = 10 ft
b = 20 ft
c = 15 ft
336
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Engineering Mechanics - Statics
Chapter 4
d = 10 ft
e = 30 ft
Solution :
⎛d⎞
rAB = ⎜ c ⎟
⎜ ⎟
⎝ −e ⎠
Fv = F
rAB
rAB
⎛ 0 ⎞
rCA = ⎜ a + b ⎟
⎜
⎟
⎝ e ⎠
FR = Fv
⎛ 14.286 ⎞
F R = ⎜ 21.429 ⎟ lb
⎜
⎟
⎝ −42.857 ⎠
MR = rCA × Fv
⎛ −1.929 ⎞
MR = ⎜ 0.429 ⎟ kip⋅ ft
⎜
⎟
⎝ −0.429 ⎠
Problem 4-168
The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment
of this force about the z axis?
Given:
F = 30 N
a = 50 mm
b = 200 mm
c = 10 mm
θ = 45 deg
Solution:
⎛ sin ( θ ) ⎞
F v = F⎜ −cos ( θ ) ⎟
⎜
⎟
⎝ 0 ⎠
⎛ −c ⎞
rOA = ⎜ b ⎟
⎜ ⎟
⎝a⎠
⎛0⎞
k = ⎜0⎟
⎜ ⎟
⎝1⎠
337
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Mz = ( rOA × F v ) k
Chapter 4
Mz = −4.03 N⋅ m
Problem 4-169
The horizontal force F acts on the
handle of the wrench. Determine the
moment of this force about point O.
Specify the coordinate direction
angles α, β, γ of the moment axis.
Given:
F = 30 N
c = 10 mm
a = 50 mm θ = 45 deg
b = 200 mm
Solution:
⎛ sin ( θ ) ⎞
F v = F⎜ −cos ( θ ) ⎟
⎜
⎟
⎝ 0 ⎠
⎛ −c ⎞
rOA = ⎜ b ⎟
⎜ ⎟
⎝a⎠
MO = rOA × Fv
⎛ 1.06 ⎞
MO = ⎜ 1.06 ⎟ N⋅ m
⎜
⎟
⎝ −4.03 ⎠
⎛⎜ α ⎞⎟
⎛ MO ⎞
⎜ β ⎟ = acos ⎜
⎟
MO ⎠
⎝
⎜γ ⎟
⎝ ⎠
⎛⎜ α ⎞⎟ ⎛ 75.7 ⎞
⎜
⎟
⎜ β ⎟ = ⎜ 75.7 ⎟ deg
⎜ γ ⎟ ⎝ 159.6 ⎠
⎝ ⎠
Problem 4-170
If the resultant couple moment of the three couples acting on the triangular block is to be zero,
determine the magnitudes of forces F and P.
338
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Engineering Mechanics - Statics
Chapter 4
Given:
F 1 = 10 lb
a = 3 in
b = 4 in
c = 6 in
d = 3 in
θ = 30 deg
Solution:
Initial Guesses:
F = 1 lb
P = 1 lb
Given
⎛ 0 ⎞ ⎛ 0 ⎞
⎜ −F c ⎟ + ⎜ 0 ⎟ +
⎜
⎟ ⎜
⎟
⎝ 0 ⎠ ⎝ −P c ⎠
⎛0⎞
⎜a⎟ = 0
2
2⎜ ⎟
a + b ⎝b⎠
⎛F⎞
⎜ ⎟ = Find ( F , P)
⎝P⎠
F1 d
⎛F⎞ ⎛3⎞
⎜ ⎟ = ⎜ ⎟ lb
⎝P⎠ ⎝4⎠
339
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Engineering Mechanics - Statics
Chapter 5
Problem 5-1
Draw the free-body diagram of the sphere of weight W resting between the smooth inclined
planes. Explain the significance of each force on the diagram.
Given:
W = 10 lb
θ 1 = 105 deg
θ 2 = 45 deg
Solution:
NA, NB force of plane on sphere.
W force of gravity on sphere.
Problem 5-2
Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the
smooth surface at B.
Given:
F = 8 lb
a = 1.5 ft
b = 0.2 ft
c = 2 ft
340
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Engineering Mechanics - Statics
Chapter 5
Solution:
Problem 5-3
Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller.
Explain the significance of each force on the diagram.
Given:
w = 40
lb
ft
a = 3 ft
b = 4 ft
θ = 30 deg
Solution:
A x, A y, MA effect of wall on beam.
NB force of roller on beam.
wa
2
resultant force of distributed load on beam.
Problem 5-4
Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by
member (link) BC.
341
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kN = 10 N
Given:
F = 8 kN
a = 3m
b = 4m
c = 0.4 m
d = 3
e = 4
Solution:
Problem 5-5
Draw the free-body diagram of the
C-bracket supported at A, B, and C by
rollers. Explain the significance of each
forcce on the diagram.
Given:
a = 3 ft
b = 4 ft
θ 1 = 30 deg
θ 2 = 20 deg
F = 200 lb
342
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Engineering Mechanics - Statics
Chapter 5
Solution:
NA , NB , NC force of rollers on beam.
Problem 5-6
Draw the free-body diagram of the smooth
rod of mass M which rests inside the
glass. Explain the significance of each
force on the diagram.
Given:
M = 20 gm
a = 75 mm
b = 200 mm
θ = 40 deg
Solution:
A x , A y , NB force of glass on rod.
M(g) N force of gravity on rod.
Problem 5-7
Draw the free-body diagram of the “spanner wrench” subjected to the force F. The support at
A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of
343
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Engineering Mechanics - Statics
Chapter 5
p
each force on the diagram.
p
g
Given:
F = 20 lb
a = 1 in
b = 6 in
Solution:
A x, A y, NB force of cylinder on wrench.
Problem 5-8
Draw the free-body diagram of the automobile, which is being towed at constant velocity up the
incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are
free to roll. Explain the significance of each force on the diagram.
Units Used:
3
Mg = 10 kg
Given:
M = 5 Mg
d = 1.50 m
a = 0.3 m
e = 0.6 m
b = 0.75 m
θ 1 = 20 deg
c = 1m
θ 2 = 30 deg
g = 9.81
m
2
s
344
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Solution:
NA, NB force of road on car.
F force of cable on car.
Mg force of gravity on car.
Problem 5-9
Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The
supports A, B, and C are smooth.
Given:
M = 100 kg
a = 1.75 m
b = 1.25 m
c = 0.5 m
d = 0.2 m
g = 9.81
m
2
s
Solution:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 5
Problem 5-10
Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B.
Given:
F = 500 N
M = 800 N⋅ m
a = 8m
b = 4m
c = 5m
Solution:
Problem 5-11
The sphere of weight W rests between the smooth
inclined planes. Determine the reaactions at the
supports.
Given:
W = 10 lb
θ 1 = 105 deg
θ 2 = 45 deg
Solution:
Initial guesses
346
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Engineering Mechanics - Statics
NA = 1 lb
Chapter 5
NB = 1 lb
Given
NB cos ( θ 1 − 90 deg) − NA cos ( θ 2 ) = 0
NA sin ( θ 2 ) − NB sin ( θ 1 − 90 deg) − W = 0
⎛ NA ⎞
⎜ ⎟ = Find ( NA , NB)
⎝ NB ⎠
⎛ NA ⎞ ⎛ 19.3 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ NB ⎠ ⎝ 14.1 ⎠
Problem 5-12
Determine the magnitude of the resultant force acting at pin A of the handpunch.
Given:
F = 8 lb
a = 1.5 ft
b = 0.2 ft
c = 2 ft
Solution:
Σ F x = 0;
Ax − F = 0
Ax = F
Σ M = 0;
F a − Ay c = 0
Ay = F
Ax = 8 lb
a
c
Ay = 6 lb
347
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Engineering Mechanics - Statics
FA =
Chapter 5
2
Ax + Ay
2
F A = 10 lb
Problem 5-13
The C-bracket is supported at A, B, and C by rollers. Determine the reactions at the supports.
Given:
a = 3 ft
b = 4 ft
θ 1 = 30 deg
θ 2 = 20 deg
F = 200 lb
Solution:
Initial Guesses:
NA = 1 lb
NB = 1 lb
NC = 1 lb
Given
NA a − F b = 0
NB sin ( θ 1 ) − NC sin ( θ 2 ) = 0
NB cos ( θ 1 ) + NC cos ( θ 2 ) − NA − F = 0
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( NA , NB , NC)
⎜N ⎟
⎝ C⎠
⎛ NA ⎞ ⎛ 266.7 ⎞
⎜ ⎟ ⎜
⎟
⎜ NB ⎟ = ⎜ 208.4 ⎟ lb
⎜ N ⎟ ⎝ 304.6 ⎠
⎝ C⎠
348
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Engineering Mechanics - Statics
Chapter 5
Problem 5-14
The smooth rod of mass M rests inside the
glass. Determine the reactions on the rod.
Given:
M = 20 gm
a = 75 mm
b = 200 mm
θ = 40 deg
g = 9.81
m
2
s
Solution:
Initial Guesses:
Ax = 1 N
Ay = 1 N
NB = 1 N
Given
Ax − NB sin ( θ ) = 0
Ay − M g + NB cos ( θ ) = 0
−M g
a+b
cos ( θ ) + NB b = 0
2
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( Ax , Ay , NB)
⎜N ⎟
⎝ B⎠
⎛ Ax ⎞ ⎛ 0.066 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0.117 ⎟ N
⎜ N ⎟ ⎝ 0.103 ⎠
⎝ B⎠
Problem 5-15
The “spanner wrench” is subjected to the force F. The support at A can be considered a pin,
and the surface of contact at B is smooth. Determine the reactions on the spanner wrench.
349
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Engineering Mechanics - Statics
Chapter 5
Given:
F = 20 lb
a = 1 in
b = 6 in
Solution:
Initial Guesses:
Ax = 1 lb
Ay = 1 lb
NB = 1 lb
Given
− Ax + NB = 0
Ay − F = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( Ax , Ay , NB)
⎜N ⎟
⎝ B⎠
−F ( a + b) + Ax a = 0
⎛ Ax ⎞ ⎛ 140 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 20 ⎟ lb
⎜ N ⎟ ⎝ 140 ⎠
⎝ B⎠
Problem 5-16
The automobile is being towed at constant velocity up the incline using the cable at C. The
automobile has a mass M and center of mass at G. The tires are free to roll. Determine the
reactions on both wheels at A and B and the tension in the cable at C.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
350
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Engineering Mechanics - Statics
Chapter 5
Given:
M = 5 Mg
d = 1.50 m
a = 0.3 m
e = 0.6 m
b = 0.75 m
θ 1 = 20 deg
c = 1m
θ 2 = 30 deg
g = 9.81
m
2
s
Solution:
Guesses
F = 1 kN
NA = 1 kN
NB = 1 kN
Given
NA + NB + F sin ( θ 2 ) − M g cos ( θ 1 ) = 0
−F cos ( θ 2 ) + M g sin ( θ 1 ) = 0
F cos ( θ 2 ) a − F sin ( θ 2 ) b − M g cos ( θ 1 ) c − M g sin ( θ 1 ) e + NB( c + d) = 0
⎛F ⎞
⎜ ⎟
⎜ NA ⎟ = Find ( F , NA , NB)
⎜ NB ⎟
⎝ ⎠
⎛ F ⎞ ⎛ 19.37 ⎞
⎜ ⎟ ⎜
⎟
⎜ NA ⎟ = ⎜ 13.05 ⎟ kN
⎜ NB ⎟ ⎝ 23.36 ⎠
⎝ ⎠
Problem 5-17
The uniform bar has mass M and
center of mass at G. The supports
A, B, and C are smooth. Determine
the reactions at the points of contact
at A, B, and C.
Given:
M = 100 kg
a = 1.75 m
b = 1.25 m
351
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Engineering Mechanics - Statics
Chapter 5
c = 0.5 m
d = 0.2 m
θ = 30 deg
g = 9.81
m
2
s
Solution:
The initial guesses:
NA = 20 N
NB = 30 N
NC = 40 N
Given
ΣMA = 0;
+
↑Σ Fy = 0;
+
↑Σ Fy = 0;
−M g cos ( θ ) a − M g sin ( θ )
d
+ NB sin ( θ ) d + NC( a + b) = 0
2
NB − M g + NC cos ( θ ) = 0
NA − NC sin ( θ ) = 0
⎛ NC ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( NC , NB , NA)
⎜N ⎟
⎝ A⎠
⎛ NC ⎞ ⎛ 493 ⎞
⎜ ⎟ ⎜
⎟
⎜ NB ⎟ = ⎜ 554 ⎟ N
⎜ N ⎟ ⎝ 247 ⎠
⎝ A⎠
Problem 5-18
The beam is pin-connected at A
and rocker-supported at B.
Determine the reactions at the pin A
and at the roller at B.
Given:
F = 500 N
M = 800 N⋅ m
a = 8m
352
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Engineering Mechanics - Statics
Chapter 5
b = 4m
c = 5m
Solution:
⎞
⎟
⎝ a + b⎠
α = atan ⎛⎜
c
ΣMA = 0;
−F
a
cos ( α )
By =
− M + By a = 0
F a + M cos ( α )
cos ( α ) a
B y = 642 N
+
Σ F x = 0;
→
− Ax + F sin ( α ) = 0
Ax = F sin ( α )
Ax = 192 N
+
− Ay − F cos ( α ) + B y = 0
Ay = −F cos ( α ) + By
Ay = 180 N
↑Σ Fy = 0;
Problem 5-19
Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam.
Given:
F 1 = 600 N
F 2 = 400 N
θ = 15 deg
a = 4m
b = 8m
c = 3
d = 4
Solution:
ΣMA = 0;
B y( a + b) − F2 cos ( θ ) ( a + b) − F1 a = 0
353
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Engineering Mechanics - Statics
By =
+
Σ F x = 0;
→
Chapter 5
F 2 cos ( θ ) ( a + b) + F 1 a
a+b
Ax − F 2 sin ( θ ) = 0
Ax = F2 sin ( θ )
+
↑Σ Fy = 0;
B y = 586 N
Ax = 104 N
Ay − F 2 cos ( θ ) + B y − F1 = 0
Ay = F2 cos ( θ ) − B y + F1
FA =
2
Ax + Ay
2
Ay = 400 N
F A = 413 N
Problem 5-20
Determine the reactions at the supports.
Given:
w = 250
lb
ft
a = 6 ft
b = 6 ft
c = 6 ft
Solution:
Guesses
Ax = 1 lb
Ay = 1 lb
B y = 1 lb
Given
Ax = 0
Ay + B y − w( a + b) −
1
2
wc = 0
⎛ b⎞ 1 ⎛ c ⎞
⎟ − w b⎜ ⎟ − w c⎜b + ⎟ + By b = 0
⎝ 2⎠
⎝ 2⎠ 2 ⎝ 3⎠
w a ⎛⎜
a⎞
354
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Engineering Mechanics - Statics
Chapter 5
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( Ax , Ay , By)
⎜B ⎟
⎝ y⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 2750 ⎟ lb
⎜ B ⎟ ⎝ 1000 ⎠
⎝ y⎠
Problem 5-21
When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts
normal forces FC and F A on the radius C and ulna A as shown. Determine these forces and the
force F B that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at
G. Neglect the weight of the arm.
Given:
W = 5 lb
θ = 75 deg
a = 2 in
b = 0.8 in
c = 14 in
Solution:
ΣMB = 0;
− W ( c − a) + F A a = 0
F A = W ⎛⎜
c − a⎞
⎟
⎝ a ⎠
F A = 30 lb
+
↑Σ Fy = 0;
F B sin ( θ ) − W − FA = 0
FB =
W + FA
sin ( θ )
F B = 36.2 lb
+
Σ F x = 0;
→
F C − F B cos ( θ ) = 0
355
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Engineering Mechanics - Statics
Chapter 5
F C = FB cos ( θ )
F C = 9.378 lb
Problem 5-22
The uniform door has a weight W and a center of gravity at G. Determine the reactions at the
hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B
exerts both horizontal and vertical reactions.
Given:
W = 100 lb
a = 3 ft
b = 3 ft
c = 0.5 ft
d = 2 ft
Solution:
ΣMB = 0;
W d − A x ( a + b) = 0
⎞
⎟
⎝ a + b⎠
Ax = W ⎛⎜
d
Ax = 33.3 lb
ΣF x = 0;
Bx = Ax
B x = 33.3 lb
ΣF y = 0;
By = W
B y = 100 lb
Problem 5-23
The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD
needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the
horizontal and vertical components of force at the hinge (pin) at A.
356
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Engineering Mechanics - Statics
Chapter 5
Given:
W = 200 lb
a = 4 ft
θ = 30 deg
b = 3 ft
φ = 20 deg
c = 6 ft
Solution:
ΣMA = 0;
−F CD cos ( θ ) ( b + c) cos ( φ ) + FCD sin ( θ ) ( b + c) sin ( φ ) + W c cos ( φ ) = 0
F CD =
W c cos ( φ )
( b + c) ( cos ( θ ) cos ( φ ) − sin ( θ ) sin ( φ ) )
+
Σ F x = 0;
→
F CD sin ( θ ) − A x = 0
Ax = FCD sin ( θ )
+
↑Σ Fy = 0;
F CD = 195 lb
Ax = 97.5 lb
Ay − W + F CD cos ( θ ) = 0
Ay = W − FCD cos ( θ )
Ay = 31.2 lb
Problem 5-24
The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A
and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa)
for 0 ≤ θ ≤ 90 deg.
357
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Engineering Mechanics - Statics
Chapter 5
Units used:
3
Mg = 10 kg
Given:
M = 1.4 Mg
a = 0.4 m
g = 9.81
m
2
s
Solution:
θ = 0 .. 90
NA ( θ ) =
M g sin ( θ deg)
3
10
NB ( θ ) =
M g cos ( θ deg)
3
10
Force in kN
15
NA( θ )10
NB( θ )
5
0
0
20
40
60
80
100
θ
Angle in Degrees
Problem 5-25
While slowly walking, a man having a total mass M places all his weight on one foot. Assuming
that the normal force NC of the ground acts on his foot at C, determine the resultant vertical
compressive force F B which the tibia T exerts on the astragalus B, and the vertical tension FA in
the achilles tendon A at the instant shown.
358
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kN = 10 N
Given:
M = 80 kg
a = 15 mm
b = 5 mm
c = 20 mm
d = 100 mm
Solution:
NC = M g
NC = 785 N
ΣMA = 0;
−F B c + NC ( c + d) = 0
F B = NC ⎛⎜
c + d⎞
⎟
⎝ c ⎠
F B = 4.71 kN
ΣF y = 0;
F A − FB + NC = 0
F A = F B − NC
F A = 3.92 kN
Problem 5-26
Determine the reactions at the roller A and pin B.
359
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Engineering Mechanics - Statics
Chapter 5
Given:
M = 800 lb ft
c = 3 ft
F = 390 lb
d = 5
a = 8 ft
e = 12
b = 4 ft
θ = 30 deg
Solution:
Guesses
R A = 1 lb B x = 1 lb
B y = 1 lb
Given
⎛
R A sin ( θ ) + B x − ⎜
d
2
⎞F = 0
2⎟
⎝ e +d ⎠
⎛
R A cos ( θ ) + By − ⎜
e
2
⎞F = 0
2⎟
⎝ e +d ⎠
M − R A cos ( θ ) ( a + b) + Bx c = 0
⎛ RA ⎞
⎜ ⎟
⎜ Bx ⎟ = Find ( RA , Bx , By)
⎜B ⎟
⎝ y⎠
R A = 105.1 lb
⎛ Bx ⎞ ⎛ 97.4 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ By ⎠ ⎝ 269 ⎠
Problem 5-27
The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a
maximum load W2 placed at point G2,,determine the smallest counterweight W that should be
placed at B in order to prevent the platform from tipping over.
Given:
W1 = 250 lb
a = 1 ft
c = 1 ft
e = 6 ft
W2 = 400 lb
b = 6 ft
d = 8 ft
f = 2 ft
360
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Engineering Mechanics - Statics
Chapter 5
Solution:
When tipping occurs, R c = 0
ΣMD = 0;
−W2 f + W1 c + WB ( b + c) = 0
WB =
W2 f − W1 c
b+c
WB = 78.6 lb
Problem 5-28
The articulated crane boom has a weight W and mass center at G. If it supports a load L,
determine the force acting at the pin A and the compression in the hydraulic cylinder BC when
the boom is in the position shown.
Units Used:
3
kip = 10 lb
Given:
W = 125 lb
361
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Engineering Mechanics - Statics
Chapter 5
L = 600 lb
a = 4 ft
b = 1 ft
c = 1 ft
d = 8 ft
θ = 40 deg
Solution:
Guesses
Ax = 1 lb
Ay = 1 lb
Given
− Ax + F B cos ( θ ) = 0
F B = 1 lb
− Ay + F B sin ( θ ) − W − L = 0
F B cos ( θ ) b + F B sin ( θ ) c − W a − L( d + c) = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( Ax , Ay , FB)
⎜F ⎟
⎝ B⎠
F B = 4.19 kip
⎛ Ax ⎞ ⎛ 3.208 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ Ay ⎠ ⎝ 1.967 ⎠
Problem 5-29
The device is used to hold an elevator
door open. If the spring has stiffness k
and it is compressed a distnace δ,
determine the horizontal and vertical
components of reaction at the pin A and
the resultant force at the wheel bearing B.
Given:
N
m
b = 125 mm
δ = 0.2 m
c = 100 mm
a = 150 mm
θ = 30 deg
k = 40
362
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Engineering Mechanics - Statics
Solution:
ΣMA = 0;
Chapter 5
F s = kδ
−F s a + F B cos ( θ ) ( a + b) − F B sin ( θ ) c = 0
FB = Fs
a
cos ( θ ) ( a + b) − sin ( θ ) c
F B = 6.378 N
+
Σ F x = 0;
→
Ax − F B sin ( θ ) = 0
Ax = FB sin ( θ )
Ax = 3.189 N
+
↑Σ Fy = 0;
Ay − F s + F B cos ( θ ) = 0
Ay = Fs − FB cos ( θ )
Ay = 2.477 N
Problem 5-30
Determine the reactions on the bent rod which is supported by a smooth surface at B and by a
collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.
Given:
F = 100 lb
M = 200 lb ft
a = 3 ft
b = 3 ft
c = 2ft
363
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Engineering Mechanics - Statics
Chapter 5
d = 3
e = 4
f = 12
g = 5
Solution:
Initial Guesses:
NA = 20 lb
NB = 10 lb
MA = 30 lb ft
Given
f
g
⎞
⎞
⎛
⎜ 2 2 ⎟ ( a + b) − NB ⎜ 2 2 ⎟ c = 0
⎝ f +g ⎠
⎝ f +g ⎠
ΣMA = 0;
MA − F a − M + NB ⎛
ΣF x = 0;
NA ⎛
ΣF y = 0;
NA ⎛
e
g
⎞
⎞
⎛
⎜ 2 2 ⎟ − NB ⎜ 2 2 ⎟ = 0
⎝ e +d ⎠
⎝ f +g ⎠
d
f
⎞
⎞
⎛
⎜ 2 2 ⎟ + NB ⎜ 2 2 ⎟ − F = 0
⎝ e +d ⎠
⎝ f +g ⎠
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( NA , NB , MA)
⎜M ⎟
⎝ A⎠
⎛ NA ⎞ ⎛ 39.7 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ NB ⎠ ⎝ 82.5 ⎠
MA = 106 lb⋅ ft
Problem 5-31
The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in
the range x1 ≤ x ≤ x2, determine the maximum magnitude of reaction at the supports A and B.
Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The
collar at B supports a force in the vertical direction, whereas the one at A does not.
Units Used:
kip = 1000 lb
Given:
F = 780 lb
364
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Engineering Mechanics - Statics
Chapter 5
a = 4 ft
b = 8 ft
x1 = 1.5 ft
x2 = 7.5 ft
Solution:
The maximum occurs when x = x2
ΣMA = 0;
−F x2 + B x a = 0
Bx = F
x2
a
3
B x = 1.462 × 10 lb
+
Σ F x = 0;
→
Ax − Bx = 0
Ax = Bx
Ax = 1.462 × 10 lb
+
By − F = 0
By = F
B y = 780 lb
↑Σ Fy = 0;
FB =
2
Bx + By
2
3
F B = 1.657 kip
Problem 5-32
The uniform rod AB has weight W. Determine the force in the cable when the rod is in the
position shown.
Given:
W = 15 lb
L = 5 ft
365
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Engineering Mechanics - Statics
Chapter 5
θ 1 = 30 deg
θ 2 = 10 deg
Solution:
ΣMA = 0;
⎛ L ⎞ cos ( θ + θ ) = 0
⎟
1
2
⎝2⎠
NB L sin ( θ 1 + θ 2 ) − W ⎜
NB =
W cos ( θ 1 + θ 2 )
2 sin ( θ 1 + θ 2 )
NB = 8.938 lb
ΣF x = 0;
T cos ( θ 2 ) − NB
T =
NB
cos ( θ 2 )
T = 9.08 lb
Problem 5-33
The power pole supports the three lines, each line exerting a vertical force on the pole due to its
weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice
to snap the lines, determine which line(s) when removed create(s) a condition for the greatest
moment reaction at D.
Units Used:
3
kip = 10 lb
366
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Engineering Mechanics - Statics
Chapter 5
Given:
W1 = 800 lb
W2 = 450 lb
W3 = 400 lb
a = 2ft
b = 4 ft
c = 3 ft
Solution:
+
Σ F x = 0;
→
Dx = 0
+
Dy − ( W1 + W2 + W3 ) = 0
↑Σ Fy = 0;
Dy = W1 + W2 + W3
ΣMD = 0;
Dy = 1.65 kip
−W2 b − W3 c + W1 a + MD = 0
MD = W2 b + W3 c − W1 a
MD = 1.4 kip⋅ ft
Examine all cases. For these numbers we require line 1 to snap.
MDmax = W2 b + W3 c
MDmax = 3 kip⋅ ft
Problem 5-34
The picnic table has a weight WT and a center of gravity at GT . If a man weighing WM has a
center of gravity at GM and sits down in the centered position shown, determine the vertical
reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude
from the results?
Given:
WT = 50 lb
367
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Engineering Mechanics - Statics
Chapter 5
WM = 225 lb
a = 6 in
b = 20 in
c = 20 in
Solution:
ΣMA = 0;
2 NB( b + c) + WM a − WT b = 0
NB =
WT b − WM a
2 ( b + c)
NB = −4.37 lb
Since NB has a negative sign, the table will tip over.
Problem 5-35
If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the
magnitude of the resultant force which the man must exert on each of the two handles in order to
hold the wheelbarrow in equilibrium.
Given:
M = 60 kg
a = 0.6 m
b = 0.5 m
c = 0.9 m
d = 0.5 m
g = 9.81
m
2
s
368
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Engineering Mechanics - Statics
Chapter 5
Solution:
ΣMB = 0;
− Ay ( b + c) + M g c = 0
Ay =
Mgc
b+c
Ay = 378.386 N
+
Σ F x = 0;
→
B x = 0N
+
Ay − M g + 2 By = 0
↑Σ Fy = 0;
By =
Bx = 0
M g − Ay
B y = 105.107 N
2
Problem 5-36
The man has weight W and stands at the center of the plank. If the planes at A and B are
smooth, determine the tension in the cord in terms of W and θ.
Solution:
ΣMB = 0;
W
L
cos ( φ ) − NA L cos ( φ ) = 0
2
NA =
ΣF x = 0;
W
2
T cos ( θ ) − NB sin ( θ ) = 0
(1)
369
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Engineering Mechanics - Statics
Chapter 5
T sin ( θ ) + NB cos ( θ ) + NA − W = 0
ΣF y = 0;
(2)
Solving Eqs. (1) and (2) yields:
T=
W
sin ( θ )
2
NB =
W
cos ( θ )
2
Problem 5-37
When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB
keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is
F , determine the unstretched length of the spring if the stiffness of the spring is k.
Given:
F = 3N
k = 80
N
m
a = 100 mm
b = 50 mm
c = 40 mm
d = 10 mm
θ = 30 deg
Solution:
ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0
Fs = F
cos ( θ ) c + sin ( θ ) d
b
Fs = k x
x =
L0 = a − x
Fs
k
F s = 2.378 N
x = 29.73 mm
L0 = 70.3 mm
370
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Engineering Mechanics - Statics
Chapter 5
Problem 5-38
The telephone pole of negligible thickness is subjected to the force F directed as shown. It is
supported by the cable BCD and can be assumed pinned at its base A. In order to provide
clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as
shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice
the tension in BCD, determine the height h for placement of the strut CE.
Given:
F = 80 lb
θ = 30 deg
a = 30 ft
b = 10 ft
Solution:
+ ΣMA = 0;
⎛
−F cos ( θ ) a + ⎜
⎞T
b
2⎟
2
⎝ a +b ⎠
TBCD = F cos ( θ )
2
BCD a = 0
2
a +b
b
TBCD = 219.089 lb
Require TCD' = 2 TBCD
+ ΣMA = 0;
TCD' = 438.178 lb
TCD' d − F cos ( θ ) a = 0
⎛ cos ( θ ) ⎞
⎟
⎝ TCD' ⎠
d = Fa⎜
d = 4.7434 ft
Geometry:
a−h
a
=
d
b
⎛ d⎞
⎟
⎝ b⎠
h = a−a⎜
h = 15.8 ft
371
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Engineering Mechanics - Statics
Chapter 5
Problem 5-39
The worker uses the hand truck to move material down the ramp. If the truck and its contents
are held in the position shown and have weight W with center of gravity at G, determine the
resultant normal force of both wheels on the ground A and the magnitude of the force required
at the grip B.
Given:
W = 100 lb
e = 1.5 ft
a = 1 ft
f = 0.5 ft
b = 1.5 ft
θ = 60 deg
c = 2 ft
φ = 30 deg
d = 1.75 ft
Solution:
ΣMB = 0;
NA cos ( θ − φ ) ( b + c + d) + NA sin ( θ − φ ) ( a − f) − W cos ( θ ) ( b + c) − W sin ( θ ) ( e + a) = 0
NA =
W cos ( θ ) ( b + c) + W sin ( θ ) ( e + a)
cos ( θ − φ ) ( b + c + d) + sin ( θ − φ ) ( a − f)
NA = 81.621 lb
ΣF x = 0;
−B x + NA sin ( φ ) = 0
B x = NA sin ( φ )
B x = 40.811 lb
ΣF y = 0;
B y + NA ( cos ( φ ) − W = 0)
B y = W − NA cos ( φ )
B y = 29.314 lb
FB =
2
Bx + By
2
F B = 50.2 lb
372
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Engineering Mechanics - Statics
Chapter 5
T sin ( θ ) + NB cos ( θ ) + NA − W = 0
ΣF y = 0;
(2)
Solving Eqs. (1) and (2) yields:
T=
W
sin ( θ )
2
NB =
W
cos ( θ )
2
Problem 5-37
When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB
keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is
F , determine the unstretched length of the spring if the stiffness of the spring is k.
Given:
F = 3N
k = 80
N
m
a = 100 mm
b = 50 mm
c = 40 mm
d = 10 mm
θ = 30 deg
Solution:
ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0
Fs = F
cos ( θ ) c + sin ( θ ) d
b
Fs = k x
x =
L0 = a − x
Fs
k
F s = 2.378 N
x = 29.73 mm
L0 = 70.3 mm
373
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Engineering Mechanics - Statics
Chapter 5
The shelf supports the electric motor which has mass m1 and mass center at Gm. The platform
upon which it rests has mass m2 and mass center at Gp. Assuming that a single bolt B holds the
shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and
the horizontal and vertical components of reaction of the bolt B on the bracket.
Given:
m1 = 15 kg
c = 50 mm
m2 = 4 kg
d = 200 mm
a = 60 mm
e = 150 mm
b = 40 mm
g = 9.81
m
2
s
Solution:
ΣMA = 0;
B x a − m2 g d − m1 g( d + e) = 0
Bx = g
+
Σ F x = 0;
→
m2 d + m1 ( d + e)
B x = 989 N
a
Ax − Bx = 0
Ax = Bx
+
↑Σ Fy = 0;
Ax = 989 N
B y − m2 g − m1 g = 0
B y = m2 g + m1 g
B y = 186 N
Problem 5-42
A cantilever beam, having an extended length L, is subjected to a vertical force F. Assuming that
the wall resists this load with linearly varying distributed loads over the length a of the beam
portion inside the wall, determine the intensities w1 and w2 for equilibrium.
374
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kN = 10 N
Given:
F = 500 N
a = 0.15 m
L = 3m
Solution:
The initial guesses
w1 = 1
kN
m
w2 = 1
kN
m
Given
+
↑Σ Fy = 0;
ΣMA = 0;
⎛ w1 ⎞
⎜ ⎟ = Find ( w1 , w2)
⎝ w2 ⎠
1
1
w1 a − w2 a − F = 0
2
2
−F L −
1
2
1
⎛ 2 a⎞
⎟ + w2 a⎜ ⎟ = 0
⎝ 3⎠ 2
⎝ 3 ⎠
w1 a⎛⎜
a⎞
⎛ w1 ⎞ ⎛ 413 ⎞ kN
⎜ ⎟=⎜
⎟
⎝ w2 ⎠ ⎝ 407 ⎠ m
Problem 5-43
The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A,
the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If
the load F is supported by the hoist line, which passes over the pulley at B, determine the
magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the
guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has
a radius of r.
375
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kN = 10 N
Given:
F = 5 kN
r = 0.1 m
a = r
b = 1.5 m
c = 5m
Solution:
From pulley, tension in the hoist line is
ΣMB = 0;
T( a) − F( r) = 0
T = F
r
T = 5 kN
a
From the jib,
ΣMA = 0;
b+a
−F ( c) + TBC
c=0
2
c + ( b + a)
2
c + ( b + a)
TBC = F
+
↑Σ Fy = 0;
2
2
TBC = 16.406 kN
b+a
b+a
⎤−F=0
⎢ 2
2⎥
⎣ c + ( b + a) ⎦
− Ay + TBC ⎡
b+a
⎤−F
⎢ 2
2⎥
⎣ c + ( b + a) ⎦
Ay = TBC⎡
+
Σ F x = 0;
→
Ay = 0 kN
c
⎤−F=0
⎢ 2
2⎥
⎣ c + ( b + a) ⎦
Ax − TBC ⎡
376
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Engineering Mechanics - Statics
Chapter 5
c
Ax = TBC
2
c + ( b + a)
FA =
2
Ax + Ay
+F
2
2
Ax = 20.6 kN
F A = 20.6 kN
Problem 5-44
The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center
of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to
overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B.
Given:
W1 = 120000 lb
W2 = 30000 lb
W = 40000 lb
a = 4 ft
b = 6 ft
c = 3 ft
d = 12 ft
e = 15 ft
Solution:
When tipping occurs, R A = 0
ΣMB = 0;
−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c) = 0
⎡W2 c + W c + W1 ( b + c) ⎤
⎥
⎣ W2 d + W ( d + e) ⎦
θ = acos ⎢
θ = 26.4 deg
377
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Problem 5-45
The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center
of gravity at G2. If the suspended load has weight W determine the normal reactions at the tracks
A and B. For the calculation, neglect the thickness of the tracks .
Units Used:
3
kip = 10 lb
Given:
W1 = 120000 lb
a = 4 ft
W2 = 30000 lb
b = 6 ft
W = 16000 lb
c = 3 ft
θ = 30 deg
d = 12 ft
e = 15 ft
Solution:
ΣMB = 0;
−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ − RA( a + b + c) + W1 ( b + c) = 0
RA =
−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c)
+
↑ Σ Fy = 0;
a+b+c
R A = 40.9 kip
R A + RB − W1 − W2 − W = 0
R B = −R A + W1 + W2 + W
R B = 125 kip
Problem 5-46
The man attempts to support the load of boards having a weight W and a center of gravity at G.
If he is standing on a smooth floor, determine the smallest angle θ at which he can hold them up
in the position shown. Neglect his weight .
378
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Engineering Mechanics - Statics
Chapter 5
Given:
a = 0.5 ft
b = 3 ft
c = 4 ft
d = 4 ft
Solution:
ΣMB = 0;
−NA( a + b) + W( b − c cos ( θ ) ) = 0
As θ becomes smaller, NA goes to 0 so that,
cos ( θ ) =
b
c
b
θ = acos ⎛⎜ ⎟⎞
⎝c⎠
θ = 41.4 deg
Problem 5-47
The motor has a weight W. Determine the force that each of the chains exerts on the
supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam.
379
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Engineering Mechanics - Statics
Chapter 5
Given:
W = 850 lb
a = 0.5 ft
b = 1 ft
c = 1.5 ft
θ 1 = 10 deg
θ 2 = 30 deg
θ 3 = 10 deg
Solution:
Guesses
F A = 1 lb
F B = 1 lb
F C = 1 lb
Given
ΣMB = 0;
F A cos ( θ 3 ) b + W a − FC cos ( θ 1 ) ( a + c) = 0
ΣF x = 0;
F C sin ( θ 1 ) − F B sin ( θ 2 ) − FA sin ( θ 3 ) = 0
ΣF y = 0;
W − F A cos ( θ 3 ) − F B cos ( θ 2 ) − F C cos ( θ 1 ) = 0
⎛ FA ⎞
⎜ ⎟
⎜ FB ⎟ = Find ( FA , FB , FC)
⎜F ⎟
⎝ C⎠
⎛ FA ⎞ ⎛ 432 ⎞
⎜ ⎟ ⎜
⎟
⎜ FB ⎟ = ⎜ −0 ⎟ lb
⎜ F ⎟ ⎝ 432 ⎠
⎝ C⎠
380
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Engineering Mechanics - Statics
Chapter 5
Problem 5-48
The boom supports the two vertical loads. Neglect the size of the collars at D and B and the
thickness of the boom, and compute the horizontal and vertical components of force at the pin
A and the force in cable CB.
Given:
F 1 = 800 N
F 2 = 350 N
a = 1.5 m
b = 1m
c = 3
d = 4
θ = 30 deg
Solution:
ΣMA = 0;
−F 1 a cos ( θ ) − F2 ( a + b) cos ( θ ) +
d
2
2
FCB ( a + b) sin ( θ ) +
c +d
F CB
d
2
c +d
Ax =
2
FCB ( a + b) cos ( θ ) = 0
F CB = 0
d
2
2
F CB = 782 N
d sin ( θ ) ( a + b) + c cos ( θ ) ( a + b)
Ax −
2
c +d
2
2
⎡⎣F1 a + F2 ( a + b)⎤⎦ cos ( θ ) c + d
=
+
Σ F x = 0;
→
c
2
Ax = 625 N
F CB
c +d
+
↑
Σ F y = 0;
Ay − F1 − F2 +
c
2
c +d
Ay = F1 + F2 −
2
F CB = 0
c
2
2
FCB
Ay = 681 N
c +d
Problem 5-49
The boom is intended to support two vertical loads F1 and F 2. If the cable CB can sustain a
381
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Engineering Mechanics - Statics
Chapter 5
maximum load Tmax before it fails, determine the critical loads if F1 = 2F2. Also, what is the
magnitude of the maximum reaction at pin A?
Units Used:
3
kN = 10 N
Given:
Tmax = 1500 N
a = 1.5 m
b = 1m
c = 3
d = 4
θ = 30deg
Solution:
ΣMA = 0;
F1 = 2 F2
−2 F 2 a cos ( θ ) − F2 ( a + b) cos ( θ ) +
d
2
2
Tmax( a + b) sin ( θ ) +
c +d
F2 =
( a + b)Tmax( d sin ( θ ) + c cos ( θ ) )
2
F1 = 2 F2
+
ΣF x = 0;
→
Ax −
d
2
c +d
Ax =
2
2
Tmax( a + b) cos ( θ ) = 0
F 1 = 1.448 kN
Tmax = 0
d
2
2
c +d
F 2 = 724 N
c + d cos ( θ ) ( 3 a + b)
2
c
2
Ax = 1.20 kN
Tmax
c +d
+
↑Σ Fy = 0;
Ay − F2 − F1 +
c
2
c +d
2
Tmax = 0
382
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Engineering Mechanics - Statics
Chapter 5
c
Ay = F2 + F1 −
2
2
Ay = 1.27 kN
Tmax
c +d
2
FA =
Ax + Ay
2
F A = 1.749 kN
Problem 5-50
The uniform rod of length L and weight W is supported on the smooth planes. Determine its
position θ for equilibrium. Neglect the thickness of the rod.
Solution:
cos ( θ ) + NA cos ( φ − θ ) L = 0
NA =
cos ( θ ) − NB cos ( ψ + θ ) L = 0
NB =
L
ΣMB = 0;
−W
ΣMA = 0;
W
ΣF x = 0;
NB sin ( ψ) − NA sin ( φ ) = 0
W cos ( θ )
2 cos ( ψ + θ )
L
2
2
sin ( ψ) −
W cos ( θ )
2 cos ( φ − θ )
W cos ( θ )
2 cos ( φ − θ )
W cos ( θ )
2 cos ( ψ + θ )
sin ( φ ) = 0
sin ( ψ) cos ( φ − θ ) − sin ( φ ) cos ( ψ + θ ) = 0
sin ( ψ) ( cos ( φ ) cos ( θ ) + sin ( φ ) sin ( θ ) ) − sin ( φ ) ( cos ( ψ) cos ( θ ) − sin ( ψ) sin ( θ ) ) = 0
2 sin ( ψ) sin ( φ ) sin ( θ ) = ( sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) ) cos ( θ )
tan ( θ ) =
sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ )
2 sin ( ψ) sin ( φ )
=
cot ( ψ) − cot ( φ )
2
⎛ cot ( ψ) − cot ( φ ) ⎞
⎟
2
⎝
⎠
θ = atan ⎜
383
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Engineering Mechanics - Statics
Chapter 5
Problem 5-51
The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in
place by the spring which has unstretched length δ. Determine the magnitude of the resultant
force at A and the normal force on the peg at B when the lever is in the position shown.
Given:
δ = 200 mm
k = 5
N
m
a = 100 mm
b = 300 mm
c = 300 mm
θ = 30 deg
Solution:
Using the law of cosines and the law of sines
c + ( a + b) − 2 c( a + b) cos ( 180 deg − θ )
2
l =
sin ( φ )
c
=
2
sin ( 180 deg − θ )
l
Fs = k s = k (l − δ )
⎛ sin (180 deg − θ ) ⎞
⎟
l
⎝
⎠
φ = asin ⎜c
φ = 12.808 deg
F s = k( l − δ )
F s = 2.3832 N
a+b
ΣMA = 0;
−F s sin ( φ ) ( a + b) + NB a = 0
NB = F s sin ( φ )
ΣF x = 0;
Ax − F s cos ( φ ) = 0
Ax = Fs cos ( φ )
Ax = 2.3239 N
ΣF y = 0;
Ay + NB − Fs sin ( φ ) = 0
Ay = Fs sin ( φ ) − NB
Ay = −1.5850 N
2
F A = 2.813 N
FA =
Ax + Ay
a
2
NB = 2.11 N
384
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Engineering Mechanics - Statics
Chapter 5
Problem 5-52
The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the
springs are uncompressed when the load is removed, determine the force in each spring when the
load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is
large enough so that only small deflections occur. Hint: The beam rotates about A so the
deflections in the springs can be related.
Solution:
ΣMA = 0;
3
F B L + FC2 L − P L = 0
2
F B + 2 FC = 1.5 P
ΔC = 2 ΔB
FC
k
=
2 FB
k
F C = 2FB
5 F B = 1.5 P
F B = 0.3 P
F c = 0.6 P
ΔC =
0.6 P
k
Problem 5-53
The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple
moment of M, determine the angle θ for equilibrium.The spring has an unstretched length δ and
a stiffness k.
385
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Engineering Mechanics - Statics
Chapter 5
Given:
W = 200 lb
M = 100 lb ft
δ = 2 ft
k = 50
lb
ft
a = 3 ft
b = 3 ft
c = 2 ft
Solution:
Initial Guess:
θ = 10 deg
Given
k⎡⎣( a + b)sin ( θ ) + c − δ⎤⎦ ( a + b) cos ( θ ) − W a cos ( θ ) − M = 0
θ = Find ( θ )
θ = 23.2 deg
Problem 5-54
The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the
reactions at these points needed to support the vertical force F . Neglect the pipe's thickness in
the calculation.
Given:
F = 45 lb
386
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Engineering Mechanics - Statics
Chapter 5
θ = 30 deg
a = 16 in
b = 20 in
c = 8 in
Solution:
R A = 1 lb
Initial Guesses:
R B = 1 lb
R C = 1 lb
Given
F cos ( θ ) ( a + b) − F sin ( θ ) c − RC b + R B c tan ( θ ) = 0
ΣMA = 0;
↑Σ Fy = 0;
R C cos ( θ ) − R B cos ( θ ) − F = 0
+
Σ F x = 0;
→
R A + RB sin ( θ ) − RC sin ( θ ) = 0
+
⎛ RA ⎞
⎜ ⎟
⎜ RB ⎟ = Find ( RA , RB , RC)
⎜R ⎟
⎝ C⎠
⎛ RA ⎞ ⎛ 25.981 ⎞
⎜ ⎟ ⎜
⎟
⎜ RB ⎟ = ⎜ 11.945 ⎟ lb
⎜ R ⎟ ⎝ 63.907 ⎠
⎝ C⎠
Problem 5-55
The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the
stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are
unstretched, determine the smallest force needed to close the contact gap at C.
Units Used:
mN = 10
−3
N
387
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Engineering Mechanics - Statics
Chapter 5
Given:
a = 50 mm
b = 50 mm
c = 10 mm
k = 5
N
m
Solution:
Initial Guesses:
F = 0.5 N
yA = 1 mm
yB = 1 mm
Given
c − yA
a+b
=
yB − yA
a
⎛ yA ⎞
⎜ ⎟
⎜ yB ⎟ = Find ( yA , yB , F)
⎜F⎟
⎝ ⎠
k yA + k yB − F = 0
⎛ yA ⎞ ⎛ −2 ⎞
⎜ ⎟ = ⎜ ⎟ mm
⎝ yB ⎠ ⎝ 4 ⎠
k yB a − F( a + b) = 0
F = 10 mN
Problem 5-56
The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine
the maximum stiffness k of the springs at A and B so that the contact at C closes when the
vertical force developed there is F . Originally the strip is horizontal as shown.
Units Used:
mN = 10
−3
N
Given:
a = 50 mm
b = 50 mm
c = 10 mm
F = 0.5 N
388
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Solution:
Initial Guesses: k = 1
N
yA = 1 mm
m
yB = 1 mm
Given
c − yA
a+b
=
yB − yA
k yA + k yB − F = 0
a
⎛ yA ⎞
⎜ ⎟
⎜ yB ⎟ = Find ( yA , yB , k)
⎜k ⎟
⎝ ⎠
⎛ yA ⎞ ⎛ −2 ⎞
⎜ ⎟ = ⎜ ⎟ mm
⎝ yB ⎠ ⎝ 4 ⎠
k yB a − F( a + b) = 0
k = 250
N
m
Problem 5-57
Determine the distance d for placement of the load P for equilibrium of the smooth bar in the
position θ as shown. Neglect the weight of the bar.
Solution:
↑Σ Fy = 0;
R cos ( θ ) − P = 0
Σ MA = 0;
−P d cos ( θ ) + R
+
R d cos ( θ ) = R
2
d=
a
cos ( θ )
=0
a
cos ( θ )
a
cos ( θ )
3
Problem 5-58
The wheelbarrow and its contents have mass m and center of mass at G. Determine the greatest
389
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Engineering Mechanics - Statics
Chapter 5
angle of tilt θ without causing the wheelbarrow to tip over.
g
Solution:
Require point G to be over the wheel axle for tipping. Thus
b cos ( θ ) = a sin ( θ )
− 1⎛ b ⎞
θ = tan
⎜ ⎟
⎝ a⎠
Problem 5-59
Determine the force P needed to pull the roller of mass M over the smooth step.
Given:
M = 50 kg
a = 0.6 m
b = 0.1 m
θ = 60 deg
θ 1 = 20 deg
g = 9.81
m
2
s
390
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Engineering Mechanics - Statics
Chapter 5
Solution:
φ = acos ⎛⎜
a − b⎞
⎟
⎝ a ⎠
φ = 33.56 deg
ΣMB = 0,
M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0
+ P cos ( θ ) ( a − b) − P sin ( θ ) a sin ( φ )
⎡sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤
⎥
⎣ cos ( θ ) ( a − b) + sin ( θ ) a sin ( φ ) ⎦
P = M g⎢
P = 441 N
Problem 5-60
Determine the magnitude and direction θ of the minimum force P needed to pull the roller of
mass M over the smooth step.
Given:
a = 0.6 m
b = 0.1 m
θ 1 = 20 deg
M = 50 kg
g = 9.81
m
2
s
Solution:
For Pmin, NA tends to 0
φ = acos ⎛⎜
a − b⎞
⎟
⎝ a ⎠
φ = 33.56 deg
ΣMB = 0 M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0
+ ⎡⎣−P cos ( θ ) ( a − b)⎤⎦ − P sin ( θ ) a sin ( φ )
391
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Engineering Mechanics - Statics
P=
Chapter 5
M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦
cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )
For Pmin :
dP
dθ
=
M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦
⎡⎣cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )⎤⎦
θ = atan ⎛⎜ sin ( φ ) ⋅
which gives,
P =
⎝
a
2
⎡⎣a sin ( φ ) cos ( θ ) − ( a − b)sin ( θ )⎤⎦ = 0
⎞
⎟
θ = 33.6 deg
a − b⎠
M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦
cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )
P = 395 N
Problem 5-61
A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a
radius r. Determine the angle of inclination θ for equilibrium.
Solution:
By Observation φ = θ.
Equilibirium :
ΣMA = 0; NB2 r cos ( θ ) − W
NB =
L
2
cos ( θ ) = 0
WL
4r
ΣF x = 0; NA cos ( θ ) − W sin ( θ ) = 0
NA = W tan ( θ )
392
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Engineering Mechanics - Statics
Chapter 5
ΣF y = 0; W tan ( θ ) sin ( θ ) +
WL
4r
− W cos ( θ ) = 0
sin ( θ ) − cos ( θ ) = 1 − 2 cos ( θ ) =
2
2
2 cos ( θ ) −
2
cos ( θ ) =
L
4r
L+
2
−L
4r
cos ( θ )
cos ( θ ) − 1 = 0
2
⎛ L + L2 + 128 r2 ⎞
⎟
16 r
⎝
⎠
2
L + 128 r
θ = acos ⎜
16 r
Problem 5-62
The disk has mass M and is supported on the smooth cylindrical surface by a spring having
stiffness k and unstretched length l0. The spring remains in the horizontal position since its end A
is attached to the small roller guide which has negligible weight. Determine the angle θ to the
nearest degree for equilibrium of the roller.
Given:
M = 20 kg
k = 400
N
m
l0 = 1 m
r = 2m
g = 9.81
m
2
s
a = 0.2 m
Guesses
F = 10 N
Solution:
Given
R = 10 N θ = 30 deg
+
Σ F y = 0;
→
R sin ( θ ) − M g = 0
+
R cos ( θ ) − F = 0
Spring
F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦
↑Σ Fx = 0;
393
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Engineering Mechanics - Statics
⎛F⎞
⎜ R ⎟ = Find ( F , R , θ )
⎜ ⎟
⎝θ ⎠
Chapter 5
⎛ F ⎞ ⎛ 163.633 ⎞
⎜ ⎟=⎜
⎟N
⎝ R ⎠ ⎝ 255.481 ⎠
θ = 50.171 deg
There is also another answer that we can find by choosing different starting guesses.
Guesses
F = 200 N R = 200 N θ = 20 deg
Solution:
Given
+
Σ F y = 0;
→
R sin ( θ ) − M g = 0
+
R cos ( θ ) − F = 0
Spring
F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦
↑Σ Fx = 0;
⎛F⎞
⎜ R ⎟ = Find ( F , R , θ )
⎜ ⎟
⎝θ ⎠
⎛ F ⎞ ⎛ 383.372 ⎞
⎜ ⎟=⎜
⎟N
⎝ R ⎠ ⎝ 430.66 ⎠
θ = 27.102 deg
Problem 5-63
Determine the x, y, z components of reaction at the fixed wall A.The force F2 is parallel to the z
axis and the force F1 is parallel to the y axis.
Given:
a = 2m
d = 2m
b = 1m
F 1 = 200 N
c = 2.5 m
F 2 = 150 N
394
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Engineering Mechanics - Statics
Chapter 5
Solution:
ΣF x = 0;
Ax = 0
ΣF x = 0;
Ay = −F1
Ay = −200 N
ΣF x = 0;
Az = F2
Az = 150 N
ΣΜx = 0;
MAx = −F2 a + F 1 d
MAx = 100 N⋅ m
ΣΜy = 0;
MAy = 0
ΣMz = 0;
MAz = F 1 c
MAz = 500 N⋅ m
Problem 5-64
The wing of the jet aircraft is
subjected to thrust T from its engine
and the resultant lift force L. If the
mass of the wing is M and the mass
center is at G, determine the x, y, z
components of reaction where the
wing is fixed to the fuselage at A.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
g = 9.81
m
2
s
395
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Given:
T = 8 kN
L = 45 kN
M = 2.1 Mg
a = 2.5 m
b = 5m
395
c = 3m
d = 7m
Solution:
ΣF x = 0;
− Ax + T = 0
Ax = T
Ax = 8 kN
ΣF y = 0;
Ay = 0
Ay = 0
ΣF z = 0;
− Az − M g + L = 0
Az = L − M g
ΣMy = 0;
Az = 24.4 kN
M y − T ( a) = 0
My = T a
ΣMx = 0;
My = 20.0 kN⋅ m
L( b + c + d) − M g b − Mx = 0
M x = L ( b + c + d) − M g b
ΣMz = 0;
Mx = 572 kN⋅ m
Mz − T( b + c) = 0
Mz = T( b + c)
Mz = 64.0 kN⋅ m
Problem 5-65
The uniform concrete slab has weight W. Determine the tension in each of the three parallel
supporting cables when the slab is held in the horizontal plane as shown.
396
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kip = 10 lb
Given:
W = 5500 lb
a = 6 ft
b = 3 ft
c = 3 ft
d = 6 ft
Solution:
Equations of Equilibrium : The
cable tension TB can be obtained
directly by summing moments
about the y axis.
ΣMy = 0;
ΣMx = 0;
d
TB =
TC a + TB( a + b) − W⎛⎜
a + b + c⎞
2
⎝
TC =
ΣF z = 0;
W
− TB d = 0
W
2
TB = 2.75 kip
⎟=0
⎠
2
1⎡ a + b + c
− TB( a + b)⎥⎤
⎢W
a⎣
2
⎦
TA + TB + TC − W = 0
TA = − TB − TC + W
TC = 1.375 kip
TA = 1.375 kip
397
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Problem 5-66
The air-conditioning unit is hoisted to the roof of a building using the three cables. If the tensions
in the cables are TA, TB and TC, determine the weight of the unit and the location (x, y) of its
center of gravity G.
Given:
TA = 250 lb
TB = 300 lb
TC = 200 lb
a = 5 ft
b = 4 ft
c = 3 ft
d = 7 ft
e = 6 ft
Solution:
ΣF z = 0; TA + TB + TC − W = 0
W = TA + TB + TC
ΣMy = 0;
W x − T A ( c + d) − T C d = 0
x =
ΣMx = 0;
W = 750 lb
TA( c + d) + TC d
W
x = 5.2 ft
TA a + TB( a + b − e) + TC( a + b) − W y = 0
y =
TA a + TB( a + b − e) + TC( a + b)
W
y = 5.267 ft
Problem 5-67
The platform truck supports the three loadings shown. Determine the normal reactions on each
of its three wheels.
398
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Engineering Mechanics - Statics
Chapter 5
Given:
F 1 = 380 lb
F 2 = 500 lb
F 3 = 800 lb
a = 8 in
b = 12 in
c = 10 in
d = 5 in
e = 12 in
f = 12 in
Solution:
The initail guesses are
F A = 1 lb
F B = 1 lb
F C = 1 lb
Given
ΣMx = 0;
F 1 ( c + d) + F2 ( b + c + d) + F3 d − F A( a + b + c + d) = 0
ΣMy = 0;
F1 e − FB e − F2 f + FC f = 0
ΣF y = 0;
FB + FC − F2 + FA − F1 − F3 = 0
⎛ FA ⎞
⎜ ⎟
⎜ FB ⎟ = Find ( FA , FB , FC)
⎜F ⎟
⎝ C⎠
⎛ FA ⎞ ⎛ 663 ⎞
⎜ ⎟ ⎜
⎟
⎜ FB ⎟ = ⎜ 449 ⎟ lb
⎜ F ⎟ ⎝ 569 ⎠
⎝ C⎠
Problem 5-68
Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane
fuselage A and wings B and C are located as shown. If these components have weights WA, WB
and WC, determine the normal reactions of the wheels D, E, and F on the ground.
399
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kip = 10 lb
Given:
WA = 45000 lb
WB = 8000 lb
WC = 6000 lb
a = 8 ft
e = 20 ft
b = 6 ft
f = 4 ft
c = 8 ft
g = 3 ft
d = 6 ft
Solution:
Initial guesses:
R D = 1 kip
R E = 1 kip
R F = 1 kip
Given
ΣMx = 0;
WB b − RD( a + b) − WC c + R E( c + d) = 0
ΣMy = 0;
W B f + W A( g + f ) + W C f − R F( e + g + f ) = 0
ΣF z = 0;
R D + R E + RF − WA − WB − WC = 0
⎛ RD ⎞
⎜ ⎟
⎜ RE ⎟ = Find ( RD , RE , RF)
⎜R ⎟
⎝ F⎠
⎛ RD ⎞ ⎛ 22.6 ⎞
⎜ ⎟ ⎜
⎟
⎜ RE ⎟ = ⎜ 22.6 ⎟ kip
⎜ R ⎟ ⎝ 13.7 ⎠
⎝ F⎠
Problem 5-69
If the cable can be subjected to a maximum tension T, determine the maximum force F which
may be applied to the plate. Compute the x, y, z components of reaction at the hinge A for this
loading.
400
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Engineering Mechanics - Statics
Chapter 5
Given:
a = 3 ft
b = 2 ft
c = 1 ft
d = 3 ft
e = 9 ft
T = 300 lb
Solution:
Initial guesses:
F = 10 lb
MAx = 10 lb ft
MAz = 10 lb ft
Ax = 10 lb
Ay = 10 lb
Az = 10 lb
Given
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ + ⎜ 0 ⎟ = 0
⎜ A ⎟ ⎝T − F⎠
⎝ z⎠
⎛ MAx ⎞ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ e ⎞ ⎛ 0 ⎞
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
⎜ 0 ⎟ + ⎜ −c ⎟ × ⎜ 0 ⎟ + ⎜ −b − c ⎟ × ⎜ 0 ⎟ = 0
⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ T ⎠
⎝
⎠
⎛ F ⎞
⎜ A ⎟
⎜ x ⎟
⎜ Ay ⎟
⎜
⎟ = Find ( F , Ax , Ay , Az , MAx , MAz)
Az
⎜
⎟
⎜ MAx ⎟
⎟
⎜
⎝ MAz ⎠
⎛ MAx ⎞ ⎛ 0 ⎞
⎜
⎟ = ⎜ ⎟ lb ft
⎝ MAz ⎠ ⎝ 0 ⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0 ⎟ lb
⎜ A ⎟ ⎝ 600 ⎠
⎝ z⎠
F = 900 lb
Problem 5-70
The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as
shown. Determine the x, y, z components of reaction at A and the tension in cable DEC.
401
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Given:
⎛ 0 ⎞
⎜ 0 ⎟ lb
F =
⎜
⎟
⎝ −1500 ⎠
a = 5 ft
b = 4 ft
c = b
d = 5 ft
e = 5 ft
f = 2 ft
Solution:
α = atan ⎛⎜
a
⎞
⎟
⎝ d + e⎠
L =
2
a + ( d + e)
β = atan ⎛⎜
2
⎞
⎟
fL
⎜L −
⎟
d+ e⎠
⎝
Guesses
b
TBE = 1 lb TDEC = 1 lb
Ax = 1 lb Ay = 1 lb
Given
Az = 1 lb
2 TDEC cos ( β ) = TBE
0
⎞
⎛0⎞
⎛ 0 ⎞ ⎛⎜
⎜ d ⎟ × F + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0
⎟
⎜ ⎟
⎜
⎟ ⎜
⎜
⎝0⎠
⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠
⎛ Ax ⎞
⎛ 0 ⎞
⎜ ⎟
⎜
⎟
⎜ Ay ⎟ + F + TBE⎜ −cos ( α ) ⎟ = 0
⎜A ⎟
⎝ sin ( α ) ⎠
⎝ z⎠
402
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜ Az ⎟ = Find ( A , A , A , T , T
x y z BE DEC) TDEC = 919 lb
⎜
⎟
⎜ TBE ⎟
⎜T
⎟
⎝ DEC ⎠
⎛ Ax ⎞ ⎛ 0
⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 1.5 × 103 ⎟ lb
⎜A ⎟ ⎜
⎟
⎝ z ⎠ ⎝ 750 ⎠
Problem 5-71
The cable CED can sustain a maximum tension Tmax before it fails. Determine the greatest
vertical force F that can be applied to the boom. Also, what are the x, y, z components of
reaction at the ball-and-socket joint A?
Given:
Tmax = 800 lb
a = 5 ft
b = 4 ft
c = b
d = 5 ft
e = 5 ft
f = 2 ft
Solution:
a ⎞
⎟
⎝ d + e⎠
α = atan ⎛⎜
L =
2
a + ( d + e)
β = atan ⎛⎜
2
⎞
fL ⎟
⎜L −
⎟
d+ e⎠
⎝
b
TDEC = Tmax
TBE = 1 lb
F = 1 lb
Ax = 1 lb Ay = 1 lb
Az = 1 lb
Guesses
Given
2 TDEC cos ( β ) = TBE
403
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Engineering Mechanics - Statics
Chapter 5
0
⎞
⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜
⎜ d ⎟ × ⎜ 0 ⎟ + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0
⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜
⎜
⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜ A ⎟ = Find ( A , A , A , T , F)
x y z BE
⎜ z ⎟
⎜ TBE ⎟
⎜
⎟
⎝ F ⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎛ 0 ⎞
⎜ ⎟ ⎜ ⎟
⎜
⎟
⎜ Ay ⎟ + ⎜ 0 ⎟ + TBE⎜ −cos ( α ) ⎟ = 0
⎜ A ⎟ ⎝ −F ⎠
⎝ sin ( α ) ⎠
⎝ z⎠
F = 1306 lb
0
⎛ Ax ⎞ ⎛
⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 1.306 × 103 ⎟ lb
⎜A ⎟ ⎜
⎟
⎝ z ⎠ ⎝ 653.197 ⎠
Problem 5-72
The uniform table has a weight W and is supported by the framework shown. Determine the
smallest vertical force P that can be applied to its surface that will cause it to tip over. Where
should this force be applied?
Given:
W = 20 lb
a = 3.5 ft
b = 2.5 ft
c = 3 ft
e = 1.5 ft
f = 1 ft
Solution:
θ = atan ⎛⎜ ⎟⎞
θ = 33.69 deg
d = e sin ( θ )
d = 0.832 ft
⎛ a −e⎞
⎜ 2
⎟
φ = atan ⎜
⎟
⎜ b ⎟
⎝ 2 ⎠
φ = 11.31 deg
f
⎝ e⎠
404
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Engineering Mechanics - Statics
2
d' =
⎛ a − e⎞ + ⎛ b ⎞
⎜
⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 2⎠
Chapter 5
2
d' = 1.275 ft
Tipping will occur about the g - g axis.
Require P to be applied at the corner of the
table for Pmin.
W d = P d' sin ( 90 deg − φ + θ )
P = W
d
P = 14.1 lb
d' sin ( 90 deg − φ + θ )
Problem 5-73
The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle
in the position shown, and the components of reaction at the ball-and-socket joint A and the
smooth journal bearing B. The bearing at B is in proper alignment and exerts only force
reactions perpendicular to the shaft on the windlass.
Given:
W = 150 lb
a = 2 ft
b = 2 ft
c = 1 ft
d = 1 ft
e = 1 ft
f = 0.5 ft
Solution:
ΣMy = 0;
W f−P d=0
P =
W f
d
ΣF y = 0;
Ay = 0 lb
ΣMx = 0;
−W a + B z( a + b) = 0
P = 75 lb
Ay = 0
405
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Engineering Mechanics - Statics
Bz =
ΣF z = 0;
Chapter 5
Wa
B z = 75 lb
a+b
Az + Bz − W = 0
Az = W − Bz
ΣMz = 0;
B x( a + b) − ( a + b + c + e)P = 0
Bx =
ΣF x = 0;
Az = 75 lb
P ( a + b + c + e)
a+b
B x = 112 lb
Ax − Bx + P = 0
Ax = Bx − P
Ax = 37.5 lb
Problem 5-74
A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at
C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the
ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as
shown.
Given:
M = 2 kg
θ 1 = 10 deg
θ 2 = 45 deg
Solution:
ΣF x = 0; F c cos ( θ 1 ) − M g sin ( θ 1 ) = 0
F c = M g⋅ tan ( θ 1 )
F c = 0.32 kg⋅ m
406
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Engineering Mechanics - Statics
Chapter 5
ΣF y = 0; NA cos ( θ 2 ) − NB cos ( θ 2 ) = 0
NA = NB
ΣF z = 0; 2 NA sin ( θ 2 ) − M g cos ( θ 1 ) − F c sin ( θ 1 ) = 0
NA =
1 M g⋅ cos ( θ 1 ) + F c⋅ sin ( θ 1 )
⋅
2
sin ( θ 2 )
NA = 1.3 kg⋅ m
N = NA = NB
Problem 5-75
Member AB is supported by cable BC and at A by a square rod which fits loosely through the
square hole at the end joint of the member as shown. Determine the components of reaction at
A and the tension in the cable needed to hold the cylinder of weight W in equilibrium.
Units Used:
3
kip = 10 lb
Given:
W = 800 lb
a = 2 ft
b = 6 ft
c = 3 ft
Solution:
⎛
ΣF x = 0
F BC ⎜
ΣF y = 0
Ay = 0
⎞=0
c
2
2⎟
F BC = 0 lb
Ay = 0lb
Ay = 0 lb
2
⎝ c +b +a ⎠
407
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
ΣF z = 0
Az − W = 0
ΣMx = 0
MAx − W b = 0
ΣMy = 0
ΣMz = 0
Az = W
Az = 800 lb
MAx = W b
MAx = 4.80 kip⋅ ft
MAy = 0 lb ft
MAy = 0 lb⋅ ft
MAz = 0 lb ft
MAz = 0 lb⋅ ft
Problem 5-76
The pipe assembly supports the vertical loads shown. Determine the components of reaction at
the ball-and-socket joint A and the tension in the supporting cables BC and BD.
Units Used:
3
kN = 10 N
Given:
F 1 = 3 kN d = 2 m
F 2 = 4 kN e = 1.5 m
a = 1m
g = 1m
b = 1.5 m
h = 3m
c = 3m
i = 2m
f = c−e
j = 2m
Solution:
The initial guesses are:
TBD = 1 kN
TBC = 1 kN
Ax = 1 kN
Ay = 1 kN
Az = 1 kN
The vectors
408
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Engineering Mechanics - Statics
Chapter 5
⎛ 0 ⎞
r1 = ⎜ a + b + f ⎟
⎜
⎟
⎝ d ⎠
⎛ 0 ⎞
r2 = ⎜ a + b + c ⎟
⎜
⎟
⎝ d ⎠
⎛ i ⎞
rBC = ⎜ −a ⎟
⎜
⎟
⎝h − g⎠
⎛ −j ⎞
rBD = ⎜ −a ⎟
⎜
⎟
⎝h − g⎠
rBC
rBD
uBC =
Given
rBC
uBD =
rBD
⎛0⎞
rAB = ⎜ a ⎟
⎜ ⎟
⎝g⎠
⎛1⎞
i = ⎜0⎟
⎜ ⎟
⎝0⎠
⎛0⎞
j = ⎜1⎟
⎜ ⎟
⎝0⎠
⎛0⎞
k = ⎜0⎟
⎜ ⎟
⎝1⎠
Ax i + Ay j + Az k − F1 k − F2 k + TBDuBD + TBC uBC = 0
rAB × ( TBDuBD + TBCuBC) + r1 × ( −F 1 k) + r2 × ( −F 2 k) = 0
⎛ TBD ⎞
⎜
⎟
⎜ TBC ⎟
⎜ Ax ⎟ = Find ( T , T , A , A , A )
BD BC x y z
⎜
⎟
⎜ Ay ⎟
⎜ A ⎟
⎝ z ⎠
⎛ TBD ⎞ ⎛ 17 ⎞
⎜
⎟ = ⎜ ⎟ kN
⎝ TBC ⎠ ⎝ 17 ⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 11.333 ⎟ kN
⎜ A ⎟ ⎝ −15.667 ⎠
⎝ z⎠
Problem 5-77
The hatch door has a weight W and center of gravity at G. If the force F applied to the handle
at C has coordinate direction angles of α, β and γ, determine the magnitude of F needed to
hold the door slightly open as shown. The hinges are in proper alignment and exert only force
reactions on the door. Determine the components of these reactions if A exerts only x and z
components of force and B exerts x, y, z force components.
Given:
W = 80 lb
α = 60 deg
409
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Engineering Mechanics - Statics
Chapter 5
β = 45 deg
γ = 60 deg
a = 3 ft
b = 2 ft
c = 4 ft
d = 3 ft
Solution:
Initial Guesses:
Ax = 1 lb
Az = 1 lb
F = 1 lb
B x = 1 lb
B y = 1 lb
B z = 1 lb
Given
⎛ Ax ⎞ ⎛ Bx ⎞
⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎜ 0 ⎟ + ⎜ By ⎟ + F⎜ cos ( β ) ⎟ + ⎜ 0 ⎟ = 0
⎜ Az ⎟ ⎜ B ⎟
⎜ cos ( γ ) ⎟ ⎝ −W ⎠
⎝
⎠
⎝ ⎠ ⎝ z⎠
⎛ a + b ⎞ ⎢⎡ ⎛⎜ cos ( α ) ⎟⎞⎥⎤ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞
⎜ 0 ⎟ × F cos ( β ) + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ c + d ⎟ × B = 0
⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎜
⎜
⎟ ⎢ ⎜
⎟ ⎜ y⎟
⎢
⎜
⎟
⎥
⎝ 0 ⎠ ⎣ ⎝ cos ( γ ) ⎠⎦ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠
⎛ Ax ⎞
⎜ ⎟
⎜ Az ⎟
⎜ Bx ⎟
⎜ ⎟ = Find ( Ax , Az , Bx , By , Bz , F)
⎜ By ⎟
⎜ Bz ⎟
⎜ ⎟
⎝F ⎠
⎛ Ax ⎞ ⎛ −96.5 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Az ⎠ ⎝ −13.7 ⎠
⎛ Bx ⎞ ⎛ 48.5 ⎞
⎜ ⎟ ⎜
⎟
⎜ By ⎟ = ⎜ −67.9 ⎟ lb
⎜ B ⎟ ⎝ 45.7 ⎠
⎝ z⎠
F = 96 lb
410
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Problem 5-78
The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at
C has coordinate direction angles α, β, γ determine the magnitude of F needed to hold the door
slightly open as shown. If the hinge at A becomes loose from its attachment and is ineffective,
what are the x, y, z components of reaction at hinge B?
Given:
W = 80 lb
α = 60 deg
β = 45 deg
γ = 60 deg
a = 3 ft
b = 2 ft
c = 4 ft
d = 3 ft
Solution:
a
ΣMy = 0;
F = W
ΣF x = 0;
B x + F cos ( α ) = 0
cos ( γ ) ( a + b)
B x = −F cos ( α )
ΣF y = 0;
B y = −67.9 lb
B z − W + F cos ( γ ) = 0
B z = W − F cos ( γ )
ΣMx = 0;
B x = −48 lb
B y + F cos ( β ) = 0
B y = −F cos ( β )
ΣF z = 0;
F = 96 lb
B z = 32 lb
MBx + W d − F cos ( γ ) ( c + d) = 0
MBx = −W d + F cos ( γ ) ( c + d)
ΣMz = 0;
MBx = 96 lb⋅ ft
MBz + F cos ( α ) ( c + d) + F cos ( β ) ( a + b) = 0
MBz = −F cos ( α ) ( c + d) − F cos ( β ) ( a + b)
MBz = −675 lb⋅ ft
411
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Engineering Mechanics - Statics
Chapter 5
Problem 5-79
The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z
components of reaction at the bearings if the rod is subjected to forces F 1 and F 2. F1 lies in the
y-z plane. The bearings are in proper alignment and exert only force reactions on the rod.
Given:
F 1 = 300 lb
d = 3 ft
F 2 = 250 lb
e = 5 ft
a = 1 ft
α = 30 deg
b = 4 ft
β = 45 deg
c = 2 ft
θ = 45 deg
Solution:
The initial guesses:
Ax = 100 lb
Ay = 200 lb
B x = 300 lb
B z = 400 lb
Cy = 500 lb
Cz = 600 lb
Given
Ax + B x + F2 cos ( β ) sin ( α ) = 0
Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0
B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0
F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0
Ax b + Cz e = 0
Ax( c + d) + Bx d − Cy e = 0
412
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Engineering Mechanics - Statics
Chapter 5
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜ ⎟
⎜ Bx ⎟ = Find ( A , A , B , B , C , C )
x y x z y z
⎜ Bz ⎟
⎜ ⎟
⎜ Cy ⎟
⎜ Cz ⎟
⎝ ⎠
⎛ Ax ⎞ ⎛ 632.883 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Ay ⎠ ⎝ −141.081 ⎠
⎛ Bx ⎞ ⎛ −721.271 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Bz ⎠ ⎝ 895.215 ⎠
⎛ Cy ⎞ ⎛ 200.12 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Cz ⎠ ⎝ −506.306 ⎠
Problem 5-80
The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude
of F2 which will cause the reaction Cy at the bearing C to be equal to zero. The bearings are in
proper alignment and exert only force reactions on the rod.
Given:
F 1 = 300 lb
d = 3 ft
Cy = 0 lb
e = 5 ft
a = 1 ft
α = 30 deg
b = 4 ft
β = 45 deg
c = 2 ft
θ = 45 deg
Solution:
The initial guesses:
Ax = 100 lb
Ay = 200 lb
B x = 300 lb
B z = 400 lb
F 2 = 500 lb
Cz = 600 lb
Given
Ax + B x + F2 cos ( β ) sin ( α ) = 0
413
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0
B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0
F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0
Ax b + Cz e = 0
Ax( c + d) + Bx d − Cy e = 0
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜ ⎟
⎜ Bx ⎟ = Find ( A , A , B , B , C , F )
x y x z z 2
⎜ Bz ⎟
⎜ ⎟
⎜ Cz ⎟
⎜ F2 ⎟
⎝ ⎠
F 2 = 673.704 lb
Problem 5-81
Determine the tension in cables BD and CD and the x, y, z components of reaction at the
ball-and-socket joint at A.
Given:
F = 300 N
a = 3m
b = 1m
c = 0.5 m
d = 1.5 m
Solution:
⎛ −b ⎞
⎜ ⎟
rBD = d
⎜ ⎟
⎝a⎠
414
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
⎛ −b ⎞
rCD = ⎜ −d ⎟
⎜ ⎟
⎝a⎠
Initial Guesses:
TBD = 1 N
TCD = 1 N
Ax = 1 N
Ay = 1 N
Az = 1 N
Given
⎛ Ax ⎞
⎛ 0 ⎞
⎜ ⎟
rBD
rCD
+ TCD
+⎜ 0 ⎟ =0
⎜ Ay ⎟ + TBD
⎜ ⎟
rBD
rCD
⎜A ⎟
⎝ −F ⎠
⎝ z⎠
⎛d⎞
⎛d⎞
⎛d − c⎞ ⎛ 0 ⎞
rBD ⎞ ⎜ ⎟ ⎛
rCD ⎞ ⎜
⎜ −d ⎟ × ⎛ T
+ d × ⎜ TCD
+
0 ⎟×⎜ 0 ⎟ = 0
⎜
⎟
⎟
BD
⎜ ⎟ ⎝
⎟ ⎜ ⎟
rBD ⎠ ⎜ ⎟ ⎝
rCD ⎠ ⎜
⎝0⎠
⎝0⎠
⎝ 0 ⎠ ⎝ −F ⎠
⎛ TBD ⎞
⎜
⎟
⎜ TCD ⎟
⎜ Ax ⎟ = Find ( T , T , A , A , A )
BD CD x y z
⎜
⎟
⎜ Ay ⎟
⎜ A ⎟
⎝ z ⎠
⎛ TBD ⎞ ⎛ 116.7 ⎞
⎜
⎟=⎜
⎟N
⎝ TCD ⎠ ⎝ 116.7 ⎠
⎛ Ax ⎞ ⎛ 66.7 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0 ⎟ N
⎜ A ⎟ ⎝ 100 ⎠
⎝ z⎠
Problem 5-82
Determine the tensions in the cables and the components of reaction acting on the smooth collar at
A necessary to hold the sign of weight W in equilibrium. The center of gravity for the sign is at G.
Given:
W = 50 lb
f = 2.5 ft
a = 4 ft
g = 1 ft
b = 3 ft
h = 1 ft
415
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Engineering Mechanics - Statics
Chapter 5
c = 2 ft
i = 2 ft
d = 2 ft
j = 2 ft
e = 2.5 ft
k = 3 ft
Solution:
The initial guesses are:
TBC = 10 lb
Ax = 10 lb
MAx = 10 lb⋅ ft
TDE = 10 lb
Ay = 10 lb
MAy = 10 lb⋅ ft
Given
TDE
( a − k)
2
2
( a − k) + i + j
−i
TDE
2
( a − k) + i + j
j
TBC
2
2
2
2
2
2
TBC
2
2
2
+ Ax = 0
+ Ay = 0
−W=0
( d − b) + i + c
i
2
2
( a − k) + i + j
MAy − TDE k
2
( d − b) + i + c
+c
( a − k) + i + j
MAx + TDE j
2
( d − b) + i + c
2
TDE
2
2
−i
2
TBC
+ ( −b + d)
2
j
2
2
( a − k) + i + j
2
i
+ c TBC
2
− Wi = 0
2
2
2
2
( d − b) + i + c
d
+ TBC c
2
+ W ( k − f) = 0
( d − b) + i + c
416
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Engineering Mechanics - Statics
Chapter 5
i
−TDE a
2
2
( a − k) + i + j
2
+ TBC b
i
2
=0
2
2
( d − b) + i + c
⎛⎜ MAx ⎞⎟
⎜ MAy ⎟
⎜
⎟
⎜ TBC ⎟ = Find ( M , M , T , T , A , A )
Ax
Ay BC DE x y
⎜ TDE ⎟
⎜
⎟
⎜ Ax ⎟
⎜ Ay ⎟
⎝
⎠
⎛ TBC ⎞ ⎛ 42.857 ⎞
⎜
⎟=⎜
⎟ lb
⎝ TDE ⎠ ⎝ 32.143 ⎠
⎛ Ax ⎞ ⎛ 3.571 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Ay ⎠ ⎝ 50 ⎠
⎛ MAx ⎞ ⎛ 2.698 × 10− 13 ⎞
⎜
⎟=⎜
⎟ lb⋅ ft
⎝ MAy ⎠ ⎝ −17.857 ⎠
Problem 5-83
The member is supported by a pin at A and a cable BC. If the load at D is W, determine the x, y,
z components of reaction at these supports.
Units Used:
3
kip = 10 lb
Given:
W = 300 lb
a = 1 ft
b = 2 ft
c = 6 ft
d = 2 ft
e = 2 ft
f = 2 ft
417
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Solution:
Initial Guesses:
TBC = 1 lb
Ax = 1 lb
Ay = 1 lb
Az = 1 lb
MAy = 1 lb ft
MAz = 1 lb ft
Given
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ +
⎜A ⎟
⎝ z⎠
⎛e + f − a⎞ ⎛ 0 ⎞
⎜ −c ⎟ + ⎜ 0 ⎟ = 0
⎟ ⎜ ⎟
2
2
2⎜
b + c + ( e + f − a) ⎝
b
⎠ ⎝ −W ⎠
TBC
⎛ 0 ⎞ ⎛ −e ⎞ ⎛ 0 ⎞ ⎛ −a ⎞ ⎡
⎛ e + f − a ⎞⎤
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢
TBC
⎜ −c ⎟⎥ = 0
⎜ MAy ⎟ + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎢
⎜
⎟⎥
⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ b ⎠ ⎣ b2 + c2 + ( e + f − a) 2 ⎝ b ⎠⎦
⎝
⎠
⎛⎜ TBC ⎞⎟
⎜ Ax ⎟
⎜
⎟
⎜ Ay ⎟ = Find ( T , A , A , A , M , M )
BC x y z
Ay
Az
⎜ Az ⎟
⎜
⎟
⎜ MAy ⎟
⎜ MAz ⎟
⎝
⎠
TBC = 1.05 kip
⎛ Ax ⎞ ⎛ −450 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 900 ⎟ lb
⎜A ⎟ ⎝ 0 ⎠
⎝ z⎠
⎛ MAy ⎞ ⎛ −600 ⎞
⎜
⎟=⎜
⎟ lb⋅ ft
⎝ MAz ⎠ ⎝ −900 ⎠
Problem 5-84
Determine the x, y, z components of reaction at the pin A and the tension in the cable BC
necessary for equilibrium of the rod.
418
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Engineering Mechanics - Statics
Chapter 5
Given:
F = 350 lb
e = 12 ft
a = 4 ft
f = 4 ft
b = 5 ft
g = 10 ft
c = 4 ft
h = 4 ft
d = 2 ft
i = 10 ft
Solution:
Initial Guesses:
F BC = 1 lb
Ay = 1 lb
MAy = 1 lb⋅ ft
Ax = 1 lb
Az = 1 lb
MAz = 1 lb⋅ ft
Given
⎛ 0 ⎞ ⎛d⎞ ⎡
⎛ g − d ⎞⎤
⎜
⎟ ⎜ ⎟ ⎢
F
⎜ e − c ⎟⎥ ... = 0
⎜ MAy ⎟ + ⎜ c ⎟ × ⎢
⎟⎥
2
2
2⎜
⎜ MAz ⎟ ⎝ 0 ⎠ ⎣ ( g − d) + ( e − c) + f ⎝ − f ⎠⎦
⎝
⎠
⎛ −a ⎞ ⎡
⎛ a − h ⎞⎤
F BC
⎜
⎟
⎢
⎜ −e ⎟⎥
+ 0 ×
⎜ ⎟ ⎢
⎟⎥
2
2
2⎜
⎝ b ⎠ ⎣ ( a − h) + e + b ⎝ b ⎠⎦
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ +
⎜A ⎟
⎝ z⎠
⎛g − d⎞
⎜e− c⎟ +
⎟
2
2
2⎜
( g − d) + ( e − c) + f ⎝ − f ⎠
F
⎛a − h⎞
⎜ −e ⎟ = 0
⎟
2
2
2⎜
( a − h) + e + b ⎝ b ⎠
F BC
419
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Engineering Mechanics - Statics
Chapter 5
⎛⎜ FBC ⎞⎟
⎜ Ax ⎟
⎜
⎟
⎜ Ay ⎟ = Find ( F , A , A , A , M , M )
BC x y z
Ay
Az
⎜ Az ⎟
⎜
⎟
⎜ MAy ⎟
⎜ MAz ⎟
⎝
⎠
F BC = 101 lb
⎛ Ax ⎞ ⎛ −233.3 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ −140 ⎟ lb
⎜ A ⎟ ⎝ 77.8 ⎠
⎝ z⎠
⎛ MAy ⎞ ⎛ −388.9 ⎞
⎜
⎟=⎜
⎟ lb⋅ ft
⎝ MAz ⎠ ⎝ 93.3 ⎠
Problem 5-85
Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z
components of reaction at these supports if the rod is subjected to a vertical force F as shown.
Given:
F = 50 lb
a = 2 ft
c = 2 ft
b = 4 ft
d = 2 ft
Solution:
TB = 10 lb
Ax = 10 lb
Ay = 10 lb
Az = 10 lb
B y = 10 lb
Given
ΣF x = 0;
− TB + A x = 0
ΣF y = 0;
Ay + By = 0
420
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Engineering Mechanics - Statics
ΣF z = 0;
−F + A z = 0
ΣMAx = 0;
F ( c) − By( b) = 0
ΣMAy = 0;
F ( a) − TB( b) = 0
ΣMAz = 0;
B y( a) − TB( c) = 0
Chapter 5
Solving,
⎛ TB ⎞
⎜ ⎟
⎜ Ax ⎟
⎜ Ay ⎟ = Find ( T , A , A , A , B )
B x y z y
⎜ ⎟
⎜ Az ⎟
⎜B ⎟
⎝ y⎠
⎛ TB ⎞ ⎛ 25 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ax ⎟ ⎜ 25 ⎟
⎜ Ay ⎟ = ⎜ −25 ⎟ lb
⎜ ⎟ ⎜
⎟
⎜ Az ⎟ ⎜ 50 ⎟
⎜ B ⎟ ⎝ 25 ⎠
⎝ y⎠
Problem 5-86
The member is supported by a square rod which fits loosely through a smooth square hole of
the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at
these supports when the member is subjected to the loading shown.
Given:
M = 50 lb⋅ ft
⎛ 20 ⎞
⎜
⎟
F = −40 lb
⎜
⎟
⎝ −30 ⎠
a = 2 ft
b = 1 ft
c = 2 ft
Solution:
Initial Guesses
Ax = 1 lb
Ay = 1 lb
421
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Engineering Mechanics - Statics
Chapter 5
MAx = 1 lb ft
MAy = 1 lb ft
MAz = 1 lb ft
B z = 1 lb
Given
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜ ⎟
⎜ Ay ⎟ + ⎜ 0 ⎟ + F = 0
⎜ 0 ⎟ ⎜ Bz ⎟
⎝ ⎠ ⎝ ⎠
⎛ MAx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎡⎛ 0 ⎞
⎛ 0 ⎞⎤
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎢⎜
⎟
⎜ ⎟⎥
⎜ MAy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ + ⎢⎜ a + b ⎟ × F + ⎜ 0 ⎟⎥ = 0
⎜ M ⎟ ⎝ 0 ⎠ ⎜ Bz ⎟ ⎣⎝ −c ⎠
⎝ −M ⎠⎦
⎝ ⎠
⎝ Az ⎠
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜
⎟
⎜ MAx ⎟ = Find ( A , A , M , M , M , B )
x y
Ax
Ay
Az z
⎜ MAy ⎟
⎜
⎟
⎜ MAz ⎟
⎜ Bz ⎟
⎝
⎠
⎛ Ax ⎞ ⎛ −20 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ Ay ⎠ ⎝ 40 ⎠
B z = 30 lb
⎛ MAx ⎞ ⎛ 110 ⎞
⎜
⎟ ⎜
⎟
⎜ MAy ⎟ = ⎜ 40 ⎟ lb⋅ ft
⎜ M ⎟ ⎝ 110 ⎠
⎝ Az ⎠
Problem 5-87
The platform has mass M and center of mass located at G. If it is lifted using the three cables,
determine the force in each of these cables.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
g = 9.81
m
2
s
422
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Engineering Mechanics - Statics
Chapter 5
Given:
M = 3 Mg
a = 4m
b = 3m
c = 3m
d = 4m
e = 2m
Solution:
The initial guesses are:
F AC = 10 N
F BC = 10 N
F DE = 10 N
Given
b ( F AC)
2
2
a +b
−
c ( F BC)
2
=0
2
a +c
M g e − ( F AC) a
d+e
2
2
− FBC
a +b
a
2
2
a( d + e)
2
=0
2
a +c
F BC( b + c) − M g b + FDE b = 0
a +c
⎛ FAC ⎞
⎜
⎟
F
⎜ BC ⎟ = Find ( FAC , FBC , FDE)
⎜F ⎟
⎝ DE ⎠
⎛ FAC ⎞
⎜
⎟
F
⎜ BC ⎟ = kN
⎜F ⎟
⎝ DE ⎠
423
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Engineering Mechanics - Statics
Chapter 5
Problem 5-88
The platform has a mass of M and center of mass located at G. If it is lifted using the three
cables, determine the force in each of the cables. Solve for each force by using a single
moment equation of equilibrium.
Units Used:
Mg = 1000 kg
3
kN = 10 N
g = 9.81
m
2
s
Given:
M = 2 Mg
c = 3m
a = 4m
d = 4m
b = 3m
e = 2m
Solution:
⎛0⎞
rBC = ⎜ −c ⎟
⎜ ⎟
⎝a⎠
⎛0⎞
rAC = ⎜ b ⎟
⎜ ⎟
⎝a⎠
⎛ −e − d ⎞
rAD = ⎜ b ⎟
⎜
⎟
⎝ 0 ⎠
⎛ −d − e ⎞
rBD = ⎜ −c ⎟
⎜
⎟
⎝ 0 ⎠
First find FDE.
ΣMy' = 0;
F DE( d + e) − M g d = 0
Next find FBC.
Guess
F DE =
Mgd
d+e
2
F DE = 4.1 s kN
F BC = 1 kN
⎡⎛ e ⎞ ⎛ 0 ⎞ ⎛ e + d ⎞
⎤
rBC ⎞⎥
⎛
⎢
⎜
⎟
⎜
⎟
⎜
⎟
0 ×
0
Given
+
c
× F
r
=0
⎢⎜ ⎟ ⎜
⎟ ⎜
⎟ ⎜⎝ BC rBC ⎟⎠⎥ AD
⎣⎝ 0 ⎠ ⎝ −M g ⎠ ⎝ 0 ⎠
⎦
F BC = Find ( FBC )
F BC = kN
Now find FAC.
Guess
F AC = 1 kN
424
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Engineering Mechanics - Statics
Chapter 5
⎡⎛ e ⎞ ⎛ 0 ⎞ ⎛ e + d ⎞
⎤
rAC ⎞⎥
⎛
⎢
⎜
⎟
⎜
⎟
⎜
⎟
0 ×
0
Given
+ −b × ⎜ FAC
⎟ rBD = 0
⎢⎜ ⎟ ⎜
⎟ ⎜
⎟ ⎝
rAC ⎠⎥
⎣⎝ 0 ⎠ ⎝ −M g ⎠ ⎝ 0 ⎠
⎦
F AC = Find ( FAC )
F AC = kN
Problem 5-89
The cables exert the forces shown on the pole. Assuming the pole is supported by a
ball-and-socket joint at its base, determine the components of reaction at A. The forces F 1 and F2
lie in a horizontal plane.
Given:
F 1 = 140 lb
F 2 = 75 lb
θ = 30 deg
a = 5 ft
b = 10 ft
c = 15 ft
Solution:
The initial guesses are
TBC = 100 lb
TBD = 100 lb
Ax = 100 lb
Ay = 100 lb
Az = 100 lb
Given
c
⎞ − T a⎛
⎞=0
BD ⎜
⎟
⎟
2
2
2
2
2
⎝ a +b +c ⎠
⎝ a +c ⎠
(F1 cos (θ ) + F2)c − TBC a⎛⎜
c
c
⎞
⎜ 2 2 2⎟ = 0
⎝ a +b +c ⎠
F 1 sin ( θ ) c − b TBC ⎛
425
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛
Chapter 5
⎞
⎟=0
2
2
2
a
+
b
+
c
⎝
⎠
Ax + F 1 sin ( θ ) − b⎜
TBC
TBC
⎞
⎞ + a⎛
=0
⎜
⎟
⎟
2
2
2
2
2
+
c
a
+
b
+
c
a
⎝
⎠ ⎝
⎠
⎛
Ay − F 1 cos ( θ ) − F 2 + TBD⎜
a
TBC
⎛ TBD ⎞ ⎛
⎞
−
c
=0
⎟
⎜
⎟
2
2
2
2
2
a
+
c
a
+
b
+
c
⎝
⎠ ⎝
⎠
Az − c⎜
⎛ TBC ⎞
⎜
⎟
⎜ TBD ⎟
⎜ Ax ⎟ = Find ( T , T , A , A , A )
BC BD x y z
⎜
⎟
⎜ Ay ⎟
⎜ A ⎟
⎝ z ⎠
⎛ TBC ⎞ ⎛ 131.0 ⎞
⎜
⎟=⎜
⎟ lb
⎝ TBD ⎠ ⎝ 509.9 ⎠
⎛ Ax ⎞ ⎛ −0.0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0.0 ⎟ lb
⎜ A ⎟ ⎝ 588.7 ⎠
⎝ z⎠
Problem 5-90
The silo has a weight W, a center of gravity at G and a radius r. Determine the vertical
component of force that each of the three struts at A, B, and C exerts on the silo if it is
subjected to a resultant wind loading of F which acts in the direction shown.
Given:
W = 3500 lb
F = 250 lb
θ 1 = 30 deg
θ 2 = 120 deg
θ 3 = 30 deg
r = 5 ft
b = 12 ft
c = 15 ft
426
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Solution:
Initial Guesses:
Az = 1 lb
B z = 2 lb
Cz = 31 lb
Given
ΣMy = 0; B z r cos ( θ 1 ) − Cz r cos ( θ 1 ) − F sin ( θ 3 ) c = 0
[1]
ΣMx = 0; −B z r sin ( θ 1 ) − Cz r sin ( θ 1 ) + Az r − F cos ( θ 3 ) c = 0
[2]
ΣF z = 0; Az + Bz + Cz = W
[3]
Solving Eqs.[1], [2] and [3] yields:
⎛ Az ⎞
⎜ ⎟
⎜ Bz ⎟ = Find ( Az , Bz , Cz)
⎜C ⎟
⎝ z⎠
⎛ Az ⎞ ⎛ 1600 ⎞
⎜ ⎟ ⎜
⎟
⎜ Bz ⎟ = ⎜ 1167 ⎟ lb
⎜ C ⎟ ⎝ 734 ⎠
⎝ z⎠
Problem 5-91
The shaft assembly is supported by two
smooth journal bearings A and B and a
short link DC. If a couple moment is
applied to the shaft as shown, determine
the components of force reaction at the
bearings and the force in the link. The
link lies in a plane parallel to the y-z plane
and the bearings are properly aligned on
the shaft.
Units Used:
3
kN = 10 N
Given:
M = 250 N⋅ m
a = 400 mm
b = 300 mm
c = 250 mm
d = 120 mm
θ = 30 deg
427
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
φ = 20 deg
Solution:
Initial Guesses:
Ay = 1 kN
Az = 1 kN
B z = 1 kN
F CD = 1 kN
B y = 1 kN
Given
0
⎞
⎛0 ⎞ ⎛0⎞ ⎛
⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ + ⎜ By ⎟ + ⎜ −FCD cos ( φ ) ⎟ = 0
⎜ Az ⎟ ⎜ Bz ⎟ ⎜ F sin ( φ ) ⎟
⎝ ⎠ ⎝ ⎠ ⎝ CD
⎠
0
⎞ ⎛ −a − b ⎞ ⎛ 0 ⎞ ⎛ −M ⎞
⎛ d − a ⎞ ⎛⎜
⎜ c sin ( θ ) ⎟ × −FCD cos ( φ ) ⎟ + ⎜ 0 ⎟ × ⎜ By ⎟ + ⎜ 0 ⎟ = 0
⎟ ⎜
⎜
⎟ ⎜
⎟ ⎜ ⎟ ⎜ ⎟
⎜
⎟
⎝ c cos ( θ ) ⎠ ⎝ FCD sin ( φ ) ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎝ 0 ⎠
⎛ Ay ⎞
⎜
⎟
A
z
⎜
⎟
⎜ By ⎟ = Find ( A , A , B , B , F )
y z y z CD
⎜
⎟
⎜ Bz ⎟
⎜F ⎟
⎝ CD ⎠
⎛ Ay ⎞ ⎛ 573 ⎞
⎜ ⎟=⎜
⎟N
⎝ Az ⎠ ⎝ −208 ⎠
⎛ By ⎞ ⎛ 382 ⎞
⎜ ⎟=⎜
⎟N
⎝ Bz ⎠ ⎝ −139 ⎠
F CD = 1.015 kN
Problem 5-92
If neither the pin at A nor the roller at B can support a load no greater than F max, determine the
maximum intensity of the distributed load w, so that failure of a support does not occur.
428
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Engineering Mechanics - Statics
Chapter 5
Units Used:
3
kN = 10 N
Given:
F max = 6 kN
a = 3m
b = 3m
Solution:
The greatest reaction is at A. Require
ΣMB = 0;
−F max( a + b) + w a ⎛⎜
a
⎝2
w =
F max( a + b)
a⎛⎜
a
⎝2
+ b⎞⎟ +
⎠
2
+ b⎞⎟ +
⎠
1
2
wb b=0
2
3
w = 2.18
b
kN
m
3
Problem 5-93
If the maximum intensity of the distributed load acting on the beam is w, determine the reactions
at the pin A and roller B.
Units Used:
3
kN = 10 N
Given:
F = 6 kN
a = 3m
b = 3m
429
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
w = 4
Chapter 5
kN
m
Solution:
ΣF x = 0;
Ax = 0
ΣMA = 0;
−w a
a
By =
1
2
6
−
1
2
w b⎛⎜ a +
⎝
⎟ + By( a + b) = 0
3⎠
2
w
3a + 3 a b + b
Ay + By − w a −
ΣF y = 0;
b⎞
2
B y = 7 kN
a+b
1
2
Ay = −By + w a +
wb=0
1
2
Ay = 11 kN
wb
Problem 5-94
Determine the normal reaction at the roller A and horizontal and vertical components at pin B for
equilibrium of the member.
Units Used:
3
kN = 10 N
Given:
F 1 = 10 kN
F 2 = 6 kN
a = 0.6 m
b = 0.6 m
c = 0.8 m
d = 0.4 m
θ = 60 deg
430
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Solution:
Initial Guesses:
NA = 1 kN
B x = 1 kN
B y = 1 kN
Given
B x − F2 sin ( θ ) = 0
B y + NA − F 1 − F 2 cos ( θ ) = 0
F 2 d + F1 ⎡⎣b + ( c + d)cos ( θ )⎤⎦ − NA⎡⎣a + b + ( c + d)cos ( θ )⎤⎦ = 0
⎛ NA ⎞
⎜ ⎟
⎜ Bx ⎟ = Find ( NA , Bx , By)
⎜B ⎟
⎝ y⎠
⎛ NA ⎞ ⎛ 8 ⎞
⎜ ⎟ ⎜
⎟
⎜ Bx ⎟ = ⎜ 5.196 ⎟ kN
⎜B ⎟ ⎝ 5 ⎠
⎝ y⎠
Problem 5-95
The symmetrical shelf is subjected to uniform pressure P. Support is provided by a bolt (or pin)
located at each end A and A' and by the symmetrical brace arms, which bear against the smooth
wall on both sides at B and B'. Determine the force resisted by each bolt at the wall and the
normal force at B for equilibrium.
Units Used:
3
kPa = 10 Pa
Given:
P = 4 kPa
a = 0.15 m
b = 0.2 m
c = 1.5 m
Solution:
ΣMA = 0;
431
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Engineering Mechanics - Statics
NB a − P ⎛⎜ b
Chapter 5
c⎞ b
⎟ =0
⎝ 2⎠ 2
2
NB = P
b c
NB = 400 N
4a
ΣF x = 0;
Ax = NB
Ax = 400 N
ΣF y = 0;
Ay = P b
c
FA =
2
Ay = 600 N
2
Ax + Ay
2
F A = 721 N
Problem 5-96
A uniform beam having a weight W
supports a vertical load F . If the ground
pressure varies linearly as shown,
determine the load intensities w1 and w2
measured in lb/ft, necessary for
equilibrium.
Given:
W = 200 lb
F = 800 lb
a = 7 ft
b = 6 ft
Solution:
Initial Guesses:
w1 = 1
lb
ft
w2 = 1
lb
ft
432
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
Given
1
w1 ( a + b) +
w1 ( a + b)
2
(w2 − w1)( a + b) − F − W = 0
a+b
2
+
(w2 − w1)( a + b) 3 ( a + b) − W
2
1
⎛ w1 ⎞
⎜ ⎟ = Find ( w1 , w2)
⎝ w2 ⎠
2
a+b
2
− Fa = 0
⎛ w1 ⎞ ⎛ 62.7 ⎞ lb
⎜ ⎟=⎜
⎟
⎝ w2 ⎠ ⎝ 91.1 ⎠ ft
Problem 5-97
The uniform ladder rests along the wall of a building at A and on the roof at B. If the ladder has
a weight W and the surfaces at A and B are assumed smooth, determine the angle θ for
equilibrium.
Given:
a = 18 ft
W = 25 lb
θ 1 = 40 deg
Solution:
Initial guesses:
R A = 10 lb
R B = 10 lb
θ = 10 deg
Given
ΣMB = 0;
−R A a sin ( θ ) + W
ΣF x = 0;
R A − RB sin ( θ 1 ) = 0
ΣF y = 0;
R B cos ( θ 1 ) − W = 0
a
2
cos ( θ ) = 0
Solving,
433
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
⎛ RB ⎞
⎜ ⎟
⎜ RA ⎟ = Find ( RB , RA , θ )
⎜ ⎟
⎝θ ⎠
⎛ RA ⎞ ⎛ 21 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ RB ⎠ ⎝ 32.6 ⎠
θ = 30.8 deg
Problem 5-98
Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket
A (not shown) for the uniformly loaded plate.
Given:
P = 2
lb
ft
2
a = 4 ft
b = 1 ft
c = 2 ft
d = 2 ft
Solution:
The initial guesses are
Ax = 1 lb
Ay = 1 lb
Az = 1 lb
B z = 1 lb
Cz = 1 lb
Given
ΣF x = 0;
Ax = 0
ΣF y = 0;
Ay = 0
ΣF z = 0; Az + Bz + Cz − P a c = 0
c⎞
⎟ + Cz b = 0
⎝ 2⎠
ΣMx = 0; c B z − P a c⎛⎜
a⎞
⎟ − Cz a = 0
⎝ 2⎠
ΣMy = 0; −B z ( a − d) + P a c⎛⎜
434
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟
⎜ Az ⎟ = Find ( A , A , A , B , C )
x y z z z
⎜ ⎟
⎜ Bz ⎟
⎜C ⎟
⎝ z⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0 ⎟ lb
⎜ A ⎟ ⎝ 5.333 ⎠
⎝ z⎠
⎛ Bz ⎞ ⎛ 5.333 ⎞
⎜ ⎟=⎜
⎟ lb
5.333
C
⎝
⎠
z
⎝ ⎠
Problem 5-99
A vertical force F acts on the crankshaft. Determine the horizontal equilibrium force P that must
be applied to the handle and the x, y, z components of force at the smooth journal bearing A and
the thrust bearing B. The bearings are properly aligned and exert the force reactions on the
shaft.
Given:
F = 80 lb
a = 10 in
b = 14 in
c = 14 in
d = 8 in
e = 6 in
f = 4 in
Solution:
ΣMy = 0;
P d−F a= 0
⎛ a⎞
⎟
⎝ d⎠
P = F⎜
ΣMx = 0;
B z( b + c) − F c = 0
⎛ c ⎞
⎟
⎝ b + c⎠
Bz = F ⎜
ΣMz = 0;
P = 100 lb
B z = 40 lb
−B x( b + c) − P ( e + f) = 0
435
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
B x = −P ⎛⎜
Chapter 5
e + f⎞
⎟ Bx = −35.7 lb
⎝ b + c⎠
ΣF x = 0;
Ax + Bx − P = 0
Ax = −Bx + P
Ax = 135.7 lb
By = 0
ΣF y = 0;
ΣF z = 0;
Az + Bz − F = 0
Az = −B z + F
Az = 40 lb
Problem 5-100
The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is kA ,
determine the required stiffness of the spring at B so that if the beam is loaded with the force F , it
remains in the horizontal position both before and after loading.
Units Used:
3
kN = 10 N
Given:
kA = 5
kN
m
F = 800 N
a = 1m
b = 2m
Solution:
Equilibrium:
ΣMA = 0;
F B ( a + b) − F a = 0
⎞
⎟
⎝ a + b⎠
F B = F ⎛⎜
a
F B = 266.667 N
ΣMB = 0;
F b − FA( a + b) = 0
436
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 5
F A = F ⎛⎜
⎞
⎟
⎝ a + b⎠
b
F A = 533.333 N
Spring force formula:
xA = xB
FA
kA
=
FB
kB
kB =
FB
FA
kA
kB = 2.5
kN
m
437
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-1
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 7 kN
P 2 = 7 kN
Solution:
θ = 45 deg
Initial Guesses:
F AB = 1 kN
F AD = 1 kN
F DC = 1 kN
F CB = 1 kN
F DB = 1 kN
Given
Joint A:
F AB + F AD cos ( θ ) = 0
−P 1 − F AD sin ( θ ) = 0
Joint D:
F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0
(FAD + FDB − FDC)sin(θ ) − P2 = 0
Joint C:
F CB + FDC sin ( θ ) = 0
⎛⎜ FAB ⎞⎟
⎜ FAD ⎟
⎜
⎟
⎜ FDB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB)
⎜F ⎟
⎜ DC ⎟
⎜ FCB ⎟
⎝
⎠
⎛⎜ FAB ⎞⎟
⎛ 7 ⎞
⎜ FAD ⎟ ⎜ −9.9 ⎟
⎟
⎜
⎟ ⎜
⎜ FDB ⎟ = ⎜ 4.95 ⎟ kN
⎜ F ⎟ ⎜ −14.85 ⎟
⎜ DC ⎟ ⎜
⎟
10.5
⎝
⎠
⎜ FCB ⎟
⎝
⎠
Positive
means
Tension,
Negative
means
Compression
438
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-2
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 8 kN
P 2 = 10 kN
Solution:
θ = 45 deg
Initial Guesses:
F AB = 1 kN
F AD = 1 kN
F DC = 1 kN
F CB = 1 kN
F DB = 1 kN
Given
Joint A:
F AB + F AD cos ( θ ) = 0
−P 1 − F AD sin ( θ ) = 0
Joint D:
F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0
(FAD + FDB − FDC)sin(θ ) − P2 = 0
Joint C:
F CB + FDC sin ( θ ) = 0
⎛⎜ FAB ⎞⎟
⎜ FAD ⎟
⎜
⎟
F
⎜ DB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB)
⎜F ⎟
⎜ DC ⎟
⎜ FCB ⎟
⎝
⎠
⎛⎜ FAB ⎞⎟
⎛ 8 ⎞
⎜ FAD ⎟ ⎜ −11.31 ⎟
⎟
⎜
⎟ ⎜
F
7.07
=
⎟ kN
⎜ DB ⎟ ⎜
⎜ F ⎟ ⎜ −18.38 ⎟
⎜ DC ⎟ ⎜
⎟
⎜ FCB ⎟ ⎝ 13 ⎠
⎝
⎠
Positive
means
Tension,
Negative
means
Compression
439
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-3
The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint
as a pin and determine the force in each member. State whether the members are in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
P 1 = 600 lb
P 2 = 400 lb
a = 4 ft
θ = 45 deg
Solution:
Initial Guesses
F AB = 1 lb
F AD = 1 lb
F DC = 1 lb
F BC = 1 lb
F BD = 1 lb
F DE = 1 lb
Given
Joint A:
F AB + F AD cos ( θ ) = 0
−P 1 − F AD sin ( θ ) = 0
Joint B:
F BC − F AB = 0
−P 2 − F BD = 0
Joint D:
(FDC − FAD)cos (θ ) + FDE = 0
(FDC + FAD)sin(θ ) + FBD = 0
⎛ FAB ⎞
⎜
⎟
⎜ FAD ⎟
⎜F ⎟
⎜ BC ⎟ = Find F , F , F , F , F , F
( AB AD BC BD DC DE)
⎜ FBD ⎟
⎜
⎟
⎜ FDC ⎟
⎜
⎟
⎝ FDE ⎠
440
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛ FAB ⎞
⎜
⎟ ⎛ 600 ⎞
F
AD
⎜
⎟ ⎜ −849 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ BC ⎟ = 600 ⎟ lb
⎜ FBD ⎟ ⎜ −400 ⎟
⎟
⎜
⎟ ⎜
1414
⎜
⎟
⎜ FDC ⎟
⎜
⎜
⎟ ⎝ −1600 ⎟⎠
F
⎝ DE ⎠
Chapter 6
Positive means Tension,
Negative means Compression
Problem 6-4
The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint
as a pin and determine the force in each member. State whether the members are in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
P 1 = 800 lb
P 2 = 0 lb
a = 4 ft
θ = 45 deg
Solution:
Initial Guesses
F AB = 1 lb
F AD = 1 lb
F DC = 1 lb
F BC = 1 lb
F BD = 1 lb
F DE = 1 lb
Given
Joint A:
F AB + F AD cos ( θ ) = 0
−P 1 − F AD sin ( θ ) = 0
Joint B:
F BC − F AB = 0
−P 2 − F BD = 0
441
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Joint D:
Chapter 6
(FDC − FAD)cos (θ ) + FDE = 0
(FDC + FAD)sin(θ ) + FBD = 0
⎛ FAB ⎞
⎜
⎟
F
⎜ AD ⎟
⎜F ⎟
⎜ BC ⎟ = Find F , F , F , F , F , F
( AB AD BC BD DC DE)
⎜ FBD ⎟
⎜
⎟
F
⎜ DC ⎟
⎜
⎟
⎝ FDE ⎠
⎛ FAB ⎞
⎜
⎟ ⎛ 800 ⎞
⎜ FAD ⎟ ⎜ −1131 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ BC ⎟ = 800 ⎟ lb
⎜ FBD ⎟ ⎜ 0 ⎟
⎟
⎜
⎟ ⎜
⎜ FDC ⎟ ⎜ 1131 ⎟
⎜
⎟ ⎜⎝ −1600 ⎟⎠
F
⎝ DE ⎠
Positive means Tension,
Negative means Compression
Problem 6-5
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 20 kN
P 2 = 10 kN
a = 1.5 m
e = 2m
Solution:
e
θ = atan ⎛⎜ ⎟⎞
⎝ a⎠
442
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Initial Guesses:
F AB = 1 kN F AG = 1 kN F CF = 1 kN
F BC = 1 kN F BG = 1 kN F DE = 1 kN
F CG = 1 kN F FG = 1 kN F EF = 1 kN
F CD = 1 kN F DF = 1 kN
Given
Joint B
F BC − F AB cos ( θ ) = 0
−F BG − F AB sin ( θ ) = 0
Joint G
F FG + F CG cos ( θ ) − F AG = 0
F CG sin ( θ ) + FBG − P1 = 0
Joint C
(
)
−F BC + F CD + FCF − F CG cos ( θ ) = 0
(
)
− FCG + FCF sin ( θ ) = 0
Joint D
−F CD + F DE cos ( θ ) = 0
−F DF − F DE sin ( θ ) = 0
Joint F
F EF − F FG − F CF cos ( θ ) = 0
F DF + F CF sin ( θ ) − P 2 = 0
Joint E
−F DE cos ( θ ) − F EF = 0
443
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCG ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FAG ⎟
⎜
⎟
⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF)
⎜F ⎟
⎜ FG ⎟
⎜ FDF ⎟
⎜
⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎝ EF ⎠
⎛ FAB ⎞
⎜
⎟ ⎛⎜ −21.88 ⎟⎞
⎜ FBC ⎟ ⎜ −13.13 ⎟
kN
⎜
⎟=⎜
⎟
F
3.13
CG
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎝ −9.37 ⎠
⎝ CD ⎠
⎛ FAG ⎞
⎜
⎟ ⎛⎜ 13.13 ⎞⎟
⎜ FBG ⎟ ⎜ 17.5 ⎟
kN
⎜
⎟=⎜
⎟
F
11.25
FG
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎝ 12.5 ⎠
⎝ DF ⎠
⎛ FCF ⎞ ⎛ −3.13 ⎞
⎜
⎟ ⎜
⎟
F
⎜ DE ⎟ = ⎜ −15.62 ⎟ kN
⎜ F ⎟ ⎝ 9.37 ⎠
⎝ EF ⎠
Positive means Tension, Negative means Compression
Problem 6-6
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 40 kN
P 2 = 20 kN
a = 1.5 m
e = 2m
444
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
e
θ = atan ⎛⎜ ⎞⎟
Solution:
⎝ a⎠
Initial Guesses:
F AB = 1 kN F AG = 1 kN F CF = 1 kN
F BC = 1 kN F BG = 1 kN F DE = 1 kN
F CG = 1 kN F FG = 1 kN F EF = 1 kN
F CD = 1 kN F DF = 1 kN
Given
Joint B
F BC − F AB cos ( θ ) = 0
−F BG − F AB sin ( θ ) = 0
Joint G
F FG + F CG cos ( θ ) − F AG = 0
F CG sin ( θ ) + FBG − P1 = 0
Joint C
(
)
−F BC + F CD + FCF − F CG cos ( θ ) = 0
(
)
− FCG + FCF sin ( θ ) = 0
Joint D
−F CD + F DE cos ( θ ) = 0
−F DF − F DE sin ( θ ) = 0
Joint F
F EF − F FG − F CF cos ( θ ) = 0
F DF + F CF sin ( θ ) − P 2 = 0
Joint E
−F DE cos ( θ ) − F EF = 0
445
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCG ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FAG ⎟
⎜
⎟
⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF)
⎜F ⎟
⎜ FG ⎟
⎜ FDF ⎟
⎜
⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎝ EF ⎠
⎛ FAB ⎞
⎜
⎟ ⎛⎜ −43.75 ⎟⎞
F
⎜ BC ⎟ ⎜ −26.25 ⎟
kN
⎜
⎟=⎜
⎟
⎜ FCG ⎟ ⎜ 6.25 ⎟
⎜ F ⎟ ⎝ −18.75 ⎠
⎝ CD ⎠
⎛ FAG ⎞
⎜
⎟ ⎛⎜ 26.25 ⎞⎟
F
⎜ BG ⎟ ⎜ 35 ⎟
kN
⎜
⎟=⎜
⎟
⎜ FFG ⎟ ⎜ 22.5 ⎟
⎜ F ⎟ ⎝ 25 ⎠
⎝ DF ⎠
⎛ FCF ⎞ ⎛ −6.25 ⎞
⎜
⎟ ⎜
⎟
⎜ FDE ⎟ = ⎜ −31.25 ⎟ kN
⎜ F ⎟ ⎝ 18.75 ⎠
⎝ EF ⎠
Positive means Tension, Negative means Compression
Problem 6-7
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 3 kN
F 2 = 8 kN
F 3 = 4 kN
F 4 = 10 kN
446
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
a = 2m
b = 1.5 m
b
θ = atan ⎛⎜ ⎟⎞
Solution:
⎝ a⎠
Initial Guesses
F BA = 1 kN
F BC = 1 kN
F AF = 1 kN
F CD = 1 kN
F DF = 1 kN
F ED = 1 kN
F AC = 1 kN
F CF = 1 kN
F EF = 1 kN
Given
Joint B
F 1 + F BC = 0
−F 2 − F BA = 0
Joint C
F CD − F BC − F AC cos ( θ ) = 0
−F 3 − F AC sin ( θ ) − FCF = 0
Joint E
−F EF = 0
Joint D
−F CD − F DF cos ( θ ) = 0
−F 4 − F DF sin ( θ ) − FED = 0
Joint F
−F AF + F EF + F DF cos ( θ ) = 0
F CF + FDF sin ( θ ) = 0
⎛ FBA ⎞
⎜
⎟
⎜ FAF ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCD ⎟ = Find ( FBA , FAF , FDF , FBC , FCD , FED , FAC , FCF , FEF)
⎜F ⎟
⎜ ED ⎟
⎜ FAC ⎟
⎜
⎟
⎜ FCF ⎟
⎟
⎜
⎝ FEF ⎠
447
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FBA ⎞
⎜
⎟ ⎛ −8 ⎞
⎜ FAF ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 4.167 ⎟
⎜ DF ⎟ ⎜ 5.208 ⎟
⎜ FBC ⎟ ⎜ −3 ⎟
⎟
⎜
⎟ ⎜
F
−
4.167
=
⎟ kN
⎜ CD ⎟ ⎜
⎜ F ⎟ ⎜ −13.125 ⎟
⎟
⎜ ED ⎟ ⎜
⎜ FAC ⎟ ⎜ −1.458 ⎟
⎜
⎟ ⎜ −3.125 ⎟
F
⎜ CF ⎟ ⎜
⎟
⎟ ⎝ 0 ⎠
⎜
⎝ FEF ⎠
Positive means tension,
Negative means compression.
Problem 6-8
Determine the force in each member of the truss in terms of the external loading and state if the
members are in tension or compression.
Solution:
ΣMA = 0;
−P a + Cy2a − P a = 0
Cy = P
Joint C:
ΣF x = 0;
1
ΣF y = 0;
P+
2
F BC −
1
17
4
17
F CD −
FCD = 0
1
2
F BC = 0
448
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F BC =
4 2P
F CD =
17 P
1
FCD +
3
3
= 1.886 P (C)
= 1.374 P (T)
Joint B:
ΣF x = 0;
P−
ΣFy = 0;
1
2
2
F CD +
2
2P
F AB =
F BD =
1
3
5P
1
2
FAB = 0
F AB − F BD = 0
= 0.471P
( C)
= 1.667P
( T)
F DA = F CD = 1.374P
(T)
3
Joint D:
ΣF x = 0;
Problem 6-9
The maximum allowable tensile force in the members of the truss is Tmax, and the maximum
allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads
that can be applied to the truss.
Given:
Tmax = 1500 lb
Cmax = 800 lb
Solution:
Set
P = 1 lb
Initial Guesses
F AB = 1 lb
F AD = 1 lb
F BC = 1 lb
F CD = 1 lb
F BD = 1 lb
449
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
Joint B
(FBC − FAB) 2 + P = 0
1
(
−F BD − FAB + FBC
Joint D
) 2 =0
1
(FCD − FAD) 17 = 0
4
(
) 17 = 0
F BD − P − F AD + F CD
Joint C
−F BC
1
2
− F CD
4
1
=0
17
⎛⎜ FAB ⎞⎟
⎜ FBC ⎟
⎜
⎟
⎜ FAD ⎟ = Find ( FAB , FBC , FAD , FCD , FBD)
⎜F ⎟
⎜ CD ⎟
⎜ FBD ⎟
⎝
⎠
⎛⎜ FAB ⎞⎟
⎛ −0.471 ⎞
⎜ FBC ⎟ ⎜ −1.886 ⎟
⎜
⎟ ⎜
⎟
⎜ FAD ⎟ = ⎜ 1.374 ⎟ lb
⎜ F ⎟ ⎜ 1.374 ⎟
⎜ CD ⎟ ⎜
⎟
⎜ FBD ⎟ ⎝ 1.667 ⎠
⎝
⎠
Now find the critical load
P1 = P
Tmax
(
max F AB , F BC , F AD , F CD , F BD
P2 = P
)
P 1 = 900 lb
)
P 2 = 424.264 lb
Cmax
(
min F AB , F BC , F AD , F CD , F BD
(
)
P = min P 1 , P 2
P = 424.3 lb
Problem 6-10
Determine the force in each member of the truss and state if the members are in tension or
compression.
Given:
P 1 = 0 lb
450
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
P 2 = 1000 lb
a = 10 ft
b = 10 ft
Solution:
b
θ = atan ⎛⎜ ⎟⎞
⎝ a⎠
Initial Guesses:
F AB = 1 lb
F AG = 1 lb
F BG = 1 lb
F BC = 1 lb
F DC = 1 lb
F DE = 1 lb
F EG = 1 lb
F EC = 1 lb
F CG = 1 lb
Given
Joint B
F BC − F AB = 0
F BG − P 1 = 0
Joint G
(FCG − FAG)cos (θ ) + FEG = 0
(
)
− FCG + FAG sin ( θ ) − FBG = 0
Joint C
F DC − FBC − FCG cos ( θ ) = 0
F EC + F CG sin ( θ ) − P2 = 0
Joint E
F DE cos ( θ ) − F EG = 0
−F EC − F DE sin ( θ ) = 0
Joint D
−F DE cos ( θ ) − F DC = 0
451
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBC ⎟
⎜F ⎟
⎜ EG ⎟
⎜ FAG ⎟
⎜
⎟
F
⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG)
⎜F ⎟
⎜ EC ⎟
⎜ FBG ⎟
⎜
⎟
F
⎜ DE ⎟
⎟
⎜
⎝ FCG ⎠
⎛ FAB ⎞
⎜
⎟ ⎛ 333 ⎞
⎜ FBC ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 333 ⎟
⎜ EG ⎟ ⎜ −667 ⎟
⎜ FAG ⎟ ⎜ −471 ⎟
⎟
⎜
⎟ ⎜
⎜ FDC ⎟ = ⎜ 667 ⎟ lb
⎜ F ⎟ ⎜ 667 ⎟
⎟
⎜ EC ⎟ ⎜
⎜ FBG ⎟ ⎜ 0 ⎟
⎜
⎟ ⎜ −943 ⎟
⎜ FDE ⎟ ⎜
⎟
⎟ ⎝ 471 ⎠
⎜
⎝ FCG ⎠
Positive means tension,
Negative means compression.
Problem 6-11
Determine the force in each member of the truss and state if the members are in tension or
compression.
Given:
P 1 = 500 lb
P 2 = 1500 lb
452
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
a = 10 ft
b = 10 ft
Solution:
b
θ = atan ⎛⎜ ⎟⎞
⎝ a⎠
Initial Guesses:
F AB = 1 lb
F AG = 1 lb
F BG = 1 lb
F BC = 1 lb
F DC = 1 lb
F DE = 1 lb
F EG = 1 lb
F EC = 1 lb
F CG = 1 lb
Given
Joint B
F BC − F AB = 0
F BG − P 1 = 0
Joint G
(FCG − FAG)cos (θ ) + FEG = 0
(
)
− FCG + FAG sin ( θ ) − FBG = 0
Joint C
F DC − FBC − FCG cos ( θ ) = 0
F EC + F CG sin ( θ ) − P2 = 0
Joint E
F DE cos ( θ ) − F EG = 0
−F EC − F DE sin ( θ ) = 0
Joint D
−F DE cos ( θ ) − F DC = 0
453
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBC ⎟
⎜F ⎟
⎜ EG ⎟
⎜ FAG ⎟
⎜
⎟
F
⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG)
⎜F ⎟
⎜ EC ⎟
⎜ FBG ⎟
⎜
⎟
F
⎜ DE ⎟
⎟
⎜
⎝ FCG ⎠
⎛ FAB ⎞
⎜
⎟ ⎛ 833 ⎞
⎜ FBC ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 833 ⎟
⎜ EG ⎟ ⎜ −1167 ⎟
⎜ FAG ⎟ ⎜ −1179 ⎟
⎟
⎜
⎟ ⎜
F
1167
=
⎟ lb
⎜ DC ⎟ ⎜
⎜ F ⎟ ⎜ 1167 ⎟
⎟
⎜ EC ⎟ ⎜
⎜ FBG ⎟ ⎜ 500 ⎟
⎜
⎟ ⎜ −1650 ⎟
F
⎜ DE ⎟ ⎜
⎟
⎟ ⎝ 471 ⎠
⎜
⎝ FCG ⎠
Positive means tension,
Negative means compression.
Problem 6-12
Determine the force in each member of
the truss and state if the members are in
tension or compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 10 kN
P 2 = 15 kN
454
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
a = 2m
b = 4m
c = 4m
Solution:
c
α = atan ⎛⎜ ⎟⎞
⎝ a⎠
c
β = atan ⎛⎜ ⎟⎞
⎝ b⎠
Initial Guesses:
F AB = 1 kN
F AF = 1 kN
F GB = 1 kN
F BF = 1 kN
F FC = 1 kN
F FE = 1 kN
F BC = 1 kN
F EC = 1 kN
F CD = 1 kN
F ED = 1 kN
Given
Joint B
−F GB + F BC − F AB cos ( α ) = 0
−F AB sin ( α ) − FBF = 0
Joint F
−F AF + F FE + F FC cos ( β ) = 0
F BF + F FC sin ( β ) − P 1 = 0
Joint C
−F BC − F FC cos ( β ) + FCD cos ( α ) = 0
−F FC sin ( β ) − F CD sin ( α ) − FEC = 0
Joint E
−F FE + F ED = 0
F EC − P 2 = 0
Joint D
−F CD cos ( α ) − F ED = 0
F CD sin ( α ) = 0
455
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBF ⎟
⎜F ⎟
⎜ BC ⎟
⎜ FED ⎟
⎜
⎟
⎜ FAF ⎟
⎜
⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD)
F
FC
⎜
⎟
⎜F ⎟
⎜ EC ⎟
⎜ FGB ⎟
⎜
⎟
F
⎜ FE ⎟
⎜F ⎟
⎝ CD ⎠
⎛⎜ FAB ⎞⎟
⎛ −27.951 ⎞
⎜ FBF ⎟ ⎜ 25 ⎟
⎟
⎜
⎟ ⎜
⎜ FBC ⎟ = ⎜ 15 ⎟ kN
⎜F ⎟ ⎜ 0 ⎟
⎜ ED ⎟ ⎜
⎟
⎜ FAF ⎟ ⎝ −15 ⎠
⎝
⎠
⎛⎜ FFC ⎞⎟
⎛ −21.213 ⎞
⎜ FEC ⎟ ⎜ 15 ⎟
⎟
⎜
⎟ ⎜
⎜ FGB ⎟ = ⎜ 27.5 ⎟ kN Positive means Tension,
Negative means Compression
⎜F ⎟ ⎜ 0 ⎟
FE
⎜
⎟ ⎜
⎟
⎜ FCD ⎟ ⎝ 0 ⎠
⎝
⎠
Problem 6-13
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 0 kN
P 2 = 20 kN
a = 2m
b = 4m
c = 4m
456
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Solution:
c
α = atan ⎛⎜ ⎞⎟
⎝ a⎠
Chapter 6
c
β = atan ⎛⎜ ⎞⎟
⎝ b⎠
Initial Guesses:
F AB = 1 kN
F AF = 1 kN
F GB = 1 kN
F BF = 1 kN
F FC = 1 kN
F FE = 1 kN
F BC = 1 kN
F EC = 1 kN
F CD = 1 kN
F ED = 1 kN
Given
Joint B
−F GB + F BC − F AB cos ( α ) = 0
−F AB sin ( α ) − FBF = 0
Joint F
−F AF + F FE + F FC cos ( β ) = 0
F BF + F FC sin ( β ) − P 1 = 0
Joint C
−F BC − F FC cos ( β ) + FCD cos ( α ) = 0
−F FC sin ( β ) − F CD sin ( α ) − FEC = 0
Joint E
−F FE + F ED = 0
F EC − P 2 = 0
Joint D
−F CD cos ( α ) − F ED = 0
F CD sin ( α ) = 0
⎛ FAB ⎞
⎜
⎟
⎜ FBF ⎟
⎜F ⎟
⎜ BC ⎟
⎜ FED ⎟
⎜
⎟
⎜ FAF ⎟
⎜
⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD)
FFC
⎜
⎟
⎜F ⎟
⎜ EC ⎟
⎜ FGB ⎟
⎜
⎟
⎜ FFE ⎟
⎜F ⎟
⎝ CD ⎠
457
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ FAB ⎞⎟
⎛ −22.361 ⎞
⎜ FBF ⎟ ⎜ 20 ⎟
⎟
⎜
⎟ ⎜
⎜ FBC ⎟ = ⎜ 20 ⎟ kN
⎜F ⎟ ⎜ 0 ⎟
⎜ ED ⎟ ⎜
⎟
−
20
⎝
⎠
⎜ FAF ⎟
⎝
⎠
⎛⎜ FFC ⎞⎟
⎛ −28.284 ⎞
⎜ FEC ⎟ ⎜ 20 ⎟
⎟
⎜
⎟ ⎜
⎜ FGB ⎟ = ⎜ 30 ⎟ kN
⎜F ⎟ ⎜ 0 ⎟
⎜ FE ⎟ ⎜
⎟
0
⎝
⎠
⎜ FCD ⎟
⎝
⎠
Positive means Tension,
Negative means Compression
Problem 6-14
Determine the force in each member of the truss and state if the members are in tension or
compression.
Given:
P 1 = 100 lb
P 2 = 200 lb
P 3 = 300 lb
a = 10 ft
b = 10 ft
θ = 30 deg
Solution:
b
φ = atan ⎛⎜ ⎟⎞
⎝ a⎠
Initial Guesses:
F AB = 1 lb
F AF = 1 lb
F BC = 1 lb
F BF = 1 lb
F FC = 1 lb
F FE = 1 lb
F ED = 1 lb
F EC = 1 lb
F CD = 1 lb
458
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
Joint B
F BC − F AB cos ( φ ) = 0
−F BF − F AB sin ( φ ) = 0
Joint F
−F AF + F FE + F FC cos ( φ ) = 0
−P 2 + F BF + F FC sin ( φ ) = 0
Joint C
−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0
−F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0
Joint E
−F FE + F ED = 0
F EC − P 3 = 0
Joint D
−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0
⎛ FAB ⎞
⎜
⎟
F
AF
⎜
⎟
⎜F ⎟
⎜ BC ⎟
⎜ FBF ⎟
⎜
⎟
F
⎜ FC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD)
⎜F ⎟
⎜ FE ⎟
⎜ FED ⎟
⎜
⎟
⎜ FEC ⎟
⎜
⎟
⎝ FCD ⎠
459
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟ ⎛ −330.0 ⎞
⎜ FAF ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 79.4 ⎟
⎜ BC ⎟ ⎜ −233.3 ⎟
⎜ FBF ⎟ ⎜ 233.3 ⎟
⎟
⎜
⎟ ⎜
F
−
47.1
=
⎟ lb
⎜ FC ⎟ ⎜
⎜ F ⎟ ⎜ 112.7 ⎟
⎟
⎜ FE ⎟ ⎜
⎜ FED ⎟ ⎜ 112.7 ⎟
⎜
⎟ ⎜ 300.0 ⎟
F
⎜ EC ⎟ ⎜
⎟
⎟ ⎝ −377.1 ⎠
⎜
⎝ FCD ⎠
Positive means Tension,
Negative means Compression
Problem 6-15
Determine the force in each member of the truss and state if the members are in tension or
compression.
Given:
P 1 = 400 lb
P 2 = 400 lb
P 3 = 0 lb
a = 10 ft
b = 10 ft
θ = 30 deg
Solution:
b
φ = atan ⎛⎜ ⎟⎞
⎝ a⎠
Initial Guesses:
F AB = 1 lb
F AF = 1 lb
F BC = 1 lb
F BF = 1 lb
F FC = 1 lb
F FE = 1lb
F ED = 1 lb
F EC = 1 lb
F CD = 1 lb
460
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
Joint B
F BC − F AB cos ( φ ) = 0
−F BF − F AB sin ( φ ) = 0
Joint F
−F AF + F FE + F FC cos ( φ ) = 0
−P 2 + F BF + F FC sin ( φ ) = 0
Joint C
−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0
−F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0
Joint E
−F FE + F ED = 0
F EC − P 3 = 0
Joint D
−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0
⎛ FAB ⎞
⎜
⎟
⎜ FAF ⎟
⎜F ⎟
⎜ BC ⎟
⎜ FBF ⎟
⎜
⎟
⎜ FFC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD)
⎜F ⎟
⎜ FE ⎟
⎜ FED ⎟
⎜
⎟
⎜ FEC ⎟
⎜
⎟
⎝ FCD ⎠
461
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛ FAB ⎞
⎜
⎟ ⎛ −377.1 ⎞
⎜ FAF ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 189.7 ⎟
⎜ BC ⎟ ⎜ −266.7 ⎟
⎜ FBF ⎟ ⎜ 266.7 ⎟
⎟
⎜
⎟ ⎜
F
188.6
=
⎟ lb
⎜ FC ⎟ ⎜
⎜ F ⎟ ⎜ 56.4 ⎟
⎟
⎜ FE ⎟ ⎜
⎜ FED ⎟ ⎜ 56.4 ⎟
⎜
⎟ ⎜ 0.0 ⎟
F
⎜ EC ⎟ ⎜
⎟
⎟ ⎝ −188.6 ⎠
⎜
⎝ FCD ⎠
Chapter 6
Positive means Tension,
Negative means Compression
Problem 6-16
Determine the force in each member of the truss in terms of the load P and state if the members
are in tension or compression.
Solution:
Support reactions:
ΣME = 0;
3
Ax d − P d = 0
2
Ax =
2P
3
462
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
2P
ΣF x = 0;
Ax − Ex = 0
Ex =
ΣF y = 0;
Ey − P = 0
Ey = P
3
Joint E:
2
ΣF x = 0;
F EC
ΣF y = 0;
P − FED − FEC
3
1
− F AD
1
2
=0
13
− Ex = 0
13
F EC =
=0
13
3
P = 1.20P (T)
F ED = 0
Joint A:
ΣF y = 0;
F AB
ΣF x = 0;
Ax − 2FAB
5
=0
5
F AB = F AD
F AB = F AD =
5
Joint D:
ΣF x = 0;
F AD
ΣF y = 0;
2FAD
2
5
− F DC
1
5
2
=0
5
F DC =
− FDB = 0
F DB =
1
F BC =
5
6
P
5
6
P = 0.373P (C)
P = 0.373P (C)
(T)
3
Joint B:
ΣF x = 0; F AB
1
5
− F BC
5
=0
5
6
P = 0.373P (C)
463
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-17
The maximum allowable tensile force in the members of the truss is Tmax and the maximum
allowable compressive force is Cmax. Determine the maximum magnitude of the load P that can
be applied to the truss.
Units Used:
3
kN = 10 N
Given:
Tmax = 5 kN
Cmax = 3 kN
d = 2m
Solution:
P = 1 kN
Set
Initial Guesses:
F AD = 1 kN
F AB = 1 kN
F BC = 1 kN
F BD = 1 kN
F CD = 1 kN
F CE = 1 kN
F DE = 1 kN
Given
Joint A
Joint B
F AD
F BC
1
5
2
5
− F AB
1
− F AB
2
=0
5
=0
5
(FBC + FAB) 5 + FBD = 0
1
Joint D
(FCD − FAD) 5 = 0
2
(
) 5 =0
1
F DE − F BD − FAD + FCD
Joint C
(
− FCD + FBC
) 5 − FCE 13 = 0
2
2
(FCD − FBC ) 5 + FCE 13 − P = 0
1
3
464
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAD ⎞
⎜
⎟
⎜ FAB ⎟
⎜
⎟
FBC
⎜
⎟
⎜ F ⎟ = Find F , F , F , F , F , F , F
( AD AB BC BD CD CE DE)
⎜ BD ⎟
⎜ FCD ⎟
⎜
⎟
⎜ FCE ⎟
⎜F ⎟
⎝ DE ⎠
⎛ FAD ⎞
⎜
⎟ ⎛ −0.373 ⎞
⎟
⎜ FAB ⎟ ⎜
⎜
⎟ ⎜ −0.373 ⎟
⎜ FBC ⎟ ⎜ −0.373 ⎟
⎜ F ⎟ = ⎜ 0.333 ⎟ kN
⎟
⎜ BD ⎟ ⎜
⎜ FCD ⎟ ⎜ −0.373 ⎟
⎜
⎟ ⎜ 1.202 ⎟
⎟
⎜ FCE ⎟ ⎜
0
⎝
⎠
⎜F ⎟
⎝ DE ⎠
Now Scale the answer
P1 = P
P2 = P
Tmax
(
max F AD , F AB , F BC , F BD , F CD , F CE , FDE
)
Cmax
min F AD , F AB , F BC , F BD , F CD , F CE , FDE
(
(
)
P = min P 1 , P 2
)
P = 4.16 kN
Problem 6-18
Determine the force in each member of the truss and state if the members are in tension or
compression. Hint: The horizontal force component at A must be zero. Why?
465
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Engineering Mechanics - Statics
Chapter 6
Units Used:
3
kip = 10 lb
Given:
F 1 = 600 lb
F 2 = 800 lb
a = 4 ft
b = 3 ft
θ = 60 deg
Solution:
Initial Guesses
F BA = 1 lb
F BD = 1 lb
F CB = 1 lb
F CD = 1 lb
Given
Joint C
−F CB − F2 cos ( θ ) = 0
Joint B
F CB + FBD
b
2
a +b
−F CD − F 2 sin ( θ ) = 0
=0
2
⎛ FBA ⎞
⎜
⎟
⎜ FBD ⎟
⎜
⎟ = Find ( FBA , FBD , FCB , FCD)
⎜ FCB ⎟
⎜F ⎟
⎝ CD ⎠
−F BA − F BD
a
2
2
a +b
− F1 = 0
⎛ FBA ⎞ ⎛
3
⎜
⎟
−1.133 × 10 ⎞
⎜
⎟ Positive means Tension
⎜ FBD ⎟
666.667
⎟ lb Negative means
⎜
⎟=⎜
⎜
⎟ Compression
F
−400
⎜ CB ⎟
⎜
⎜ F ⎟ ⎝ −692.82 ⎟⎠
⎝ CD ⎠
Problem 6-19
Determine the force in each member of the truss and state if the members are in tension or
compression. Hint: The resultant force at the pin E acts along member ED. Why?
Units Used:
3
kN = 10 N
466
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 3 kN
F 2 = 2 kN
a = 3m
b = 4m
Solution:
Initial Guesses:
F CB = 1 kN
F CD = 1 kN
F BA = 1 kN
F BD = 1 kN
F DA = 1 kN
F DE = 1 kN
Given
Joint C
−F 2 − F CD
Joint B
2a
−F CB − FCD
=0
2
( 2 a) + b
b
2
2
=0
2
( 2 a) + b
−F BA + F CB = 0
−F 1 − F BD = 0
Joint D
(FCD − FDA − FDE)
(
2a
2
=0
2
( 2 a) + b
F BD + FCD + FDA − FDE
)
b
2
=0
2
( 2 a) + b
467
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FCB ⎞
⎜
⎟
F
CD
⎜
⎟
⎜F ⎟
⎜ BA ⎟ = Find F , F , F , F , F , F
( CB CD BA BD DA DE)
⎜ FBD ⎟
⎜
⎟
⎜ FDA ⎟
⎜
⎟
⎝ FDE ⎠
⎛ FCB ⎞
⎜
⎟ ⎛ 3 ⎞
F
⎜ CD ⎟ ⎜ −3.606 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎟
BA
3
⎜
⎟=
⎜
⎟ kN
⎜ FBD ⎟
−3 ⎟
⎜
⎜
⎟
⎜ FDA ⎟ ⎜ 2.704 ⎟
⎜
⎟ ⎜⎝ −6.31 ⎟⎠
F
⎝ DE ⎠
Positive means Tension,
Negative means Compression
Problem 6-20
Each member of the truss is uniform and has a mass density ρ. Determine the approximate force
in each member due to the weight of the truss. State if the members are in tension or
compression. Solve the problem by assuming the weight of each member can be represented as
a vertical force, half of which is applied at each end of the member.
Given:
ρ = 8
kg
m
g = 9.81
m
2
s
F1 = 0 N
F2 = 0 N
a = 3m
b = 4m
468
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Initial Guesses:
F CB = 1 N
F CD = 1 N
F BA = 1 N
F BD = 1 N
F DA = 1 N
F DE = 1 N
Given
Joint C
−F CB − FCD
−F 2 − F CD
Joint B
2a
2
( 2 a) + b
2
2⎤
⎛ a⎞ + ⎛ b⎞ ⎥ = 0
⎜ ⎟ ⎜ ⎟
⎝ 2⎠ ⎝ 4⎠ ⎦
⎡a
− ρ g⎢ +
2
2
⎣2
( 2 a) + b
b
−F BA + F CB = 0
⎛
⎝
−F 1 − F BD − ρ g⎜ a +
Joint D
=0
2
b⎞
⎟=0
4⎠
(
F BD + FCD + FDA − FDE
(FCD − FDA − FDE)
⎡b
− ρ g⎢ + 3
2
2
⎣4
( 2 a) + b
)
b
2a
2
2
2⎤
⎛ a⎞ + ⎛ b⎞ ⎥ = 0
⎜ ⎟ ⎜ ⎟
⎝ 2⎠ ⎝ 4⎠ ⎦
=0
2
( 2 a) + b
⎛ FCB ⎞
⎜
⎟
F
CD
⎜
⎟
⎜F ⎟
⎜ BA ⎟ = Find F , F , F , F , F , F
( CB CD BA BD DA DE)
⎜ FBD ⎟
⎜
⎟
⎜ FDA ⎟
⎜
⎟
⎝ FDE ⎠
469
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛ FCB ⎞
⎜
⎟ ⎛ 389 ⎞
F
CD
⎜
⎟ ⎜ −467 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ BA ⎟ = 389 ⎟ N
⎜ FBD ⎟ ⎜ −314 ⎟
⎟
⎜
⎟ ⎜
736
⎜
⎟
⎜ FDA ⎟
⎜
⎜
⎟ ⎝ −1204 ⎟⎠
F
⎝ DE ⎠
Chapter 6
Positive means Tension,
Negative means Compression
Problem 6-21
Determine the force in each member of the truss in terms of the external loading and state if the
members are in tension or compression.
Solution:
Joint B:
+
↑Σ Fy = 0;
F BA sin ( 2 θ ) − P = 0
F BA = P csc ( 2 θ )
+
Σ F x = 0;
→
( C)
F BAcos( 2 θ ) − FBC = 0
F BC = Pcot( 2 θ )
( C)
Joint C:
+
Σ F x = 0;
→
P cot ( 2 θ ) + P + F CD cos ( 2 θ ) − F CA cos ( θ ) = 0
+
F CD sin ( 2 θ ) − FCA sin ( θ ) = 0
↑Σ Fy = 0;
470
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Engineering Mechanics - Statics
F CA =
Chapter 6
cot ( 2 θ ) + 1
cos ( θ ) − sin ( θ ) cot( 2 θ )
P
F CA = ( cot ( θ ) csc ( θ ) − sin ( θ ) + 2 cos ( θ ) ) P
( T)
F CD = ( cot ( 2 θ ) + 1) P
( C)
Joint D:
+
Σ F x = 0;
→
F DA − ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P = 0
F DA = ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P
( C)
Problem 6-22
The maximum allowable tensile force in the members of the truss is Tmax, and the maximum
allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that
can be applied to the truss.
Units Used:
3
kN = 10 N
Given:
Tmax = 2 kN
Cmax = 1.2 kN
L = 2m
θ = 30 deg
Solution:
Initial guesses (assume all bars are in tension). Use a unit load
for P and then scale the answer later.
F BA = 1 kN
F BC = 1 kN
F CA = 1 kN
F CD = 1 kN
F DA = 1 kN
P = 1 kN
471
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
Joint B
↑ Σ Fy = 0; −FBA sin(2 θ) − P = 0
+
Σ F x = 0; −F BA cos ( 2 θ ) + F BC = 0
→
+
Joint C
+
↑Σ Fy = 0;
−F CA sin ( θ ) − F CD sin ( 2 θ ) = 0
+
Σ F x = 0;
→
Joint D
−F BC + P − FCD cos ( 2 θ ) − FCA cos ( θ ) = 0
+
Σ F x = 0;
→
−F DA + F CD cos ( 2 θ ) = 0
⎛⎜ FBA ⎞⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCA ⎟ = Find ( FBA , FBC , FCA , FCD , FDA)
⎜F ⎟
⎜ CD ⎟
⎜ FDA ⎟
⎝
⎠
⎛⎜ FBA ⎞⎟
⎜ FBC ⎟
⎜
⎟
ans = ⎜ FCA ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FDA ⎟
⎝
⎠
⎛ −1.155 ⎞
⎜ −0.577 ⎟
⎜
⎟
ans = ⎜ 2.732 ⎟ kN
⎜ −1.577 ⎟
⎜
⎟
⎝ −0.789 ⎠
Now find the biggest tension and the biggest compression.
T = max ( ans)
T = 2.732 kN
C = min ( ans)
C = −1.577 kN
Decide which is more important and scale the answer
⎡⎛ Tmax ⎞ ⎤
⎢⎜
⎟ ⎥
T
⎢
⎜
⎟ P⎥
P = min
⎢⎜ −Cmax ⎟ ⎥
⎢⎜
⎟ ⎥
⎣⎝ C ⎠ ⎦
P = 732.051 N
Problem 6-23
472
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
The Fink truss supports the loads shown. Determine the force in each member and state if the
members are in tension or compression. Approximate each joint as a pin.
Units Used:
3
kip = 10 lb
Given:
F 1 = 500 lb
a = 2.5 ft
F 2 = 1 kip
θ = 30 deg
F 3 = 1 kip
Solution:
Entire truss:
(
)
ΣF x = 0;
E x = F1 + F2 + F3 + F2 + F1 sin ( θ )
ΣME = 0;
− Ay 4a cos ( θ ) + F 14a + F 23a + F 32a + F 2 a = 0
Ay =
ΣF y = 0;
2 F1 + 2 F2 + F3
E x = 2000 lb
Ay = 2309.4 lb
2 cos ( θ )
E y = − Ay + 2 cos ( θ ) F1 + 2 cos ( θ ) F 2 + cos ( θ ) F3
E y = 1154.7 lb
Joint A:
ΣF y = 0;
F AB =
−cos ( θ ) F 1 + Ay
sin ( θ )
F AB = 3.75 kip
ΣF x = 0;
(C)
F AH = −sin ( θ ) F 1 + F AB cos ( θ )
F AH = 3 kip
(T)
473
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Joint B:
ΣF x = 0;
F BC = FAB
F BC = 3.75 kip
ΣF y = 0;
F BH = F 2
F BH = 1 kip
(C)
Σ F y = 0;
F HC = F 2
F HC = 1 kip
(T)
ΣF x = 0;
F GH = −F2 cos ( −90 deg + θ ) − FHC cos ( −90 deg + θ ) + F AH
(C)
Joint H:
F GH = 2 kip
(T)
Joint E:
ΣF y = 0;
F EF =
(
− F 1 − E x sin ( θ ) − Ey cos ( θ )
F EF = 3 kip
ΣF x = 0;
sin ( θ )
)
(T)
F ED = −Ey sin ( θ ) + Ex cos ( θ ) + F EF cos ( θ )
F ED = 3.75 kip
(C)
Joint D:
ΣF x = 0;
F DC = F ED
F DC = 3.75 kip
ΣF y = 0;
(C)
F DF = F2
F DF = 1 kip
(C)
Joint C:
ΣF x = 0;
F CF = F HC
F CF = 1 kip
ΣF y = 0;
(T)
F CG = F3 + FHC cos ( 90 deg − θ ) ( 2)
F CG = 2 kip
(C)
F FG = FEF − FCF cos ( 90 deg − θ ) ( 2)
F FG = 2 kip
(T)
Joint F:
ΣF x = 0;
474
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-24
Determine the force in each member of the double scissors truss in terms of the load P and state
if the members are in tension or compression.
Solution:
ΣΜ A = 0;
+
↑ ΣF y = 0;
P
L
2L
+P
− Dy L = 0
3
3
Ay + Dy − 2 P = 0
Joint F:
+
1
↑ ΣF y = 0;
F FB
+
ΣF x = 0;
→
F FD − F FE − F FB
−P=0
2
1
=0
2
475
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Joint E:
+
↑ ΣF y = 0;
+
ΣF x = 0;
→
1
F EC
−P=0
2
1
F EF − F EA + F EC
=0
2
Joint B:
+
↑ ΣF y = 0;
F BA
+
ΣF x = 0;
→
F BA
1
2
1
2
+ F BD
1
+ F FB
1
1
− F FB
5
=0
2
⎛ 2 ⎞=0
⎟
⎝ 5⎠
− F BD ⎜
2
Joint C:
+
↑ ΣF y = 0;
F CA
+
ΣF x = 0;
→
F CA
1
5
2
5
+ FCD
1
− FEC
1
2
2
− FEC
1
− FCD
1
=0
2
=0
2
476
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Joint A:
+
ΣF x = 0;
→
F AE − F BA
1
2
− F CA
2
=0
5
Solving we find
F EF = 0.667 P ( T )
F FD = 1.67 P ( T)
F AB = 0.471 P ( C)
F AE = 1.67 P ( T)
F AC = 1.49 P ( C)
F BF = 1.41 P ( T)
F BD = 1.49 P ( C)
F EC = 1.41 P ( T)
F CD = 0.471 P ( C)
Problem 6-25
Determine the force in each member of the truss and state if the members are in tension or
compression. Hint: The vertical component of force at C must equal zero. Why?
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
477
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F 2 = 8 kN
a = 1.5 m
b = 2m
c = 2m
Solution:
Initial Guesses:
F AB = 1 kN
F AE = 1 kN
F EB = 1 kN
F BC = 1 kN
F BD = 1 kN
F ED = 1 kN
Given
Joint A
F AB
F AB
Joint E
a
2
2
a +c
c
2
2
a +c
+ F AE = 0
− F1 = 0
F ED − F AE = 0
F EB − F 2 = 0
Joint B
F BC + F BD
−F EB − F BD
b
2
2
b +c
c
2
2
b +c
− F AB
− F AB
a
2
=0
2
a +c
c
2
=0
2
a +c
⎛ FAB ⎞
⎜
⎟
F
AE
⎜
⎟
⎜F ⎟
⎜ EB ⎟ = Find F , F , F , F , F , F
( AB AE EB BC BD ED)
⎜ FBC ⎟
⎜
⎟
⎜ FBD ⎟
⎜
⎟
⎝ FED ⎠
478
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟ ⎛ 7.5 ⎞
F
AE
⎜
⎟ ⎜ −4.5 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎟
8
⎜ EB ⎟ =
kN
⎜ FBC ⎟ ⎜ 18.5 ⎟
⎟
⎜
⎟ ⎜
−
19.799
⎜
⎟
⎜ FBD ⎟
⎜
⎜
⎟ ⎝ −4.5 ⎟⎠
F
⎝ ED ⎠
Positive means Tension,
Negative means Compresson.
Problem 6-26
Each member of the truss is uniform and has a mass density ρ. Remove the external loads F1
and F 2 and determine the approximate force in each member due to the weight of the truss.
State if the members are in tension or compression. Solve the problem by assuming the weight
of each member can be represented as a vertical force, half of which is applied at each end of
the member.
Given:
F1 = 0
F2 = 0
ρ = 8
kg
m
a = 1.5 m
b = 2m
c = 2m
g = 9.81
m
2
s
Solution:
Find the weights of each bar.
2
2
WAB = ρ g a + c
WBC = ρ g b
WBE = ρ g c
WAE = ρ g a
WBD = ρ g b + c
2
2
WDE = ρ g b
479
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Guesses
Chapter 6
F AB = 1 N
F AE = 1 N
F BE = 1 N
F BC = 1 N
F BD = 1 N
F DE = 1 N
Given
Joint A
a
F AE +
2
a +c
c
2
2
a +c
Joint E
F AB −
F AB = 0
WAB + WAE
2
=0
F DE − F AE = 0
WAE + WBE + WDE
F BE −
Joint B
2
2
b
F BC +
2
−c
2
2
b +c
2
a +c
F BD −
F AB − F BE −
=0
a
F AB = 0
2
2
c
F BD −
a +c
2
2
b +c
WAB + WBE + WBD + WBC
2
=0
⎛ FAB ⎞
⎜
⎟
⎜ FAE ⎟
⎜F ⎟
⎜ BC ⎟ = Find F , F , F , F , F , F
( AB AE BC BD BE DE)
⎜ FBD ⎟
⎜
⎟
⎜ FBE ⎟
⎜
⎟
⎝ FDE ⎠
⎛ FAB ⎞
⎜
⎟ ⎛ 196 ⎞
F
⎜ AE ⎟ ⎜ −118 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎟
BC
857
⎜
⎟=
⎜
⎟N
⎜ FBD ⎟
−1045
⎜
⎟
⎜
⎟
216
⎜
⎟
⎜ FBE ⎟
⎜
⎟ ⎜⎝ −118 ⎟⎠
FDE
⎝
⎠
Positive means tension,
Negative means Compression.
480
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-27
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 4 kN
P 2 = 0 kN
a = 2m
θ = 15 deg
Solution:
Take advantage of the symetry.
Initial Guesses:
F BD = 1 kN
F CD = 1 kN
F CA = 1 kN
F BC = 1 kN
Given
Joint D
Joint B
−P 1
2
F AB = 1 kN
− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0
−P 2 cos ( 2 θ ) − F BC = 0
F BD − F AB − P 2 sin ( 2 θ ) = 0
Joint C
F CD cos ( θ ) − F CA cos ( θ ) = 0
(FCD + FCA)sin(θ ) + FBC = 0
⎛⎜ FBD ⎞⎟
⎜ FCD ⎟
⎜
⎟
F
⎜ AB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC )
⎜F ⎟
⎜ CA ⎟
⎜ FBC ⎟
⎝
⎠
⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟
⎜ FED ⎟ ⎜ FCD ⎟
⎜
⎟ ⎜
⎟
F
F
=
⎜ GF ⎟ ⎜ AB ⎟
⎜F ⎟ ⎜F ⎟
⎜ EG ⎟ ⎜ CA ⎟
⎜ FFE ⎟ ⎜ FBC ⎟
⎝
⎠ ⎝
⎠
481
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Engineering Mechanics - Statics
⎛⎜ FBD ⎞⎟
⎛ −4 ⎞
⎜ FCD ⎟ ⎜ 0 ⎟
⎜
⎟ ⎜ ⎟
⎜ FAB ⎟ = ⎜ −4 ⎟ kN
⎜F ⎟ ⎜ 0 ⎟
⎜ CA ⎟ ⎜ ⎟
⎜ FBC ⎟ ⎝ 0 ⎠
⎝
⎠
Chapter 6
⎛⎜ FFD ⎞⎟
⎛ −4 ⎞
⎜ FED ⎟ ⎜ 0 ⎟
⎜
⎟ ⎜ ⎟
⎜ FGF ⎟ = ⎜ −4 ⎟ kN
⎜F ⎟ ⎜ 0 ⎟
⎜ EG ⎟ ⎜ ⎟
⎜ FFE ⎟ ⎝ 0 ⎠
⎝
⎠
Positvive means Tension,
Negative means Compression
Problem 6-28
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
P 1 = 2 kN
P 2 = 4 kN
a = 2m
θ = 15 deg
Solution:
Take advantage of the symmetry.
Initial Guesses:
F BD = 1 kN
F CD = 1 kN
F CA = 1 kN
F BC = 1 kN
Given
Joint D
Joint B
−P 1
2
F AB = 1 kN
− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0
−P 2 cos ( 2 θ ) − F BC = 0
F BD − F AB − P 2 sin ( 2θ ) = 0
Joint C
F CD cos ( θ ) − F CA cos ( θ ) = 0
(FCD + FCA)sin(θ ) + FBC = 0
482
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ FBD ⎞⎟
⎜ FCD ⎟
⎜
⎟
⎜ FAB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC )
⎜F ⎟
⎜ CA ⎟
⎜ FBC ⎟
⎝
⎠
⎛⎜ FBD ⎞⎟
⎛ −11.46 ⎞
⎜ FCD ⎟ ⎜ 6.69 ⎟
⎜
⎟ ⎜
⎟
⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN
⎜ F ⎟ ⎜ 6.69 ⎟
⎜ CA ⎟ ⎜
⎟
⎜ FBC ⎟ ⎝ −3.46 ⎠
⎝
⎠
⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟
⎜ FED ⎟ ⎜ FCD ⎟
⎜
⎟ ⎜
⎟
⎜ FGF ⎟ = ⎜ FAB ⎟
⎜F ⎟ ⎜F ⎟
⎜ EG ⎟ ⎜ CA ⎟
⎜ FFE ⎟ ⎜ FBC ⎟
⎝
⎠ ⎝
⎠
⎛⎜ FFD ⎞⎟
⎛ −11.46 ⎞
⎜ FED ⎟ ⎜ 6.69 ⎟
⎜
⎟ ⎜
⎟
⎜ FGF ⎟ = ⎜ −13.46 ⎟ kN
⎜ F ⎟ ⎜ 6.69 ⎟
⎜ EG ⎟ ⎜
⎟
⎜ FFE ⎟ ⎝ −3.46 ⎠
⎝
⎠
Positvive means Tension,
Negative means Compression
Problem 6-29
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
F 1 = 2 kip
F 2 = 1.5 kip
F 3 = 3 kip
F 4 = 3 kip
a = 4 ft
b = 10 ft
Solution:
a
θ = atan ⎛⎜ ⎟⎞
⎝ b⎠
Initial Guesses
483
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Engineering Mechanics - Statics
Chapter 6
F AB = 1 lb
F BC = 1 lb
F CD = 1 lb
F DE = 1 lb
F AI = 1 lb
F BI = 1 lb
F CI = 1 lb
F CG = 1 lb
F CF = 1 lb
F DF = 1 lb
F EF = 1 lb
F HI = 1 lb
F GI = 1 lb
F GH = 1 lb
F FG = 1 lb
Given
Joint A
F AI cos ( θ ) + FAB = 0
Joint B
F BC − F AB = 0
F BI = 0
Joint C
(
)
F CD − F BC + FCF − F CI cos ( θ ) = 0
(
)
F CG + FCF + F CI sin ( θ ) = 0
Joint D
F DE − F CD = 0
F DF = 0
Joint I
(
)
F 2 + FGI + F CI − F AI cos ( θ ) = 0
(
)
F HI − FBI + FGI − F AI − FCI sin ( θ ) = 0
Joint H
F GH cos ( θ ) + F 1 = 0
−F GH sin ( θ ) − FHI = 0
Joint G
Joint F
(FFG − FGH − FGI )cos (θ ) = 0
−F 3 − F CG + ( FGH − FFG − FGI ) sin ( θ ) = 0
(FEF − FFG − FCF)cos (θ ) = 0
(FFG − FCF − FEF)sin(θ ) − F4 − FDF = 0
484
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Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FBC ⎟
⎜
⎟
⎜ FCD ⎟
⎜F ⎟
⎜ DE ⎟
⎜ FAI ⎟
⎜
⎟
⎜ FBI ⎟
⎜
⎟
⎜ FCI ⎟
⎜ F ⎟ = Find F , F , F , F , F , F , F , F , F , F , F , F , F , F , F
( AB BC CD DE AI BI CI CG CF DF EF HI GI GH
⎜ CG ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDF ⎟
⎜F ⎟
⎜ EF ⎟
⎜ FHI ⎟
⎜
⎟
⎜ FGI ⎟
⎜F ⎟
⎜ GH ⎟
⎜ FFG ⎟
⎝
⎠
⎛⎜ FAB ⎞⎟
⎛ 3.75 ⎞
⎜ FBC ⎟ ⎜ 3.75 ⎟
⎟
⎜
⎟ ⎜
⎜ FCD ⎟ = ⎜ 7.75 ⎟ kip
⎜ F ⎟ ⎜ 7.75 ⎟
⎜ DE ⎟ ⎜
⎟
⎜ FAI ⎟ ⎝ −4.04 ⎠
⎝
⎠
⎛⎜ FBI ⎞⎟
⎛ 0 ⎞
⎜ FCI ⎟ ⎜ 0.27 ⎟
⎟
⎜
⎟ ⎜
⎜ FCG ⎟ = ⎜ 1.4 ⎟ kip
⎜ F ⎟ ⎜ −4.04 ⎟
⎜ CF ⎟ ⎜
⎟
⎜ FDF ⎟ ⎝ 0 ⎠
⎝
⎠
⎛⎜ FEF ⎟⎞
⎛ −12.12 ⎞
⎜ FHI ⎟ ⎜ 0.8 ⎟
⎟
⎜
⎟ ⎜
⎜ FGI ⎟ = ⎜ −5.92 ⎟ kip
⎜ F ⎟ ⎜ −2.15 ⎟
⎜ GH ⎟ ⎜
⎟
⎜ FFG ⎟ ⎝ −8.08 ⎠
⎝
⎠
Positive means Tension, Negative means Compression
Problem 6-30
The Howe bridge truss is subjected to the loading shown. Determine the force in members DE,
EH, and HG, and state if the members are in tension or compression.
Units Used:
3
kN = 10 N
485
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 30 kN
F 2 = 20 kN
F 3 = 20 kN
F 4 = 40 kN
a = 4m
b = 4m
Solution:
−F 2 a − F 3( 2a) − F 4( 3a) + Gy( 4a) = 0
Gy =
F2 + 2F 3 + 3F4
4
Gy = 45 kN
Guesses
F DE = 1 kN
F EH = 1 kN
F HG = 1 kN
Given
−F DE − F HG = 0
Gy − F 4 − F EH = 0
F DE b + Gy a = 0
⎛ FDE ⎞
⎜
⎟
F
⎜ EH ⎟ = Find ( FDE , FEH , FHG)
⎜F ⎟
⎝ HG ⎠
⎛ FDE ⎞ ⎛ −45 ⎞
⎜
⎟ ⎜
⎟
F
⎜ EH ⎟ = ⎜ 5 ⎟ kN
⎜ F ⎟ ⎝ 45 ⎠
⎝ HG ⎠
Positive (T)
Negative (C)
Problem 6-31
The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD,
LK, CD, and KD, and state if the members are in tension or compression.
Units Used:
3
kN = 10 N
486
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Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 50 kN
F 2 = 50 kN
F 3 = 50 kN
a = 4m
b = 3m
Solution:
Ax = 0
Ay =
3F 3 + 4F2 + 5F 1
6
Guesses
F LD = 1 kN
F LK = 1 kN
F CD = 1 kN
F KD = 1 kN
Given
F 2 b + F 1( 2b) − Ay( 3b) − F LK a = 0
F CD a + F1 b − Ay( 2b) = 0
a
⎞
⎜ 2 2 ⎟ FLD = 0
⎝ a +b ⎠
Ay − F1 − F2 − ⎛
−F 3 − F KD = 0
⎛ FLD ⎞
⎜
⎟
⎜ FLK ⎟
⎜
⎟ = Find ( FLD , FLK , FCD , FKD)
⎜ FCD ⎟
⎜F ⎟
⎝ KD ⎠
⎛ FLD ⎞
⎜
⎟ ⎛⎜ 0 ⎟⎞
F
⎜ LK ⎟ ⎜ −112.5 ⎟
kN
⎜
⎟=⎜
⎟
⎜ FCD ⎟ ⎜ 112.5 ⎟
⎜ F ⎟ ⎝ −50 ⎠
⎝ KD ⎠
Positive (T)
Negative (C)
487
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Engineering Mechanics - Statics
Chapter 6
Problem 6-32
The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI,
JE, and DE, and state if the members are in tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 50 kN
F 2 = 50 kN
F 3 = 50 kN
a = 4m
b = 3m
Solution:
Initial Guesses
Gy = 1 kN
F JI = 1 kN
F JE = 1 kN F DE = 1 kN
Given
Entire Truss
−F 1 b − F2( 2b) − F3( 3b) + Gy( 6b) = 0
Section
−F DE − F JI = 0
F JE + Gy = 0
Gy( 2b) − F DE a = 0
488
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ Gy ⎞
⎜
⎟
⎜ FJI ⎟
⎜
⎟ = Find ( Gy , FJI , FJE , FDE) Gy = 50 kN
F
JE
⎜
⎟
⎜F ⎟
⎝ DE ⎠
⎛ FJI ⎞ ⎛ −75 ⎞
⎜
⎟ ⎜
⎟
F
⎜ JE ⎟ = ⎜ −50 ⎟ kN
⎜ F ⎟ ⎝ 75 ⎠
⎝ DE ⎠
Positive means Tension,
Negative means Compression
Problem 6-33
The roof truss supports the vertical loading shown. Determine the force in members BC, CK,
and KJ and state if these members are in tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 4 kN
F 2 = 8 kN
a = 2m
b = 3m
489
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Initial Guesses
Ax = 1 kN
Ay = 1 kN
F BC = 1 kN
F CK = 1 kN
F KJ = 1 kN
Given
Ax = 0
F 2( 3a) + F 1( 4a) − Ay( 6a) = 0
⎛ 2b ⎞
⎟ + Ax⎜ ⎟ − Ay( 2a) = 0
⎝3⎠
⎝3⎠
F KJ ⎛⎜
2b ⎞
3a
⎞F = 0
BC
⎜ 2
⎟
2
⎝ b + 9a ⎠
F KJ + A x + ⎛
b
⎞F = 0
BC
⎜ 2
2⎟
⎝ b + 9a ⎠
F CK + A y + ⎛
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜
⎟
F
⎜ KJ ⎟ = Find ( Ax , Ay , FKJ , FCK , FBC )
⎜F ⎟
⎜ CK ⎟
⎜ FBC ⎟
⎝
⎠
⎛⎜ Ax ⎞⎟
⎛ 0 ⎞
⎜ Ay ⎟ ⎜ 6.667 ⎟
⎜
⎟ ⎜
⎟
F
13.333
=
⎜ KJ ⎟ ⎜
⎟ kN Positive (T)
Negative (C)
⎜F ⎟ ⎜ 0 ⎟
CK
⎜
⎟ ⎜
⎟
⎜ FBC ⎟ ⎝ −14.907 ⎠
⎝
⎠
Problem 6-34
Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the
deck of a bridge. State if these members are in tension or compression.
Units Used:
3
kip = 10 lb
Given:
F 1 = 4000 lb
F 2 = 8000 lb
490
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F 3 = 5000 lb
a = 9 ft
b = 12 ft
Solution
Initial Guesses:
F DJ = 1 kip
Ay = 1 kip
F CD = 1 kip
F CJ = 1 kip
F KJ = 1 kip
Given
F 3 a + F 2( 4a) + F 1( 5a) − Ay( 6a) = 0
− Ay( 2a) + F 1 a + FKJ b = 0
⎛
⎞F
a
F CD + F KJ + ⎜
2⎟
2
⎝ a +b ⎠
⎛
CJ = 0
⎞F = 0
CJ
2
2⎟
⎝ a +b ⎠
Ay − F1 − F2 − ⎜
b
−F DJ = 0
⎛⎜ Ay ⎞⎟
⎜ FKJ ⎟
⎜
⎟
⎜ FCJ ⎟ = Find ( Ay , FKJ , FCJ , FDJ , FCD)
⎜F ⎟
⎜ DJ ⎟
⎜ FCD ⎟
⎝
⎠
⎛⎜ Ay ⎞⎟
⎛ 9.5 ⎞
⎜ FKJ ⎟ ⎜ 11.25 ⎟
⎟
⎜
⎟ ⎜
⎜ FCJ ⎟ = ⎜ −3.125 ⎟ kip
⎜F ⎟ ⎜ 0 ⎟
⎜ DJ ⎟ ⎜
⎟
−
9.375
⎝
⎠
⎜ FCD ⎟
⎝
⎠
Positive (T)
Negative (C)
491
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-35
Determine the force in members EI and JI of the truss which serves to support the deck of a
bridge. State if these members are in tension or compression.
Units Used:
3
kip = 10 lb
Given:
F 1 = 4000 lb
F 2 = 8000 lb
F 3 = 5000 lb
a = 9 ft
b = 12 ft
Solution:
Initial Guesses:
Gy = 1 kip
F EI = 1 kip
F JI = 1 kip
Given
−F 1 a − F22a − F35a + Gy6a = 0
Gy2a − F 3 a − FJI b = 0
F EI − F3 + Gy = 0
⎛ Gy ⎞
⎜
⎟
F
⎜ JI ⎟ = Find ( Gy , FJI , FEI )
⎜F ⎟
⎝ EI ⎠
⎛ Gy ⎞ ⎛ 7.5 ⎞
⎜
⎟ ⎜
⎟
F
⎜ JI ⎟ = ⎜ 7.5 ⎟ kip
⎜ F ⎟ ⎝ −2.5 ⎠
⎝ EI ⎠
Positive (T)
Negative (C)
Problem 6-36
Determine the force in members BE, EF, and CB, and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
492
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Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 5 kN
F 4 = 10 kN
F 2 = 10 kN
a = 4m
F 3 = 5 kN
b = 4m
a
θ = atan ⎛⎜ ⎟⎞
Solution:
⎝ b⎠
Inital Guesses
F CB = 1 kN F BE = 1 kN
F EF = 1 kN
Given
F 1 + F 2 − F BE cos ( θ ) = 0
−F CB − FEF − FBE sin ( θ ) − F3 = 0
−F 1 a + F CB b = 0
⎛ FCB ⎞
⎜
⎟
F
⎜ BE ⎟ = Find ( FCB , FBE , FEF)
⎜F ⎟
⎝ EF ⎠
⎛ FCB ⎞ ⎛ 5 ⎞
⎜
⎟ ⎜
⎟
⎜ FBE ⎟ = ⎜ 21.2 ⎟ kN
⎜ F ⎟ ⎝ −25 ⎠
⎝ EF ⎠
Positive (T)
Negative (C)
Problem 6-37
Determine the force in members BF, BG, and AB, and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 5 kN
F 4 = 10 kN
493
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F 2 = 10 kN
a = 4m
F 3 = 5 kN
b = 4m
Solution:
Chapter 6
a
θ = atan ⎛⎜ ⎟⎞
⎝ b⎠
Inital Guesses
F AB = 1 kN F BG = 1 kN
F BF = 1 kN
Given
F 1 + F 2 + F 4 + F BG cos ( θ ) = 0
−F 1 3a − F 22a − F 4 a + FAB b = 0
−F BF = 0
⎛ FAB ⎞
⎜
⎟
⎜ FBG ⎟ = Find ( FAB , FBG , FBF)
⎜F ⎟
⎝ BF ⎠
⎛ FAB ⎞ ⎛ 45 ⎞
⎜
⎟ ⎜
⎟
F
⎜ BG ⎟ = ⎜ −35.4 ⎟ kN Positive (T)
Negative (C)
⎜F ⎟ ⎝ 0 ⎠
BF
⎝
⎠
Problem 6-38
Determine the force developed in members GB and GF of the bridge truss and state if these
members are in tension or compression.
Given:
F 1 = 600 lb
494
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F 2 = 800 lb
a = 10 ft
b = 10 ft
c = 4 ft
Solution:
Initial Guesses
Ax = 1 lb
Ay = 1 lb
F GB = 1 lb
F GF = 1 lb
Given
F 2 b + F1( b + 2c) − A y2( b + c) = 0
Ax = 0
Ay − F GB = 0
− Ay b − FGF a = 0
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜
⎟ = Find ( Ax , Ay , FGB , FGF)
⎜ FGB ⎟
⎜F ⎟
⎝ GF ⎠
⎛ Ax ⎞
⎞
⎜
⎟ ⎛⎜ 0
⎟
⎜ Ay ⎟ ⎜ 671.429 ⎟
lb Positive (T)
⎜
⎟=⎜
FGB
671.429 ⎟
Negative (C)
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎝ −671.429 ⎠
⎝ GF ⎠
Problem 6-39
Determine the force members BC, FC, and FE, and state if the members are in tension or
compression.
Units Used:
3
kN = 10 N
495
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 6 kN
F 2 = 6 kN
a = 3m
b = 3m
Solution:
a
θ = atan ⎛⎜ ⎟⎞
⎝ b⎠
Initial Guesses
Dy = 1 kN
F BC = 1 kN
F FC = 1 kN
F FE = 1 kN
Given
−F 1 b − F2( 2b) + Dy( 3b) = 0
Dy b − FFE cos ( θ ) a = 0
(
)
−F FC − FBC + FFE cos ( θ ) = 0
(
)
−F 2 + Dy + FFE + FBC sin ( θ ) = 0
⎛ Dy ⎞
⎜
⎟
⎜ FBC ⎟
⎜
⎟ = Find ( Dy , FBC , FFC , FFE)
⎜ FFC ⎟
⎜F ⎟
⎝ FE ⎠
⎛ Dy ⎞
⎜
⎟ ⎛⎜ 6 ⎟⎞
F
⎜ BC ⎟ ⎜ −8.49 ⎟
kN
⎜
⎟=⎜
⎟
⎜ FFC ⎟ ⎜ 0 ⎟
⎜ F ⎟ ⎝ 8.49 ⎠
⎝ FE ⎠
Positive (T)
Negative (C)
496
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Engineering Mechanics - Statics
Chapter 6
Problem 6-40
Determine the force in members IC and CG of the truss and state if these members are in tension
or compression. Also, indicate all zero-force members.
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 6 kN
a = 1.5 m
b = 2m
Solution:
By inspection of joints B, D, H and I.
AB, BC, CD, DE, HI, and GI are all zero-force members.
Guesses
Ay = 1 kN
Given
− Ay( 4a) + F 1( 2a) + F 2 a = 0
− Ay( 2a) −
−a
2
2
a +b
−b
2
2
a +b
F IC = 1 kN
b
2
2
a +b
FIC +
FIC −
FIC a −
a
2
a +b
2
b
2
a +b
2
⎛ Ay ⎞
⎜
⎟
⎜ FIC ⎟
⎜
⎟ = Find ( Ay , FIC , FCG , FCJ )
⎜ FCG ⎟
⎜F ⎟
⎝ CJ ⎠
F CG = 1 kN
a
2
2
a +b
F CJ = 1 kN
FIC b = 0
F CJ = 0
F CJ − FCG = 0
⎛ Ay ⎞
⎜
⎟ ⎛⎜ 4.5 ⎟⎞
F
⎜ IC ⎟ ⎜ −5.625 ⎟
kN
⎜
⎟=⎜
⎟
⎜ FCG ⎟ ⎜ 9 ⎟
⎜ F ⎟ ⎝ −5.625 ⎠
⎝ CJ ⎠
Positive (T)
Negative (C)
Problem 6-41
Determine the force in members JE and GF of the truss and state if these members are in
tension or compression. Also, indicate all zero-force members.
497
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Engineering Mechanics - Statics
Chapter 6
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 6 kN
a = 1.5 m
b = 2m
Solution:
By inspection of joints B, D, H and I.
AB, BC, CD, DE, HI, and GI are all zero-force members.
E y = 1 kN
Guesses
Given
F JE = 1 kN
F GF = 1 kN
−F 1( 2a) − F 2( 3a) + E y( 4a) = 0
Ey +
b
2
a +b
−a
2
2
a +b
2
F JE = 0
FJE − FGF = 0
⎛ Ey ⎞
⎜
⎟
F
⎜ JE ⎟ = Find ( Ey , FJE , FGF)
⎜F ⎟
⎝ GF ⎠
⎛ Ey ⎞ ⎛ 7.5 ⎞
⎜
⎟ ⎜
⎟
F
⎜ JE ⎟ = ⎜ −9.375 ⎟ kN Positive (T)
Negative (C)
⎜ F ⎟ ⎝ 5.625 ⎠
⎝ GF ⎠
Problem 6-42
Determine the force in members BC, HC, and HG. After the truss is sectioned use a single
equation of equilibrium for the calculation of each force. State if these members are in tension
or compression.
Units Used:
3
kN = 10 N
498
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Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 2 kN F 4 = 5 kN a = 5 m
F 2 = 4 kN F 5 = 3 kN b = 2 m
F 3 = 4 kN
c = 3m
Solution:
Guesses
Ax = 1 kN
Ay = 1 kN
F BC = 1 kN
F HC = 1 kN
F HG = 1 kN
d = 1m
Given
c
b
=
a+d
a
− Ax = 0
(F1 − Ay)( 4a) + F2( 3a) + F3( 2a) + F4( a) = 0
(F1 − Ay)( a) + Ax( c) − FBC ( c) = 0
(F1 − Ay)( 2a) + F2( a) +
( A y − F 1) ( d) − F 2( a + d ) +
a
2
a +b
2
F HG( c) +
c
2
2
a +c
b
2
a +b
FHC ( a + d) +
2
F HG( a) = 0
a
2
2
a +c
F HC ( c) = 0
⎛ Ay ⎞
⎜
⎟
⎜ Ax ⎟
⎜
⎟
A
FBC
⎜
⎟ = Find A , A , F , F , F , d ⎛⎜ x ⎞⎟ = ⎛⎜ 0 ⎞⎟ kN
( y x BC HC HG ) ⎜ A ⎟ ⎝ 8.25 ⎠
⎜F ⎟
⎝ y⎠
HC
⎜
⎟
⎜ FHG ⎟
⎜
⎟
d = 2.5 m
⎝ d ⎠
⎛ FBC ⎞ ⎛ −10.417 ⎞
⎜
⎟ ⎜
⎟
⎜ FHC ⎟ = ⎜ 2.235 ⎟ kN
⎜ F ⎟ ⎝ 9.155 ⎠
⎝ HG ⎠
Positive (T)
Negative (C)
Problem 6-43
Determine the force in members CD, CF, and CG and state if these members are in tension or
499
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Engineering Mechanics - Statics
Chapter 6
compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 2 kN F 4 = 5 kN a = 5 m
F 2 = 4 kN F 5 = 3 kN b = 2 m
F 3 = 4 kN
c = 3m
Solution:
Guesses
E y = 1 kN
F CD = 1 kN
F CF = 1 kN
F CG = 1 kN
F FG = 1 kN
F GH = 1 kN
Given
(
)
−F 2( a) − F3( 2a) − F4( 3a) + E y − F5 ( 4a) = 0
(
)
(
)
F CD( c) + Ey − F 5 ( a) = 0
a
−F 4( a) − F 5 − E y ( 2a) −
a
2
2
a +b
b
2
2
a +b
(F5 − Ey)
FFG −
a
2
a +b
2
2
2
a +b
FFG( b + c) = 0
F GH = 0
(FFG + FGH ) + FCG = 0
a( c − b)
b
+ F4⎡⎢a +
⎣
a( c − b) ⎤
c
b
2
⎥−
⎦
2
a +c
F CF⎡⎢2 a +
⎣
a( c − b) ⎤
b
⎥=0
⎦
500
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Engineering Mechanics - Statics
Chapter 6
⎛ Ey ⎞
⎜
⎟
F
CD
⎜
⎟
⎜F ⎟
⎜ CF ⎟ = Find E , F , F , F , F , F
( y CD CF CG FG GH )
⎜ FCG ⎟
⎜
⎟
⎜ FFG ⎟
⎜
⎟
⎝ FGH ⎠
⎛ Ey ⎞
⎜
⎟ ⎛ 9.75 ⎞
F
CD
⎜
⎟ ⎜ −11.25 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ CF ⎟ = 3.207 ⎟ kN Positive (T)
⎜ FCG ⎟ ⎜ −6.8 ⎟
Negative (C)
⎟
⎜
⎟ ⎜
⎜ FFG ⎟ ⎜ 9.155 ⎟
⎜
⎟ ⎜⎝ 9.155 ⎟⎠
F
⎝ GH ⎠
Problem 6-44
Determine the force in members OE, LE, and LK of the Baltimore truss and state if the
members are in tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 2 kN
a = 2m
F 2 = 2 kN
b = 2m
F 3 = 5 kN
F 4 = 3 kN
Solution:
Ax = 0 kN
Initial Guesses
Ay = 1 kN
F OE = 1 kN
F DE = 1 kN F LK = 1 kN
F LE = 1 kN
Given
F LE = 0
F 4( 3b) + F 3( 4b) + F 2( 5b) + F 1( 6b) − Ay( 8b) = 0
F LK + FDE + FOE
b
2
a +b
=0
2
501
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Engineering Mechanics - Statics
Ay − F 1 − F 2 − F OE
Chapter 6
a
2
=0
2
a +b
−F LK ( 2a) + F2( b) + F 1( 2b) − Ay( 4b) = 0
⎛⎜ Ay ⎞⎟
⎜ FOE ⎟
⎜
⎟
⎜ FDE ⎟ = Find ( Ay , FOE , FDE , FLK , FLE)
⎜F ⎟
⎜ LK ⎟
⎜ FLE ⎟
⎝
⎠
⎛⎜ Ay ⎞⎟ ⎛ 6.375 ⎞
=
kN
⎜ FDE ⎟ ⎜⎝ 7.375 ⎟⎠
⎝
⎠
⎛ FOE ⎞ ⎛ 3.36 ⎞
⎜
⎟ ⎜
⎟
⎜ FLE ⎟ = ⎜ 0 ⎟ kN
⎜ F ⎟ ⎝ −9.75 ⎠
⎝ LK ⎠
Positive (T)
Negative (C)
Problem 6-45
Determine the force in member GJ of the truss and state if this member is in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
F 1 = 1000 lb
F 2 = 1000 lb
F 3 = 1000 lb
F 4 = 1000 lb
a = 10 ft
θ = 30 deg
502
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=
Chapter 6
Engineering Mechanics - Statics
Solution:
Guess
E y = 1 lb
F GJ = 1 lb
Given
−F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0
−F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0
⎛⎜ Ey ⎟⎞
= Find ( Ey , F GJ )
⎜ FGJ ⎟
⎝
⎠
⎛⎜ Ey ⎟⎞ ⎛ 1.5 ⎞
=
kip
⎜ FGJ ⎟ ⎜⎝ −2 ⎟⎠
⎝
⎠
Positive (T)
Negative (C)
Problem 6-46
Determine the force in member GC of the truss and state if this member is in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
F 1 = 1000 lb
F 2 = 1000 lb
F 3 = 1000 lb
F 4 = 1000 lb
a = 10 ft
θ = 30 deg
Solution:
Guess
E y = 1 lb
F GJ = 1 lb
F HG = 1 lb
F GC = 1 lb
Given
−F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0
−F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0
503
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
−F HG cos ( θ ) + FGJ cos ( θ ) = 0
(
)
−F 3 − F GC − F HG + FGJ sin ( θ ) = 0
⎛ Ey ⎞
⎜
⎟
⎜ FGJ ⎟
⎜
⎟ = Find ( Ey , FGJ , FGC , FHG)
F
GC
⎜
⎟
⎜F ⎟
⎝ HG ⎠
⎛ Ey ⎞
⎜
⎟ ⎛⎜ 1.5 ⎟⎞
⎜ FGJ ⎟ ⎜ −2 ⎟
⎜
⎟ = ⎜ ⎟ kip
F
GC
⎜
⎟ ⎜ 1 ⎟
⎜ F ⎟ ⎝ −2 ⎠
⎝ HG ⎠
Positive (T)
Negative (C)
Problem 6-47
Determine the force in members KJ, JN, and CD, and state if the members are in tension or
compression. Also indicate all zero-force members.
Units Used:
3
kip = 10 lb
Given:
F = 3 kip
a = 20 ft
b = 30 ft
c = 20 ft
Solution:
Ax = 0
2c ⎞
⎟
⎝ 2a + b ⎠
θ = atan ⎛⎜
2c ⎞
⎟
⎝b⎠
φ = atan ⎛⎜
Initial Guesses:
Ay = 1 lb
F CD = 1 lb
F KJ = 1 lb
F JN = 1 lb
504
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
⎛
⎝
F⎜a +
b⎞
⎟ − Ay( 2a + b) = 0
2⎠
⎛
⎝
F CD c − Ay⎜ a +
b⎞
⎟=0
2⎠
F CD + F JN cos ( φ ) + F KJ cos ( θ ) = 0
Ay + F JN sin ( φ ) + FKJ sin ( θ ) = 0
⎛ Ay ⎞
⎜
⎟
⎜ FCD ⎟
⎜
⎟ = Find ( Ay , FCD , FJN , FKJ )
⎜ FJN ⎟
⎜F ⎟
⎝ KJ ⎠
Ay = 1.5 kip
⎛ FCD ⎞ ⎛ 2.625 ⎞
⎜
⎟ ⎜
⎟
⎜ FJN ⎟ = ⎜ 0 ⎟ kip
⎜ F ⎟ ⎝ −3.023 ⎠
⎝ KJ ⎠
Positive (T), Negative (C)
Problem 6-48
Determine the force in members BG, HG, and BC of the truss and state if the members are in
tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 7 kN
F 3 = 4 kN
a = 3m
b = 3m
c = 4.5 m
505
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Initial Guesses
F BG = 1 kN
Ax = 1 kN
F HG = 1 kN
Ay = 1 kN
F BC = 1 kN
Given
− Ax = 0
⎡
⎤ F ( b) = 0
HG
2
2⎥
⎣ ( c − b) + a ⎦
a
− Ay( a) − ⎢
F 3( a) + F2( 2a) + F1( 3a) − A y( 4a) = 0
⎡
⎤F + ⎛ a ⎞F − A = 0
HG ⎜
BG
x
2
2⎥
2
2⎟
⎣ ( c − b) + a ⎦
⎝ a +c ⎠
a
F BC + ⎢
⎡
c
⎤F + ⎛
⎞F = 0
HG ⎜
BG
⎥
⎟
2
2
2
2
⎣ ( c − b) + a ⎦
⎝ a +c ⎠
Ay − F1 + ⎢
c−b
506
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜
⎟
⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC )
⎜F ⎟
⎜ BG ⎟
⎜ FBC ⎟
⎝
⎠
⎛⎜ Ax ⎞⎟
⎛ 0 ⎞
⎜ Ay ⎟ ⎜ 9 ⎟
⎟
⎜
⎟ ⎜
⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN
⎜ F ⎟ ⎜ 1.803 ⎟
⎜ BG ⎟ ⎜
⎟
8
⎝
⎠
⎜ FBC ⎟
⎝
⎠
Positive (T)
Negative (C)
Problem 6-49
The skewed truss carries the load shown. Determine the force in members CB, BE, and EF
and state if these members are in tension or compression. Assume that all joints are pinned.
Solution:
ΣMB = 0;
−P d − F EF d = 0
ΣME = 0;
−P d +
2
5
F CB d = 0
+
1
Σ F x = 0; P −
FCB − F BE = 0
→
5
F EF = −P
F EF = P
( C)
F CB =
5
P
2
F CB = 1.12P
( T)
P BE =
P
2
F BE = 0.5P
( T)
Problem 6-50
The skewed truss carries the load shown. Determine the force in members AB, BF, and EF
and state if these members are in tension or compression. Assume that all joints are pinned.
507
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Engineering Mechanics - Statics
Chapter 6
Solution:
ΣMF = 0;
−P 2d + P d + FAB d = 0
F AB = P
F AB = P
( T)
ΣMB = 0;
−P d − F EF d = 0
F EF = −P
F EF = P
( C)
F BE = − 2 P
F BF = 1.41P
( C)
+
Σ F x = 0;
→
P + FBF
1
=0
2
Problem 6-51
Determine the force developed in members BC and CH of the roof truss and state if the
members are in tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 1.5 kN
F 2 = 2 kN
508
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Engineering Mechanics - Statics
Chapter 6
a = 1.5 m
b = 1m
c = 2m
d = 0.8 m
Solution:
a
θ = atan ⎛⎜ ⎟⎞
⎝c⎠
a ⎞
⎟
⎝ c − b⎠
φ = atan ⎛⎜
Initial Guesses:
E y = 1 kN
F BC = 1 kN
F CH = 1 kN
Given
−F 2( d) − F1( c) + Ey( 2c) = 0
F BC sin ( θ ) ( c) + FCH sin ( φ ) ( c − b) + E y( c) = 0
−F BC sin ( θ ) − FCH sin ( φ ) − F1 + Ey = 0
⎛ Ey ⎞
⎜
⎟
⎜ FBC ⎟ = Find ( Ey , FBC , FCH ) Ey = 1.15 kN
⎜F ⎟
⎝ CH ⎠
⎛⎜ FBC ⎞⎟ ⎛ −3.25 ⎞
=
kN
⎜ FCH ⎟ ⎜⎝ 1.923 ⎟⎠
⎝
⎠
Positive (T)
Negative (C)
Problem 6-52
Determine the force in members CD
and GF of the truss and state if the
members are in tension or
compression. Also indicate all
zero-force members.
Units Used:
3
kN = 10 N
Given:
F 1 = 1.5 kN
F 2 = 2 kN
509
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
a = 1.5 m
b = 1m
c = 2m
d = 0.8 m
Solution:
a
θ = atan ⎛⎜ ⎟⎞
⎝c⎠
⎞
⎟
⎝ c − b⎠
φ = atan ⎛⎜
a
Initial Guesses:
E y = 1 kN
F CD = 1 kN
F GF = 1 kN
Given
−F 2( d) − F1( c) + Ey( 2c) = 0
E y( b) + F CD sin ( θ ) ( b) = 0
E y( c) − FGF( a) = 0
⎛ Ey ⎞
⎜
⎟
F
⎜ CD ⎟ = Find ( Ey , FCD , FGF) Ey = 1.15 kN
⎜F ⎟
⎝ GF ⎠
⎛⎜ FCD ⎞⎟ ⎛ −1.917 ⎞
Positive (T)
=
kN
Negative (C)
⎜ FGF ⎟ ⎜⎝ 1.533 ⎟⎠
⎝
⎠
DF and CF are zero force members.
Problem 6-53
Determine the force in members DE, DL,
and ML of the roof truss and state if the
members are in tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 12 kN
F 3 = 12 kN
510
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F 4 = 12 kN
a = 4m
b = 3m
c = 6m
Solution:
θ = atan ⎛⎜
c − b⎞
⎟
⎝ 3a ⎠
⎡ b + 2 ( c − b) ⎤
⎢ 3
⎥
φ = atan ⎢
⎥
a
⎣
⎦
Initial Guesses:
Ay = 1 kN
F ML = 1 kN
F DL = 1 kN
F DE = 1 kN
Given
F 2( a) + F3( 2a) + F4( 3a) + F3( 4a) + F2( 5a) + F1( 6a) − A y( 6a) = 0
F 1( 2a) + F 2( a) − A y( 2a) + FML⎡⎢b +
⎣
2
3
( c − b)⎤⎥ = 0
⎦
Ay − F 1 − F 2 − F 3 + F DE sin ( θ ) − FDL sin ( φ ) = 0
F ML + F DL cos ( φ ) + FDE cos ( θ ) = 0
⎛ Ay ⎞
⎜
⎟
⎜ FML ⎟
⎜
⎟ = Find ( Ay , FML , FDE , FDL)
⎜ FDE ⎟
⎜F ⎟
⎝ DL ⎠
Ay = 36 kN
⎛ FML ⎞ ⎛ 38.4 ⎞
⎜
⎟ ⎜
⎟
⎜ FDE ⎟ = ⎜ −37.1 ⎟ kN
⎜ F ⎟ ⎝ −3.8 ⎠
⎝ DL ⎠
Positive (T), Negative (C)
Problem 6-54
Determine the force in members EF and EL of the roof truss and state if the members are in
511
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Engineering Mechanics - Statics
Chapter 6
tension or compression.
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 12 kN
F 3 = 12 kN
F 4 = 12 kN
a = 4m
b = 3m
c = 6m
Solution:
θ = atan ⎛⎜
c − b⎞
⎟
⎝ 3a ⎠
Initial Guesses:
Iy = 1 kN
F EF = 1 kN
F EL = 1 kN
Given
−F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = 0
−F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = 0
−F 4 − F EL − 2F EF sin ( θ ) = 0
⎛ Iy ⎞
⎜
⎟
⎜ FEF ⎟ = Find ( Iy , FEF , FEL)
⎜F ⎟
⎝ EL ⎠
Iy = 36 kN
⎛⎜ FEF ⎞⎟ ⎛ −37.108 ⎞
=
kN
⎜ FEL ⎟ ⎜⎝ 6 ⎟⎠
⎝
⎠
Positive (T)
Negative (C)
Problem 6-55
Two space trusses are used to equally support the uniform sign of mass M. Determine the force
developed in members AB, AC, and BC of truss ABCD and state if the members are in tension
or compression. Horizontal short links support the truss at joints B and D and there is a ball-and512
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Engineering Mechanics - Statics
Chapter 6
p
socket joint at C.
pp
j
Given:
M = 50 kg
g = 9.81
m
2
s
a = 0.25 m
b = 0.5 m
c = 2m
Solution:
h =
2
2
b −a
⎛ −a ⎞
AB = ⎜ −c ⎟
⎜ ⎟
⎝h⎠
⎛a⎞
AD = ⎜ −c ⎟
⎜ ⎟
⎝h⎠
⎛ 2a ⎞
BD = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
⎛a⎞
BC = ⎜ 0 ⎟
⎜ ⎟
⎝ −h ⎠
⎛0⎞
AC = ⎜ −c ⎟
⎜ ⎟
⎝0⎠
Guesses
F AB = 1 N
F AD = 1 N
F AC = 1 N
F BC = 1 N
F BD = 1 N
By = 1 N
Given
⎛⎜ 0 ⎟⎞
0 ⎟
AB
AD
AC
F AB
+ FAD
+ FAC
+⎜
=0
⎜ −M g ⎟
AB
AD
AC
⎜ 2 ⎟
⎝
⎠
⎛⎜ 0 ⎟⎞
F AB
+ FBD
+ F BC
+ ⎜ −B y ⎟ = 0
AB
BD
BC
⎜ 0 ⎟
⎝
⎠
−AB
BD
BC
513
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Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
F
AD
⎜
⎟
⎜F ⎟
⎜ AC ⎟ = Find F , F , F , F , F , B
( AB AD AC BC BD y)
⎜ FBC ⎟
⎜
⎟
⎜ FBD ⎟
⎜
⎟
⎝ By ⎠
⎛ By ⎞ ⎛ 566 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AD ⎟ = ⎜ 584 ⎟ N
⎜F ⎟ ⎝ 0 ⎠
⎝ BD ⎠
⎛ FAB ⎞ ⎛ 584 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AC ⎟ = ⎜ −1133 ⎟ N
⎜ F ⎟ ⎝ −142 ⎠
⎝ BC ⎠
Positive (T), Negative (C)
Problem 6-56
Determine the force in each member of the space truss and state if the members are in tension or
compression.The truss is supported by short links at B, C, and D.
Given:
F = 600 N
a = 3m
b = 1m
c = 1.5 m
Solution:
⎛b⎞
⎜ ⎟
AB = −c
⎜ ⎟
⎝ −a ⎠
⎛0⎞
⎜ ⎟
AC = c
⎜ ⎟
⎝ −a ⎠
⎛ −b ⎞
⎜ ⎟
AD = −c
⎜ ⎟
⎝ −a ⎠
⎛ −b ⎞
⎜
⎟
CD = −2c
⎜
⎟
⎝ 0 ⎠
514
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Engineering Mechanics - Statics
⎛ b ⎞
CB = ⎜ −2c ⎟
⎜
⎟
⎝ 0 ⎠
Chapter 6
⎛ −2b ⎞
BD = ⎜ 0 ⎟
⎜
⎟
⎝ 0 ⎠
Guesses
F BA = 1 N
F BC = 1 N
F CA = 1 N
F DA = 1 N
F BD = 1 N
F DC = 1 N
By = 1 N
Bz = 1 N
Cz = 1 N
Given
⎛ 0 ⎞
F BA
+ FCA
+ F DA
+⎜ 0 ⎟ =0
⎜ ⎟
AB
AC
AD
⎝ −F ⎠
AB
AC
AD
⎛⎜ 0 ⎟⎞
F CA
+ FDC
+ F BC
+⎜0 ⎟ =0
AC
CD
CB
⎜ Cz ⎟
⎝ ⎠
−AC
CD
CB
⎛0⎞
⎜ ⎟
F BC
+ FBD
+ F BA
+ ⎜ By ⎟ = 0
CB
BD
AB
⎜B ⎟
⎝ z⎠
−CB
BD
−AB
⎛ FBA ⎞
⎜
⎟
F
BC
⎜
⎟
⎜F ⎟
⎜ CA ⎟
⎜ FDA ⎟
⎜
⎟
F
⎜ BD ⎟ = Find ( FBA , FBC , FCA , FDA , FBD , FDC , By , Bz , Cz)
⎜F ⎟
⎜ DC ⎟
⎜ By ⎟
⎜
⎟
⎜ Bz ⎟
⎟
⎜
⎝ Cz ⎠
515
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Engineering Mechanics - Statics
⎛ By ⎞ ⎛ 1.421 × 10− 14 ⎞
⎜ ⎟ ⎜
⎟
B
=
⎜ z⎟ ⎜
⎟N
150
⎜C ⎟ ⎜
⎟
300
⎠
⎝ z⎠ ⎝
Chapter 6
⎛ FBA ⎞
⎜
⎟ ⎛ −175 ⎞
F
BC
⎜
⎟ ⎜ 79.1 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ CA ⎟ = −335.4 ⎟ N
⎜ FDA ⎟ ⎜ −175 ⎟
⎟
⎜
⎟ ⎜
25
⎜
⎟
⎜ FBD ⎟
⎜
⎜
⎟ ⎝ 79.1 ⎟⎠
F
⎝ DC ⎠
Positive (T),
Negative (C)
Problem 6-57
Determine the force in each member of the space truss and state if the members are in tension
or compression.The truss is supported by short links at A, B, and C.
Given:
a = 4 ft
b = 2 ft
c = 3 ft
d = 2 ft
e = 8 ft
⎛ 0 ⎞
⎜
⎟
F = 500 lb
⎜
⎟
⎝ 0 ⎠
Solution:
⎛ −c ⎞
⎜ ⎟
AD = a
⎜ ⎟
⎝e⎠
⎛ −c ⎞
⎜ ⎟
BD = −b
⎜ ⎟
⎝e ⎠
⎛d⎞
⎜ ⎟
CD = −b
⎜ ⎟
⎝e ⎠
⎛ 0 ⎞
⎜
⎟
AB = a + b
⎜
⎟
⎝ 0 ⎠
⎛ −c − d ⎞
⎛ −c − d ⎞
⎜
⎟
⎜ 0 ⎟
AC = a + b
BC =
⎜
⎟
⎜
⎟
⎝ 0 ⎠
⎝ 0 ⎠
Guesses
516
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F BA = 1 lb
F BC = 1 lb
F BD = 1 lb
F AD = 1 lb
F AC = 1 lb
F CD = 1 lb
Ay = 1 lb
Az = 1 lb
B x = 1 lb
B z = 1 lb
Cy = 1 lb
Cz = 1 lb
Given
F + FAD
−AD
AD
+ FBD
−BD
BD
+ F CD
−CD
CD
=0
⎛0 ⎞
⎜ ⎟
F AD
+ F BA
+ FAC
+ ⎜ Ay ⎟ = 0
AD
AB
AC
⎜A ⎟
⎝ z⎠
AD
AB
AC
⎛ Bx ⎞
⎜ ⎟
−AB
BC
BD
F BA
+ FBC
+ F BD
+⎜ 0 ⎟ =0
AB
BC
BD
⎜B ⎟
⎝ z⎠
⎛0⎞
CD
−AC
−BC ⎜ C ⎟
F CD
+ F AC
+ F BC
+ ⎜ y⎟ = 0
CD
AC
BC
⎜C ⎟
⎝ z⎠
⎛⎜ FBA ⎞⎟
⎜ FBC ⎟
⎜
⎟
⎜ FBD ⎟
⎜F ⎟
⎜ AD ⎟
⎜ FAC ⎟
⎜
⎟
⎜ FCD ⎟
⎜
⎟ = Find ( FBA , FBC , FBD , FAD , FAC , FCD , Ay , Az , Bx , Bz , Cy , Cz)
⎜ Ay ⎟
⎜ A ⎟
⎜ z ⎟
⎜ Bx ⎟
⎜
⎟
⎜ Bz ⎟
⎜ C ⎟
⎜ y ⎟
⎜ Cz ⎟
⎝
⎠
517
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Engineering Mechanics - Statics
⎛ Ay ⎞
⎜ ⎟ ⎛ −200 ⎞
⎜ Az ⎟ ⎜ −667 ⎟
⎟
⎜B ⎟ ⎜
⎜
⎜ x ⎟ = 0 ⎟ lb
⎜ Bz ⎟ ⎜ 667 ⎟
⎟
⎜ ⎟ ⎜
−
300
⎜
⎟
⎜ Cy ⎟
⎜
⎜ ⎟ ⎝ 0 ⎟⎠
⎝ Cz ⎠
Chapter 6
⎛ FBA ⎞
⎜
⎟ ⎛ 167 ⎞
F
BC
⎜
⎟ ⎜ 250 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎜ BD ⎟ = −731 ⎟ lb
⎜ FAD ⎟ ⎜ 786 ⎟
⎟
⎜
⎟ ⎜
−
391
⎜
⎟
⎜ FAC ⎟
⎜
⎜
⎟ ⎝ 0 ⎟⎠
F
⎝ CD ⎠
Positive (T)
Negative (C)
Problem 6-58
The space truss is supported by a
ball-and-socket joint at D and short
links at C and E. Determine the
force in each member and state if
the members are in tension or
compression.
Given:
⎛ 0 ⎞
⎜ 0 ⎟ lb
F1 =
⎜
⎟
⎝ −500 ⎠
⎛ 0 ⎞
⎜
⎟
F 2 = 400 lb
⎜
⎟
⎝ 0 ⎠
a = 4 ft
b = 3 ft
c = 3 ft
Solution:
Find the external reactions
Guesses
E y = 1 lb
Cy = 1 lb
Cz = 1 lb
Dx = 1 lb
Dy = 1 lb
Dz = 1 lb
518
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Engineering Mechanics - Statics
Chapter 6
Given
⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0
⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟
⎝ z⎠ ⎝ ⎠ ⎝ z⎠
D
⎛0⎞
⎛ −b ⎞
⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞
⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0
⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟
⎝0⎠
⎝0⎠
⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠
⎛ Ey ⎞
⎜ ⎟
⎜ Cy ⎟
⎜C ⎟
⎜ z ⎟ = Find E , C , C , D , D , D
( y y z x y z)
⎜ Dx ⎟
⎜ ⎟
⎜ Dy ⎟
⎜ ⎟
⎝ Dz ⎠
⎛ Ey ⎞
⎜ ⎟ ⎛ 266.667 ⎞
⎜ Cy ⎟ ⎜ −400 ⎟
⎟
⎜C ⎟ ⎜
⎜
⎟
0
⎜ z⎟ =
⎟ lb
⎜ Dx ⎟ ⎜
0
⎟
⎜ ⎟ ⎜
−
266.667
⎜
⎟
⎜ Dy ⎟
⎜
⎜ ⎟ ⎝ 500 ⎟⎠
⎝ Dz ⎠
Now find the force in each member.
⎛ −b ⎞
⎛ −b ⎞
⎜
⎟
AB = 0
AC = ⎜ −a ⎟
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝c ⎠
⎛0⎞
AD = ⎜ −a ⎟
⎜ ⎟
⎝c ⎠
⎛0⎞
AE = ⎜ −a ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
⎛b⎞
⎜
⎟
BC = −a
BE = ⎜ −a ⎟
⎜ ⎟
⎜ ⎟
c
⎝ ⎠
⎝0⎠
⎛0⎞
BF = ⎜ −a ⎟
⎜ ⎟
⎝0⎠
⎛b⎞
CD = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
⎛0⎞
⎜
⎟
CF = 0
DE = ⎜ 0 ⎟
⎜ ⎟
⎜ ⎟
⎝ −c ⎠
⎝ −c ⎠
⎛ −b ⎞
DF = ⎜ 0 ⎟
⎜ ⎟
⎝ −c ⎠
⎛ −b ⎞
EF = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
Guesses
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
F AE = 1 lb
F BC = 1 lb
F BE = 1 lb
F BF = 1 lb
F CD = 1 lb
519
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Engineering Mechanics - Statics
F CF = 1 lb
Chapter 6
F DE = 1lb
F DF = 1 lb
F EF = 1 lb
Given
F 1 + F AB
AB
F 2 + F BC
BC
AB
BC
+ FAC
AC
+ FBF
BF
AC
BF
+ FAD
AD
+ FBE
BE
AD
BE
+ FAE
+ F AB
AE
AE
−AB
AB
=0
=0
⎛⎜ 0 ⎟⎞
−AE
−BE
EF
−DE
+ FBE
+ F EF
+ F DE
=0
⎜ Ey ⎟ + FAE
AE
BE
EF
DE
⎜0⎟
⎝ ⎠
F BF
−BF
BF
+ F CF
−CF
CF
+ F DF
−DF
DF
+ FEF
−EF
EF
=0
⎛0⎞
⎜ ⎟
−BC
−AC
CD
CF
+ FAC
+ FCD
+ FCF
=0
⎜ Cy ⎟ + FBC
BC
AC
CD
CF
⎜C ⎟
⎝ z⎠
⎛⎜ FAB ⎞⎟
⎜ FAC ⎟
⎜
⎟
⎜ FAD ⎟
⎜F ⎟
⎜ AE ⎟
⎜ FBC ⎟
⎜
⎟
⎜ FBE ⎟
⎜
⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF)
⎜ FBF ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FEF ⎟
⎝
⎠
520
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎛ FBF ⎞
⎜
⎟ ⎛ −300 ⎞ ⎜
⎟ ⎛ 0 ⎞
⎜
⎟
F
F
AC
CD
⎜
⎟
⎜
⎟ ⎜ −300 ⎟
583.095 ⎟
⎜
⎟
⎜F ⎟
⎜F ⎟ ⎜
Positive (T)
⎜
⎟
⎜
⎜ AD ⎟ = 333.333 lb ⎜ CF ⎟ = −300 ⎟ lb Negative (C)
⎜ FAE ⎟ ⎜ −666.667 ⎟ ⎜ FDE ⎟ ⎜ 0 ⎟
⎟ ⎜
⎟
⎜
⎟ ⎜
⎟ ⎜
0
424.264
⎜
⎟
⎜
⎟
⎜ FBC ⎟
⎜ FDF ⎟
⎜
⎟
⎜
⎜
⎟ ⎝ 500 ⎠ ⎜
⎟ ⎝ −300 ⎟⎠
F
F
⎝ BE ⎠
⎝ EF ⎠
Problem 6-59
The space truss is supported by
a ball-and-socket joint at D and
short links at C and E. Determine
the force in each member and
state if the members are in
tension or compression.
Given:
⎛ 200 ⎞
⎜
⎟
F 1 = 300 lb
⎜
⎟
⎝ −500 ⎠
⎛ 0 ⎞
⎜
⎟
F 2 = 400 lb
⎜
⎟
⎝ 0 ⎠
a = 4 ft
b = 3 ft
c = 3 ft
Solution:
Find the external reactions
Guesses
E y = 1 lb
Cy = 1 lb
Cz = 1 lb
Dx = 1 lb
Dy = 1 lb
Dz = 1 lb
521
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0
⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟
⎝ z⎠ ⎝ ⎠ ⎝ z⎠
D
⎛0⎞
⎛ −b ⎞
⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞
⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0
⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟
⎝0⎠
⎝0⎠
⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠
⎛ Ey ⎞
⎜ ⎟
⎜ Cy ⎟
⎜C ⎟
⎜ z ⎟ = Find E , C , C , D , D , D
( y y z x y z)
⎜ Dx ⎟
⎜ ⎟
⎜ Dy ⎟
⎜ ⎟
⎝ Dz ⎠
⎛ Ey ⎞
⎞
⎜ ⎟ ⎛ −33.333
⎜
⎟
C
⎜ y⎟
−666.667
⎟
⎜C ⎟ ⎜
200
⎜
⎟
z
⎜ ⎟=
⎜
⎟ lb
−200
⎜ Dx ⎟
⎜
⎟
⎜ ⎟
− 13 ⎟
⎜
−1.253 × 10
⎜ Dy ⎟
⎜
⎟
⎜ ⎟ ⎝
300
⎠
Dz
⎝ ⎠
Now find the force in each member.
⎛ −b ⎞
⎛ −b ⎞
⎜
⎟
AB = 0
AC = ⎜ −a ⎟
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝c ⎠
⎛0⎞
AD = ⎜ −a ⎟
⎜ ⎟
⎝c ⎠
⎛0⎞
AE = ⎜ −a ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
⎛b⎞
⎜
⎟
BC = −a
BE = ⎜ −a ⎟
⎜ ⎟
⎜ ⎟
⎝c ⎠
⎝0⎠
⎛0⎞
BF = ⎜ −a ⎟
⎜ ⎟
⎝0⎠
⎛b⎞
CD = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
⎛0⎞
⎜
⎟
CF = 0
DE = ⎜ 0 ⎟
⎜ ⎟
⎜ ⎟
⎝ −c ⎠
⎝ −c ⎠
⎛ −b ⎞
DF = ⎜ 0 ⎟
⎜ ⎟
⎝ −c ⎠
⎛ −b ⎞
EF = ⎜ 0 ⎟
⎜ ⎟
⎝0⎠
Guesses
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
F AE = 1 lb
F BC = 1 lb
F BE = 1 lb
F BF = 1 lb
F CD = 1 lb
F CF = 1 lb
F DE = 1lb
F DF = 1 lb
F EF = 1 lb
Given
522
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Engineering Mechanics - Statics
F 1 + F AB
AB
F 2 + F BC
BC
AB
BC
Chapter 6
+ FAC
AC
+ FBF
BF
AC
BF
+ FAD
AD
+ FBE
BE
AD
BE
+ FAE
+ F AB
AE
AE
−AB
AB
=0
=0
⎛⎜ 0 ⎟⎞
−AE
−BE
EF
−DE
+ FBE
+ F EF
+ F DE
=0
⎜ Ey ⎟ + FAE
AE
BE
EF
DE
⎜0⎟
⎝ ⎠
F BF
−BF
BF
+ F CF
−CF
CF
+ F DF
−DF
DF
+ FEF
−EF
EF
=0
⎛0⎞
⎜ ⎟
−BC
−AC
CD
CF
+ FAC
+ FCD
+ FCF
=0
⎜ Cy ⎟ + FBC
BC
AC
CD
CF
⎜C ⎟
⎝ z⎠
⎛⎜ FAB ⎞⎟
⎜ FAC ⎟
⎜
⎟
⎜ FAD ⎟
⎜F ⎟
⎜ AE ⎟
⎜ FBC ⎟
⎜
⎟
⎜ FBE ⎟
⎜
⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF)
F BF
⎜
⎟
⎜F ⎟
⎜ CD ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FEF ⎟
⎝
⎠
523
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎛ FBF ⎞
−300
⎜
⎟ ⎛
⎜⎞
⎟ ⎛ 0 ⎞
⎜
⎟
F
F
AC
CD
⎜
⎟
⎜
⎟ ⎜ −500 ⎟
971.825
⎜
⎟
⎟
⎜F ⎟
⎜F ⎟ ⎜
Positive (T)
−
11
⎜
⎟
⎜
⎜ AD ⎟ = 1.121 × 10 ⎜ lbCF ⎟ = −300 ⎟ lb Negative (C)
⎜ FAE ⎟ ⎜ −366.667 ⎜⎟ FDE ⎟ ⎜ 0 ⎟
⎟
⎜
⎟ ⎜
⎜⎟
⎟ ⎜
424.264
⎜
⎟
⎜
⎟
0
⎜ FBC ⎟
⎜ FDF ⎟
⎜
⎟
⎜
⎜
⎟ ⎝
⎜⎠
⎟ ⎝ −300 ⎟⎠
500
F
F
⎝ BE ⎠
⎝ EF ⎠
Problem 6-60
Determine the force in each member of the space truss and state if the members are in tension or
compression. The truss is supported by a ball-and-socket joints at A, B, and E. Hint: The
support reaction at E acts along member EC. Why?
Given:
⎛ −200 ⎞
⎜
⎟
F = 400 N
⎜
⎟
⎝ 0 ⎠
a = 2m
b = 1.5 m
c = 5m
d = 1m
e = 2m
Solution:
⎛ 0 ⎞
⎜
⎟
AC = a + b
⎜
⎟
⎝ 0 ⎠
⎛d⎞
⎜ ⎟
AD = a
⎜ ⎟
⎝e ⎠
⎛ −c − d ⎞
⎜ 0 ⎟
BC =
⎜
⎟
⎝ 0 ⎠
⎛ −c ⎞
⎜ ⎟
BD = −b
⎜ ⎟
⎝e ⎠
Guesses
Given
⎛d⎞
⎜ ⎟
CD = −b
⎜ ⎟
⎝e ⎠
F AC = 1 N
F AD = 1 N
F BC = 1 N
F CD = 1 N
F EC = 1 N
F BD = 1 N
F + FAD
−AD
AD
+ FBD
−BD
BD
+ F CD
−CD
CD
=0
524
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ 0 ⎞⎟
F CD
+ F BC
+ FAC
+⎜ 0 ⎟ =0
CD
BC
AC
⎜ −FEC ⎟
⎝
⎠
CD
−BC
−AC
⎛ FAC ⎞
⎜
⎟
F
⎜ AD ⎟
⎜F ⎟
⎜ BC ⎟ = Find F , F , F , F , F , F
( AC AD BC BD CD EC )
⎜ FBD ⎟
⎜
⎟
F
⎜ CD ⎟
⎜
⎟
⎝ FEC ⎠
⎛ FAC ⎞
⎜
⎟ ⎛ 221 ⎞
F
⎜ AD ⎟ ⎜ 343 ⎟
⎟
⎜F ⎟ ⎜
⎜ BC ⎟ = ⎜ 148 ⎟ N
⎜ FBD ⎟ ⎜ 186 ⎟
⎟
⎜
⎟ ⎜
−
397
⎜
⎟
⎜ FCD ⎟
⎜
⎜
⎟ ⎝ −295 ⎟⎠
F
⎝ EC ⎠
Positive (T)
Negative (C)
Problem 6-61
Determine the force in each member of the space truss and state if the members are in tension or
compression. The truss is supported by ball-and-socket joints at C, D, E, and G.
Units Used:
3
kN = 10 N
Given:
F = 3 kN
a = 2m
b = 1.5 m
c = 2m
d = 1m
e = 1m
Solution:
Fv =
⎛0⎞
⎜b⎟
2
2⎜ ⎟
b + c ⎝ −c ⎠
F
525
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
uAG =
⎛ −e ⎞
⎜ −a ⎟
⎟
2
2⎜
a +e ⎝ 0 ⎠
uAE =
⎛d⎞
⎜ −a ⎟
⎟
2
2⎜
a +d ⎝ 0 ⎠
1
1
⎛0⎞
⎜ ⎟
uAB = 0
⎜ ⎟
⎝ −1 ⎠
uBC = uAE
uBD = uAG
uBE =
⎛d⎞
⎜ −a ⎟
⎟
2
2
2⎜
a +c +d ⎝ c ⎠
uBG =
⎛ −e ⎞
⎜ −a ⎟
⎟
2
2
2⎜
a +e +c ⎝ c ⎠
1
1
Guesses
F AB = 1 kN
F AE = 1 kN
F BC = 1 kN
F BD = 1 kN
F BE = 1 kN
F BG = 1 kN
F AG = 1 kN
Given
−c
2
c +b
2
e
2
a
F ( a) −
2
a +e
2
2
a +d
F BD( a) −
FBC ( c) −
d
2
a +d
2
a
2
2
a +e
FBD( c) = 0
F BC ( a) = 0
F v + F AEuAE + FAGuAG + F ABuAB = 0
−F AB uAB + FBGuBG + FBEuBE + F BC uBC + FBDuBD = 0
526
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟
⎜ FAE ⎟
⎜
⎟
FAG
⎜
⎟
⎜ F ⎟ = Find F , F , F , F , F , F , F
( AB AE AG BC BD BE BG)
⎜ BC ⎟
⎜ FBD ⎟
⎜
⎟
⎜ FBE ⎟
⎜F ⎟
⎝ BG ⎠
⎛ FAB ⎞
⎜
⎟ ⎛ −2.4 ⎞
⎟
⎜ FAE ⎟ ⎜
⎜
⎟ ⎜ 1.006 ⎟
⎜ FAG ⎟ ⎜ 1.006 ⎟
⎜ F ⎟ = ⎜ −1.342 ⎟ kN
⎟
⎜ BC ⎟ ⎜
⎜ FBD ⎟ ⎜ −1.342 ⎟
⎜
⎟ ⎜ 1.8 ⎟
⎟
⎜ FBE ⎟ ⎜
1.8
⎝
⎠
⎜F ⎟
⎝ BG ⎠
Positive (T)
Negative (C)
Problem 6-62
Determine the force in
members BD, AD, and AF of
the space truss and state if the
members are in tension or
compression. The truss is
supported by short links at A,
B, D, and F.
Given:
⎛ 0 ⎞
⎜
⎟
F = 250 lb
⎜
⎟
⎝ −250 ⎠
a = 6 ft
b = 6 ft
θ = 60 deg
Solution:
Find the external reactions
h = b sin ( θ )
Guesses
Ax = 1 lb
B y = 1 lb
B z = 1 lb
Dy = 1 lb
F y = 1 lb
F z = 1 lb
⎞
⎛
Given
⎛
⎞
527
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟
F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0
⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟
⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠
⎛ 0.5b ⎞
⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞
⎜ a ⎟ × F + ⎜a⎟ ×
⎜ ⎟ B
⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0
⎜
0 ⎟ + 0 × ⎜ y⎟ +
⎜
⎟
⎜ ⎟
⎜ ⎟
⎜
⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟
⎜
⎟
⎜
⎟
⎝ h ⎠
⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠
⎛ Ax ⎞
⎜ ⎟
⎜ By ⎟
⎜B ⎟
⎜ z ⎟ = Find A , B , B , D , F , F
( x y z y y z)
⎜ Dy ⎟
⎜ ⎟
⎜ Fy ⎟
⎜ ⎟
⎝ Fz ⎠
⎛ Ax ⎞
⎜ ⎟ ⎛ 0 ⎞
⎜ By ⎟ ⎜ 72 ⎟
⎟
⎜B ⎟ ⎜
⎜
⎜ z ⎟ = 125 ⎟ lb
⎜ Dy ⎟ ⎜ −394 ⎟
⎟
⎜ ⎟ ⎜
72
⎜
⎟
⎜ Fy ⎟
⎜
⎜ ⎟ ⎝ 125 ⎟⎠
⎝ Fz ⎠
Now find the forces in the members
⎛0⎞
⎜ ⎟
AB = −a
⎜ ⎟
⎝0⎠
⎛ −b ⎞
⎜ ⎟
AC = −a
⎜ ⎟
⎝0⎠
⎛ −0.5b ⎞
⎜ −a ⎟
AD =
⎜
⎟
⎝ h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
AE =
⎜
⎟
⎝ h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
BD =
⎜
⎟
⎝ h ⎠
⎛ −b ⎞
⎜ ⎟
AF = 0
⎜ ⎟
⎝0⎠
⎛ 0.5b ⎞
⎜ 0 ⎟
CD =
⎜
⎟
⎝ h ⎠
⎛ −b ⎞
⎜ ⎟
BC = 0
⎜ ⎟
⎝0⎠
⎛0⎞
⎜ ⎟
CF = a
⎜ ⎟
⎝0⎠
⎛0⎞
⎜ ⎟
DE = a
⎜ ⎟
⎝0⎠
⎛ −0.5b ⎞
⎜ a ⎟
DF =
⎜
⎟
⎝ −h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
EF =
⎜
⎟
⎝ −h ⎠
Guesses
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
F AE = 1 lb
F AF = 1 lb
F BC = 1 lb
F BD = 1 lb
F CD = 1 lb
F CF = 1 lb
528
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F DE = 1 lb
Chapter 6
F DF = 1 lb
F EF = 1 lb
Given
F + FDE
F CF
CF
CF
−DE
DE
+ F AE
+ F CD
−AE
CD
CD
AE
+ FEF
+ F BC
EF
EF
−BC
BC
=0
+ FAC
−AC
AC
=0
⎛⎜ 0 ⎞⎟
F DE
+ FDF
+ F AD
+ F BD
+ FCD
+ ⎜ Dy ⎟ = 0
DE
DF
AD
BD
CD
⎜0 ⎟
⎝ ⎠
⎛0⎞
⎜ ⎟
−AB
BC
BD
F AB
+ FBC
+ F BD
+ ⎜ By ⎟ = 0
AB
BC
BD
⎜B ⎟
⎝ z⎠
DE
DF
−AD
−BD
−CD
⎛⎜ Ax ⎞⎟
F AB
+ FAC
+ FAF
+ F AD
+ F AE
+⎜ 0 ⎟ =0
AB
AC
AF
AD
AE
⎜0 ⎟
⎝ ⎠
AB
AC
AF
AD
AE
⎛⎜ FAB ⎞⎟
⎜ FAC ⎟
⎜
⎟
⎜ FAD ⎟
⎜F ⎟
⎜ AE ⎟
⎜ FAF ⎟
⎜
⎟
⎜ FBC ⎟
⎜
⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF)
⎜ FBD ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FEF ⎟
⎝
⎠
529
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛ FBD ⎞ ⎛ −144.3 ⎞
⎜
⎟ ⎜
⎟
⎜ FAD ⎟ = ⎜ 204.1 ⎟ lb
⎜ F ⎟ ⎝ 72.2 ⎠
⎝ AF ⎠
Chapter 6
Positive (T)
Negative (C)
Problem 6-63
Determine the force in members CF, EF, and DF of the space truss and state if the members
are in tension or compression. The truss is supported by short links at A, B, D, and F.
Given:
⎛ 0 ⎞
⎜
⎟
F = 250 lb
⎜
⎟
⎝ −250 ⎠
a = 6 ft
b = 6 ft
θ = 60 deg
Solution:
Find the external reactions
h = b sin ( θ )
Guesses
Ax = 1 lb
B y = 1 lb
B z = 1 lb
Dy = 1 lb
F y = 1 lb
F z = 1 lb
Given
⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞
F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0
⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟
⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠
⎛ 0.5b ⎞
⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞
⎜ a ⎟ × F + ⎜a⎟ ×
⎜ ⎟
⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0
+ 0 × By +
⎜
⎟
⎜ ⎟ ⎜0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟
⎜
⎟
⎜
⎟
⎝ h ⎠
⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠
530
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ Ax ⎞
⎜ ⎟
⎜ By ⎟
⎜B ⎟
⎜ z ⎟ = Find A , B , B , D , F , F
( x y z y y z)
⎜ Dy ⎟
⎜ ⎟
⎜ Fy ⎟
⎜ ⎟
⎝ Fz ⎠
⎛ Ax ⎞
⎜ ⎟ ⎛ 0 ⎞
⎜ By ⎟ ⎜ 72 ⎟
⎟
⎜B ⎟ ⎜
⎜
⎜ z ⎟ = 125 ⎟ lb
⎜ Dy ⎟ ⎜ −394 ⎟
⎟
⎜ ⎟ ⎜
72
⎜
⎟
⎜ Fy ⎟
⎜
⎜ ⎟ ⎝ 125 ⎟⎠
⎝ Fz ⎠
Now find the forces in the members
⎛0⎞
⎜ ⎟
AB = −a
⎜ ⎟
⎝0⎠
⎛ −b ⎞
⎜ ⎟
AC = −a
⎜ ⎟
⎝0⎠
⎛ −0.5b ⎞
⎜ −a ⎟
AD =
⎜
⎟
⎝ h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
AE =
⎜
⎟
⎝ h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
BD =
⎜
⎟
⎝ h ⎠
⎛ −b ⎞
⎜ ⎟
AF = 0
⎜ ⎟
⎝0⎠
⎛ 0.5b ⎞
⎜ 0 ⎟
CD =
⎜
⎟
⎝ h ⎠
⎛ −b ⎞
⎜ ⎟
BC = 0
⎜ ⎟
⎝0⎠
⎛0⎞
⎜ ⎟
CF = a
⎜ ⎟
⎝0⎠
⎛0⎞
⎜ ⎟
DE = a
⎜ ⎟
⎝0⎠
⎛ −0.5b ⎞
⎜ a ⎟
DF =
⎜
⎟
⎝ −h ⎠
⎛ −0.5b ⎞
⎜ 0 ⎟
EF =
⎜
⎟
⎝ −h ⎠
Guesses
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
F AE = 1 lb
F AF = 1 lb
F BC = 1 lb
F BD = 1 lb
F CD = 1 lb
F CF = 1 lb
F DE = 1 lb
F DF = 1 lb
F EF = 1 lb
Given
F + FDE
F CF
CF
CF
−DE
DE
+ F AE
+ F CD
−AE
CD
CD
AE
+ FEF
+ F BC
EF
EF
−BC
BC
=0
+ FAC
−AC
AC
=0
531
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Engineering Mechanics - Statics
Chapter 6
⎛⎜ 0 ⎞⎟
F DE
+ FDF
+ F AD
+ F BD
+ FCD
+ ⎜ Dy ⎟ = 0
DE
DF
AD
BD
CD
⎜0 ⎟
⎝ ⎠
DE
DF
−AD
−BD
−CD
⎛0⎞
⎜ ⎟
F AB
+ FBC
+ F BD
+ ⎜ By ⎟ = 0
AB
BC
BD
⎜B ⎟
⎝ z⎠
−AB
BC
BD
⎛⎜ Ax ⎞⎟
F AB
+ FAC
+ FAF
+ F AD
+ F AE
+⎜ 0 ⎟ =0
AB
AC
AF
AD
AE
⎜0 ⎟
⎝ ⎠
AB
AC
AF
AD
AE
⎛⎜ FAB ⎞⎟
⎜ FAC ⎟
⎜
⎟
⎜ FAD ⎟
⎜F ⎟
⎜ AE ⎟
⎜ FAF ⎟
⎜
⎟
⎜ FBC ⎟
⎜
⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF)
⎜ FBD ⎟
⎜F ⎟
⎜ CD ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎜ DF ⎟
⎜ FEF ⎟
⎝
⎠
⎛ FCF ⎞ ⎛ 72.2 ⎞
⎜
⎟ ⎜
⎟
⎜ FEF ⎟ = ⎜ −144.3 ⎟ lb
⎜F ⎟ ⎝ 0 ⎠
⎝ DF ⎠
Positive (T)
Negative (C)
532
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-64
Determine the force developed in each member of the space truss and state if the members are
in tension or compression. The crate has weight W.
Given:
W = 150 lb
a = 6 ft
b = 6 ft
c = 6 ft
Solution:
h =
Unit Vectors
c − ⎛⎜
2
a⎞
2
⎟
⎝ 2⎠
uAD =
⎛⎜ −a ⎟⎞
1
⎜ 2 ⎟
2⎜ 0 ⎟
2 ⎛ a⎞ ⎜
⎟
h +⎜ ⎟
⎝ 2⎠ ⎝ h ⎠
uBD =
⎛⎜ a ⎟⎞
1
⎜2⎟
2⎜ 0 ⎟
a
2
h + ⎛⎜ ⎟⎞ ⎜ h ⎟
⎝ 2⎠ ⎝ ⎠
uAC =
⎛⎜ − a ⎟⎞
1
⎜ 2⎟
2⎜ b ⎟
a
2
2
h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟
⎝ 2⎠ ⎝ ⎠
uBC =
⎛⎜ a ⎟⎞
1
⎜2⎟
2⎜ b ⎟
a
2
2
h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟
⎝ 2⎠ ⎝ ⎠
Guesses
F AB = 1 lb
B y = 1 lb
Ax = 1 lb
F AC = 1 lb
Ay = 1 lb
F AD = 1 lb
F BC = 1 lb
F BD = 1 lb
F CD = 1 lb
533
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
⎛⎜ 0 ⎟⎞
⎜ −FCD ⎟ − FAC uAC − FBC uBC = 0
⎜ −W ⎟
⎝
⎠
⎛ FAB ⎞
⎜
⎟
B
⎜ y ⎟ + FBC uBC + FBDuBD = 0
⎜
⎟
⎝ 0 ⎠
⎛ Ax − FAB ⎞
⎜
⎟
A
⎜
⎟ + FAC uAC + FADuAD = 0
y
⎜
⎟
0
⎝
⎠
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜ B ⎟
⎜ y ⎟
⎜ FAB ⎟
⎜
⎟
⎜ FAC ⎟ = Find ( Ax , Ay , By , FAB , FAC , FAD , FBC , FBD , FCD)
⎜F ⎟
⎜ AD ⎟
⎜ FBC ⎟
⎜
⎟
⎜ FBD ⎟
⎟
⎛ FAB ⎞
⎜
⎜
⎟ ⎛ 0.0 ⎞
⎝ FCD ⎠
F
⎜ AC ⎟ ⎜ −122.5 ⎟
⎟
⎜F ⎟ ⎜
⎜
⎟
AD
86.6
Positive (T)
⎜
⎟=
lb
⎜
⎟
Negative (C)
⎜ FBC ⎟
−122.5
⎜
⎟
⎜
⎟
⎜ FBD ⎟ ⎜ 86.6 ⎟
⎜
⎟ ⎜⎝ 173.2 ⎟⎠
FCD
⎝
⎠
Problem 6-65
The space truss is used to support vertical forces at joints B, C, and D. Determine the force in
each member and state if the members are in tension or compression.
534
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Engineering Mechanics - Statics
Chapter 6
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
a = 0.75 m
F 2 = 8 kN
b = 1.00 m
F 3 = 9 kN
c = 1.5 m
Solution:
Assume that the connections at A, E,
and F are rollers
Guesses
F BC = 1 kN
F CF = 1 kN
F CD = 1 kN
F AD = 1 kN
F DF = 1 kN
F DE = 1 kN
F BD = 1 kN
F BA = 1 kN
F EF = 1 kN
F AE = 1 kN
F AF = 1 kN
Given
Joint C
F BC = 0
F CD = 0
−F 2 − F CF = 0
Joint D
a
2
2
a +b
FBD +
a
2
2
2
a +b +c
FAD = 0
535
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Engineering Mechanics - Statics
b
−F CD −
2
−b
+
2
a +b
2
2
2
2
2
b +c
−c
2
F AD −
c
−F 3 − F DE −
+
FBD ...
2
a +b +c
2
a +b +c
Chapter 6
=0
b
2
2
b +c
F DF
F DF ... = 0
F AD
Joint B
a
−F BC −
2
2
a +b
FBD = 0
b
2
2
a +b
FBD = 0
−F 1 − F BA = 0
Joint E
a
2
2
a +b
FAE = 0
−F EF −
b
2
2
a +b
FAE = 0
⎛ FBC ⎞
⎜
⎟
⎜ FCF ⎟
⎜
⎟
⎜ FCD ⎟
⎜F ⎟
⎜ AD ⎟
⎜ FDF ⎟
⎜
⎟
⎜ FDE ⎟ = Find ( FBC , FCF , FCD , FAD , FDF , FDE , FBD , FBA , FEF , FAE , FAF)
⎜F ⎟
⎜ BD ⎟
⎜ FBA ⎟
⎜
⎟
⎜ FEF ⎟
⎜
⎟
⎜ FAE ⎟
⎜F ⎟
⎝ AF ⎠
536
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Engineering Mechanics - Statics
⎛ FBC ⎞
⎜
⎟
⎜ FCF ⎟ ⎛⎜ 0.00 ⎟⎞
⎜
⎟ ⎜ −8.00 ⎟
⎜ FCD ⎟ ⎜
⎟
⎜ F ⎟ ⎜ 0.00 ⎟
AD
⎜
⎟ ⎜ 0.00 ⎟
⎜ FDF ⎟ ⎜ 0.00 ⎟
⎜
⎟ ⎜
⎟
⎜ FDE ⎟ = ⎜ −9.00 ⎟ kN
⎜ F ⎟ ⎜ 0.00 ⎟
⎜ BD ⎟ ⎜
⎟
⎜ FBA ⎟ ⎜ −6.00 ⎟
⎜
⎟ ⎜ 0.00 ⎟
⎜ FEF ⎟ ⎜
⎟
⎜
⎟ ⎜ 0.00 ⎟
F AE
⎜
⎟ ⎝ 0.00 ⎠
⎜F ⎟
⎝ AF ⎠
Chapter 6
Positive (T)
Negative (C)
Problem 6-66
A force P is applied to the handles of the pliers. Determine the force developed on the smooth
bolt B and the reaction that pin A exerts on its attached members.
Given:
P = 8 lb
a = 1.25 in
b = 5 in
c = 1.5 in
Solution:
ΣMA = 0;
−R B c + P b = 0
RB = P
b
c
R B = 26.7 lb
ΣF x = 0;
Ax = 0
ΣF y = 0;
Ay − P − RB = 0
537
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Ay = P + RB
Ay = 34.7 lb
Problem 6-67
The eye hook has a positive locking latch when it supports the load because its two parts are
pin-connected at A and they bear against one another along the smooth surface at B. Determine
the resultant force at the pin and the normal force at B when the eye hook supports load F .
Given:
F = 800 lb
a = 0.25 in
b = 3 in
c = 2 in
θ = 30 deg
Solution:
Σ MA = 0;
−F B cos ( 90 deg − θ ) ( b) − F B sin ( 90 deg − θ ) ( c) + F a = 0
FB = F
+
↑Σ Fy = 0;
+
Σ F x = 0;
→
a
cos ( 90 deg − θ ) b + sin ( 90 deg − θ ) c
F B = 61.9 lb
−F − FB sin ( 90 deg − θ ) + A y = 0
Ay = F + F B sin ( 90 deg − θ )
Ay = 854 lb
Ax − F B cos ( 90 deg − θ ) = 0
Ax = FB cos ( 90 deg − θ )
Ax = 30.9 lb
2
F A = 854 lb
FA =
2
Ax + Ay
Problem 6-68
Determine the force P needed to hold the block of mass F in equilibrium.
538
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Engineering Mechanics - Statics
Chapter 6
Given:
F = 20 lb
Solution:
Pulley B:
ΣF y = 0;
2P − T = 0
Pulley A:
ΣF y = 0;
2T − F = 0
T =
1
F
2
2P = T
T = 10 lb
P =
1
T
2
P = 5 lb
Problem 6-69
The link is used to hold the rod in place. Determine the required axial force on the screw at E if
the largest force to be exerted on the rod at B, C or D is to be F max. Also, find the magnitude of
the force reaction at pin A. Assume all surfaces of contact are smooth.
Given:
F max = 100 lb
a = 100 mm
b = 80 mm
c = 50 mm
θ = 45 deg
539
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Assign an initial value for R E. This will be scaled
at the end of the problem.
Guesses
Given
Ax = 1 lb
Ay = 1 lb
R B = 1 lb
R C = 1 lb
R D = 1 lb
R E = 1 lb
− Ax + R E − R B cos ( θ ) = 0
Ay − R B sin ( θ ) = 0
R E a − RB cos ( θ ) ( a + b + c cos ( θ ) ) − RB sin ( θ ) c sin ( θ ) = 0
R B cos ( θ ) − R D = 0
R B sin ( θ ) − RC = 0
⎛⎜ Ax ⎟⎞
⎜ Ay ⎟
⎜ ⎟
⎜ RB ⎟ = Find ( Ax , Ay , RB , RC , RD)
⎜R ⎟
⎜ C⎟
⎜ RD ⎟
⎝ ⎠
⎛⎜ Ax ⎟⎞
⎛ 0.601 ⎞
⎜ Ay ⎟ ⎜ 0.399 ⎟
⎜ ⎟ ⎜
⎟
⎜ RB ⎟ = ⎜ 0.564 ⎟ lb
⎜ R ⎟ ⎜ 0.399 ⎟
⎜ C⎟ ⎜
⎟
0.399
⎝
⎠
⎜ RD ⎟
⎝ ⎠
Now find the critical load and scale the problem
⎛ RB ⎞
⎜ ⎟
ans = ⎜ R C ⎟
⎜R ⎟
⎝ D⎠
F scale =
Fmax
R E = Fscale RE
max ( ans )
F A = Fscale
2
2
Ax + Ay
R E = 177.3 lb
F A = 127.9 lb
Problem 6-70
The man of weight W1 attempts to lift himself and the seat of weight W2 using the rope and
pulley system shown. Determine the force at A needed to do so, and also find his reaction on
the seat.
540
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
W1 = 150 lb
W2 = 10 lb
Solution:
Pulley C:
ΣF y = 0;
3T − R = 0
Pulley B:
ΣF y = 0;
3R − P = 0
Thus,
P = 9T
Man and seat:
ΣF y = 0;
T + P − W1 − W2 = 0
10T = W1 + W2
T =
W1 + W2
10
P = 9T
T = 16 lb
P = 144 lb
Seat:
ΣF y = 0;
P − N − W2 = 0
N = P − W2
N = 134 lb
Problem 6-71
Determine the horizontal and vertical components of force that pins A and C exert on the frame.
Given:
F = 500 N
a = 0.8 m
d = 0.4 m
541
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
b = 0.9 m
e = 1.2 m
c = 0.5 m
θ = 45deg
Solution:
BC is a two-force member
Member AB :
ΣMA = 0;
e
−F c + FBC
2
2
a +e
b + F BC
2
a
2
( c + d) = 0
2
a +e
2
a +e
F BC = F c
eb+a c+a d
F BC = 200.3 N
Thus,
Cx = F BC
Cy = F BC
ΣF x = 0;
ΣF y = 0;
Ax − F BC
e
2
Cx = 167 N
2
a +e
a
2
Cy = 111 N
2
a +e
e
=0
2
2
Ay − F + FBC
a
Ax = FBC
a +e
2
=0
2
a +e
e
2
Ax = 167 N
2
a +e
Ay = F − F BC
a
2
2
a +e
Ay = 389 N
Problem 6-72
Determine the horizontal and vertical components of force that pins A and C exert on the
frame.
Units Used:
3
kN = 10 N
Given:
F 1 = 1 kN
542
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Engineering Mechanics - Statics
Chapter 6
F 2 = 500 N
θ = 45 deg
a = 0.2 m
b = 0.2 m
c = 0.4 m
d = 0.4 m
Solution:
Guesses
Ax = 1 N
Ay = 1 N
Cx = 1 N
Cy = 1 N
Given
Ax − Cx = 0
Ay + Cy − F1 − F2 = 0
F1 a − Ay 2 a + Ax d = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟
⎜ ⎟ = Find ( Ax , Ay , Cx , Cy)
⎜ Cx ⎟
⎜C ⎟
⎝ y⎠
−F 2 b + Cy( b + c) − Cx d = 0
⎛ Ax ⎞
⎜ ⎟ ⎛⎜ 500 ⎞⎟
⎜ Ay ⎟ ⎜ 1000 ⎟
N
⎜ ⎟=⎜
⎟
C
500
x
⎜ ⎟ ⎜
⎟
⎜ C ⎟ ⎝ 500 ⎠
⎝ y⎠
Problem 6-73
The truck exerts the three forces shown on the girders of the bridge. Determine the reactions at
the supports when the truck is in the position shown. The girders are connected together by a
short vertical link DC.
Units Used:
3
kip = 10 lb
543
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
a = 55 ft
f = 12 ft
b = 10 ft
F 1 = 5 kip
c = 48 ft
F 2 = 4 kip
d = 5 ft
F 3 = 2 kip
e = 2 ft
Solution:
Member CE:
e
ΣMC = 0;
−F 3 e + E y( e + c) = 0
Ey = F3
ΣF y = 0;
Cy − F3 + Ey = 0
Cy = F 3 − E y
e+c
E y = 80 lb
Cy = 1920 lb
Member ABD:
ΣMA = 0;
−F 1 a − F2( d + a) − Cy( a + d + b) + By( a + d) = 0
By =
ΣF y = 0;
F 1 a + F2( d + a) + Cy( a + d + b)
d+a
B y = 10.8 kip
Ay − F 1 + B y − F2 − Cy = 0
Ay = Cy + F 1 − B y + F2
Ay = 96.7 lb
Problem 6-74
Determine the greatest force P that can be applied to the frame if the largest force resultant
acting at A can have a magnitude F max.
Units Used:
3
kN = 10 N
544
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F max = 2 kN
a = 0.75 m
b = 0.75 m
c = 0.5 m
d = 0.1 m
Solution:
Σ MA = 0;
T( c + d) − P( a + b) = 0
+
Σ F x = 0;
→
Ax − T = 0
+
↑Σ Fy = 0;
Ay − P = 0
Thus,
T=
a+b
c+d
P
Ay = P
Ax =
a+b
c+d
P
Require,
F max =
P =
2
2
Ax + Ay
F max
2
⎛ a + b⎞ + 1
⎜
⎟
⎝c + d⎠
P = 743 N
Problem 6-75
The compound beam is pin supported at
B and supported by rockers at A and C.
There is a hinge (pin) at D. Determine
the reactions at the supports.
Units Used:
3
kN = 10 N
545
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 7 kN
a = 4m
F 2 = 6 kN
b = 2m
F 3 = 16 kN
c = 3m
θ = 60 deg
d = 4m
Solution:
Member DC :
ΣMD = 0;
−F 1 sin ( θ ) ( a − c) + Cy a = 0
Cy = F 1 sin ( θ )
ΣF y = 0;
a−c
Cy = 1.52 kN
a
Dy − F 1 sin ( θ ) + Cy = 0
Dy = F1 sin ( θ ) − Cy
ΣF x = 0;
Dy = 4.55 kN
Dx − F 1 cos ( θ ) = 0
Dx = F1 cos ( θ )
Dx = 3.5 kN
Member ABD :
ΣMA = 0;
−F 3 a − F2( 2 a + b) − Dy( 3 a + b) + By2 a = 0
By =
ΣF y = 0;
F 3 a + F2( 2a + b) + Dy ( 3a + b)
2a
Ay − F 3 + B y − F2 − Dy = 0
Ay = Dy + F3 − By + F 2
ΣF x = 0;
B y = 23.5 kN
Ay = 3.09 kN
B x − F1 cos ( θ ) = 0
B x = F 1 cos ( θ )
B x = 3.5 kN
Problem 6-76
The compound beam is fixed supported at A and supported by rockers at B and C. If there are
hinges (pins) at D and E, determine the reactions at the supports A, B, and C.
546
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Units Used:
3
kN = 10 N
Given:
a = 2 m M = 48 kN⋅ m
b = 4m
kN
w1 = 8
m
c = 2m
kN
w2 = 6
d = 6m
m
e = 3m
Solution:
Guesses
Ax = 1 N
Ay = 1 N
MA = 1 N m
Dx = 1 N
Dy = 1 N
By = 1 N
Ey = 1 N
Ex = 1 N
Cy = 1 N
Given
Ay − w2 a − Dy = 0
MA − w2 a
−w1
a
2
( b + c)
2
2
E y − w1
−w1⎛⎜
− Dy a = 0
d+e
2
+ B y b − E y( b + c) = 0
+ Cy = 0
− Ax − Dx = 0
Dy − w1( b + c) + By − E y = 0
Dx + E x = 0
−E x = 0
d + e⎞⎛ d + e⎞
⎟⎜
⎟ + Cy d − M = 0
⎝ 2 ⎠⎝ 3 ⎠
547
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜M ⎟
⎜ A⎟
⎜ Dx ⎟
⎜
⎟
D
⎜ y ⎟ = Find ( Ax , Ay , MA , Dx , Dy , By , Ey , Ex , Cy)
⎜B ⎟
⎜ y⎟
⎜ Ey ⎟
⎜
⎟
E
⎜ x⎟
⎟
⎜
⎝ Cy ⎠
⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞
=
kN
⎜ Ay ⎟ ⎜⎝ 19 ⎟⎠
⎝ ⎠
MA = 26 kN m
B y = 51 kN
Cy = 26 kN
Problem 6-77
Determine the reactions at supports A and B.
Units Used:
3
kip = 10 lb
Given:
lb
w1 = 500
ft
lb
w2 = 700
ft
a = 6 ft
b = 8 ft
c = 9 ft
d = 6 ft
Solution:
Guesses
Ax = 1 lb
Ay = 1 lb
B x = 1 lb
B y = 1 lb
F CD = 1 lb MA = 1 lb⋅ ft
548
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Engineering Mechanics - Statics
Chapter 6
Given
Ay − w1 a −
MA − w1 a a −
d
2
2
b +d
d
2
b +d
2
F CD = 0
d
2
d +b
FCD − w2
c
2
2
Ax −
b
2
b +d
b
F CD2 a = 0
2
2
b +d
+ By = 0
B y c − w2
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜ B ⎟
⎜ x ⎟ = Find A , A , B , B , F , M
( x y x y CD A)
⎜ By ⎟
⎜
⎟
⎜ FCD ⎟
⎜
⎟
⎝ MA ⎠
2
FCD = 0
FCD − Bx = 0
c c
2 3
=0
⎛⎜ Ax ⎞⎟ ⎛ 2.8 ⎞
=
kip
⎜ Ay ⎟ ⎜⎝ 5.1 ⎟⎠
⎝ ⎠
MA = 43.2 kip⋅ ft
⎛⎜ Bx ⎟⎞ ⎛ 2.8 ⎞
=
kip
⎜ By ⎟ ⎜⎝ 1.05 ⎟⎠
⎝ ⎠
Problem 6-78
Determine the horizontal and vertical components of force at C which member ABC exerts on
member CEF.
Given:
F = 300 lb
a = 4 ft
b = 6 ft
c = 3 ft
r = 1 ft
Solution:
Guesses
Ax = 1 lb
Ay = 1 lb
F y = 1 lb
549
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Engineering Mechanics - Statics
Cx = 1 lb
Chapter 6
Cy = 1 lb
Given
Ax a − Ay
b
2
− Cx a − Cy
b
2
=0
Cx a − F r = 0
Ax = 0
Ay + Fy − F = 0
F y b − F ( b + c + r) = 0
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜ ⎟
⎜ Fy ⎟ = Find ( Ax , Ay , Fy , Cx , Cy)
⎜C ⎟
⎜ x⎟
⎜ Cy ⎟
⎝ ⎠
⎛ Ax ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ −200 ⎟ lb
⎜ F ⎟ ⎝ 500 ⎠
⎝ y⎠
⎛⎜ Cx ⎟⎞ ⎛ 75 ⎞
=
lb
⎜ Cy ⎟ ⎜⎝ 100 ⎟⎠
⎝ ⎠
Problem 6-79
Determine the horizontal and vertical components of force that the pins at A, B, and C exert on
their connecting members.
Units Used:
3
kN = 10 N
Given:
F = 800 N
a = 1m
r = 50 mm
b = 0.2 m
Solution:
−F ( a + r) + A x b = 0
550
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Engineering Mechanics - Statics
Ax = F
a+r
b
Chapter 6
Ax = 4.2 kN
− Ax + Bx = 0
Bx = Ax
B x = 4.2 kN
−F r − Ay b + A x b = 0
Ay =
−F r + A x b
b
Ay = 4 kN
Ay − By − F = 0
By = Ay − F
B y = 3.2 kN
− Ax + F + Cx = 0
Cx = Ax − F
Cx = 3.4 kN
Ay − Cy = 0
Cy = Ay
Cy = 4 kN
Problem 6-80
Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod
DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the spring is
compressed a distance δ when the valve is open as shown, determine the normal force acting on
the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth.The spring has a
stiffness k.
Given:
a = 25 mm
b = 40 mm
δ = 20 mm
k = 300
N
m
551
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Engineering Mechanics - Statics
Chapter 6
Solution:
F s = kδ
Fs = 6 N
ΣF y = 0;
−F G + Fs = 0
FG = Fs
FG = 6 N
ΣMF = 0;
FG b + T a = 0
T = FG
b
a
T = 9.60 N
552
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Engineering Mechanics - Statics
Chapter 6
Problem 6-81
Determine the force P on the cord, and the angle θ that the pulley-supporting link AB makes
with the vertical. Neglect the mass of the pulleys and the link. The block has weight W and the
cord is attached to the pin at B. The pulleys have radii of r1 and r2.
Given:
W = 200 lb
r1 = 2 in
r2 = 1 in
φ = 45 deg
Solution:
The initial guesses are
θ = 30 deg
F AB = 30 lb
+
2P − W = 0
↑Σ Fy = 0;
P =
1
2
W
P = 100 lb
Given
+
Σ F x = 0;
→
P cos ( φ ) − FAB sin ( θ ) = 0
+
F AB cos ( θ ) − P − P − P sin ( φ ) = 0
↑Σ Fy = 0;
⎛ θ ⎞
⎜ F ⎟ = Find ( θ , FAB)
⎝ AB ⎠
θ = 14.6 deg
F AB = 280 lb
Problem 6-82
The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin
connected at B and the surface at D is smooth, determine the normal force on the fingernail
when a force F is applied to the handles as shown.The pin AC slides through a smooth hole at
A and is attached to the bottom member at C.
553
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F = 1 lb
a = 0.25 in
b = 1.5 in
Solution:
Handle :
ΣMD = 0;
FA a − F b = 0
F A = F ⎛⎜
b⎞
⎟
⎝ a⎠
ΣF y = 0;
F A = 6 lb
ND − FA − F = 0
ND = F A + F
ND = 7 lb
Top blade :
ΣMB = 0;
ND b − F N( 2 a + b) = 0
F N = ND ⎛⎜
b
⎞
⎟
⎝ 2 a + b⎠
F N = 5.25 lb
Problem 6-83
The wall crane supports load F. Determine the horizontal and vertical components of
reaction at the pins A and D. Also, what is the force in the cable at the winch W?
Units Used:
3
kip = 10 lb
Given:
F = 700 lb
554
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
a = 4 ft
b = 4 ft
c = 4 ft
θ = 60 deg
Solution:
Pulley E:
+
↑Σ Fy = 0;
T =
1
2
2T − F = 0
T = 350 lb
F
This is the force in the cable at the winch W
a
φ = atan ⎛⎜ ⎟⎞
Member ABC:
⎝ b⎠
Σ MA = 0;
(
)
−F ( b + c) + TBD sin ( φ ) − T sin ( θ ) b = 0
⎛ b + c ⎞ + T sin ( θ )
⎟
⎝ b ⎠
F⎜
TBD =
sin ( φ )
3
TBD = 2.409 × 10 lb
+
↑Σ Fy = 0;
− Ay + TBD sin ( φ ) − T sin ( θ ) − F = 0
Ay = TBD sin ( φ ) − T sin ( θ ) − F
Ay = 700 lb
+
Σ F x = 0;
→
Ax − TBD cos ( φ ) − T cos ( θ ) = 0
Ax = TBD cos ( φ ) + T cos ( θ )
Ax = 1.878 kip
At D:
555
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Dx = TBD cos ( φ )
Dx = 1.703 kip
Dy = TBD sin ( φ )
Dy = 1.703 kip
Problem 6-84
Determine the force that the smooth roller C exerts on beam AB. Also, what are the horizontal and
vertical components of reaction at pin A? Neglect the weight of the frame and roller.
Given:
M = 60 lb⋅ ft
a = 3 ft
b = 4 ft
c = 0.5 ft
Solution:
Σ MA = 0;
−M + Dx c = 0
M
Dx =
c
Dx = 120 lb
+
Σ F x = 0;
→
Ax − Dx = 0
Ax = Dx
Ax = 120 lb
↑Σ Fy = 0;
Ay = 0
Ay = 0 lb
Σ MB = 0;
−Nc b + Dx c = 0
+
c
Nc = Dx
b
Nc = 15.0 lb
Problem 6-85
Determine the horizontal and vertical components of force which the pins exert on member ABC.
556
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Engineering Mechanics - Statics
Chapter 6
Given:
W = 80 lb
a = 6 ft
b = 9 ft
c = 3 ft
r = 0.5 ft
Solution:
+
Σ F x = 0;
→
− Ax + W = 0
Ax = W
+
↑Σ Fy = 0;
Ax = 80 lb
Ay − W = 0
Ay = W
Σ MC = 0;
Ay = 80 lb
Ay( a + b) − B y b = 0
By = Ay
a+b
b
B y = 133 lb
Σ MD = 0;
−W( c − r) + By b − Bx c = 0
Bx =
+
Σ F x = 0;
→
B y b − W ( c − r)
B x = 333 lb
c
Ax + B x − Cx = 0
Cx = Ax + B x
+
↑Σ Fy = 0;
Cx = 413 lb
− Ay + B y − Cy = 0
Cy = B y − A y
Cy = 53.3 lb
Problem 6-86
The floor beams AB and BC are stiffened using the two tie rods CD and AD. Determine the
force along each rod when the floor beams are subjected to a uniform load w. Assume the
three contacting members at B are smooth and the joints at A, C, and D are pins. Hint:
557
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
g
j
Members AD, CD, and BD are two-force members
p
3
kip = 10 lb
Units Used:
Given:
w = 80
lb
ft
b = 5 ft
a = 12 ft
Solution:
Due to summetry:
Cy =
w( 2a)
2
Cy = 960 lb
Member BC :
ΣMB = 0;
a
Cy( a) − w a⎛⎜ ⎟⎞ − T⎛
⎜
2
b
2
2
⎝ ⎠
T = ⎛⎜ Cy − w
⎝
a⎞
⎟
2⎠
⎞a = 0
2
2⎟
⎝ a +b ⎠
a +b
T = 1.248 kip
b
Problem 6-87
Determine the horizontal and vertical components of force at pins B and C.
Given:
F = 50 lb
c = 6 ft
a = 4 ft
d = 1.5 ft
558
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
b = 4 ft
Chapter 6
r = 0.5 ft
Solution:
Guesses
Cx = 1 lb
Cy = 1 lb
B x = 1 lb
B y = 1 lb
Given
F ( a − r) + Cx c − Cy( a + b) = 0
−F ( a − r) − F( d + r) + Cy( a + b) = 0
−B x + F + Cx = 0
B y − F + Cy = 0
⎛ Bx ⎞
⎜ ⎟
⎜ By ⎟
⎜ ⎟ = Find ( Bx , By , Cx , Cy)
⎜ Cx ⎟
⎜C ⎟
⎝ y⎠
⎛ Bx ⎞
⎜ ⎟ ⎛⎜ 66.667 ⎞⎟
⎜ By ⎟ ⎜ 15.625 ⎟
lb
⎜ ⎟=⎜
⎟
⎜ Cx ⎟ ⎜ 16.667 ⎟
⎜ C ⎟ ⎝ 34.375 ⎠
⎝ y⎠
559
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-88
The skid steer loader has a mass M1, and in the position shown the center of mass is at G1. If
there is a stone of mass M2 in the bucket, with center of mass at G2 determine the reactions
of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and
at the pin E. There is a similar linkage on each side of the loader.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
Given:
M1 = 1.18 Mg
M2 = 300 kg
a = 1.25 m
d = 0.15 m
b = 1.5 m
e = 0.5 m
c = 0.75 m
θ = 30 deg
Solution:
Entire System:
ΣMA = 0;
M2 g b − M1 g( c − d) + NB c = 0
NB =
ΣF y = 0;
M1 g( c − d) − M2 g b
c
NB = 3.37 kN
(Both wheels)
NA = 11.1 kN
(Both wheels)
NB − M2 g − M1 g + NA = 0
NA = −NB + M2 g + M1 g
Upper member:
ΣME = 0;
M2 g( a + b) − 2 F CD sin ( θ ) a = 0
560
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
M 2 g( a + b)
F CD =
ΣF x = 0;
E x = F CD( cos ( θ ) )
ΣF y = 0;
Ey −
Ey =
FR =
F CD = 6.5 kN
2 sin ( θ ) a
M2 g
E x = 5607 N
+ FCD sin ( θ ) = 0
2
M2 g
2
− F CD sin ( θ )
E y = −1766 N
2
F R = 5.879 kN
2
Ex + Ey
Problem 6-89
Determine the horizontal and vertical components of force at each pin. The suspended cylinder
has a weight W.
Given:
W = 80 lb
d = 6 ft
a = 3 ft
e = 2 ft
b = 4 ft
r = 1 ft
c = 4 ft
Solution:
Guesses
Ax = 1 lb
B x = 1 lb
B y = 1 lb
F CD = 1 lb
E x = 1 lb
E y = 1 lb
Given
Ex − Bx + W = 0
−E y + By = 0
561
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
−W r + Ey a − Ex d = 0
−B y +
−B y a +
c
2
FCD − W = 0
2
c +d
c
2
c +d
− Ax + Bx +
2
F CD d − W( d + e − r) = 0
d
2
2
c +d
FCD − W = 0
⎛ Ax ⎞
⎜
⎟
B
x
⎜
⎟
⎜ B ⎟
⎜ y ⎟ = Find A , B , B , F , E , E
( x x y CD x y)
⎜ FCD ⎟
⎜
⎟
⎜ Ex ⎟
⎜
⎟
⎝ Ey ⎠
Cx = F CD
d
2
c +d
2
Cy = F CD
c
2
c +d
2
Dx = −Cx
Dy = −Cy
⎛ Ax ⎞
⎜ ⎟ ⎛
160
⎞
⎜ Bx ⎟ ⎜
⎟
80
⎜B ⎟ ⎜
⎟
26.667
⎜ y⎟ ⎜
⎟
⎜ Cx ⎟ ⎜
⎟
160
⎜ ⎟ ⎜
⎟
106.667
C
=
⎜ y⎟ ⎜
⎟ lb
⎜D ⎟ ⎜
⎟
−160
⎜ x⎟ ⎜
⎟
⎜ Dy ⎟ ⎜ −106.667 ⎟
⎜ ⎟ ⎜ −8.694 × 10− 13 ⎟
⎜ Ex ⎟ ⎜
⎟
26.667
⎠
⎜ ⎟ ⎝
Ey
⎝ ⎠
562
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Engineering Mechanics - Statics
Chapter 6
Problem 6-90
The two-member frame is pin connected at C, D, and E. The cable is attached to A, passes
over the smooth peg at B, and is attached to a load W. Determine the horizontal and vertical
reactions at each pin.
Given:
a = 2 ft
b = 1 ft
c = 0.75 ft
W = 100 lb
Solution:
d =
c
( a + 2 b)
b
Initial guesses:
Cx = 1 lb
Cy = 1 lb
Dx = 1 lb
Dy = 1 lb
E x = 1 lb
E y = 1 lb
Given
−Dx + Cx − W = 0
−Dy + Cy − W = 0
−Cx c + Cy b + W d − W( a + 2 b) = 0
W − Cx + Ex = 0
E y − Cy = 0
−W d + Cx c + Cy b = 0
562
563
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ Cx ⎞
⎜ ⎟
⎜ Cy ⎟
⎜D ⎟
⎜ x ⎟ = Find C , C , D , D , E , E
( x y x y x y)
⎜ Dy ⎟
⎜ ⎟
⎜ Ex ⎟
⎜ ⎟
⎝ Ey ⎠
⎛ Cx ⎞
⎜ ⎟ ⎛ 133 ⎞
⎜ Cy ⎟ ⎜ 200 ⎟
⎟
⎜D ⎟ ⎜
⎜
⎜ x ⎟ = 33 ⎟ lb
⎜ Dy ⎟ ⎜ 100 ⎟
⎟
⎜ ⎟ ⎜
33
⎜
⎟
⎜ Ex ⎟
⎜
⎜ ⎟ ⎝ 200 ⎟⎠
⎝ Ey ⎠
Problem 6-91
Determine the horizontal and vertical components of force which the pins at A, B, and C exert
on member ABC of the frame.
Given:
F 1 = 400 N
F 2 = 300 N
F 3 = 300 N
a = 1.5 m
b = 2m
c = 1.5 m
d = 2.5 m
f = 1.5 m
g = 2m
e = a+b+c−d
Solution:
Guesses
Ay = 1 N
Cx = 1 N
F BD = 1 N
F BE = 1 N
Cy = 1 N
564
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
F 1 g + F2( a + b) + F3 a − Ay( f + g) = 0
F 1 g − Cy( f + g) = 0
Cx e = 0
−Cx −
f+g
2
e + ( f + g)
Ay − Cy −
2
FBD +
e
2
e + ( f + g)
2
f+g
2
d + ( f + g)
FBD −
2
F BE = 0
d
2
d + ( f + g)
2
F BE = 0
⎛⎜ Ay ⎞⎟
⎜ Cx ⎟
⎜
⎟
C
⎜ y ⎟ = Find ( Ay , Cx , Cy , FBD , FBE)
⎜F ⎟
⎜ BD ⎟
⎜ FBE ⎟
⎝
⎠
Bx = −
By =
f+g
2
e + ( f + g)
e
2
e + ( f + g)
2
2
F BD +
F BD +
f+g
2
d + ( f + g)
d
2
d + ( f + g)
2
2
FBE
FBE
Ay = 657 N
⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞
=
N
⎜ By ⎟ ⎜⎝ 429 ⎟⎠
⎝ ⎠
⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞
=
N
⎜ Cy ⎟ ⎜⎝ 229 ⎟⎠
⎝ ⎠
565
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-92
The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported
by the derrick if the maximum force that can be sustained by the pin at A is Fmax.
Units Used:
3
kN = 10 N
m
g = 9.81
2
s
3
Mg = 10 kg
Given:
F max = 18 kN
L = 5m
θ = 60 deg
Solution:
AB is a two-force member.
Require F AB = Fmax
+
↑
Σ F y = 0;
F AB sin ( θ ) −
Mg
2
sin ( θ ) − W = 0
⎛ FAB ⎞ ⎛ sin ( θ ) ⎞
⎟⎜
⎟
⎝ g ⎠ ⎝ sin ( θ ) + 2 ⎠
M = 2⎜
M = 5.439
1
2
Mg
s
Problem 6-93
Determine the required mass of the suspended cylinder if the tension in the chain wrapped
around the freely turning gear is T. Also, what is the magnitude of the resultant force on pin A?
Units Used:
3
kN = 10 N
g = 9.8
m
2
s
566
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
T = 2 kN
L = 2 ft
θ = 30 deg
φ = 45 deg
Solution:
Σ MA = 0;
−2 T L cos ( θ ) + M g cos ( φ ) L cos ( θ ) + M g sin ( φ ) L sin ( θ ) = 0
M =
2 T cos ( θ )
(cos ( φ ) cos ( θ ) + sin (φ ) sin ( θ )) g
M = 1793
+
→ Σ Fx = 0;
1
2
kg
s
2 T − M g cos ( φ ) − A x = 0
Ax = 2 T − M g cos ( φ )
+
↑Σ Fy = 0;
M g sin ( φ ) − Ay = 0
Ay = M g sin ( φ )
FA =
2
2
Ax + Ay
F A = 2.928 kN
567
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-94
The tongs consist of two jaws pinned to links at A, B, C, and D. Determine the horizontal and
vertical components of force exerted on the stone of weight W at F and G in order to lift it.
Given:
a = 1 ft
b = 2 ft
c = 1.5 ft
d = 1 ft
W = 500 lb
Solution:
Guesses
F x = 1 lb
F y = 1 lb
F AD = 1 lb
F BE = 1 lb
Given
2 Fy − W = 0
F AD b − Fx( b + c) = 0
a
⎞
⎜ 2 2 ⎟ FBE = 0
⎝ a +d ⎠
F AD − F x − ⎛
⎛ Fx ⎞
⎜
⎟
⎜ Fy ⎟
⎜
⎟ = Find ( Fx , Fy , FAD , FBE)
⎜ FAD ⎟
⎜F ⎟
⎝ BE ⎠
d
⎞
⎜ 2 2 ⎟ FBE = 0
⎝ a +d ⎠
−F y + ⎛
⎛⎜ Gx ⎞⎟ ⎛⎜ Fx ⎟⎞
=
⎜ Gy ⎟ ⎜ Fy ⎟
⎝ ⎠ ⎝ ⎠
⎛ Fx ⎞
⎜ ⎟ ⎛⎜ 333 ⎟⎞
⎜ Fy ⎟ ⎜ 250 ⎟
lb
⎜ ⎟=⎜
⎟
⎜ Gx ⎟ ⎜ 333 ⎟
⎜ G ⎟ ⎝ 250 ⎠
⎝ y⎠
Problem 6-95
Determine the force P on the cable if the spring is compressed a distance δ when the mechanism
is in the position shown. The spring has a stiffness k.
568
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
δ = 0.5 in
c = 6 in
lb
d = 6 in
k = 800
ft
a = 24 in
e = 4 in
b = 6 in
θ = 30 deg
Solution:
F E = kδ
F E = 33.333 lb
The initial guesses are
P = 20 lb
B x = 11 lb
B y = 34 lb
F CD = 34 lb
Given
ΣMA = 0;
B x b + B y c − F E ( a + b) = 0
ΣMD = 0;
By d − P e = 0
+
Σ F x = 0;
→
ΣMB = 0;
−B x + FCD cos ( θ ) = 0
F CD sin ( θ ) d − P ( d + e) = 0
⎛⎜ FCD ⎞⎟
⎜ Bx ⎟
⎜
⎟ = Find ( FCD , Bx , By , P)
B
⎜ y ⎟
⎜ P ⎟
⎝
⎠
B x = 135.398 lb
B y = 31.269 lb
F CD = 156.344 lb
P = 46.903 lb
569
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-96
The scale consists of five pin-connected members. Determine the load W on the pan EG if a
weight F is suspended from the hook at A.
Given:
F = 3 lb
b = 3 in
a = 5 in
c = 4 in
d = 6 in
f = 2 in
e = 8 in
Solution:
Guesses
TC = 10 lb
TD = 10 lb
TG = 10 lb
W = 10 lb
Given
Member ABCD:
ΣMB = 0;
F a − TC b − TD( b + e − d) = 0
Member EG:
ΣMG = 0;
− TC e + W c = 0
ΣF y = 0;
TG − W + TC = 0
Member FH:
ΣMH = 0;
−TD( d + f) + TG f = 0
⎛⎜ TC ⎟⎞
⎜ TD ⎟
⎜ ⎟ = Find ( TC , TD , TG , W)
⎜ TG ⎟
⎜W⎟
⎝ ⎠
W = 7.06 lb
Problem 6-97
The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF,
570
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
g
p
gg
which are operated by the hydraulic cylinder H. The toggles push the movable bar G forward,
pressing the plate p into the cavity. If the force which the plate exerts on the head is P,
determine the force F in the hydraulic cylinder for the given angle θ.
Units Used:
3
kN = 10 N
Given:
P = 12 kN
a = 200 mm
θ = 30 deg
Solution:
Member EF:
ΣME = 0;
ΣF x = 0;
ΣF y = 0;
−F y a cos ( θ ) +
Fy =
P
Ex −
P
Ex =
P
2
a sin ( θ ) = 0
tan ( θ )
2
2
P
F y = 3.464 kN
=0
E x = 6 kN
2
Ey − Fy = 0
Ey = Fy
E y = 3.464 kN
Joint E:
ΣF x = 0;
−F DE cos ( θ ) − E x = 0
F DE =
−E x
cos ( θ )
F DE = −6.928 kN
F − Ey + F DE sin ( θ ) = 0
ΣF y = 0;
F = E y − FDE sin ( θ )
F = 6.93 kN
Problem 6-98
Determine the horizontal and vertical components of force at pins A and C of the two-member
frame.
571
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Engineering Mechanics - Statics
Chapter 6
Given:
N
w1 = 500
m
N
w2 = 400
m
N
w3 = 600
m
a = 3m
b = 3m
Solution:
Guesses
Ax = 1 N
Ay = 1 N Cx = 1 N
Cy = 1 N
Given
Ay + Cy −
1
w a − w2 a = 0
2 1
Ax a −
1
w b=0
2 3
− Ay a +
− Ax + Cx +
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟
⎜ ⎟ = Find ( Ax , Ay , Cx , Cy)
⎜ Cx ⎟
⎜C ⎟
⎝ y⎠
1
2a 1
b
a
w1 a
− w3 b − w2 a = 0
2
3
2
3
2
1
a
w1 a = 0
2
3
⎛ Ax ⎞
⎜ ⎟ ⎛⎜ 1400 ⎞⎟
⎜ Ay ⎟ ⎜ 250 ⎟
N
⎜ ⎟=⎜
⎟
C
500
x
⎜ ⎟ ⎜
⎟
⎜ C ⎟ ⎝ 1700 ⎠
⎝ y⎠
572
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-99
The truck rests on the scale, which consists of a series of compound levers. If a mass M1 is
placed on the pan P and it is required that the weight is located at a distance x to balance the
“beam” ABC, determine the mass of the truck. There are pins at all lettered points. Is it
necessary for the truck to be symmetrically placed on the scale? Explain.
Units Used:
3
Mg = 10 kg
g = 9.81
m
2
s
Given:
M1 = 15 kg
FD = 3 m
x = 0.480 m
EF = 0.2 m
a = 0.2 m
HI = 0.1 m
GH = 2.5 m
KJ = HI
KG = GH
Solution:
Member ABC :
ΣMB = 0;
−M1 g x + F AD a = 0
F AD = M1 g
x
a
2
F AD = 72 s N
Member EFD :
ΣME = 0;
−F y EF + F AD( FD + EF) = 0
⎛ FD + EF ⎞
⎟
⎝ EF ⎠
F y = F AD⎜
2
F y = 1152 s N
573
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Member GHI :
ΣMI = 0;
Hy HI − GY( GH + HI ) = 0
Member JKG :
ΣMJ = 0;
(Fy − Gy)( KJ + GH ) − Ky( KJ ) = 0
Ky + Hy = Fy
KJ + KG
HI
Scale Platform :
ΣF y = 0;
Ky + Hy = W
W = Fy⎛⎜
⎝
M =
W
g
KJ + KG ⎞
HI
⎟
⎠
M = 14.98 Mg
Because KJ = HI and KG = GH it doesn't matter where the truck is on the scale.
Problem 6-100
By squeezing on the hand brake of the bicycle, the
rider subjects the brake cable to a tension T If the
caliper mechanism is pin-connected to the bicycle
frame at B, determine the normal force each brake
pad exerts on the rim of the wheel. Is this the
force that stops the wheel from turning? Explain.
Given:
T = 50 lb
a = 2.5 in
b = 3 in
574
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Σ MB = 0;
−N b + T a = 0
N = T
a
b
N = 41.7 lb
This normal force does not stop the wheel from turning. A frictional force (see Chapter 8),
which acts along the wheel's rim stops the wheel.
Problem 6-101
If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine
the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D.
Given:
P = 6 lb
a = 25 in
b = 4 in
c = 5 in
d = 4 in
e = 5 in
f = 5 in
g = 30 in
Solution:
Σ MA = 0;
F BC b − P a = 0
F BC =
Pa
b
575
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
F BC = 37.5 lb
+
Σ F x = 0;
→
− Ax + P = 0
Ax = P
+
↑Σ Fy = 0;
Ax = 6 lb
− Ay + F BC = 0
Ay = FBC
Σ MD = 0;
Ay = 37.5 lb
−e Ax − Ay( b + c) + ( g + b + c)F = 0
F =
e Ax + Ay( b + c)
F = 9.423 lb
g+b+c
Problem 6-102
The pillar crane is subjected to the load having a mass M. Determine the force developed in the
tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied
in the position shown.
Units Used:
3
kN = 10 N
Given:
M = 500 kg
a = 1.8 m
b = 2.4 m
θ 1 = 10 deg
θ 2 = 20 deg
g = 9.81
m
2
s
Solution:
initial guesses: F CB = 10 kN
F AB = 10 kN
576
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
−M
2
−M
2
( )
( )
( )
( )
g cos θ 1 − F AB cos θ 2 + FCB
g sin θ 1 − FAB sin θ 2 + FCB
⎛ FAB ⎞
⎜
⎟ = Find F , F
( AB CB)
⎜ FCB ⎟
⎝
⎠
⎛⎜ Cx ⎟⎞
⎜ Cy ⎟
⎝ ⎠
b
=0
2
2
a +b
a
− Mg = 0
2
2
=
⎛b⎞
⎜ ⎟
2
2 a
a +b ⎝ ⎠
a +b
FCB
⎛ FAB ⎞ ⎛ 9.7 ⎞
⎜
⎟ ⎜
⎟
C
⎜ x ⎟ = ⎜ 11.53 ⎟ kN
⎜ C ⎟ ⎝ 8.65 ⎠
⎝ y ⎠
Problem 6-103
The tower truss has a weight W and a center of gravity at G. The rope system is used to hoist it
into the vertical position. If rope CB is attached to the top of the shear leg AC and a second
rope CD is attached to the truss, determine the required tension in BC to hold the truss in the
position shown. The base of the truss and the shear leg bears against the stake at A, which can
be considered as a pin. Also, compute the compressive force acting along the shear leg.
Given:
W = 575 lb
θ = 40 deg
a = 5 ft
b = 3 ft
c = 10 ft
d = 4 ft
e = 8 ft
Solution: Entire system:
ΣMA = 0;
TBC cos ( θ ) ( d + e) − TBC sin ( θ ) a − W( a + b) = 0
TBC =
W( a + b)
cos ( θ ) ( d + e) − sin ( θ ) a
TBC = 769 lb
577
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
CA is a two-force member.
At C:
e ⎞
⎟
⎝ b + c⎠
φ = atan ⎛⎜
initial guesses:
F CA = 500 lb
TCD = 300 lb
Given
ΣF x = 0;
ΣF y = 0;
F CA
F CA
a
2
a + ( d + e)
2
d+e
2
a + ( d + e)
⎛⎜ FCA ⎟⎞
= Find ( FCA , TCD)
⎜ TCD ⎟
⎝
⎠
2
+ TCD cos ( φ ) − TBC cos ( θ ) = 0
− TCD sin ( φ ) − TBC sin ( θ ) = 0
TCD = 358 lb
F CA = 739 lb
Problem 6-104
The constant moment M is applied to the crank shaft. Determine the compressive force P that is
exerted on the piston for equilibrium as a function of θ. Plot the results of P (ordinate) versus
θ (abscissa) for 0 deg ≤ θ ≤ 90 deg.
Given:
a = 0.2 m
b = 0.45 m
M = 50 N⋅ m
Solution:
a cos ( θ ) = b sin ( φ )
φ = asin ⎛⎜ cos ( θ )⎟⎞
a
⎝b
⎠
−M + F BC cos ( θ − φ ) a = 0
578
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F BC =
Chapter 6
M
a cos ( θ − φ )
P = FBC cos ( φ ) =
M cos ( φ )
a cos ( θ − φ )
This function goes to infinity at θ = 90 deg,
so we will only plot it to θ = 80 deg.
θ = 0 , 0.1 .. 80
φ ( θ ) = asin ⎛⎜ cos ( θ deg)⎟⎞
P (θ) =
a
⎝b
⎠
M cos ( φ ( θ ) )
a cos ( θ deg − φ ( θ ) )
Newtons
1500
P( θ )
1000
500
0
0
20
40
60
80
θ
Degrees
Problem 6-105
Five coins are stacked in the smooth plastic
container shown. If each coin has weight W,
determine the normal reactions of the bottom
coin on the container at points A and B.
Given:
W = 0.0235 lb
a = 3
b = 4
579
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
All coins :
ΣF y = 0;
NB = 5W
NB = 0.1175 lb
Bottom coin :
ΣF y = 0;
⎛
⎞=0
b
NB − W − N⎜
2⎟
2
⎝ a +b ⎠
⎛ a2 + b2 ⎞
)⎝ b ⎟⎠
(
N = NB − W ⎜
N = 0.1175 lb
ΣF x = 0;
⎛
NA = N⎜
a
2
⎞
2⎟
⎝ a +b ⎠
NA = 0.0705 lb
Problem 6-106
Determine the horizontal and vertical components of force at pin B and the normal force the pin
at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions
of force at A. There is a pulley at E.
Given:
F = 50 lb
a = 4 ft
b = 3 ft
Solution:
Guesses
B x = 1 lb
B y = 1 lb
NC = 1 lb
580
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ax = 1 lb
Chapter 6
Ay = 1 lb
MA = 1 lb⋅ ft
Given
⎛
⎞N − F = 0
C
⎝ a +b ⎠
a
Bx + ⎜
2
⎛
b
2⎟
⎞N − F = 0
C
⎝ a +b ⎠
By − ⎜
2
2⎟
(F + Bx)a − (F + By)b = 0
⎛
⎞N − A = 0
C
x
2
2⎟
⎝ a +b ⎠
F−⎜
a
b
⎛
⎞
⎜ 2 2 ⎟ NC − Ay = 0
⎝ a +b ⎠
⎛
⎞N a + M = 0
C
A
⎝ a +b ⎠
−F 2 a + ⎜
a
2
2⎟
⎛ Bx ⎞
⎜
⎟
B
y
⎜
⎟
⎜N ⎟
⎜ C ⎟ = Find B , B , N , A , A , M
( x y C x y A)
⎜ Ax ⎟
⎜
⎟
⎜ Ay ⎟
⎜
⎟
⎝ MA ⎠
NC = 20 lb
⎛⎜ Bx ⎟⎞ ⎛ 34 ⎞
=
lb
⎜ By ⎟ ⎜⎝ 62 ⎟⎠
⎝ ⎠
⎛⎜ Ax ⎞⎟ ⎛ 34 ⎞
=
lb
⎜ Ay ⎟ ⎜⎝ 12 ⎟⎠
⎝ ⎠
MA = 336 lb⋅ ft
Problem 6-107
A force F is applied to the handles of the vise grip. Determine the compressive force developed
on the smooth bolt shank A at the jaws.
581
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F = 5 lb
b = 1 in
a = 1.5 in
c = 3 in
d = 0.75 in
e = 1 in
θ = 20 deg
Solution:
From FBD (a)
ΣME = 0;
⎡
F ( b + c) − F CD⎢
d+e
2
⎤b = 0
2⎥
⎣ c + ( d + e) ⎦
⎡ c2 + ( d + e) 2⎤
⎥
⎣ b( d + e) ⎦
F CD = F( b + c) ⎢
ΣF x = 0;
⎡
F CD = 39.693 lb
⎤
⎥
2
2
⎣ c + ( d + e) ⎦
E x = F CD⎢
c
E x = 34.286 lb
From FBD (b)
ΣMB = 0;
NA sin ( θ ) d + NA cos ( θ ) a − Ex( d + e) = 0
d+e
⎞
⎛
⎟
(
)
(
)
sin
θ
d
+
cos
θ
a
⎝
⎠
NA = Ex ⎜
NA = 36.0 lb
Problem 6-108
If a force of magnitude P is applied to the grip of the clamp, determine the compressive force
F that the wood block exerts on the clamp.
582
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 6
Given:
P = 10 lb
a = 2 in
b = 2 in
c = 0.5 in
d = 0.75 in
e = 1.5 in
Solution:
Define
b
φ = atan ⎛⎜ ⎟⎞
φ = 69.444 deg
⎝ d⎠
From FBD (a),
ΣMB = 0;
F CD cos ( φ ) c − P ( a + b + c) = 0
F CD =
+
↑Σ Fy = 0;
P( a + b + c)
F CD = 256.32 lb
cos ( φ ) ( c)
F CD sin ( φ ) − By = 0
B y = F CD sin ( φ )
B y = 240 lb
From FBD (b),
ΣMA = 0;
By d − F e = 0
F =
By d
F = 120 lb
e
583
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-109
The hoist supports the engine of
mass M. Determine the force in
member DB and in the hydraulic
cylinder H of member FB.
Units Used:
3
kN = 10 N
Given:
M = 125 kg
d = 1m
a = 1m
e = 1m
b = 2m
f = 2m
c = 2m
g = 9.81
m
2
s
Solution:
Member GFE:
ΣME = 0; −F FB⎡
⎤ b + M g ( a + b) = 0
⎢
2
2⎥
⎣ ( c + d) + ( b − e) ⎦
c+d
⎡ a + b ⎤ ( c + d) 2 + ( b − e) 2
⎥
⎣ b( c + d) ⎦
F FB = M g ⎢
F FB = 1.94 kN
ΣF x = 0;
⎡
⎤=0
b−e
E x − FFB⎢
2⎥
2
⎣ ( c + d) + ( b − e) ⎦
⎡
⎤
b−e
E x = F FB⎢
2⎥
2
⎣ ( c + d) + ( b − e) ⎦
Member EDC:
ΣΜc = 0;
⎛
⎞d = 0
⎟
2
2
⎝ e +d ⎠
E x( c + d) − F DB⎜
⎛ c + d⎞
⎟
⎝ ed ⎠
F DB = Ex ⎜
e
2
2
e +d
F DB = 2.601 kN
584
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Engineering Mechanics - Statics
Chapter 6
Problem 6-110
The flat-bed trailer has weight W1 and center of gravity at GT . It is pin-connected to the cab at
D. The cab has a weight W2 and center of gravity at GC. Determine the range of values x for the
position of the load L of weight W3 so that no axle is subjected to a force greater than FMax. The
load has a center of gravity at GL .
Given:
W1 = 7000 lb
a = 4 ft
W2 = 6000 lb
b = 6 ft
W3 = 2000 lb
c = 3 ft
F max = 5500 lb d = 10 ft
e = 12 ft
Solution:
Case 1:
Assume
Guesses
Ay = Fmax
x = 1 ft
Given
Ay = Fmax
B y = F max
Cy = F max
Dy = Fmax
Ay + B y − W2 − Dy = 0
−W2 a − Dy( a + b) + B y( a + b + c) = 0
Dy − W1 − W3 + Cy = 0
W3 x + W1 e − Dy( c + d + e) = 0
⎛⎜ By ⎞⎟
⎜ Cy ⎟
⎜ ⎟ = Find ( By , Cy , Dy , x)
⎜ Dy ⎟
⎜ x ⎟
⎝ ⎠
⎛ Ay ⎞ ⎛⎜ 5.5 × 103 ⎟⎞
⎜ ⎟
x1 = 30.917 ft
⎜ By ⎟ = ⎜ 6.333 × 103 ⎟ lb x1 = x
⎟
⎜C ⎟ ⎜
Since By > Fmax then this solution is no good.
⎝ y ⎠ ⎜⎝ 3.167 × 103 ⎟⎠
Case 2:
Assume
Guesses
Ay = Fmax
B y = F max
B y = F max
Cy = F max
585
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Engineering Mechanics - Statics
x = 1 ft
Given
Chapter 6
Dy = Fmax
Ay + B y − W2 − Dy = 0
−W2 a − Dy( a + b) + B y( a + b + c) = 0
Dy − W1 − W3 + Cy = 0
W3 x + W1 e − Dy( c + d + e) = 0
⎛⎜ Ay ⎞⎟
⎜ Cy ⎟
⎜ ⎟ = Find ( Ay , Cy , Dy , x)
⎜ Dy ⎟
⎜ x ⎟
⎝ ⎠
⎛ Ay ⎞ ⎛⎜ 5.25 × 103 ⎟⎞
⎜ ⎟
⎜ By ⎟ = ⎜ 5.5 × 103 ⎟ lb x2 = x
⎟
⎜C ⎟ ⎜
3
⎜
⎝ y ⎠ ⎝ 4.25 × 10 ⎟⎠
x2 = 17.375 ft
Since Ay < Fmax and Cy < F max then this solution is good.
Cy = F max
Case 3:
Assume
Guesses
Ay = Fmax
B y = F max
x = 1 ft
Dy = Fmax
Given
Cy = F max
Ay + B y − W2 − Dy = 0
−W2 a − Dy( a + b) + B y( a + b + c) = 0
Dy − W1 − W3 + Cy = 0
W3 x + W1 e − Dy( c + d + e) = 0
⎛⎜ Ay ⎞⎟
⎜ By ⎟
⎜ ⎟ = Find ( Ay , By , Dy , x)
⎜ Dy ⎟
⎜ x ⎟
⎝ ⎠
⎛ Ay ⎞ ⎛⎜ 4.962 × 103 ⎟⎞
⎜ ⎟
⎜ By ⎟ = ⎜ 4.538 × 103 ⎟ lb x3 = x
⎟
⎜C ⎟ ⎜
3
⎜
⎝ y ⎠ ⎝ 5.5 × 10 ⎟⎠
x3 = 1.75 ft
Since Ay < Fmax and B y < F max then this solution is good.
We conclude that x3 = 1.75 ft < x < x2 = 17.375 ft
586
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Engineering Mechanics - Statics
Chapter 6
Problem 6-111
Determine the force created in
the hydraulic cylinders EF and
AD in order to hold the shovel
in equilibrium. The shovel load
has a mass W and a center of
gravity at G. All joints are pin
connected.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
Given:
a = 0.25 m θ 1 = 30 deg
b = 0.25 m θ 2 = 10 deg
c = 1.5 m
θ 3 = 60 deg
d = 2m
W = 1.25 Mg
e = 0.5 m
Solution:
Assembly FHG :
(
( )) = 0
ΣMH = 0; −[ W g( e) ] + FEF c sin θ 1
⎛
⎞ F = 8.175 kN
(T)
EF
⎟
c
sin
θ
(
)
1
⎝
⎠
e
F EF = W g ⎜
Assembly CEFHG:
(
)
( )
ΣMC = 0; F AD cos θ 1 + θ 2 b − W g⎡( a + b + c)cos θ 2 + e⎤ = 0
⎣
⎦
⎛ cos ( θ 2) a + cos ( θ 2) b + cos ( θ 2) c + e ⎞
⎟
cos ( θ 1 + θ 2) b
⎝
⎠
F AD = W g ⎜
F AD = 158 kN (C)
587
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-112
The aircraft-hangar door opens and closes slowly by means of a motor which draws in the
cable AB. If the door is made in two sections (bifold) and each section has a uniform weight
W and length L, determine the force in the cable as a function of the door's position θ. The
sections are pin-connected at C and D and the bottom is attached to a roller that travels along
the vertical track.
587
Solution:
L
⎛ θ ⎞ − 2L sin ⎛ θ ⎞ N = 0
⎟
⎜ ⎟ A
⎝2⎠
⎝2⎠
cos ⎜
ΣMD = 0;
2W
ΣΜC = 0;
T L cos ⎜
2
⎛ θ ⎞ − N L sin ⎛ θ ⎞ − W L cos ⎛ θ ⎞ = 0
⎟
⎜ ⎟
⎜ ⎟
A
2
⎝2⎠
⎝2⎠
⎝2⎠
W
⎛θ⎞
NA =
cot ⎜ ⎟
2
2
⎝ ⎠
T=W
Problem 6-113
A man having weight W attempts to lift himself using one of the two methods shown.
Determine the total force he must exert on bar AB in each case and the normal reaction he
exerts on the platform at C. Neglect the weight of the platform.
Given:
W = 175 lb
588
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Engineering Mechanics - Statics
Chapter 6
Solution:
(a)
Bar:
+
↑Σ Fy = 0;
⎛ F⎞ − 2 ⎛ W⎞ = 0
⎟
⎜ ⎟
⎝ 2⎠
⎝2⎠
2⎜
F = W
F = 175 lb
Man:
+
↑Σ Fy = 0;
⎛ F⎞
NC − W − 2 ⎜ ⎟ = 0
2
⎝ ⎠
NC = W + F
NC = 350 lb
( b)
Bar:
+
↑
Σ F y = 0;
⎛ W⎞ − 2 F = 0
⎟
2
⎝4⎠
2⎜
F =
Man:
+
↑Σ Fy = 0;
W
2
F = 87.5 lb
⎛ F⎞
NC − W + 2 ⎜ ⎟ = 0
2
⎝ ⎠
NC = W − F
NC = 87.5 lb
589
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Problem 6-114
A man having weight W1 attempts to lift himself using one of the two methods shown.
Determine the total force he must exert on bar AB in each case and the normal reaction he
exerts on the platform at C. The platform has weight W2.
Given:
W1 = 175 lb W2 = 30 lb
Solution:
(a)
Bar:
+
↑Σ Fy = 0;
2
F
2
(
)
− W1 + W2 = 0
F = W1 + W2
F = 205 lb
Man:
+
↑
Σ F y = 0;
F
NC − W1 − 2
=0
2
NC = F + W1
NC = 380 lb
( b)
Bar:
+
↑
Σ F y = 0;
⎛ W1 + W2 ⎞
⎟=0
⎟+2⎜
⎝ 2⎠
⎝ 4 ⎠
−2 ⎛⎜
F⎞
590
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F =
Chapter 6
W1 + W2
2
F = 102 lb
Man:
+
↑
⎛F⎞
NC − W1 + 2 ⎜ ⎟ = 0
2
Σ F y = 0;
⎝ ⎠
NC = W1 − F
NC = 72.5 lb
Problem 6-115
The piston C moves vertically between the two smooth walls. If the spring has stiffness k and
is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the
mechanism in equilibrium.
Given:
k = 15
lb
in
θ = 30 deg
a = 8 in
b = 12 in
Solution:
Geometry:
b sin ( ψ) = a sin ( θ )
a⎞
⎟
b⎠
ψ = 19.471 deg
φ = 180 deg − ψ − θ
φ = 130.529 deg
ψ = asin ⎛⎜ sin ( θ )
⎝
591
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
rAC
sin ( φ )
=
Chapter 6
⎛ sin ( φ ) ⎞
⎟
⎝ sin ( θ ) ⎠
b
rAC = b ⎜
sin ( θ )
rAC = 18.242 in
Free Body Diagram: The solution for this problem will be simplified if one realizes that
member CB is a two force member. Since the spring stretches
x = ( a + b) − rAC
x = 1.758 in
F sp = k x
the spring force is
F sp = 26.371 lb
Equations of Equilibrium: Using the method of joints
+
↑
F CB cos ( ψ) − Fsp = 0
Σ F y = 0;
F CB =
Fsp
cos ( ψ)
F CB = 27.971 lb
From FBD of bar AB
F CB sin ( φ ) a − M = 0
+ ΣMA = 0;
M = FCB sin ( φ ) a
M = 14.2 lb⋅ ft
Problem 6-116
The compound shears are used to cut metal parts. Determine the vertical cutting force exerted
on the rod R if a force F is applied at the grip G. The lobe CDE is in smooth contact with the
head of the shear blade at E.
Given:
F = 20 lb
e = 0.5 ft
a = 1.4 ft
f = 0.5 ft
b = 0.2 ft
g = 0.5 ft
c = 2 ft
h = 2.5 ft
d = 0.75 ft
θ = 60 deg
Solution:
Member AG:
ΣMA = 0;
F ( a + b) − FBC b sin ( θ ) = 0
592
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F BC = F ⎛⎜
Chapter 6
a+b
⎞
⎟
⎝ b sin ( θ ) ⎠
F BC = 184.75 lb
Lobe:
ΣMD = 0;
F BC f − NE e = 0
f
NE = FBC
e
NE = 184.75 lb
Head:
ΣMF = 0;
−NE( h + g) + h NR = 0
h + g⎞
NR = NE ⎛⎜
⎟
⎝ h ⎠
NR = 222 lb
Problem 6-117
The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear
C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is
allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential
forces between them, determine the compressive force developed on the cylinder S when a
vertical force F is applied to the handle of the press.
Given:
F = 40 N
a = 0.5 m
b = 0.2 m
c = 1.2 m
d = 0.35 m
e = 0.65 m
Solution:
Member GD:
ΣMG = 0;
−F a + F CG b = 0
593
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F CG = F
Chapter 6
a
F CG = 100 N
b
Sector gear :
⎞d = 0
2
2⎟
⎝ c +d ⎠
ΣMH = 0; F CG( d + e) − FAB⎛
⎜
c
⎛ c2 + d2 ⎞
⎟ F = 297.62 N
F AB = FCG ( d + e) ⎜
⎝ c d ⎠ AB
Table:
ΣF y = 0;
c
⎞
⎜ 2 2 ⎟ − Fs = 0
⎝ c +d ⎠
F AB⎛
c
⎞
⎜ 2 2⎟
⎝ c +d ⎠
F s = FAB ⎛
F s = 286 N
Problem 6-118
The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to
rotate downward. If the mixer has weight W, is centered on the shelf, and has a mass center at
G, determine the stretch in the spring necessary to hold the shelf in the equilibrium position
shown. There is a similar mechanism on each side of the shelf, so that each mechanism
supports half of the load W. The springs each have stiffness k.
Given:
W = 10 lb
k = 4
a = 2 in
lb
b = 4 in
in
φ = 30 deg
c = 15 in
θ = 30 deg
d = 6 in
Solution:
ΣMF = 0;
W
2
b − a F ED cos ( φ ) = 0
594
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
F ED =
Chapter 6
Wb
2 a cos ( φ )
+
Σ F x = 0;
→
−F x + FED cos ( φ ) = 0
F x = F ED cos ( φ )
+
↑Σ Fy = 0;
Fy =
W
2
F ED = 11.547 lb
−W
2
F x = 10 lb
+ F y − FED sin ( φ ) = 0
+ F ED⋅ sin ( φ )
F y = 10.774 lb
Member FBA:
ΣMA = 0;
F y( c + d) cos ( φ ) − F x( c + d) sin ( φ ) − Fs sin ( θ + φ ) d = 0
Fs =
F s = k s;
Fy( c + d) cos ( φ ) − Fx( c + d) sin ( φ )
Fs = k x
d sin ( θ + φ )
x =
Fs
F s = 17.5 lb
x = 4.375 in
k
Problem 6-119
If each of the three links of the mechanism has a weight W, determine the angle θ for
equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°.
595
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
W = 25 lb
k = 60
lb
ft
a = 4 ft
b = 4 ft
Solution:
Guesses
θ = 30 deg
B x = 10 lb
B y = 10 lb
Cx = 10 lb
Cy = 10 lb
Given
⎛ a ⎞ sin ( θ ) − C a sin ( θ ) + C a cos ( θ ) = 0
⎟
y
x
⎝ 2⎠
−W⎜
⎛ b⎞ + C b = 0
⎟
y
⎝ 2⎠
−W⎜
B y + Cy − W = 0
−B x + Cx = 0
⎛ a ⎞ sin ( θ ) + k⎛ a ⎞ sin ( θ ) ⎛ a ⎞ cos ( θ ) = 0
⎟
⎜ ⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
−B x a cos ( θ ) − B y a sin ( θ ) − W⎜
⎛ Bx ⎞
⎜ ⎟
⎜ By ⎟
⎜ C ⎟ = Find B , B , C , C , θ
(x y x y )
⎜ x⎟
⎜ Cy ⎟
⎜ ⎟
⎝θ ⎠
⎛ Bx ⎞
⎜ ⎟ ⎛⎜ 16.583 ⎞⎟
⎜ By ⎟ ⎜ 12.5 ⎟
lb
⎜ ⎟=⎜
⎟
⎜ Cx ⎟ ⎜ 16.583 ⎟
⎜ C ⎟ ⎝ 12.5 ⎠
⎝ y⎠
θ = 33.6 deg
Problem 6-120
Determine the required force P that must be applied at the blade of the pruning shears so
that the blade exerts a normal force F on the twig at E.
596
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F = 20 lb
a = 0.5 in
b = 4 in
c = 0.75 in
d = 0.75 in
e = 1 in
Solution:
initial guesses:
Ax = 1 lb
Ay = 1 lb
Dy = 10 lb P = 20 lb
Dx = 10 lb
F CB = 20 lb
Given
− P ( b + c + d) − A x a + F e = 0
Dy − P − A y − F = 0
Dx − Ax = 0
− Ay( d) − A x( a) + ( b + c)P = 0
c
⎞
⎜ 2 2⎟ = 0
⎝ c +a ⎠
Ax − F CB⎛
a
⎞
⎜ 2 2⎟ = 0
⎝ a +c ⎠
Ay + P − FCB⎛
597
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟
⎜
⎟
Dx
⎜
⎟ = Find A , A , D , D , P , F
( x y x y CB)
⎜ D ⎟
y
⎜
⎟
⎜ P ⎟
⎜F ⎟
⎝ CB ⎠
⎛⎜ Ax ⎟⎞
⎛ 13.333 ⎞
⎜ Ay ⎟ ⎜ 6.465 ⎟
⎜
⎟ ⎜
⎟
⎜ Dx ⎟ = ⎜ 13.333 ⎟ lb
⎜ D ⎟ ⎜ 28.889 ⎟
⎜ y ⎟ ⎜
⎟
16.025
⎝
⎠
⎜ FCB ⎟
⎝
⎠
P = 2.424 lb
Problem 6-121
The three power lines exert the forces shown on the truss joints, which in turn are
pin-connected to the poles AH and EG. Determine the force in the guy cable AI and the pin
reaction at the support H.
Units Used:
3
kip = 10 lb
Given:
F 1 = 800 lb
d = 125 ft
F 2 = 800 lb
e = 50 ft
a = 40 ft
f = 30 ft
b = 20 ft
g = 30 ft
c = 20 ft
Solution:
AH is a two-force member.
c
θ = atan ⎛⎜ ⎟⎞
φ = atan ⎛⎜
d
β = atan ⎛⎜ ⎟⎞
e
γ = atan ⎛⎜ ⎟⎞
⎝ b⎠
⎝ f⎠
⎞
⎟
⎝ a + b⎠
c
⎝ d⎠
α = 90 deg − β + γ
Guesses
F AB = 1 lb
F BC = 1 lb
F CA = 1 lb
F AI = 1 lb
F H = 1 lb
598
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given
F BC − F AB cos ( θ ) = 0
F AB sin ( θ ) − F1 = 0
2 F CA sin ( φ ) − F 2 = 0
−F AI sin ( α ) + F AB sin ( β − θ ) + F CA sin ( β − φ ) = 0
−F AI cos ( α ) − FAB cos ( β − θ ) − F CA cos ( β − φ ) + F H = 0
⎛⎜ FAB ⎞⎟
⎜ FCA ⎟
⎜
⎟
F
⎜ BC ⎟ = Find ( FAB , FCA , FBC , FAI , FH)
⎜F ⎟
⎜ AI ⎟
⎜ FH ⎟
⎝
⎠
⎛ FAB ⎞ ⎛ 1.131 ⎞
⎜
⎟ ⎜
⎟
⎜ FCA ⎟ = ⎜ 1.265 ⎟ kip
⎜ F ⎟ ⎝ 0.8 ⎠
⎝ BC ⎠
⎛⎜ FAI ⎞⎟ ⎛ 2.881 ⎞
=
kip
⎜ FH ⎟ ⎜⎝ 3.985 ⎟⎠
⎝
⎠
Problem 6-122
The hydraulic crane is used to lift the
load of weight W. Determine the force
in the hydraulic cylinder AB and the
force in links AC and AD when the
load is held in the position shown.
Units Used:
3
kip = 10 lb
Given:
W = 1400 lb
a = 8 ft
c = 1 ft
b = 7 ft
γ = 70 deg
599
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
ΣMD = 0;
F CA sin ( 60 deg)c − W a = 0
F CA =
Wa
sin ( 60 deg) c
F CA = 12.9 kip
↑Σ Fy = 0; FCA sin( 60 deg) − FAB sin(γ ) = 0
+
F AB = FCA
sin ( 60 deg)
sin ( γ )
F AB = 11.9 kip
+
Σ F x = 0; −F AB cos ( γ ) + FCA cos ( 60 deg) − FAD = 0
→
F AD = −FAB cos ( γ ) + F CA cos ( 60 deg)
F AD = 2.39 kip
Problem 6-123
The kinetic sculpture requires that each of the three pinned beams be in perfect balance at all
times during its slow motion. If each member has a uniform weight density γ and length L,
determine the necessary counterweights W1, W2 and W3 which must be added to the ends of
each member to keep the system in balance for any position. Neglect the size of the
counterweights.
Given:
γ = 2
lb
ft
L = 3 ft
a = 1 ft
Solution:
ΣMA = 0;
W1 a cos ( θ ) − γ L cos ( θ ) ⎛⎜
L
⎝2
− a⎟⎞ = 0
⎠
600
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Engineering Mechanics - Statics
Chapter 6
γ L⎛⎜
⎝2
W1 =
+
↑Σ Fy = 0;
L
− a⎞⎟
⎠
W1 = 3 lb
a
R A − W1 − γ L = 0
R A = W1 + γ L
R A = 9 lb
ΣMB = 0;
⎛ L − a⎞ − R ( L − a) cos ( φ ) = 0
⎟
A
⎝2 ⎠
W2 a cos ( φ ) − γ L cos ( φ ) ⎜
γ L⎛⎜
W2 =
L
⎝2
+
↑Σ Fy = 0;
− a⎟⎞ + R A( L − a)
⎠
W2 = 21 lb
a
R B − W2 − RA − γ L = 0
R B = W2 + R A + γ L
R B = 36 lb
L
⎞
− a⎟ cos ( φ ) − W3 a cos ( φ ) = 0
⎝2 ⎠
ΣMC = 0; R B( L − a) cos ( φ ) + γ L⎛⎜
W3 =
R B( L − a) + γL⎛⎜
L
⎝2
− a⎟⎞
⎠
W3 = 75 lb
a
Problem 6-124
The three-member frame is connected at its ends using ball-and-socket joints. Determine the x,
y, z components of reaction at B and the tension in member ED. The force acting at D is F.
Given:
⎛ 135 ⎞
⎜
⎟
F = 200 lb
⎜
⎟
⎝ −180 ⎠
a = 6 ft
e = 3 ft
b = 4 ft
f = 1 ft
601
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
d = 6 ft
Chapter 6
g = 2 ft
c = g+ f
Solution:
AC and DE are two-force members.
Define some vectors
⎛ −e ⎞
⎜
⎟
rDE = −b − g
⎜
⎟
⎝ a ⎠
uDE =
⎛ −d − e ⎞
⎜ −b ⎟
rAC =
⎜
⎟
⎝ 0 ⎠
uAC =
⎛e⎞
⎜ ⎟
rBD = − f
⎜ ⎟
⎝0⎠
⎛e + d⎞
⎜ −c ⎟
rBA =
⎜
⎟
⎝ 0 ⎠
rDE
rDE
rAC
rAC
Guesses
B x = 1 lb
B y = 1 lb
B z = 1 lb
F DE = 1 lb
F AC = 1 lb
Given
⎛ Bx ⎞
⎜ ⎟
⎜ By ⎟ + FDE uDE + FAC uAC + F = 0
⎜B ⎟
⎝ z⎠
rBD × FDEuDE + F + rBA × FAC uAC = 0
⎛⎜ Bx ⎞⎟
⎜ By ⎟
⎜
⎟
⎜ Bz ⎟ = Find ( Bx , By , Bz , FDE , FAC )
⎜F ⎟
⎜ DE ⎟
⎜ FAC ⎟
⎝
⎠
⎛ Bx ⎞ ⎛
−30
⎞ F
⎜ ⎟ ⎜
⎟ ⎛⎜ DE ⎞⎟ ⎛ 270 ⎞
=
lb
⎜ By ⎟ = ⎜ −13.333 ⎟ lb⎜
⎟ ⎜ 16.415 ⎟⎠
F
⎜B ⎟ ⎜
− 12 ⎟ ⎝ AC ⎠ ⎝
⎝ z ⎠ ⎝ 3.039 × 10 ⎠
(
)
(
)
602
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Engineering Mechanics - Statics
Chapter 6
Problem 6-125
The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All
the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is
F max, determine the largest vertical force P that can be supported by the frame. Also, what
are the x, y, z force components which member BD exerts on members EDC and ABC? The
collars at A and E and the pon at G only exert force components on the frame.
Given:
F max = 800 N
a = 300 mm
b = 600 mm
c = 600 mm
Solution:
ΣMx = 0;
b
−P 2 c +
2
2
b +c
Fmax c = 0
F max b
P =
2
2
P = 282.843 N
2
b +c
c
B z + Dz − Fmax
2
b +c
F max c
Bz =
2
2
=0
2
Dz = B z
2
b +c
Dz = Bz
B z = 283 N
Dz = 283 N
b
B y + Dy − Fmax
2
b +c
F max b
By =
2
2
=0
2
2
b +c
Dy = By
Dy = B y
B y = 283 N
Dy = 283 N
B x = Dx = 0
Problem 6-126
The structure is subjected to the loading shown. Member AD is supported by a cable AB and a
roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at
603
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Engineering Mechanics - Statics
Chapter 6
g
pp
y
D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of
reaction at E and the tension in cable AB.
Units Used:
3
kN = 10 N
Given:
⎛ 0 ⎞
F = ⎜ 0 ⎟ kN
⎜
⎟
⎝ −2.5 ⎠
a = 0.5 m
d = 0.3 m
b = 0.4 m
e = 0.8 m
c = 0.3 m
Solution:
Guesses
⎛ −c − d ⎞
AB = ⎜ 0 ⎟
⎜
⎟
⎝ e ⎠
F AB = 1 kN
Dx = 1 kN
Dz = 1 kN
Dz2 = 1 kN
E x = 1 kN
E y = 1 kN
MDx = 1 kN⋅ m
MDz = 1 kN⋅ m
Cx = 1 kN
MEx = 1 kN⋅ m
MEy = 1⋅ kN m
Given
⎛⎜ Cx ⎟⎞ ⎛⎜ Dx ⎞⎟
F + FAB
+⎜ 0 ⎟+⎜ 0 ⎟ =0
AB
⎜ 0 ⎟ ⎜D ⎟
⎝ ⎠ ⎝ z⎠
AB
⎛ MDx ⎞ ⎛ 0 ⎞ ⎛ Cx ⎞ ⎛ d ⎞
⎛c + d⎞
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
AB ⎞
⎜
⎟ ⎛
⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × F + ⎜ b ⎟ × ⎜ FAB
⎟=0
AB ⎠
⎝
⎜M ⎟ ⎝0⎠ ⎜ 0 ⎟ ⎝0⎠
⎝ 0 ⎠
⎝ ⎠
⎝ Dz ⎠
⎛ −Dx ⎞ ⎛ Ex ⎞
⎜
⎟ ⎜ ⎟
⎜ 0 ⎟ + ⎜ Ey ⎟ = 0
⎜D − D ⎟ ⎜ ⎟
z⎠ ⎝ 0 ⎠
⎝ z2
⎛ −MDx ⎞ ⎛ MEx ⎞ ⎛ 0 ⎞ ⎛ Dx ⎞
⎜
⎟ ⎜
⎟ ⎜ ⎟ ⎜
⎟
⎜ 0 ⎟ + ⎜ MEy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ = 0
⎜ −M ⎟ ⎜
⎟
⎜
⎟
⎝ Dz ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Dz − Dz2 ⎠
604
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Engineering Mechanics - Statics
Chapter 6
⎛ Cx ⎞
⎜
⎟
⎜ Dx ⎟
⎜
⎟
D
z
⎜
⎟
⎜D ⎟
⎜ z2 ⎟
⎜ Ex ⎟
⎜
⎟
E
⎜ y ⎟ = Find ( Cx , Dx , Dz , Dz2 , Ex , Ey , FAB , MDx , MDz , MEx , MEy)
⎜F ⎟
⎜ AB ⎟
⎜ MDx ⎟
⎜
⎟
⎜ MDz ⎟
⎜
⎟
⎜ MEx ⎟
⎜M ⎟
⎝ Ey ⎠
⎛ Cx ⎞
⎜
⎟ ⎛⎜ 0.937 ⎞⎟
D
⎜ x⎟ ⎜ 0 ⎟
kN
⎜
⎟=⎜
⎟
⎜ Dz ⎟ ⎜ 1.25 ⎟
⎜ D ⎟ ⎝ 1.25 ⎠
⎝ z2 ⎠
⎛⎜ MDx ⎞⎟ ⎛ 0.5 ⎞
=
kN⋅ m
⎜ MDz ⎟ ⎜⎝ 0 ⎟⎠
⎝
⎠
⎛⎜ Ex ⎟⎞ ⎛ 0 ⎞
=
kN
⎜ Ey ⎟ ⎜⎝ 0 ⎟⎠
⎝ ⎠
⎛⎜ MEx ⎟⎞ ⎛ 0.5 ⎞
=
kN⋅ m
⎜ MEy ⎟ ⎜⎝ 0 ⎟⎠
⎝
⎠
F AB = 1.562 kN
Problem 6-127
The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket
at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z
components of reaction at A and C.
605
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Engineering Mechanics - Statics
Chapter 6
Given:
a = 2m
M = 800 N⋅ m
b = 1.5 m
F = 250 N
c = 3m
θ 1 = 60 deg
d = 4m
θ 2 = 45 deg
Solution:
θ 3 = 60 deg
Guesses
Bx = 1 N
By = 1 N
Ax = 1 N
Ay = 1 N
Az = 1 N
Cx = 1 N
Cy = 1 N
Cz = 1 N
MBx = 1 N⋅ m
MBy = 1 N⋅ m
MCy = 1 N⋅ m
MCz = 1 N⋅ m
Given
⎛ Ax ⎞ ⎛ Bx ⎞
⎜ ⎟ ⎜ ⎟
⎜ Ay ⎟ + ⎜ By ⎟ = 0
⎜A ⎟ ⎜ ⎟
⎝ z⎠ ⎝ 0 ⎠
⎛ c ⎞ ⎛⎜ Bx ⎟⎞ ⎛ M ⎞ ⎛⎜ −MBx ⎞⎟
⎜ a ⎟ × B + ⎜ 0 ⎟ + −M
=0
⎜ ⎟ ⎜ y ⎟ ⎜ ⎟ ⎜ By ⎟
⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠
⎛ cos ( θ 1) ⎞ ⎛ −Bx ⎞ ⎛ Cx ⎞
⎜
⎟ ⎜
⎟ ⎜ ⎟
F ⎜ cos ( θ 2) ⎟ + ⎜ −B y ⎟ + ⎜ Cy ⎟ = 0
⎜
⎟ ⎜
⎟ ⎜ ⎟
⎝ cos ( θ 3) ⎠ ⎝ 0 ⎠ ⎝ Cz ⎠
⎡ ⎛ cos θ ⎞⎤
⎛ 0 ⎞ ⎢ ⎜ ( 1) ⎟⎥ ⎛ 0 ⎞ ⎛⎜ −Bx ⎞⎟ ⎛⎜ MBx ⎟⎞ ⎛⎜ 0 ⎞⎟
⎜ 0 ⎟ × F cos ( θ ) + ⎜ 0 ⎟ × −B + M
+ MCy ⎟ = 0
2 ⎟⎥
⎜
⎟ ⎢ ⎜
⎜ ⎟ ⎜ y ⎟ ⎜ By ⎟ ⎜
⎝ b + d ⎠ ⎢ ⎜ cos ( θ 3) ⎟⎥ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ MCz ⎟⎠
⎣ ⎝
⎠⎦
606
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
⎛⎜ Ax ⎞⎟
⎜ Ay ⎟
⎜
⎟
⎜ Az ⎟
⎜ C ⎟
⎜ x ⎟
⎜ Cy ⎟
⎜
⎟
⎜ Cz ⎟
⎜
⎟ = Find ( Ax , Ay , Az , Cx , Cy , Cz , Bx , By , MBx , MBy , MCy , MCz )
⎜ Bx ⎟
⎜ B ⎟
⎜ y ⎟
⎜ MBx ⎟
⎜
⎟
⎜ MBy ⎟
⎜M ⎟
⎜ Cy ⎟
⎜ MCz ⎟
⎝
⎠
⎛ Ax ⎞
⎜ ⎟ ⎛ −172.3 ⎞
⎜ Ay ⎟ ⎜ −114.8 ⎟
⎟
⎜A ⎟ ⎜
⎜
⎟
z
0
⎜ ⎟=
⎜
⎟N
⎜ Cx ⎟
47.3
⎜
⎟
⎜ ⎟
−
61.9
⎜
⎟
⎜ Cy ⎟
⎜
⎜ ⎟ ⎝ −125 ⎟⎠
⎝ Cz ⎠
⎛⎜ MCy ⎞⎟ ⎛ −429 ⎞
=
N⋅ m
⎜ MCz ⎟ ⎜⎝ 0 ⎟⎠
⎝
⎠
Problem 6-128
Determine the resultant forces at pins B and C on member ABC of the four-member frame.
Given:
w = 150
lb
ft
a = 5 ft
b = 2 ft
607
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Engineering Mechanics - Statics
Chapter 6
c = 2 ft
e = 4 ft
d = a+b−c
Solution:
The initial guesses are
F CD = 20 lb
F BE = 40 lb
Given
F CD( c + d) − FBE
⎛ a + b⎞ +
⎟
⎝ 2 ⎠
−w( a + b) ⎜
ec
=0
2
2
( d − b) + e
F BE e
2
2
( d − b) + e
⎛⎜ FCD ⎞⎟
= Find ( F CD , F BE)
⎜ FBE ⎟
⎝
⎠
a − FCD( a + b) = 0
⎛⎜ FCD ⎞⎟ ⎛ 350 ⎞
=
lb
⎜ FBE ⎟ ⎜⎝ 1531 ⎟⎠
⎝
⎠
Problem 6-129
The mechanism consists of identical meshed gears A and B and arms which are fixed to the
gears. The spring attached to the ends of the arms has an unstretched length δ and a stiffness
k. If a torque M is applied to gear A, determine the angle θ through which each arm rotates.
The gears are each pinned to fixed supports at their centers.
608
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
δ = 100 mm
k = 250
N
m
M = 6 N⋅ m
r =
δ
2
a = 150 mm
Solution:
ΣMA = 0;
−F r − P a cos ( θ ) + M = 0
ΣMB = 0;
P a cos ( θ ) − F r = 0
2P a cos ( θ ) = M
2k( 2a) sin ( θ ) a cos ( θ ) = M
2k a sin ( 2θ ) = M
2
θ =
1
2
⎞
⎟
2
⎝ 2ka ⎠
asin ⎛⎜
M
θ = 16.1 deg
Problem 6-130
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
kN = 1000 N
609
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
F 1 = 20 kN
F 2 = 10 kN
a = 1.5 m
b = 2m
Solution:
b
θ = atan ⎛⎜ ⎟⎞
⎝ a⎠
Guesses
F AG = 1 kN
F BG = 1 kN
F GC = 1 kN
F GF = 1 kN
F AB = 1 kN
F BC = 1 kN
F CD = 1 kN
F CF = 1 kN
F DF = 1 kN
F DE = 1 kN
F EF = 1 kN
Given
−F AB cos ( θ ) + F BC = 0
−F AB sin ( θ ) − FBG = 0
F GC cos ( θ ) + FGF − FAG = 0
F GC sin ( θ ) + F BG − F 1 = 0
−F BC + F CD − F GC cos ( θ ) + FCF cos ( θ ) = 0
−F GC sin ( θ ) − F CF sin ( θ ) = 0
−F CD + F DE cos ( θ ) = 0
−F DF − F DE sin ( θ ) = 0
−F GF − F CF cos ( θ ) + FEF = 0
610
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Engineering Mechanics - Statics
Chapter 6
F DF + F CF sin ( θ ) − F 2 = 0
−F DE cos ( θ ) − F EF = 0
611
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Engineering Mechanics - Statics
Chapter 6
⎛ FAG ⎞
⎜
⎟
⎜ FBG ⎟
⎜
⎟
⎜ FGC ⎟
⎜F ⎟
⎜ GF ⎟
⎜ FAB ⎟
⎜
⎟
⎜ FBC ⎟ = Find ( FAG , FBG , FGC , FGF , FAB , FBC , FCD , FCF , FDF , FDE , FEF)
⎜F ⎟
⎜ CD ⎟
⎜ FCF ⎟
⎜
⎟
⎜ FDF ⎟
⎜
⎟
⎜ FDE ⎟
⎜F ⎟
⎝ EF ⎠
⎛ FAG ⎞
⎜
⎟
⎜ FBG ⎟ ⎛⎜ 13.13 ⎟⎞
⎜
⎟ ⎜ 17.50 ⎟
F
GC
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎜ 3.13 ⎟
⎜ GF ⎟ ⎜ 11.25 ⎟
⎜ FAB ⎟ ⎜ −21.88 ⎟
⎜
⎟ ⎜
⎟
F
⎜ BC ⎟ = ⎜ −13.13 ⎟ kN
⎜ F ⎟ ⎜ −9.37 ⎟
⎜ CD ⎟ ⎜
⎟
⎜ FCF ⎟ ⎜ −3.13 ⎟
⎜
⎟ ⎜ 12.50 ⎟
F
⎜ DF ⎟ ⎜
⎟
⎜
⎟ ⎜ −15.62 ⎟
⎜ FDE ⎟ ⎝ 9.37 ⎠
⎜F ⎟
⎝ EF ⎠
Positive (T)
Negative (C)
Problem 6-131
The spring has an unstretched length δ. Determine the angle θ for equilibrium if the uniform
links each have a mass mlink.
612
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Given:
mlink = 5 kg
δ = 0.3 m
N
k = 400
m
a = 0.1 m
b = 0.6 m
g = 9.81
m
2
s
Solution:
Guesses
θ = 10 deg
F BD = 1 N
Ex = 1 N
Given
mlink g
a+b
cos ( θ ) − FBD b cos ( θ ) + E x b sin ( θ ) = 0
2
−2 mlink g
a+b
2
cos ( θ ) + Ex 2 b sin ( θ ) = 0
F BD = k( 2 b sin ( θ ) − δ )
⎛ FBD ⎞
⎜
⎟
⎜ Ex ⎟ = Find ( FBD , Ex , θ )
⎜
⎟
⎝ θ ⎠
θ = 21.7 deg
Problem 6-132
The spring has an unstretched length δ. Determine the mass mlink of each uniform link if the
angle for equilibrium is θ.
Given:
δ = 0.3 m
613
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Engineering Mechanics - Statics
Chapter 6
θ = 20 deg
k = 400
N
m
a = 0.1 m
b = 0.6 m
g = 9.81
m
2
s
Solution:
Guesses
Ey = 1 N
mlink = 1 kg
Fs = 1 N
Given
F s = ( 2 b sin ( θ ) − δ ) k
mlink g
a+b
cos ( θ ) − Fs b cos ( θ ) + E y b sin ( θ ) = 0
2
−2mlink g
a+b
cos ( θ ) + E y2b sin ( θ ) = 0
2
⎛ mlink ⎞
⎜
⎟
⎜ Fs ⎟ = Find ( mlink , Fs , Ey)
⎜ E ⎟
⎝ y ⎠
mlink = 3.859 kg
y
= sin ( θ )
2( b)
y = 2 b sin ( θ )
F s = ( y − δ ) ( k)
F s = 44.17 N
614
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Engineering Mechanics - Statics
ΣMA = 0;
Chapter 6
a+b
( cos (θ ) ) = 0
⎡⎣Ey ( 2) ( a + b) sin ( θ ) − 2( M) ( g)⎤⎦
2
⎡E ( 2) sin ( θ ) − 2 ( m) g⎤ ( cos ( θ ) = 0
⎢ y
⎥
2⎦
⎣
⇒ E y ( 2 sin ( θ ) ) = m ( g) ( cos ( θ ) )
Ey =
ΣMC = 0;
m( g) ( cos ( θ ) )
2 sin ( θ )
m( g) ( cos ( θ ) )
2 sin ( θ )
m = Fs
( a + b) sin ( θ ) + m( g) ⎛⎜
a + b⎞
⎟ cos ( θ ) − Fs( b cos ( θ ) ) = 0
⎝ 2 ⎠
b
g ( a + b)
m = 3.859 kg
Problem 6-133
Determine the horizontal and vertical components of force that the pins A and B exert on the
two-member frame.
Given:
w = 400
N
m
a = 1.5 m
b = 1m
c = 1m
F = 0N
θ = 60 deg
Solution:
Guesses
Ax = 1 N
Ay = 1 N
Bx = 1 N
615
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Engineering Mechanics - Statics
By = 1 N
Cx = 1 N
Chapter 6
Cy = 1 N
Given
−w a
a
+ Cx a sin ( θ ) − Cy a cos ( θ ) = 0
2
F c − Cx c − Cy b = 0
Ay − Cy − w a cos ( θ ) = 0
− Ax − Cx + w a sin ( θ ) = 0
Cx − Bx − F = 0
Cy + By = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟
⎜B ⎟
⎜ x ⎟ = Find A , A , B , B , C , C
( x y x y x y)
⎜ By ⎟
⎜ ⎟
⎜ Cx ⎟
⎜ ⎟
⎝ Cy ⎠
⎛⎜ Ax ⎞⎟ ⎛ 300.0 ⎞
=
N
⎜ Ay ⎟ ⎜⎝ 80.4 ⎟⎠
⎝ ⎠
⎛⎜ Bx ⎟⎞ ⎛ 220 ⎞
=
N
⎜ By ⎟ ⎜⎝ 220 ⎟⎠
⎝ ⎠
Problem 6-134
Determine the horizontal and vertical components of force that the pins A and B exert on the
two-member frame.
Given:
w = 400
N
m
a = 1.5 m
b = 1m
c = 1m
F = 500 N
θ = 60 deg
616
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 6
Solution:
Guesses
Ax = 1 N
Ay = 1 N
Bx = 1 N
By = 1 N
Cx = 1 N
Cy = 1 N
Given
−w a
a
2
+ Cx a sin ( θ ) − Cy a cos ( θ ) = 0
Ay − Cy − w a cos ( θ ) = 0
Cx + Bx − F = 0
F c − Cx c − Cy b = 0
− Ax − Cx + w a sin ( θ ) = 0
Cy − By = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟
⎜B ⎟
⎜ x ⎟ = Find A , A , B , B , C , C
( x y x y x y)
⎜ By ⎟
⎜ ⎟
⎜ Cx ⎟
⎜ ⎟
⎝ Cy ⎠
⎛⎜ Ax ⎞⎟ ⎛ 117.0 ⎞
=
N
⎜ Ay ⎟ ⎜⎝ 397.4 ⎟⎠
⎝ ⎠
⎛⎜ Bx ⎟⎞ ⎛ 97.4 ⎞
=
N
⎜ By ⎟ ⎜⎝ 97.4 ⎟⎠
⎝ ⎠
Problem 6-135
Determine the force in each member of the truss and
indicate whether the members are in tension or
compression.
Units Used:
kip = 1000 lb
Given:
F 1 = 1000 lb
b = 8 ft
F 2 = 500 lb
c = 4 ft
a = 4 ft
Solution:
Joint B:
617
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Engineering Mechanics - Statics
Chapter 6
Initial Guesses:
Given
F BC = 100 lb
F BA = 150 lb
ΣF x = 0;
F 1 − F BC cos ⎛⎜ atan ⎛⎜
⎛ ⎛ c ⎞⎞
⎟⎟ − FBA cos ⎜atan ⎜ ⎟ ⎟ = 0
⎝ a ⎠⎠
⎝ ⎝ a ⎠⎠
ΣF y = 0;
−F BC sin ⎜⎛ atan ⎛⎜
⎝
b ⎞⎞
⎛ ⎛ c ⎞⎞
⎟ ⎟ + FBA sin ⎜atan ⎜ ⎟ ⎟ − F2 = 0
⎝ a ⎠⎠
⎝ ⎝ a ⎠⎠
⎝
b ⎞⎞
⎛⎜ FBC ⎞⎟
= Find ( F BC , F BA)
⎜ FBA ⎟
⎝
⎠
F BC = 373 lb(C)
F BA = 1178.51 lb
F BA = 1.179 kip (C)
Joint A:
ΣF y = 0;
F AC − FBA sin ⎜⎛ atan ⎛⎜
(
)
⎝
c ⎞⎞
c
F AC = FBA
⎟⎟ = 0
⎝ a ⎠⎠
2
a
2
a +c
2
a
F AC = 833 lb (T)
Problem 6-136
Determine the force in each
member of the truss and state if
the members are in tension or
compression.
Units Used:
3
kip = 10 lb
Given:
F = 1000 lb
a = 10 ft
b = 10 ft
618
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Engineering Mechanics - Statics
Solution:
Chapter 6
b
θ = atan ⎛⎜ ⎞⎟
⎝ a⎠
Guesses
F AB = 1 lb
F AG = 1 lb
F BC = 1 lb
F BG = 1 lb
F CD = 1 lb
F CE = 1 lb
F CG = 1 lb
F DE = 1 lb
F EG = 1 lb
Given
F AB + F AG cos ( θ ) = 0
−F AB + F BC = 0
F BG = 0
F CD − F BC − F CG cos ( θ ) = 0
F CE − F + F CG sin ( θ ) = 0
−F CD − F DE cos ( θ ) = 0
F DE cos ( θ ) − F EG = 0
−F CE − FDE sin ( θ ) = 0
F EG − F AG cos ( θ ) + F CG cos ( θ ) = 0
−F AG sin ( θ ) − FBG − FCG sin ( θ ) = 0
⎛ FAB ⎞
⎜
⎟
⎜ FAG ⎟
⎜F ⎟
⎜ BC ⎟
⎜ FBG ⎟
⎜
⎟
⎜ FCD ⎟ = Find ( FAB , FAG , FBC , FBG , FCD , FCE , FCG , FDE , FEG)
⎜F ⎟
⎜ CE ⎟
⎜ FCG ⎟
⎜
⎟
⎜ FDE ⎟
⎟
⎜
⎝ FEG ⎠
619
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Engineering Mechanics - Statics
Chapter 6
⎛ FAB ⎞
⎜
⎟ ⎛ 333 ⎞
⎜ FAG ⎟ ⎜
⎟
⎜ F ⎟ ⎜ −471 ⎟
⎜ BC ⎟ ⎜ 333 ⎟
⎜ FBG ⎟ ⎜ 0 ⎟
⎟
⎜
⎟ ⎜
F
667
=
⎟ lb
⎜ CD ⎟ ⎜
⎜ F ⎟ ⎜ 667 ⎟
⎟
⎜ CE ⎟ ⎜
⎜ FCG ⎟ ⎜ 471 ⎟
⎜
⎟ ⎜ −943 ⎟
F
⎜ DE ⎟ ⎜
⎟
⎟ ⎝ −667 ⎠
⎜
⎝ FEG ⎠
Positive (T), Negative (C)
Problem 6-137
Determine the force in members AB, AD, and AC of the space truss and state if the members are
in tension or compression. The force F is vertical.
Units Used:
3
kip = 10 lb
Given:
F = 600 lb
a = 1.5 ft
b = 2 ft
c = 8 ft
Solution:
⎛a⎞
AB = ⎜ −c ⎟
⎜ ⎟
⎝0⎠
⎛ −a ⎞
AC = ⎜ −c ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
AD = ⎜ −c ⎟
⎜ ⎟
⎝b⎠
Guesses
F AB = 1 lb
F AC = 1 lb
F AD = 1 lb
620
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Given
Chapter 6
⎛ 0 ⎞
F AB
+ FAC
+ FAD
+⎜ 0 ⎟ =0
⎜ ⎟
AB
AC
AD
⎝ −F ⎠
AB
AC
AD
⎛ FAB ⎞
⎜
⎟
F
⎜ AC ⎟ = Find ( FAB , FAC , FAD)
⎜F ⎟
⎝ AD ⎠
⎛ FAB ⎞ ⎛ −1.221 ⎞
⎜
⎟ ⎜
⎟
F
⎜ AC ⎟ = ⎜ −1.221 ⎟ kip
⎜ F ⎟ ⎝ 2.474 ⎠
⎝ AD ⎠
Positive (T)
Negative (C)
621
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Engineering Mechanics - Statics
Chapter 7
Problem 7-1
The column is fixed to the floor and is subjected to the loads shown. Determine the internal normal
force, shear force, and moment at points A and B.
Units Used:
3
kN = 10 N
Given:
F 1 = 6 kN
F 2 = 6 kN
F 3 = 8 kN
a = 150 mm
b = 150 mm
c = 150 mm
Solution:
Free body Diagram: The support reaction need not be computed in this case.
Internal Forces: Applying equations of equillibrium to the top segment sectioned through
point A, we have
+
Σ F x = 0; V A = 0
→
+
↑Σ Fy = 0;
ΣMA = 0;
VA = 0
NA − F1 − F2 = 0
NA = F 1 + F 2
NA = 12.0 kN
F 1 a − F2 b − MA = 0
MA = F1 a − F 2 b
MA = 0 kN⋅ m
Applying equations of equillibrium to the top segment sectioned through point B, we have
+
Σ F x = 0; V B = 0
→
+
↑Σ Fy = 0;
+ΣMB = 0;
VB = 0
NB − F1 − F2 − F3 = 0
NB = F 1 + F 2 + F 3
NB = 20.0 kN
F 1 a − F 2 b − F3 c + MB = 0
MB = −F1 a + F 2 b + F3 c
MB = 1.20 kN⋅ m
Problem 7-2
The axial forces act on the shaft as shown. Determine the internal normal forces at points A
and B.
622
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Engineering Mechanics - Statics
Chapter 7
Given:
F 1 = 20 lb
F 2 = 50 lb
F 3 = 10 lb
Solution:
Section A:
ΣF z = 0;
F 2 − 2 F1 − NA = 0
NA = F 2 − 2 F 1
NA = 10.00 lb
Section B:
ΣF z = 0;
F 2 − 2 F1 − NA + NB = 0
NB = −F 2 + 2 F1 + NA
NB = 0.00 lb
Problem 7-3
The shaft is supported by smooth bearings at A and B and subjected to the torques shown.
Determine the internal torque at points C, D, and E.
Given:
M1 = 400 N⋅ m
M2 = 150 N⋅ m
M3 = 550 N⋅ m
623
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Engineering Mechanics - Statics
Chapter 7
Solution:
Section C:
ΣMx = 0;
TC = 0
Section D:
ΣMx = 0;
TD − M 1 = 0
TD = M 1
TD = 400.00 N⋅ m
Section E:
ΣMx = 0;
M 1 + M 2 − TE = 0
TE = M 1 + M 2
TE = 550.00 N⋅ m
Problem 7-4
Three torques act on the shaft. Determine the internal torque at points A, B, C, and D.
Given:
M1 = 300 N⋅ m
M2 = 400 N⋅ m
M3 = 200 N⋅ m
Solution:
Section A:
ΣΜx = 0;
− TA + M 1 − M 2 + M 3 = 0
TA = M 1 − M 2 + M 3
TA = 100.00 N⋅ m
Section B:
ΣMx = 0;
TB + M 3 − M 2 = 0
624
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
TB = − M 3 + M 2
TB = 200.00 N⋅ m
Section C:
ΣΜx = 0;
− TC + M 3 = 0
TC = M 3
TC = 200.00 N⋅ m
Section D:
ΣΜx = 0;
TD = 0
Problem 7-5
The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal
force, shear force, and moment at a section passing through (a) point C, which is just to the
right of the bearing at A, and (b) point D, which is just to the left of the force F 2.
Units Used:
3
kip = 10 lb
Given:
F 1 = 2.5 kip
F 2 = 3 kip
w = 75
lb
ft
a = 6 ft
b = 12 ft
c = 2 ft
Solution:
Σ MB = 0;
625
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Engineering Mechanics - Statics
Chapter 7
− Ay( b + c) + F1 ( a + b + c) + w b⎛⎜ c +
⎝
F1 ( a + b + c) + w b⎛⎜ c +
⎝
Ay =
b⎞
⎟ + F2 c = 0
2⎠
b⎞
⎟ + F2 c
2⎠
b+c
Ay = 4514 lb
+
Σ F x = 0;
→
+
↑Σ Fy = 0;
B x = 0 lb
Ay − F1 − w b − F2 + By = 0
By = − Ay + F1 + w b + F2
Σ MC = 0;
F 1 a + Mc = 0
+
Σ F x = 0;
→
NC = 0 lb
+
↑ Σ Fy = 0;
−F 1 + Ay − V C = 0
Σ MD = 0;
−MD + By c = 0
+
Σ F x = 0;
→
+
↑
Σ F y = 0;
B y = 1886 lb
MC = −F 1 a
MC = −15.0 kip⋅ ft
NC = 0.00 lb
VC = Ay − F1
V C = 2.01 kip
MD = B y c
MD = 3.77 kip⋅ ft
ND = 0 lb
ND = 0.00 lb
VD − F2 + By = 0
VD = F2 − By
V D = 1.11 kip
Problem 7-6
Determine the internal normal force, shear force, and moment at point C.
Given:
M = 400 lb⋅ ft
a = 4 ft
b = 12 ft
626
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Engineering Mechanics - Statics
Chapter 7
Solution:
Beam:
ΣMB = 0;
M − Ay( a + b) = 0
Ay =
M
a+b
Ay = 25.00 lb
Segment AC:
ΣF x = 0;
NC = 0
ΣF y = 0;
Ay − VC = 0
VC = Ay
V C = 25.00 lb
ΣMC = 0;
− Ay a + MC = 0
MC = Ay a
MC = 100.00 lb⋅ ft
Problem 7-7
Determine the internal normal force, shear force, and moment at point C.
Units Used:
kN = 103 N
Given:
F 1 = 30 kN
F 2 = 50 kN
F 3 = 25 kN
a = 1.5 m
b = 3m
θ = 30 deg
Solution:
ΣF x = 0;
−NC + F 3 cos ( θ ) = 0
627
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Engineering Mechanics - Statics
Chapter 7
NC = F3 cos ( θ )
NC = 21.7 kN
V C − F 2 − F 3 sin ( θ ) = 0
ΣF y = 0;
V C = F2 + F3 sin ( θ )
V C = 62.50 kN
ΣMC = 0; −MC − F2 b − F 3 sin ( θ ) 2 b = 0
MC = −F 2 b − F3 sin ( θ ) 2b
MC = −225.00 kN⋅ m
Problem 7-8
Determine the normal force, shear force, and moment at a section passing through point
C. Assume the support at A can be approximated by a pin and B as a roller.
Units used:
3
kip = 10 lb
Given:
F 1 = 10 kip
a = 6 ft
F 2 = 8 kip
b = 12 ft
kip
ft
c = 12 ft
w = 0.8
d = 6 ft
Solution:
ΣMA = 0;
⎛ b + c ⎞ − F ( b + c + d) + B ( b + c) + F a = 0
⎟
2
y
1
⎝ 2 ⎠
−w( b + c) ⎜
w
( b + c)
By =
+
Σ F x = 0;
→
2
2
+ F 2 ( b + c + d) − F 1 a
B y = 17.1 kip
b+c
NC = 0
628
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Engineering Mechanics - Statics
+
↑Σ Fy = 0;
Chapter 7
VC − w c + By − F2 = 0
VC = w c − By + F2
V C = 0.5 kip
⎛ c ⎞ + B c − F ( c + d) = 0
⎟
y
2
⎝ 2⎠
−MC − w c⎜
ΣMC = 0;
⎛ c2 ⎞
⎟ + B y c − F 2 ( c + d)
⎝2⎠
MC = −w⎜
MC = 3.6 kip⋅ ft
Problem 7-9
The beam AB will fail if the maximum internal moment at D reaches Mmax or the normal force in
member BC becomes Pmax. Determine the largest load w it can support.
Given:
Mmax = 800 N⋅ m
P max = 1500 N
a = 4m
b = 4m
c = 4m
d = 3m
Solution:
⎛ a + b ⎞ − A ( a + b) = 0
⎟
y
⎝ 2 ⎠
w ( a + b) ⎜
Ay =
w( a + b)
2
⎛ a⎞ − A a + M = 0
⎟
y
D
⎝2⎠
w a⎜
⎛ a b⎞
⎟
⎝2 ⎠
MD = w ⎜
⎛
d
Ay − w( a + b) + ⎜
2
2
2
⎞T = 0
2⎟
⎝ c +d ⎠
T=
w( a + b)
c +d
2d
Assume the maximum moment has been reached
629
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Engineering Mechanics - Statics
Chapter 7
MD = Mmax
2 MD
w1 =
ab
w1 = 100
N
m
w2 = 225
N
m
Assume that the maximum normal force in BC has been reached
T = Pmax
T2 d
w2 =
2
( a + b) c + d
2
w = min ( w1 , w2 )
Now choose the critical load
w = 100
N
m
Problem 7-10
Determine the shear force and moment acting at a section passing through point C in the beam.
Units Used:
3
kip = 10 lb
Given:
w = 3
kip
ft
a = 6 ft
b = 18 ft
Solution:
ΣMB = 0;
− Ay b +
Ay =
ΣMC = 0;
1
6
− Ay a +
1
2
⎛ b⎞ = 0
⎟
⎝3⎠
wb ⎜
Ay = 9 kip
wb
a⎞ ⎛ a⎞
w ⎟ a ⎜ ⎟ + MC = 0
⎜
2 ⎝ b⎠ ⎝ 3 ⎠
1⎛
3
wa
MC = Ay a −
6b
MC = 48 kip⋅ ft
+
↑
Σ F y = 0;
Ay −
a⎞
w ⎟ a − VC = 0
⎜
2 ⎝ b⎠
1 ⎛
2
wa
VC = Ay −
2b
630
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Engineering Mechanics - Statics
Chapter 7
V C = 6 kip
Problem 7-11
Determine the internal normal force, shear force, and moment at points E and D of the
compound beam.
Given:
M = 200 N⋅ m
c = 4m
F = 800 N
d = 2m
a = 2m
e = 2m
b = 2m
Solution:
Segment BC :
M
d+e
−M + Cy( d + e) = 0
Cy =
−B y + Cy = 0
B y = Cy
Segment EC :
−NE = 0
NE = 0 N
NE = 0.00
V E + Cy = 0
V E = −Cy
V E = −50.00 N
−ME − M + Cy e = 0
ME = Cy e − M
ME = −100.00 N⋅ m
Segment DB :
−ND = 0
ND = 0 N
ND = 0.00
VD − F + By = 0
VD = F − By
V D = 750.00 N
−MD − F b + By( b + c) = 0
MD = −F b + B y( b + c)
MD = −1300 N⋅ m
631
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Engineering Mechanics - Statics
Chapter 7
Problem 7-12
The boom DF of the jib crane and the
column DE have a uniform weight density
γ. If the hoist and load have weight W,
determine the normal force, shear force,
and moment in the crane at sections
passing through points A, B, and C. Treat
the boom tip, beyond the hoist, as
weightless.
Given:
W = 300 lb
γ = 50
lb
ft
a = 7 ft
b = 5 ft
c = 2 ft
d = 8 ft
e = 3 ft
Solution:
+
ΣF x = 0;
→
−NA = 0
+
VA − W − γ e = 0
↑ ΣFy = 0;
NA = 0 lb
VA = W + γ e
ΣΜΑ = 0;
+
ΣF x = 0;
→
+
↑ ΣFy = 0;
ΣΜΒ = 0;
NA = 0.00 lb
V A = 450 lb
⎛e⎞ − We = 0
⎟
⎝ 2⎠
⎛ e2 ⎞
MA = γ ⎜ ⎟ + W e
⎝2⎠
MA − γ e⎜
−NB = 0
MA = 1125.00 lb⋅ ft
NB = 0 lb
V B − γ ( d + e) − W = 0
NB = 0.00 lb
V B = γ ( d + e) + W
V B = 850 lb
⎛ d + e ⎞ − W( d + e) = 0
⎟
⎝ 2 ⎠
MB − γ ( d + e) ⎜
MB =
1
2
2
γ ( d + e) + W( d + e)
MB = 6325.00 lb⋅ ft
632
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Engineering Mechanics - Statics
Chapter 7
+
ΣF x = 0;
→
VC = 0
+
NC − ( c + d + e) γ − W − γ ( b) = 0
↑ ΣFy = 0;
V C = 0 lb
V C = 0.00 lb
NC = γ ( c + d + e + b) + W
NC = 1200.00 lb
⎛ c + d + e ⎞ − W( c + d + e) = 0
⎟
⎝ 2 ⎠
MC − ( c + d + e) γ ⎜
ΣΜC = 0;
⎛ c + d + e ⎞ + W( c + d + e)
⎟
⎝ 2 ⎠
MC = ( c + d + e) γ ⎜
MC = 8125.00 lb⋅ ft
Problem 7-13
Determine the internal normal force, shear force, and moment at point C.
Units Used:
3
kip = 10 lb
Given:
a = 0.5 ft
d = 8 ft
b = 2 ft
e = 4 ft
c = 3 ft
w = 150
lb
ft
Solution:
Entire beam:
ΣMA = 0;
⎛ d + e ⎞ + T ( a + b) = 0
⎟
⎝ 2 ⎠
−w( d + e) ⎜
w ( d + e)
T =
2 ( a + b)
ΣF x = 0;
2
Ax − T = 0
Ax = T
ΣF y = 0;
T = 4.32 kip
Ax = 4.32 kip
Ay − w( d + e) = 0
Ay = w( d + e)
Ay = 1.80 kip
Segment AC:
633
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
ΣF x = 0;
Ax + NC = 0
NC = − A x
NC = −4.32 kip
ΣF y = 0;
Ay − w c − VC = 0
VC = Ay − w c
V C = 1.35 kip
ΣMC = 0;
− Ay c + w c⎜
⎛ c2 ⎞
⎟
⎝2⎠
⎛c⎞ + M = 0
⎟
C
⎝2⎠
MC = Ay c − w ⎜
MC = 4.72 kip⋅ ft
Problem 7-14
Determine the normal force, shear force, and moment at a section passing through point D of
the two-member frame.
Units Used:
3
kN = 10 N
Given:
w = 400
N
m
a = 2.5 m
b = 3m
c = 6m
Solution:
ΣMA = 0;
−1
2
⎛2 ⎞
⎛
w c ⎜ c⎟ + F BC ⎜
⎝3 ⎠
1
F BC =
+
→ Σ Fx = 0;
⎞c = 0
⎟
2
2
⎝ a +c ⎠
3
⎛ a2 + c2 ⎞
⎟
⎝ a ⎠
wc ⎜
⎛
⎞F
c
2⎟
2
⎝ a +c ⎠
↑Σ Fy = 0;
F BC = 2080 N
⎛ c ⎞F − A = 0
x
⎜ 2 2 ⎟ BC
⎝ a +c ⎠
Ax = ⎜
+
a
Ay −
1
2
⎛
wc + ⎜
a
2
Ax = 1920 N
BC
⎞F
2⎟
⎝ a +c ⎠
BC = 0
634
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Engineering Mechanics - Statics
Ay =
Chapter 7
1
2
wc − ⎛
a
⎞
⎜ 2 2 ⎟ FBC
⎝ a +c ⎠
+
Σ F x = 0;
→
ND − Ax = 0
+
Ay −
↑Σ Fy = 0;
ND = 1.920 kN
b⎞
⎜w ⎟ b − VD = 0
c⎠
1⎛
2⎝
VD = Ay −
ΣF y = 0;
ND = A x
Ay = 400 N
− Ay b +
1
2
⎛ b2 ⎞
⎟
⎝c⎠
w⎜
V D = 100 N
b
b
w ⎞⎟ b⎛⎜ ⎟⎞ + MD = 0
⎜
2⎝ c⎠ ⎝3⎠
1⎛
⎛ b3 ⎞
MD = Ay b − w ⎜ ⎟
6
⎝c⎠
1
MD = 900 N m
Problem 7-15
The beam has weight density γ. Determine the internal normal force, shear force, and
moment at point C.
Units Used:
kip = 103 lb
Given:
γ = 280
lb
ft
a = 3 ft
b = 7 ft
c = 8 ft
d = 6 ft
635
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
c
θ = atan ⎛⎜ ⎟⎞
W = γ ( a + b)
⎝ d⎠
Guesses
Ax = 1 lb
Ay = 1 lb
B x = 1 lb
NC = 1 lb
V C = 1 lb
MC = 1 lb ft
Given
Entire beam:
Ax − Bx = 0
Ay − W = 0
⎛ d⎞ = 0
⎟
⎝ 2⎠
Bx c − W ⎜
Bottom Section
Ax − NC cos ( θ ) + V C sin ( θ ) = 0
⎛ a ⎞ − N sin ( θ ) − V cos ( θ ) = 0
⎟
C
C
⎝ a + b⎠
Ay − W ⎜
⎛ a ⎞ ⎛ a ⎞ cos ( θ ) = 0
⎟⎜ ⎟
⎝ a + b⎠ ⎝ 2 ⎠
MC − VC a − W ⎜
⎛⎜ Ax ⎟⎞
⎜ Ay ⎟
⎜
⎟
⎜ Bx ⎟ = Find ( A , A , B , N , V , M )
x y x C C
C
⎜ NC ⎟
⎜
⎟
⎜ VC ⎟
⎜ MC ⎟
⎝
⎠
⎛ Ax ⎞ ⎛ 1.05 ⎞
⎜ ⎟
⎜ Ay ⎟ = ⎜ 2.80 ⎟ kip
⎜ B ⎟ ⎜⎝ 1.05 ⎟⎠
⎝ x⎠
⎛ NC ⎞ ⎛ 2.20 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ VC ⎠ ⎝ 0.34 ⎠
MC = 1.76 kip⋅ ft
Problem 7-16
Determine the internal normal force, shear force, and moment at points C and D of the
beam.
636
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Engineering Mechanics - Statics
Chapter 7
Units Used:
3
kip = 10 lb
Given:
w1 = 60
lb
ft
a = 12 ft
b = 15 ft
lb
w2 = 40
ft
c = 10 ft
d = 5 ft
F = 690 lb
e = 12
f = 5
θ = atan ⎛⎜ ⎟⎞
e
⎝ f⎠
Solution:
Guesses
B y = 1 lb
NC = 1 lb
V C = 1 lb
MC = 1 lb⋅ ft
ND = 1 lb
V D = 1 lb
MD = 1 lb⋅ ft
Given
⎛ b ⎞ ... = 0
⎟
⎝ 2⎠
B y b − F sin ( θ ) ( b + c) − w2 b⎜
+
−1
2
(w1 − w2)b⎛⎜ 3 ⎟⎞
b
⎝ ⎠
−NC − F cos ( θ ) = 0
VC −
(w1 − w2) ⎛⎜ b ⎞⎟ ( b − a) ... = 0
2
b−a
1
⎝
⎠
+ B y − w2 ( b − a) − F sin ( θ )
−ND − F cos ( θ ) = 0
V D − F sin ( θ ) = 0
−MD − F sin ( θ ) d = 0
⎛ b − a ⎞ − 1 ( w − w ) ⎛ b − a ⎞ ( b − a) ⎛ b − a ⎞ ... = 0
⎟
⎟
⎜ 3 ⎟
1
2 ⎜
⎝
⎠
⎝ 2 ⎠ 2
⎝ b ⎠
−MC − w2 ( b − a) ⎜
+ B y ( b − a) − F sin ( θ ) ( c + b − a)
637
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Engineering Mechanics - Statics
Chapter 7
⎛ By ⎞
⎜
⎟
⎜ NC ⎟
⎜ VC ⎟
⎜
⎟
M
⎜ C ⎟ = Find ( By , NC , VC , MC , ND , VD , MD)
⎜N ⎟
⎜ D⎟
⎜ VD ⎟
⎜
⎟
⎝ MD ⎠
⎛ NC ⎞ ⎛ −265 ⎞
⎜ ⎟ ⎜
⎟
⎜ VC ⎟ = ⎜ −649 ⎟ lb
⎜ ND ⎟ ⎜ −265 ⎟
⎜ ⎟ ⎜ 637 ⎟
⎠
⎝ VD ⎠ ⎝
⎛ MC ⎞ ⎛ −4.23 ⎞
⎜
⎟=⎜
⎟ kip⋅ ft
⎝ MD ⎠ ⎝ −3.18 ⎠
Problem 7-17
Determine the normal force, shear force, and moment acting at a section passing through point C.
3
kip = 10 lb
Units Used:
Given:
F 1 = 800 lb
F 2 = 700 lb
F 3 = 600 lb
θ = 30 deg
a = 1.5 ft
b = 1.5 ft
c = 3 ft
d = 2 ft
e = 1 ft
f = a+b+c−d−e
Solution:
Guesses
B y = 1 lb
Ax = 1 lb
Ay = 1 lb
NC = 1 lb
V C = 1 lb
MC = 1 lb⋅ ft
638
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given
− Ax + V C sin ( θ ) + NC cos ( θ ) = 0
Ay − V C cos ( θ ) + NC sin ( θ ) = 0
MC − A x ( a) sin ( θ ) − Ay ( a) cos ( θ ) = 0
− Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0
Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0
−F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0
+ F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )
⎛⎜ Ax ⎟⎞
⎜ Ay ⎟
⎜
⎟
⎜ By ⎟ = Find ( A , A , B , N , V , M )
x y y C C
C
⎜ NC ⎟
⎜
⎟
⎜ VC ⎟
⎜ MC ⎟
⎝
⎠
⎛ Ax ⎞ ⎛ 100 ⎞
⎜ ⎟
⎜ Ay ⎟ = ⎜ 985 ⎟ lb
⎜ B ⎟ ⎜⎝ 927 ⎟⎠
⎝ y⎠
⎛ NC ⎞ ⎛ −406 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ VC ⎠ ⎝ 903 ⎠
MC = 1.355 kip⋅ ft
Problem 7-18
Determine the normal force, shear force, and moment acting at a section passing through point D.
Units Used:
kip = 103 lb
639
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
F 1 = 800 lb
F 2 = 700 lb
F 3 = 600 lb
θ = 30 deg
a = 1.5 ft
b = 1.5 ft
c = 3 ft
d = 2 ft
e = 1 ft
f = a+b+c−d−e
Solution:
Guesses
B y = 1 lb
Ax = 1 lb
Ay = 1 lb
ND = 1 lb
V D = 1 lb
MD = 1 lb⋅ ft
Given
V D sin ( θ ) − ND cos ( θ ) − F3 sin ( θ ) = 0
B y + VD cos ( θ ) + ND sin ( θ ) − F 3 cos ( θ ) = 0
−MD − F3 e + By( e + f) cos ( θ ) = 0
Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0
− Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0
−F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0
+ F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )
⎛⎜ Ax ⎟⎞
⎜ Ay ⎟
⎜
⎟
⎜ By ⎟ = Find ( A , A , B , N , V , M )
x y y D D
D
⎜ ND ⎟
⎜
⎟
⎜ VD ⎟
⎜ MD ⎟
⎝
⎠
⎛ Ax ⎞ ⎛ 100 ⎞
⎜ ⎟
⎜ Ay ⎟ = ⎜ 985 ⎟ lb
⎜ B ⎟ ⎜⎝ 927 ⎟⎠
⎝ y⎠
⎛ ND ⎞ ⎛ −464 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ VD ⎠ ⎝ −203 ⎠
MD = 2.61 kip⋅ ft
640
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-19
Determine the normal force, shear force, and moment at a section passing through point C.
Units Used:
3
kN = 10 N
Given:
P = 8 kN
c = 0.75 m
a = 0.75m
d = 0.5 m
b = 0.75m
r = 0.1 m
Solution:
ΣMA = 0;
−T ( d + r) + P( a + b + c) = 0
⎛ a + b + c ⎞ T = 30 kN
⎟
⎝ d+r ⎠
T = P⎜
+
Σ F x = 0;
→
Ax = T
Ax = 30 kN
+
Ay = P
Ay = 8 kN
+
Σ F x = 0;
→
−NC − T = 0
↑Σ Fy = 0;
NC = −T
+
↑Σ Fy = 0;
VC + P = 0
V C = −P
ΣMC = 0;
NC = −30 kN
V C = −8 kN
−MC + P c = 0
MC = P c
MC = 6 kN⋅ m
Problem 7-20
The cable will fail when subjected to a tension Tmax. Determine the largest vertical load P the frame
will support and calculate the internal normal force, shear force, and moment at a section passing
through point C for this loading.
Units Used:
3
kN = 10 N
641
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
Tmax = 2 kN
a = 0.75 m
b = 0.75 m
c = 0.75 m
d = 0.5 m
r = 0.1 m
Solution:
ΣMA = 0;
−Tmax( r + d) + P ( a + b + c) = 0
⎛ d+r ⎞
⎟
⎝ a + b + c⎠
P = Tmax⎜
P = 0.533 kN
+
Σ F x = 0;
→
Tmax − Ax = 0
Ax = Tmax
Ax = 2 kN
+
↑Σ Fy = 0;
Ay − P = 0
Ay = P
Ay = 0.533 kN
+
Σ F x = 0;
→
−NC − Ax = 0
NC = − A x
NC = −2 kN
VC = Ay
V C = 0.533 kN
MC = Ay c
MC = 0.400 kN⋅ m
+
↑Σ Fy = 0;
ΣMC = 0;
−V C + Ay = 0
−MC + A y c = 0
Problem 7-21
Determine the internal shear force and moment acting at point C of the beam.
Units Used:
3
kip = 10 lb
Given:
w = 2
kip
ft
a = 9 ft
642
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
ΣF x = 0;
NC = 0
ΣF y = 0;
wa
ΣMC = 0;
2
NC = 0
wa
−
− VC = 0
2
MC − ⎛⎜
wa⎞
MC =
wa
VC = 0
wa⎞ ⎛ a⎞
a + ⎛⎜
⎟
⎟⎜ ⎟=0
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝3⎠
2
MC = 54.00 kip⋅ ft
3
Problem 7-22
Determine the internal shear force and moment acting at point D of the beam.
Units Used:
kip = 103 lb
Given:
w = 2
kip
ft
a = 6 ft
b = 9 ft
Solution:
ΣF x = 0;
ΣF y = 0;
ND = 0
wb
2
− w⎛⎜
VD =
a⎞ ⎛ a⎞
⎟ ⎜ ⎟ − VD = 0
⎝ b⎠ ⎝ 2 ⎠
wb
2
− w⎛⎜
a⎞ ⎛ a⎞
⎟⎜ ⎟
⎝ b⎠ ⎝ 2 ⎠
V D = 5.00 kip
ΣMD = 0;
w a ⎞⎛ a ⎞ ⎛ a ⎞
a + ⎛⎜
⎟
⎟⎜ ⎟ ⎜ ⎟ = 0
⎝ 2 ⎠ ⎝ b ⎠⎝ 2 ⎠ ⎝ 3 ⎠
⎛ w a3 ⎞
wb⎞
⎟
MD = ⎛⎜
a−⎜
⎟
⎝ 2 ⎠ ⎝ 6b ⎠
MD − ⎛⎜
wb⎞
MD = 46.00 kip⋅ ft
643
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-23
The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal
normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at
A, and (b) point D, which is just to the left of the F2 force.
Units Used:
kip = 103 lb
Given:
F 1 = 2500 lb
a = 6 ft
F 2 = 3000 lb
b = 12 ft
w = 75
lb
ft
Solution:
ΣMB = 0;
c = 2 ft
− Ay( b + c) + F1 ( a + b + c) + w b ⎛⎜
Ay =
(2
2
ΣF y = 0;
Ay − F1 − w b − F2 + By = 0
By = − Ay + F1 + w b + F2
ΣF x = 0;
ΣF y = 0;
)
⎠
b+c
B x = 0 lb
ΣMC = 0;
⎝2
+ c⎞⎟ + F2 c = 0
1 2 F 1 ( a + b + c) + w b + 2 c b + 2 F 2 c
ΣF x = 0;
Segment AC :
b
Ay = 4514 lb
B y = 1886 lb
F 1 a + MC = 0
MC = −F 1 a
MC = −15 kip⋅ ft
NC = 0
NC = 0
−F 1 + Ay − V C = 0
VC = Ay − F1
V C = 2.01 kip
Segment BD:
ΣMD = 0;
−MD + By c = 0
MD = B y c
MD = 3.77 kip⋅ ft
ΣF x = 0;
ND = 0
ND = 0
ΣF y = 0;
VD − F2 + By = 0
VD = F2 − By
V D = 1.11 kip
644
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-24
The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial
compressive force in the jack is P, determine the internal moment developed at point C of the
top beam. Neglect the weight of the beams.
Units Used:
kip = 103 lb
Given:
P = 5000 lb
a = 2 ft
b = 10 ft
Solution:
Segment:
ΣMC = 0; MC + ⎛⎜
P⎞
⎟b = 0
⎝2⎠
MC = −
P
2
b
MC = −25.00 kip⋅ ft
Problem 7-25
The jack AB is used to straighten the bent
beam DE using the arrangement shown. If
the axial compressive force in the jack is P,
determine the internal moment developed at
point C of the top beam. Assume that each
beam has a uniform weight density γ.
Units Used:
kip = 103 lb
Given:
P = 5000 lb
645
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
γ = 150
Chapter 7
lb
ft
a = 2 ft
b = 10 ft
Solution:
Beam:
+
↑Σ Fy = 0;
P − 2 γ ( a + b) − 2 R = 0
R =
P
2
− γ ( a + b)
R = 700 lb
Segment:
ΣMC = 0;
MC + R b + γ ( a + b) ⎛⎜
⎝
MC = −R b − γ
( a + b)
a + b⎞
2
⎟=0
⎠
2
2
MC = −17.8 kip⋅ ft
Problem 7-26
Determine the normal force, shear force, and moment in the beam at sections passing through
points D and E. Point E is just to the right of the F load.
Units Used:
kip = 103 lb
Given:
w = 1.5
a = 6 ft
kip
ft
F = 3 kip
b = 6 ft
c = 4 ft
d = 4 ft
646
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
Σ MB = 0;
1
2
⎛ a + b ⎞ − A ( a + b) = 0
⎟
y
⎝ 3 ⎠
w( a + b) ⎜
1
2
Ay =
⎛ a + b⎞
⎟
⎝ 3 ⎠
w( a + b) ⎜
a+b
Ay = 3 kip
+
Σ F x = 0;
→
Bx = 0
+
By + Ay −
↑
Σ F y = 0;
1
2
By = − Ay +
w( a + b) = 0
1
2
w( a + b)
B y = 6 kip
+
Σ F x = 0;
→
ND = 0
+
Ay −
↑
Σ F y = 0;
1⎛ aw ⎞
⎜
VD = Ay −
Σ MD = 0;
⎟ a − VD = 0
2 ⎝ a + b⎠
MD +
1⎛ aw ⎞
⎜
V D = 0.75 kip
1⎛ aw ⎞
⎛ a⎞
a ⎜ ⎟ − Ay a = 0
⎜
⎟
2 ⎝ a + b⎠ ⎝ 3⎠
MD =
−1 ⎛ a w ⎞ ⎛ a ⎞
⎜
⎟ a ⎜ ⎟ + Ay a
2 ⎝ a + b⎠ ⎝ 3 ⎠
+
Σ F x = 0;
→
NE = 0
+
−V E − F − B y = 0
↑Σ Fy = 0;
⎟a
2 ⎝ a + b⎠
V E = −F − By
ΣME = 0;
MD = 13.5 kip⋅ ft
V E = −9 kip
ME + B y c = 0
ME = −By c
ME = −24.0 kip⋅ ft
647
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-27
Determine the normal force, shear force, and moment at a section passing through point D of the
two-member frame.
Units Used:
kN = 103 N
Given:
w1 = 200
N
w2 = 400
N
m
m
a = 2.5 m
b = 3m
c = 6m
Solution:
Σ MA = 0;
−w1 c⎛⎜
1
2c
− ( w2 − w1 ) c ⎛⎜ ⎟⎞ +
⎟
2
⎝ 2⎠
⎝3⎠
c⎞
2
⎡ c2
c⎤
⎢
F BC = w1
+ ( w2 − w1 ) ⎥
3⎦
⎣ 2
+
Σ F x = 0;
→
Ax = ⎛
+
Ay − w1 c −
1
Ay = w1 c +
1
↑Σ Fy = 0;
+
(FBC c) = 0
F BC = 2600 N
(w2 − w1)c − ⎛⎜
2
a
⎞F = 0
BC
⎟
2
2
+
c
a
⎝
⎠
2
⎞F
2⎟
⎝ a +c ⎠
+
Ay − w1 b −
BC
ND = A x
Ay = 800 N
ND = 2.40 kN
(w2 − w1)⎛⎜ c ⎟⎞ b − VD = 0
2
b
1
⎝ ⎠
V D = A y − w1 b −
Σ MD = 0;
2
Ax = 2400 N
a
− Ax + ND = 0
↑
ac
(w2 − w1)c + ⎛⎜
2
2
a +c
2
a +c
c
⎞
⎜ 2 2 ⎟ FBC
⎝ a +c ⎠
→ Σ Fx = 0;
Σ F y = 0;
2
a
⎛ b2 ⎞
(w2 − w1)⎜ c ⎟
2
1
⎝ ⎠
V D = 50 N
1
b
b
+ ( w2 − w1 ) ⎛⎜ ⎟⎞ b ⎛⎜ ⎟⎞ + MD = 0
⎟
⎝2⎠ 2
⎝c⎠ ⎝3⎠
− Ay b + w1 b⎛⎜
b⎞
⎛ b2 ⎞ 1
⎛ b3 ⎞
⎜
⎟
MD = Ay( b) − w1
− ( w2 − w1 ) ⎜ ⎟
⎝2⎠ 2
⎝ 3c ⎠
648
MD = 1.35 kN⋅ m
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-28
Determine the normal force, shear force, and moment at sections passing through points E and F.
Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD.
Units Used:
3
kip = 10 lb
Given:
M = 350 lb⋅ ft
lb
ft
c = 2 ft
F = 500 lb
d = 3 ft
θ = 60 deg
e = 2 ft
a = 2 ft
f = 4 ft
b = 1 ft
g = 2 ft
w = 80
Solution:
Σ MB = 0;
−1
2
⎛ 2d ⎞ − F sin ( θ ) d + C ( d + e) = 0
⎟
y
⎝3⎠
wd ⎜
⎛ w d2 ⎞
⎜
⎟ + F sin ( θ ) d
3 ⎠
⎝
Cy =
Cy = 307.8 lb
d+e
+
Σ F x = 0;
→
B x − F cos ( θ ) = 0
+
By −
1
By =
1
↑
Σ F y = 0;
+
Σ F x = 0;
→
2
B x = F cos ( θ )
B x = 250 lb
w d − F sin ( θ ) + Cy = 0
2
w d + F sin ( θ ) − Cy
−NE − Bx = 0
NE = −B x
B y = 245.2 lb
NE = −250 lb
649
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
+
Chapter 7
↑Σ Fy = 0;
VE − By = 0
VE = By
V E = 245 lb
Σ ME = 0
−ME − B y c = 0
ME = −By c
ME = −490 lb⋅ ft
+
Σ F x = 0;
→
NF = 0
NF = 0 lb
NF = 0.00 lb
+
↑Σ Fy = 0;
−Cy − VF = 0
V F = −Cy
V F = −308 lb
Σ MF = 0;
Cy( f) + MF = 0
MF = − f Cy
MF = −1.23 kip⋅ ft
Problem 7-29
The bolt shank is subjected to a tension F. Determine the internal normal force, shear force,
and moment at point C.
Given:
F = 80 lb
a = 6 in
Solution:
ΣF x = 0;
NC + F = 0
NC = −F
NC = −80.00 lb
ΣF y = 0;
VC = 0
ΣMC = 0;
MC + F a = 0
MC = −F a
MC = −480.00 lb⋅ in
Problem 7-30
Determine the normal force, shear force, and moment acting at sections passing through points B
and C on the curved rod.
650
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Units Used:
kip = 103 lb
Given:
F 1 = 300 lb
F 2 = 400 lb
θ = 30 deg
r = 2 ft
φ = 45 deg
Solution:
Σ F x = 0;
F 2 sin ( θ ) − F1 cos ( θ ) + NB = 0
NB = −F 2 sin ( θ ) + F1 cos ( θ )
NB = 59.8 lb
Σ F y = 0;
V B + F2 cos ( θ ) + F1 sin ( θ ) = 0
V B = −F 2 cos ( θ ) − F 1 sin ( θ )
V B = −496 lb
Σ MB = 0;
MB + F 2 r sin ( θ ) + F 1 ( r − r cos ( θ ) ) = 0
MB = −F2 r sin ( θ ) − F1 r( 1 − cos ( θ ) )
MB = −480 lb⋅ ft
+
Σ F x = 0;
→
F2 − Ax = 0
Ax = F2
Ax = 400 lb
+
Ay − F1 = 0
Ay = F1
Ay = 300 lb
↑Σ Fy = 0;
Σ MA = 0;
−MA + F 1 2 r = 0
MA = 2 F 1 r
Σ F x = 0;
NC + Ax sin ( φ ) + A y cos ( φ ) = 0
MA = 1200 lb⋅ ft
NC = − A x sin ( φ ) − Ay cos ( φ )
Σ F y = 0;
V C − Ax cos ( φ ) + Ay sin ( φ ) = 0
V C = A x cos ( φ ) − A y sin ( φ )
Σ MC = 0;
NC = −495 lb
V C = 70.7 lb
−MC − MA − A x r sin ( φ ) + A y( r − r cos ( φ ) ) = 0
651
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
MC = −MA − Ax r sin ( φ ) + Ay r( 1 − cos ( φ ) )
MC = −1.59 kip⋅ ft
Problem 7-31
The cantilevered rack is used to support each end of a smooth pipe that has total weight W.
Determine the normal force, shear force, and moment that act in the arm at its fixed support A along
a vertical section.
Units Used:
3
kip = 10 lb
Given:
W = 300 lb
r = 6 in
θ = 30 deg
Solution:
Pipe:
+
↑
Σ F y = 0;
NB cos ( θ ) −
NB =
W
2
=0
W ⎞
⎜
⎟
2 ⎝ cos ( θ ) ⎠
1 ⎛
NB = 173.205 lb
Rack:
+
Σ F x = 0; −NA + NB sin ( θ ) = 0
→
NA = NB sin ( θ )
+
↑Σ Fy = 0;
NA = 86.6 lb
V A − NB cos ( θ ) = 0
V A = NB cos ( θ )
Σ MA = 0;
V A = 150 lb
⎛ r + r sin ( θ ) ⎞ = 0
⎟
⎝ cos ( θ ) ⎠
MA − NB ⎜
⎛ r + r sin ( θ ) ⎞
⎟
⎝ cos ( θ ) ⎠
MA = NB ⎜
MA = 1.800 kip⋅ in
652
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-32
Determine the normal force, shear force, and moment at a section passing through point D of the
two-member frame.
Units Used:
3
kN = 10 N
Given:
kN
m
w = 0.75
F = 4 kN
a = 1.5 m
d = 1.5 m
b = 1.5 m
e = 3
c = 2.5 m
f = 4
Solution:
Σ MC = 0;
⎛
⎞F d = 0
⎟
2
2
⎝ e +f ⎠
−B x( c + d) + ⎜
f
fdF
Bx =
2
2
B x = 1.2 kN
e + f ( c + d)
Σ MA = 0;
⎛ c + d ⎞ + B ( a + b) + B ( c + d) = 0
⎟
y
x
⎝ 2 ⎠
−w( c + d) ⎜
⎡ ( c + d) 2⎤
⎥ − B x ( c + d)
w⎢
2
⎣
⎦
By =
a+b
B y = 0.40 kN
+ Σ F x = 0;
→
−ND − B x = 0
ND = −Bx
ND = −1.2 kN
+
VD + By = 0
V D = −By
V D = −0.4 kN
↑Σ Fy = 0;
653
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Σ MD = 0;
−MD + By b = 0
Chapter 7
MD = B y b
MD = 0.6 kN⋅ m
Problem 7-33
Determine the internal normal force, shear force, and moment acting at point A of the smooth
hook.
Given:
θ = 45 deg
a = 2 in
F = 20 lb
Solution:
ΣF x = 0;
NA − F cos ( θ ) = 0
NA = F cos ( θ )
ΣF y = 0;
V A − F sin ( θ ) = 0
V A = F sin ( θ )
ΣMB = 0;
NA = 14.1 lb
V A = 14.1 lb
MA − NA a = 0
MA = NA a
MA = 28.3 lb⋅ in
Problem 7-34
Determine the internal normal force, shear force, and moment acting at points B and C on the
curved rod.
654
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Units Used:
kip = 103 lb
Given:
θ 2 = 30 deg
F = 500 lb
r = 2 ft
a = 3
θ 1 = 45 deg
b = 4
Solution:
ΣF N = 0;
⎛ F b ⎞ sin ( θ ) − ⎛ F a ⎞ cos ( θ ) + N = 0
2
2
B
⎜ 2 2⎟
⎜ 2 2⎟
⎝ a +b ⎠
⎝ a +b ⎠
⎡( a)cos ( θ 2 ) − b sin ( θ 2 )⎥⎤
2
2
⎢
⎥
a +b
⎣
⎦
NB = F ⎢
ΣF V = 0;
NB = 59.8 lb
VB + ⎛
Fb ⎞
⎛ Fa ⎞
⎜ 2 2 ⎟ cos ( θ 2) + ⎜ 2 2 ⎟ sin ( θ 2 ) = 0
⎝ a +b ⎠
⎝ a +b ⎠
⎡ b cos ( θ 2) + ( a)sin ( θ 2 )⎥⎤
2
2
⎢
⎥
a +b
⎣
⎦
V B = −F ⎢
ΣMB = 0;
V B = −496 lb
MB + ⎛
Fb ⎞
⎛ a ⎞
⎜ 2 2 ⎟ r sin ( θ 2 ) + F⎜ 2 2 ⎟ ( r − r cos ( θ 2 ) ) = 0
⎝ a +b ⎠
⎝ a +b ⎠
⎡−b sin ( θ 2) − a + ( a)cos ( θ 2 )⎥⎤
2
2
⎢
⎥
a +b
⎣
⎦
MB = F r ⎢
MB = −480 lb⋅ ft
Also,
ΣF x = 0;
Fb
− Ax +
=0
2
a +b
2
Ax = F ⎛
b
⎞
⎜ 2 2⎟
⎝ a +b ⎠
ΣF y = 0;
Ay −
Fa
2
a +b
Ax = 400.00 lb
=0
2
655
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
⎛
⎞
⎟
2
2
⎝ a +b ⎠
ΣMA = 0;
Ay = F ⎜
a
⎛
Fa
Ay = 300.00 lb
⎞ 2r = 0
⎟
2
2
⎝ a +b ⎠
−MA + ⎜
MA =
2F r a
2
2
MA = 1200 lb⋅ ft
a +b
ΣF x = 0;
NC + Ax sin ( θ 1 ) + Ay cos ( θ 1 ) = 0
NC = − A x sin ( θ 1 ) − A y cos ( θ 1 )
ΣF y = 0;
NC = −495 lb
V C − Ax cos ( θ 1 ) + A y sin ( θ 1 ) = 0
V C = A x cos ( θ 1 ) − Ay sin ( θ 1 )
ΣMC = 0;
V C = 70.7 lb
−MC − MA + A y( r − r cos ( θ 1 ) ) − Ax r sin ( θ 1 ) = 0
MC = −MA + Ay( r − r cos ( θ 1 ) ) − A x r sin ( θ 1 )
MC = −1.59 kip⋅ ft
Problem 7-35
Determine the ratio a/b for which the shear force will be zero at the midpoint C of the beam.
Solution:
Find Ay:
ΣMB = 0;
656
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
( 2 a + b) w ⎡⎢ ( b − a)⎤⎥ − Ay b = 0
1
1
⎣3
2
Ay =
⎦
w
( 2 a + b) ( b − a)
6b
This problem requires
V C = 0.
Summing forces vertically for the section, we have
+
↑Σ Fy = 0;
w
1 ⎛
b⎞ w
( 2a + b) ( b − a) − ⎜ a + ⎟
=0
2 ⎝
2⎠ 2
6b
w
w
( 2a + b) ( b − a) = ( 2a + b)
8
6b
4 ( b − a) = 3 b
b = 4a
a
1
=
4
b
Problem 7-36
The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length.
Determine the internal normal force, shear force, and moment in the arch at angle θ.
Given:
θ = 45 deg
Solution:
Resultants of distributed load:
657
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
θ
⌠
F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) )
⌡0
F Rx = r w0 ( 1 − cos ( θ ) )
θ
⌠
F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ )
⌡0
F Rx = r w0 ( sin ( θ ) )
θ
⌠
2
MRo = ⎮ w0 r dθ r = r w0 θ
⌡0
Σ F x = 0;
−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0
V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ )
V = w0 r( cos ( θ ) − 1 )
a = cos ( θ ) − 1
a = −0.293
Σ F y = 0;
V = a r w0
N + FRy cos ( θ ) + F Rx sin ( θ ) = 0
N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ )
N = −w0 r sin ( θ )
b = −sin ( θ )
b = −0.707
ΣMo = 0;
N = w0 r b
−M + r w0 ( θ ) + b r w0 r = 0
2
M = w0 r ( θ + b)
2
c = θ+b
c = 0.0783
2
M = c r w0
658
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-37
The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length.
Determine the internal normal force, shear force, and moment in the arch at angle θ.
Given:
θ = 120 deg
Solution:
Resultants of distributed load:
θ
⌠
F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) )
⌡0
F Rx = r w0 ( 1 − cos ( θ ) )
θ
⌠
F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ )
⌡0
F Rx = r w0 ( sin ( θ ) )
θ
⌠
2
MRo = ⎮ w0 r dθ r = r w0 θ
⌡0
Σ F x = 0;
−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0
V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ )
V = w0 r( cos ( θ ) − 1 )
a = cos ( θ ) − 1
a = −1.500
Σ F y = 0;
V = a r w0
N + FRy cos ( θ ) + F Rx sin ( θ ) = 0
N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ )
N = −w0 r sin ( θ )
b = −sin ( θ )
659
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
b = −0.866
Σ Mo = 0;
N = w0 ⋅ r⋅ b
−M + r w0 ( θ ) + b r w0 r = 0
2
M = w0 r ( θ + b)
2
c = θ+b
c = 1.2284
2
M = c r w0
Problem 7-38
Determine the x, y, z components of internal loading at a section passing through point C in the pipe
assembly. Neglect the weight of the pipe.
Units Used:
kip = 103 lb
Given:
⎛ 0 ⎞
F 1 = ⎜ 350 ⎟ lb
⎜
⎟
⎝ −400 ⎠
⎛ 150 ⎞
F 2 = ⎜ 0 ⎟ lb
⎜
⎟
⎝ −300 ⎠
a = 1.5 ft
b = 2 ft
c = 3 ft
Solution:
⎛c⎞
r1 = ⎜ b ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
r2 = ⎜ b ⎟
⎜ ⎟
⎝0⎠
F C = −F1 − F2
⎛ −150.00 ⎞
F C = ⎜ −350.00 ⎟ lb
⎜
⎟
⎝ 700.00 ⎠
660
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
MC = −r1 × F1 − r2 × F2
Chapter 7
⎛ 1400.00 ⎞
MC = ⎜ −1200.00 ⎟ lb⋅ ft
⎜
⎟
⎝ −750.00 ⎠
Problem 7-39
Determine the x, y, z components of internal loading at a section passing through point C in the pipe
assembly. Neglect the weight of the pipe.
Units Used:
kip = 103 lb
Given:
⎛ −80 ⎞
F 1 = ⎜ 200 ⎟ lb
⎜
⎟
⎝ −300 ⎠
⎛ 250 ⎞
F 2 = ⎜ −150 ⎟ lb
⎜
⎟
⎝ −200 ⎠
a = 1.5 ft
b = 2 ft
c = 3 ft
Solution:
⎛c⎞
r1 = ⎜ b ⎟
⎜ ⎟
⎝0⎠
⎛0⎞
r2 = ⎜ b ⎟
⎜ ⎟
⎝0⎠
F C = −F1 − F2
⎛ −170.00 ⎞
F C = ⎜ −50.00 ⎟ lb
⎜
⎟
⎝ 500.00 ⎠
MC = −r1 × F1 − r2 × F2
⎛ 1000.00 ⎞
MC = ⎜ −900.00 ⎟ lb⋅ ft
⎜
⎟
⎝ −260.00 ⎠
Problem 7-40
Determine the x, y, z components of internal loading in the rod at point D.
661
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Units Used:
kN = 103 N
Given:
M = 3 kN⋅ m
⎛ 7 ⎞
F = ⎜ −12 ⎟ kN
⎜
⎟
⎝ −5 ⎠
a = 0.75 m
b = 0.2 m
c = 0.2 m
d = 0.6 m
e = 1m
Solution:
Guesses
Cx = 1 N
Cy = 1 N
Bx = 1 N
Bz = 1 N
Ay = 1 N
Az = 1 N
Given
⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞
⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0
⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0
⎛ 0 ⎞
⎞
⎜ b + c + d ⎟ × ⎜ Ay ⎟ + ⎜ b + c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c + d ⎟ × F + ⎜ 0 ⎟ = 0
y
⎟
⎜ ⎟
⎜
⎟
⎜ ⎟
⎜
⎟ ⎜ ⎟ ⎜
⎜
⎟
⎜
0
a
⎠ ⎝ Bz ⎠ ⎝ ⎠ ⎝ 0 ⎟⎠ ⎝ 0
⎠
⎝ −M ⎠
⎝ 0
⎠ ⎝ Az ⎠ ⎝
662
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Engineering Mechanics - Statics
⎛⎜ Ay ⎞⎟
⎜ Az ⎟
⎜ ⎟
⎜ Bx ⎟ = Find ( A , A , B , B , C , C )
y z x z x y
⎜ Bz ⎟
⎜ ⎟
⎜ Cx ⎟
⎜ Cy ⎟
⎝ ⎠
Chapter 7
⎛⎜ Ay ⎞⎟
⎛ −53.60 ⎞
⎜ Az ⎟ ⎜ 87.00 ⎟
⎜ ⎟ ⎜
⎟
⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN
⎜ Bz ⎟ ⎜ −82.00 ⎟
⎜ ⎟ ⎜ −116.00 ⎟
⎜ Cx ⎟ ⎜
⎟
65.60
⎝
⎠
⎜ Cy ⎟
⎝ ⎠
Guesses
V Dx = 1 N
NDy = 1 N
V Dz = 1 N
MDx = 1 N⋅ m
MDy = 1 N⋅ m
MDz = 1 N⋅ m
Given
⎛⎜ Cx ⎟⎞ ⎛⎜ VDx ⎟⎞
⎜ Cy ⎟ + ⎜ NDy ⎟ = 0
⎜ 0 ⎟ ⎜V ⎟
⎝ ⎠ ⎝ Dz ⎠
⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛⎜ MDx ⎟⎞ ⎛ 0 ⎞
⎜ −b ⎟ × ⎜ C ⎟ + ⎜ MDy ⎟ + ⎜ 0 ⎟ = 0
⎜ ⎟
⎜ ⎟ ⎜ y⎟ ⎜
⎝ a ⎠ ⎝ 0 ⎠ ⎝ MDz ⎟⎠ ⎝ −M ⎠
⎛⎜ VDx ⎟⎞
⎜ NDy ⎟
⎜
⎟
⎜ VDz ⎟ = Find ( V , N , V , M , M , M )
Dx Dy Dz
Dx
Dy
Dz
⎜ MDx ⎟
⎜
⎟
⎜ MDy ⎟
⎜ MDz ⎟
⎝
⎠
⎛ VDx ⎞ ⎛ 116.00 ⎞
⎜
⎟
⎜ NDy ⎟ = ⎜ −65.60 ⎟ kN
⎜ V ⎟ ⎜⎝ 0.00 ⎟⎠
⎝ Dz ⎠
⎛ MDx ⎞ ⎛ 49.20 ⎞
⎜
⎟
⎜ MDy ⎟ = ⎜ 87.00 ⎟ kN⋅ m
⎜ M ⎟ ⎜⎝ 26.20 ⎟⎠
⎝ Dz ⎠
663
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Engineering Mechanics - Statics
Chapter 7
Problem 7-41
Determine the x, y, z components of internal loading in the rod at point E.
Units Used:
3
kN = 10 N
Given:
M = 3 kN⋅ m
⎛ 7 ⎞
F = ⎜ −12 ⎟ kN
⎜
⎟
⎝ −5 ⎠
a = 0.75 m
b = 0.4 m
c = 0.6 m
d = 0.5 m
e = 0.5 m
Solution:
Guesses
Cx = 1 N
Cy = 1 N
Bx = 1 N
Bz = 1 N
Ay = 1 N
Az = 1 N
Given
⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞
⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0
⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ −d − e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞
⎛ 0 ⎞
⎜ b + c ⎟ × ⎜ Ay ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c ⎟ × F + ⎜ 0 ⎟ = 0
⎟
⎜ ⎟
⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜
0
⎝ −M ⎠
⎝
⎠ ⎝ Az ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠
664
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Engineering Mechanics - Statics
⎛⎜ Ay ⎞⎟
⎜ Az ⎟
⎜ ⎟
⎜ Bx ⎟ = Find ( A , A , B , B , C , C )
y z x z x y
⎜ Bz ⎟
⎜ ⎟
⎜ Cx ⎟
⎜ Cy ⎟
⎝ ⎠
Chapter 7
⎛⎜ Ay ⎞⎟
⎛ −53.60 ⎞
⎜ Az ⎟ ⎜ 87.00 ⎟
⎜ ⎟ ⎜
⎟
⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN
⎜ Bz ⎟ ⎜ −82.00 ⎟
⎜ ⎟ ⎜ −116.00 ⎟
⎜ Cx ⎟ ⎜
⎟
65.60
⎝
⎠
⎜ Cy ⎟
⎝ ⎠
Guesses
NEx = 1 N
V Ey = 1 N
V Ez = 1 N
MEx = 1 N⋅ m
MEy = 1 N⋅ m
MEz = 1 N⋅ m
Given
⎛⎜ 0 ⎞⎟ ⎛⎜ NEx ⎟⎞
⎜ Ay ⎟ + ⎜ VEy ⎟ = 0
⎜ Az ⎟ ⎜ V ⎟
⎝ ⎠ ⎝ Ez ⎠
⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ MEx ⎞⎟
⎜ 0 ⎟ × Ay + M
=0
⎜ ⎟ ⎜ ⎟ ⎜ Ey ⎟
⎝ 0 ⎠ ⎜⎝ Az ⎟⎠ ⎜⎝ MEz ⎟⎠
⎛⎜ NEx ⎞⎟
⎜ VEy ⎟
⎜
⎟
⎜ VEz ⎟ = Find ( N , V , V , M , M , M )
Ex Ey Ez
Ex
Ey
Ez
⎜ MEx ⎟
⎜
⎟
⎜ MEy ⎟
⎜ MEz ⎟
⎝
⎠
⎛ NEx ⎞ ⎛ 0.00 ⎞
⎜
⎟
⎜ VEy ⎟ = ⎜ 53.60 ⎟ kN
⎜ V ⎟ ⎜⎝ −87.00 ⎟⎠
⎝ Ez ⎠
⎛ MEx ⎞ ⎛ 0.00 ⎞
⎜
⎟
⎜ MEy ⎟ = ⎜ −43.50 ⎟ kN⋅ m
⎜ M ⎟ ⎜⎝ −26.80 ⎟⎠
⎝ Ez ⎠
Problem 7-42
Draw the shear and moment diagrams for the shaft in terms of the parameters shown; There is
665
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Engineering Mechanics - Statics
Chapter 7
g
a thrust bearing at A and a journal bearing at B.
p
Units Used:
3
kN = 10 N
Given:
P = 9 kN
a = 2m
L = 6m
Solution:
P ( L − a) − Ay L = 0
Ay = P
L−a
L
x1 = 0 , 0.01a .. a
Ay − V 1 ( x) = 0
V 1 ( x) =
M1 ( x) − Ay x = 0
M1 ( x) =
Ay
kN
Ay x
kN⋅ m
x2 = a , 1.01a .. L
Ay − P − V2 ( x) = 0
V 2 ( x) =
Ay − P
kN
M2 ( x) − Ay x + P ( x − a) = 0
M2 ( x) =
A y x − P ( x − a)
kN⋅ m
666
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (kN)
10
V1( x1) 5
V2( x2)
0
5
0
1
2
3
4
5
6
5
6
x1 , x2
Distance (m)
Moment (kN-m)
15
10
M1( x1)
M2( x2)
5
0
5
0
1
2
3
4
x1 , x2
Distance (m)
Problem 7-43
Draw the shear and moment diagrams
for the beam in terms of the parameters
shown.
Given:
P = 800 lb
a = 5 ft
L = 12 ft
Solution:
667
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Engineering Mechanics - Statics
x1 = 0 , 0.01a .. a
Chapter 7
x2 = a , 1.01a .. L − a
x3 = L − a , 1.01( L − a) .. L
V 2 ( x) = 0
V 3 ( x) =
P
V 1 ( x) =
lb
Px
M1 ( x) =
M2 ( x) =
lb⋅ ft
Pa
M3 ( x) =
lb⋅ ft
−P
lb
P( L − x)
lb⋅ ft
Force (lb)
1000
V1( x1)
500
V2( x2)
0
V3( x3)
500
1000
0
2
4
6
8
10
12
8
10
12
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Moment (lb-ft)
4000
M1( x1)
M2( x2)
M3( x3)2000
0
0
2
4
6
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Problem 7-44
Draw the shear and moment diagrams for the beam (a) in terms of the parameters
668
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
g
shown; (b) set M0 and L as given.
( )
p
Given:
M0 = 500 N⋅ m
L = 8m
Solution:
For
0 ≤ x≤
L
3
+
↑Σ Fy = 0;
V1 = 0
ΣMx = 0;
M1 = 0
For
L
3
≤x≤
2L
3
+
↑Σ Fy = 0;
ΣMx = 0;
For
2L
3
V2 = 0
M2 = M0
≤x≤L
+
↑Σ Fy = 0;
V3 = 0
ΣMx = 0;
M3 = 0
( b)
x1 = 0 , 0.01L ..
L
3
x2 =
L L
2L
, 1.01 ..
3
3
3
x3 =
2L 2L
3
,
3
V 1 ( x1 ) = 0
V 2 ( x2 ) = 0
V 3 ( x3 ) = 0
M1 ( x1 ) = 0
M2 ( x2 ) = M0
M3 ( x3 ) = 0
1.01 .. L
669
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Engineering Mechanics - Statics
Chapter 7
1
Shear force in N
0.5
V1( x1)
V2( x2)
0
V3( x3)
0.5
1
0
1
2
3
4
5
6
7
8
9
x1 , x2 , x3
Distance in m
600
Moment in N - m
400
M1( x1)
M2( x2)
M3( x3)200
0
0
1
2
3
4
5
6
7
8
9
x1 , x2 , x3
Distance in m
Problem 7-45
The beam will fail when the maximum shear force is V max or the maximum bending moment is Mmax.
Determine the magnitude M0 of the largest couple moments it will support.
670
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Units Used:
3
kN = 10 N
Given:
L = 9m
V max = 5 kN
Mmax = 2 kN⋅ m
Solution:
The shear force is zero everywhere in the beam.
The moment is zero in the first third and the last third of the bam.
In the middle section of the beam the moment is
Thus the beam will fail when
M = M0
M0 = Mmax
M0 = 2.00 kN⋅ m
Problem 7-46
The shaft is supported by a thrust bearing at A and a journal bearing at B. Draw the shear and
moment diagrams for the shaft in terms of the parameters shown.
Given:
w = 500
lb
ft
L = 10 ft
Solution:
For
0 ≤ x< L
ΣF y = 0;
wL
− wx − V = 0
ΣM = 0;
2
−w L
2
⎛x⎞ + M = 0
⎟
⎝ 2⎠
x + w x⎜
V ( x) =
w
2
( L − 2 x)
M ( x) =
1
lb
1
2
(
Lx − x )
2
lb⋅ ft
w
671
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (lb)
5000
V( x )
0
5000
0
2
4
6
8
10
x
ft
Distance (ft)
1 .10
Moment (lb-ft)
4
M( x)
0
1 .10
4
0
2
4
6
8
10
x
ft
Distance (ft)
Problem 7-47
The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the
maximum moment is Mmax. Determine the largest uniformly distributed load w the shaft will support.
Units Used:
3
kip = 10 lb
Given:
L = 10 ft
Mmax = 5 kip⋅ ft
672
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
wL
2
− wx − V = 0
V = −w x +
wL
2
2
⎛x⎞
−
x + wx ⎜ ⎟ + M = 0
2
⎝2⎠
⎛ w L ⎞x − w x
M=⎜
⎟
2
⎝ 2 ⎠
wL
From the Moment Diagram,
wL
Mmax =
2
w =
8
8 Mmax
L
2
w = 400.00
lb
ft
Problem 7-48
Draw the shear and moment
diagrams for the beam.
3
kN = 10 N
Units Used:
Given:
w = 2
kN
m
L = 5m
MB = 5 kN⋅ m
Solution:
ΣF y = 0; −V ( x) + wL − wx = 0
ΣM = 0;
V ( x) = ( w L − w x)
1
kN
⎛ L ⎞ + M − w L x + w x⎛ x ⎞ = 0
⎟
⎜ 2⎟
B
⎝2⎠
⎝ ⎠
M ( x) + w L⎜
⎡
⎣
⎛ x ⎞ − w L⎛ L ⎞ − M ⎤ 1
⎟
⎜2⎟
B⎥
⎝2⎠
⎝ ⎠
⎦ kN⋅ m
M ( x) = ⎢w L x − w x⎜
673
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (kN)
10
V( x ) 5
0
0
1
2
3
4
5
4
5
x
Distance (m)
Moment (kN-m)
0
M( x) 20
40
0
1
2
3
x
Distance (m)
Problem 7-49
Draw the shear and moment diagrams
for the beam.
3
kN = 10 N
Units Used:
Given:
w = 3
kN
m
F = 10 kN
L = 6m
Solution:
V ( x) − w( L − x) − F = 0
V ( x) = [ w( L − x) + F ]
1
kN
⎛ L − x ⎞ − F( L − x) = 0
⎟
⎝ 2 ⎠
−M ( x) − w( L − x) ⎜
674
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎡
( L − x)
⎣
2
M ( x) = ⎢−w
2
Chapter 7
⎤
− F ( L − x)⎥
1
⎦ kN⋅ m
Force (kN)
40
V( x) 20
0
0
1
2
3
4
5
6
x
Distance (m)
Moment (kN-m)
0
50
M( x)
100
150
0
1
2
3
4
5
6
x
Distance (m)
Problem 7-50
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kN = 10 N
Given:
a = 2m
b = 4m
w = 1.5
kN
m
Solution:
⎛ b − a⎞ − A b = 0
⎟
y
⎝ 2 ⎠
w( b − a) ⎜
Ay =
w ( b − a)
2b
2
Ay = 0.75 kN
x1 = 0 , 0.01a .. a
Ay − V 1 ( x) = 0
V 1 ( x) = Ay
1
kN
675
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
− Ay x + M1 ( x) = 0
M1 ( x) = A y x
1
kN⋅ m
x2 = b − a , 1.01( b − a) .. b
1
Ay − w( x − a) − V 2 ( x) = 0
V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦
⎛ x − a ⎞ + M ( x) = 0
− Ay x + w( x − a) ⎜
⎟
2
⎝ 2 ⎠
2
⎡
( x − a) ⎤
⎢
⎥ 1
M2 ( x) = Ay x − w
2
⎣
⎦ kN⋅ m
kN
Force (kN)
2
V1( x1) 0
V2( x2)
2
4
0
0.5
1
1.5
2
2.5
3
3.5
4
2.5
3
3.5
4
x1 , x2
Distance (m)
Moment (kN-m)
2
M1( x1) 1
M2( x2)
0
1
0
0.5
1
1.5
2
x1 , x2
Distance (m)
676
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-51
Draw the shear and moment
diagrams for the beam.
Given:
L = 20 ft
w = 250
lb
ft
M1 = 150 lb⋅ ft
Solution:
⎛ L2 ⎞
⎟
⎝2⎠
M1 − M2 + w ⎜
⎛ L⎞ − M − A l = 0
⎟
2
y
⎝2⎠
M 1 + w L⎜
V ( x) = ( Ay − w x)
M2 = 150 lb⋅ ft
Ay =
L
Ay = 2500 lb
⎡
⎛ x2 ⎞
⎤ 1
⎢
M ( x) = A y x − w⎜ ⎟ − M1⎥
⎣
⎝2⎠
⎦ lb⋅ ft
1
lb
Force in lb
2000
V( x )
0
2000
0
5
10
15
20
x
ft
Distance in ft
677
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
1.5 .10
Moment in lb ft
4
1 .10
4
M( x)
5000
0
0
5
10
15
20
x
ft
Distance in ft
Problem 7-52
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kN = 10 N
Given:
w = 40
kN
m
F = 20 kN
Solution:
⎛ a⎞ − F b − M = 0
⎟
⎝ 2⎠
− Ay a + w a ⎜
x1 = 0 , 0.01a .. a
1
kN
⎛x
M1 ( x) = ⎢Ay x − w⎜
⎣
a = 8m
b = 3m
⎡ ⎛ a2 ⎞
⎤
⎟ − F b − M⎥ ⎛⎜ 1 ⎟⎞
⎣ ⎝2⎠
⎦⎝ a ⎠
Ay = ⎢w⎜
Ay = 133.75 kN
x2 = a , 1.01a .. a + b
V 1 ( x) = ( A y − w x)
⎡
M = 150 kN⋅ m
2 ⎞⎤
⎟⎥ 1
⎝ 2 ⎠⎦ kN⋅ m
V 2 ( x) = F
1
kN
M2 ( x) = [ −F( a + b − x) − M]
1
kN⋅ m
678
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Engineering Mechanics - Statics
Chapter 7
Force in kN
200
V1( x1)
V2( x2)
0
200
0
2
4
6
8
10
8
10
x1 , x2
Distance in m
Moment (kN-m)
400
M1( x1)
200
M2( x2)
0
200
400
0
2
4
6
x1 , x2
Distance (m)
Problem 7-53
Draw the shear and moment diagrams
for the beam.
679
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Engineering Mechanics - Statics
Chapter 7
Solution:
0 ≤ x< a
ΣF y = 0; −V − w x = 0
V = −w x
ΣM = 0;
M + w x⎛⎜
x⎞
M = −w
x
⎟=0
⎝2⎠
2
2
a < x ≤ 2a
ΣF y = 0; −V + 2 w a − w x = 0
V = w( 2 a − x)
ΣM = 0;
M + w x⎛⎜
x⎞
⎟ − 2 w a( x − a) = 0
⎝2⎠
2
M = 2 w a( x − a) −
wx
2
Problem 7-54
Draw the shear and bending-moment
diagrams for beam ABC. Note that
there is a pin at B.
680
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Engineering Mechanics - Statics
Chapter 7
Solution:
Support Reactions: From FBD (b),
wL ⎛ L⎞
L
− By⎛⎜ ⎞⎟ = 0
⎜
⎟
2 ⎝4⎠
⎝2⎠
wL
By =
4
Ay −
From FBD (a),
Ay =
3w L
−B y
L
wL
2
−
B
y
=0
4
2
− w⎛⎜
L⎞ ⎛ L⎞
⎟ ⎜ ⎟ + MA = 0
⎝2⎠ ⎝4⎠
⎛ L2 ⎞
⎟
⎝4⎠
MA = w ⎜
Shear and Moment Functions:
From FBD (c) For
0 ≤ x≤ L
V=
w
4
Ay − w x − V = 0
( 3 L − 4 x)
MA − Ay x + w x ⎛⎜
x⎞
⎟+M=0
⎝ 2⎠
M=
w
4
(3L x − 2x2 − L2)
681
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Engineering Mechanics - Statics
Chapter 7
Problem 7-55
The beam has depth a and is subjected to a uniform distributed loading w which acts at an
angle θ from the vertical as shown. Determine the internal normal force, shear force, and
moment in the beam as a function of x. Hint:The moment loading is to be determined from a
point along the centerline of the beam (x axis).
Given:
a = 2 ft
L = 10 ft
θ = 30 deg
w = 50
lb
ft
Solution:
0 ≤ x≤ L
N + w sin ( θ ) x = 0
ΣF x = 0;
N ( x) = −w sin ( θ ) x
−V − w cos ( θ ) x = 0
ΣF y = 0;
V = −w cos ( θ ) x
⎛ x ⎞ − w sin ( θ ) x⎛ a ⎞ + M = 0
⎟
⎜ 2⎟
⎝2⎠
⎝ ⎠
w cos ( θ ) x⎜
ΣM = 0;
⎛ x2 ⎞
⎟ + w sin ( θ ) ⎛⎜ x a ⎞⎟
2
⎝ ⎠
⎝2⎠
M ( x) = −w cos ( θ ) ⎜
Problem 7-56
Draw the shear and moment
diagrams for the beam.
Given:
w = 250
lb
ft
L = 12 ft
682
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Engineering Mechanics - Statics
Chapter 7
Solution:
1
2
wx
=0
L
ΣF y = 0;
−V −
ΣM = 0;
x ⎛ w x⎞ 1
M + ⎜ ⎟ ⎛⎜ x⎟⎞ = 0
2 ⎝ L ⎠⎝ 3 ⎠
2
x
V ( x) = −
wx 1
2 L lb
3
−w x
M ( x) =
6L
1
lb⋅ ft
Force (lb)
0
V( x) 1000
2000
0
2
4
6
8
10
12
x
ft
Distance (ft)
Moment (lb-ft)
0
M( x)
5000
1 .10
4
0
2
4
6
8
10
12
x
ft
Distance (ft)
Problem 7-57
The beam will fail when the maximum shear force is V max or the maximum moment is Mmax.
Determine the largest intensity w of the distributed loading it will support.
683
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Engineering Mechanics - Statics
Chapter 7
Given:
L = 18 ft
V max = 800 lb
Mmax = 1200 lb⋅ ft
Solution:
For 0 ≤ x ≤ L
2
V=
3
−w x
M=
2L
6L
wL
V max =
2
⎛ Vmax ⎞
⎟
⎝ L ⎠
w1 = 2 ⎜
Mmax =
−w x
w2 L
lb
ft
2
6
⎛ Mmax ⎞
2 ⎟
⎝ L ⎠
w2 = 6 ⎜
w1 = 88.9
w2 = 22.2
Now choose the critical case
lb
ft
w = min ( w1 , w2 )
w = 22.22
lb
ft
Problem 7-58
The beam will fail when the maximum internal moment is Mmax. Determine the position x of the
concentrated force P and its smallest magnitude that will cause failure.
684
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Engineering Mechanics - Statics
Chapter 7
Solution:
For ξ < x,
M1 =
For ξ > x,
M2 =
Pξ ( L − x)
L
P x( L − ξ )
L
Note that M1 = M2 when x = ξ
Mmax = M1 = M2 =
d
dx
P x( L − x)
L
(L x − x2) = L − 2x
Thus,
Mmax =
x=
=
P
L
(L x − x2)
L
2
P ⎛ L ⎞⎛
L⎞
P ⎛ L⎞
⎜ 2 ⎟ ⎜L − 2 ⎟ = 2 ⎜ 2 ⎟
L ⎝ ⎠⎝
⎠
⎝ ⎠
P=
4 Mmax
L
Problem 7-59
Draw the shear and moment diagrams
for the beam.
Given:
w = 30
lb
ft
a = 9 ft
MC = 180 lb⋅ ft
b = 4.5 ft
Solution:
− Ay a +
1
2
w a ⎛⎜
a⎞
⎟ − MC = 0
⎝ 3⎠
685
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Engineering Mechanics - Statics
Ay =
wa
6
Ay + By −
By =
wa
2
MC
−
1
2
Chapter 7
a
wa = 0
− Ay
x1 = 0 , 0.01a .. a
⎛x⎞
Ay − w ⎜ ⎟ x − V1 ( x) = 0
2
⎝ a⎠
2
⎛
wx ⎞ 1
⎜
⎟
V 1 ( x) = A y −
2 a ⎠ lb
⎝
⎛x⎞ ⎛x⎞
− Ay x + w⎜ ⎟ x⎜ ⎟ + M1 ( x) = 0
2 ⎝ a⎠ ⎝ 3 ⎠
3
⎛
wx ⎞ 1
⎜
⎟
M1 ( x) = Ay x −
6 a ⎠ lb⋅ ft
⎝
1
1
x2 = a , 1.01a .. a + b
Ay −
1
2
⎛
⎝
V 2 ( x) = ⎜ A y + By −
w a + By − V 2 ( x) = 0
− Ay x +
1
2
⎛
⎝
w a ⎜x −
wa⎞ 1
⎟
2 ⎠ lb
2a ⎞
⎟ − By( x − a) + M2 ( x) = 0
3 ⎠
⎡
⎣
M2 ( x) = ⎢Ay x + B y( x − a) −
2 a ⎞⎤
wa ⎛
⋅ ⎜x −
⎟⎥
2
⎝
1
3 ⎠⎦ lb⋅ ft
Force (lb)
100
V1( x1)
V2( x2)
0
100
200
0
2
4
6
8
10
12
x1 x2
,
ft ft
Distance (ft)
686
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment (lb-ft)
100
M1( x1)
M2( x2)
0
100
200
0
2
4
6
8
10
12
x1 x2
,
ft ft
Diastance (ft)
Problem 7-60
The cantilevered beam is made of material having a specific weight γ. Determine the shear and
moment in the beam as a function of x.
Solution:
By similar triangles
y
h
=
x
d
y=
h
x
d
1
⎡ 1 ⎛ h ⎞ ⎤ ⎛ γ h t ⎞ x2
W = γ V = γ ⎛⎜ y x t⎟⎞ = γ ⎢ ⎜ x⎟ x t⎥ = ⎜
⎟
⎝2 ⎠
⎣2 ⎝ d ⎠ ⎦ ⎝ 2d ⎠
ΣF y = 0;
⎛ γ h t ⎞ x2 = 0
⎟
⎝ 2d ⎠
V−⎜
⎛ γ h t ⎞ x2
⎟
⎝ 2d ⎠
V=⎜
ΣM = 0;
⎛ γ h t ⎞ x2 ⎛ x ⎞ = 0
⎟ ⎜ ⎟
⎝ 2d ⎠ ⎝ 3 ⎠
−M − ⎜
687
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Engineering Mechanics - Statics
M=−
Chapter 7
γh t 3
6d
x
Problem 7-61
Draw the shear and moment
diagrams for the beam.
3
kip = 10 lb
Given:
w1 = 30
lb
ft
w2 = 120
lb
ft
L = 12 ft
Solution:
⎛ L ⎞ + 1 (w − w )L ⎛ L ⎞ − A L = 0
⎟
2
1 ⎜ ⎟
y
⎝2⎠ 2
⎝3⎠
w1 L ⎜
⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞
⎟
⎜ ⎟
⎝2⎠ ⎝ 2 ⎠ ⎝3⎠
Ay = w1 ⎜
Ay − w1 x −
(w2 − w1)⎛⎜ L ⎞⎟ x − V ( x) = 0
2
1
x
⎝ ⎠
⎡
⎛ x2 ⎞⎤ 1
1
V ( x) = ⎢Ay − w1 x − ( w2 − w1 ) ⎜ ⎟⎥
2
⎣
⎝ L ⎠⎦ lb
⎛ x ⎞ + 1 ( w − w ) ⎛ x ⎞ x ⎛ x ⎞ + M ( x) = 0
⎟
2
1 ⎜ ⎟ ⎜ ⎟
⎝ 2⎠ 2
⎝ L⎠ ⎝ 3⎠
− Ay x + w1 x⎜
⎡
⎛ x2 ⎞
⎛ x3 ⎞⎤ 1
⎜
⎟
M ( x) = ⎢A y x − w1
− ( w2 − w1 ) ⎜ ⎟⎥
⎝2⎠
⎣
⎝ 6 L ⎠⎦ kip⋅ ft
688
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (lb)
500
0
V( x )
500
0
2
4
6
8
10
12
x
ft
Distance (ft)
Moment (kip-ft)
2
M( x) 1
0
0
2
4
6
8
10
12
x
ft
Distance (ft)
Problem 7-62
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kip = 10 lb
Given:
F 1 = 20 kip
F 2 = 20 kip
w = 4
kip
ft
a = 15 ft
b = 30 ft
c = 15 ft
Solution:
⎛ a + b⎞ + wb − F ⎛ c ⎞
⎟
2⎜ ⎟
⎝ b ⎠ 2
⎝ b⎠
⎛ b⎞ − F c − A b = 0
⎟
2
y
⎝ 2⎠
F 1 ( a + b) + w b⎜
Ay = F1 ⎜
Ay + By − F1 − F2 − w b = 0
By = F1 + F2 + w b − Ay
689
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
x1 = 0 , 0.01a .. a
1
−F 1 − V 1 ( x) = 0
V 1 ( x) = −F 1
F 1 x + M1 ( x) = 0
M1 ( x) = −F1 x
kip
1
kip⋅ ft
x2 = a , 1.01a .. a + b
1
V 2 ( x) = ⎡⎣−F1 − w( x − a) + A y⎤⎦
−F 1 − w( x − a) + Ay − V 2 ( x) = 0
kip
⎛ x − a ⎞ + M ( x) = 0
⎟
2
⎝ 2 ⎠
F 1 x − Ay( x − a) + w( x − a) ⎜
⎡
( x − a) ⎤
⎣
2
M2 ( x) = ⎢−F 1 x + Ay( x − a) − w
2
⎥ 1
⎦ kip⋅ ft
x3 = a + b , 1.01( a + b) .. a + b + c
1
V 3 ( x) − F2 = 0
V 3 ( x) = F 2
−M3 ( x) − F 2 ( a + b + c − x) = 0
M3 ( x) = −F2 ( a + b + c − x)
kip
1
kip⋅ ft
Force (kip)
100
V1( x1)
V2( x2)
V3( x3)
50
0
50
100
0
10
20
30
40
50
60
x1 x2 x3
, ,
ft ft ft
Distance (ft)
690
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
200
Moment (kip-ft)
M1( x1)
0
M2( x2)
M3( x3)
200
400
0
10
20
30
40
50
60
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Problem 7-63
Express the x, y, z components of internal loading in the rod at the specific value for y, where
0 < y< a
Units Used:
3
kip = 10 lb
Given:
y = 2.5 ft
w = 800
lb
F = 1500 lb
ft
a = 4 ft
b = 2 ft
Solution:
In general we have
⎡ F ⎤
⎥
V = ⎢
0
⎢
⎥
⎣w( a − y) ⎦
691
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be reproduced, in any form or by any means, without permission in writing from the publisher.
.
Engineering Mechanics - Statics
Chapter 7
⎡⎢ w( a − y) ⎛ a − y ⎞ ⎤⎥
⎜ 2 ⎟
⎝
⎠⎥
⎢
M =
⎢
⎥
−F b
⎢ −F( a − y) ⎥
⎣
⎦
For these values we have
⎛ 1500.00 ⎞
V = ⎜ 0.00 ⎟ lb
⎜
⎟
⎝ 1200.00 ⎠
⎛ 900.00 ⎞
M = ⎜ −3000.00 ⎟ lb⋅ ft
⎜
⎟
⎝ −2250.00 ⎠
Problem 7-64
Determine the normal
force, shear force, and
moment in the curved rod
as a function of θ.
Given:
c = 3
d = 4
Solution:
For
0 ≤ θ ≤ π
ΣF x = 0;
N−⎜
⎛
d
⎞ P cos ( θ ) − ⎛ c ⎞ P sin ( θ ) = 0
⎜ 2 2⎟
2
2⎟
⎝ c +d ⎠
⎝ c +d ⎠
⎛
P
N=⎜
2
V−⎜
⎛
d
⎛
P
⎞ ( d cos ( θ ) + c sin ( θ ) )
2⎟
⎝ c +d ⎠
ΣF y = 0;
⎞ P sin ( θ ) + ⎛ c ⎞ P cos ( θ ) = 0
⎜ 2 2⎟
2
2⎟
⎝ c +d ⎠
⎝ c +d ⎠
V=⎜
2
⎞ ( d sin ( θ ) − c cos ( θ ) )
2⎟
⎝ c +d ⎠
692
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
ΣM = 0;
Chapter 7
⎛ −d ⎞ P( r − r cos ( θ ) ) + ⎛ c ⎞ P r sin ( θ ) + M = 0
⎜ 2 2⎟
⎜ 2 2⎟
⎝ c +d ⎠
⎝ c +d ⎠
⎛
M=⎜
Pr
2
⎞ ( d − d cos ( θ ) − c sin ( θ ) )
2⎟
⎝ c +d ⎠
Problem 7-65
The quarter circular rod lies in the horizontal
plane and supports a vertical force P at its
end. Determine the magnitudes of the
components of the internal shear force,
moment, and torque acting in the rod as a
function of the angle θ.
Solution:
ΣF z = 0;
V= P
ΣMx = 0;
M + P r cos ( θ ) = 0
M = −P r cos ( θ )
M = P r cos ( θ )
ΣMy = 0;
T + P r( l − sin ( θ ) ) = 0
T = −P r ( l − sin ( θ )
T = P r ( 1 − sin ( θ )
693
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-66
Draw the shear and moment
diagrams for the beam.
Given:
MB = 800 lb⋅ ft a = 5 ft
F = 100 lb
b = 5 ft
Solution:
V 1 ( x) = F
x2 = a , 1.01a .. a + b
V 2 ( x) = F
Force (lb)
x1 = 0 , 0.01a .. a
1
M1 ( x) = ⎡⎣−F ( a + b − x) − MB⎤⎦
lb
1
M2 ( x) = −F( a + b − x)
lb
1
lb⋅ ft
1
lb⋅ ft
V1( x1) 100
V2( x2)
99.9
99.8
0
2
4
6
8
10
x1 x2
,
ft ft
Distance (ft)
Moment (lb-ft)
0
M1( x1)
M2( x2)
1000
2000
0
2
4
6
8
10
x1 x2
,
ft ft
Distane (ft)
694
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-67
Draw the shear and moment diagrams
for the beam.
kN = 103 N
Units Used:
Given:
w = 3
kN
F = 10 kN
m
L = 6m
Solution:
V ( x) = [ w( L − x) + F ]
⎡
( L − x)
⎣
2
M ( x) = ⎢−w
2
1
kN
⎤
− F ( L − x)⎥
1
⎦ kN⋅ m
Force (kN)
40
V( x) 20
0
0
1
2
3
4
5
6
x
Distance (m)
Moment (kN-m)
0
50
M( x)
100
150
0
1
2
3
4
5
6
x
Distance (m)
695
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-68
Draw the shear and moment
diagrams for the beam.
3
kN = 10 N
Units Used:
Given:
F = 7 kN
M = 12 kN⋅ m
A = 1N
Solution:
Guesses
Given
A+B−F=0
a = 2m
b = 2m
c = 4m
B = 1N
⎛A⎞
⎜ ⎟ = Find ( A , B)
⎝B ⎠
−F a − M + B ( a + b + c) = 0
x1 = 0 , 0.01a .. a
x2 = a , 1.01a .. a + b
x3 = a + b , 1.01( a + b) .. a + b + c
V 1 ( x1 ) = A
V 2 ( x2 ) = ( A − F)
V 3 ( x3 ) = −B
M1 ( x1 ) =
1
kN
A x1
kN⋅ m
M2 ( x2 ) =
1
kN
A x2 − F( x2 − a)
1
kN
M3 ( x3 ) = B ( a + b + c − x3 )
kN⋅ m
1
kN⋅ m
Force (kN)
5
V1( x1)
V2( x2)
V3( x3)
0
5
0
1
2
3
4
5
6
7
x1 , x2 , x3
Distance (m)
696
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8
Engineering Mechanics - Statics
Chapter 7
Moment (kN-m)
15
M1( x1)
10
M2( x2)
5
M3( x3)
0
5
0
1
2
3
4
5
6
7
8
x1 , x2 , x3
Distance (m)
Problem 7-69
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kN = 10 N
Given:
a = 2m
b = 4m
w = 1.5
kN
m
Solution:
⎛ b − a⎞ − A b = 0
⎟
y
⎝ 2 ⎠
w( b − a) ⎜
x1 = 0 , 0.01a .. a
V 1 ( x) = Ay
x2 = b − a , 1.01( b − a) .. b
Ay =
1
kN
w ( b − a)
2b
2
Ay = 0.75 kN
M1 ( x) = A y x
V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦
1
kN⋅ m
2
⎡
( x − a) ⎤ 1
⎢
⎥
kN M2 ( x) = Ay x − w
2
⎣
⎦ kN⋅ m
1
697
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (kN)
2
V1( x1) 0
V2( x2)
2
4
0
0.5
1
1.5
2
2.5
3
3.5
4
2.5
3
3.5
4
x1 , x2
Distance (m)
Moment (kN-m)
2
M1( x1) 1
M2( x2)
0
1
0
0.5
1
1.5
2
x1 , x2
Distance (m)
Problem 7-70
Draw the shear and moment diagrams
for the beam.
Given:
lb
ft
w = 30
MC = 180 lb⋅ ft
a = 9 ft
b = 4.5 ft
Solution:
− Ay a +
Ay =
1
2
wa
6
⎛ a⎞ − M = 0
⎟
C
⎝ 3⎠
w a⎜
−
MC
a
698
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ay + By −
By =
wa
2
1
Chapter 7
wa = 0
2
− Ay
x1 = 0 , 0.01a .. a
⎛
V 1 ( x) = ⎜ A y −
⎝
wx ⎞ 1
⎟
2 a ⎠ lb
⎝
wx ⎞ 1
⎟
6 a ⎠ lb⋅ ft
8
10
⎛
2
M1 ( x) = ⎜ Ay x −
3
x2 = a , 1.01a .. a + b
⎛
⎝
V 2 ( x) = ⎜ A y + By −
⎡
⎣
wa⎞ 1
⎟
2 ⎠ lb
M2 ( x) = ⎢Ay x + B y( x − a) −
wa⎛
2 a ⎞⎤ 1
x−
⎜
⎟⎥
2 ⎝
3 ⎠⎦ lb⋅ ft
Force (lb)
100
V1( x1)
V2( x2)
0
100
200
0
2
4
6
12
x1 x2
,
ft ft
Distance (ft)
699
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment (lb-ft)
100
M1( x1)
M2( x2)
0
100
200
0
2
4
6
8
10
12
x1 x2
,
ft ft
Diastance (ft)
Problem 7-71
Draw the shear and moment
diagrams for the beam.
3
kip = 10 lb
Given:
w1 = 30
lb
ft
w2 = 120
lb
ft
L = 12 ft
Solution:
⎛ L ⎞ + 1 ( w − w ) L⎛ L ⎞ − A L = 0
⎟
2
1 ⎜ ⎟
y
⎝2⎠ 2
⎝3⎠
w1 L⎜
⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞
⎟
⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠⎝ 3 ⎠
Ay = w1 ⎜
⎡
1
⎡
x
2
⎣
2
⎣
M ( x) = ⎢A y x − w1
x⎤1
2
(w2 − w1) L ⎥ lb
2
V ( x) = ⎢Ay − w1 x −
⎦
− ( w2 − w1 )
x ⎤ 1
⎥
6 L⎦ kip⋅ ft
3
700
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (lb)
500
0
V( x )
500
0
2
4
6
8
10
12
x
ft
Distance (ft)
Moment (kip-ft)
2
M( x) 1
0
0
2
4
6
8
10
12
x
ft
Distance (ft)
Problem 7-72
Draw the shear and moment diagrams for
the shaft. The support at A is a journal
bearing and at B it is a thrust bearing.
Given:
F 1 = 400 lb
w = 100
Solution:
lb
in
F 2 = 800 lb
a = 4 in
b = 12 in
c = 4 in
⎛ b ⎞ − F b + B( b + c) = 0
⎟
2
⎝2⎠
F 1 a − w b⎜
⎛ b2 ⎞
⎟ + F2 b − F1 a
⎝2⎠
B = 950.00 lb
w⎜
B =
b+c
701
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
x1 = 0 , 0.01a .. a
Chapter 7
V 1 ( x) = −F 1
x2 = a , 1.01a .. a + b
1
M1 ( x) = −F1 x
lb
V 2 ( x2 ) = ⎡⎣−B + F2 + w( a + b − x2 )⎤⎦
1
lb⋅ in
1
lb
2
⎡
a + b − x2 ) ⎤ 1
(
⎥
M2 ( x2 ) = ⎢B( a + b + c − x2 ) − F2 ( a + b − x2 ) − w⋅
2
⎣
⎦ lb⋅ in
V 3 ( x3 ) =
x3 = a + b , 1.01( a + b) .. a + b + c
−B
lb
M3 ( x3 ) = B ( a + b + c − x3 )
1
lb⋅ in
Force (lb)
2000
V1( x1)
V2( x2)
1000
V3( x3)
0
1000
0
5
10
15
20
15
20
x1 x2 x3
, ,
in in in
Distance (in)
Moment (lb-in)
4000
M1( x1)
M2( x2)
M3( x3)
2000
0
2000
0
5
10
x1 x2 x3
, ,
in in in
Distance (ini)
702
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-73
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kN = 10 N
Given:
F 1 = 10 kN
F 2 = 10 kN
M0 = 12 kN⋅ m
a = 2m
b = 2m
c = 2m
d = 2m
Solution:
F 1 ( b + c + d) − M0 + F2 d − Ay( a + b + c + d) = 0
Ay =
F 1 ( b + c + d) − M 0 + F 2 d
Ay = 8.50 kN
a+b+c+d
Ay + By − F1 − F2 = 0
By = F1 + F2 − Ay
B y = 11.50 kN
x1 = 0 , 0.01a .. a
V 1 ( x) = Ay
1
M1 ( x) = A y x
kN
1
kN⋅ m
x2 = a , 1.01a .. a + b
V 2 ( x) = ( A y − F1 )
1
kN
M2 ( x) = ⎡⎣Ay x − F 1 ( x − a)⎤⎦
1
kN⋅ m
x3 = a + b , 1.01( a + b) .. a + b + c
V 3 ( x) = ( A y − F1 )
1
kN
M3 ( x) = ⎡⎣Ay x − F 1 ( x − a) + M0⎤⎦
1
kN⋅ m
x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d
V 4 ( x) = −B y
1
kN
M4 ( x) = By( a + b + c + d − x)
1
kN⋅ m
703
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
10
Force (kN)
V1( x1)
5
V2( x2) 0
V3( x3)
5
V4( x4)
10
15
0
1
2
3
4
5
6
7
8
5
6
7
8
x1 , x2 , x3 , x4
Distance (m)
Moment (kN-m)
30
M1( x1)
M2( x2)
20
M3( x3)
M4( x4)10
0
0
1
2
3
4
x1 , x2 , x3 , x4
Distance (m)
Problem 7-74
Draw the shear and moment
diagrams for the shaft. The support
at A is a journal bearing and at B it
is a thrust bearing.
704
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
F = 200 lb
w = 100
lb
ft
M = 300 lb⋅ ft
Solution:
F ( a + b) − A b + w b⎛⎜
b = 4 ft
⎛ w b2 ⎞
⎟−M
⎝ 2 ⎠
c = 1 ft
F( a + b) + ⎜
b⎞
⎟−M=0
⎝ 2⎠
x1 = 0 , 0.01a .. a
a = 1 ft
V 1 ( x1 ) = −F
A =
M1 ( x1 ) = −F x1
1
lb
V 2 ( x2 ) = ⎡⎣−F + A − w( x2 − a)⎤⎦
x2 = a , 1.01a .. a + b
A = 375.00 lb
b
1
lb⋅ ft
1
lb
⎡
(x2 − a)2⎥⎤ 1
⎢
M2 ( x2 ) = −F x2 + A ( x2 − a) − w
2
⎣
⎦ lb⋅ ft
V 3 ( x3 ) = 0
x3 = a + b , 1.01( a + b) .. a + b + c
1
lb
M3 ( x3 ) = −M
1
lb⋅ ft
Force (lb)
200
V1( x1)
V2( x2)
V3( x3)
0
200
400
0
1
2
3
4
5
x1 x2 x3
, ,
ft ft ft
Distance (ft)
705
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
6
Engineering Mechanics - Statics
Chapter 7
0
Moment (lb-ft)
M1( x1)
100
M2( x2)
M3( x3)
200
300
0
1
2
3
4
5
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Problem 7-75
Draw the shear and moment diagrams for
the beam.
Units Used:
3
kN = 10 N
Given:
F = 8 kN
M = 20 kN⋅ m
c = 2m
d = 3m
w = 15
kN
m
a = 2m
Solution:
F c − M − w⎜
⎛ d⎞ = 0
⎟
⎝2⎠
− A( a + b + c) − M + F c − w d⎜
x1 = 0 , 0.01a .. a
V 1 ( x1 ) = A
x2 = a , 1.01a .. a + b
V 2 ( x2 ) = A
x3 = a + b , 1.01( a + b) .. a + b + c
⎛ d2 ⎞
⎟
⎝2⎠
b = 1m
A =
1
kN
1
kN
A = −14.30 kN
a+b+c
M1 ( x1 ) = A x1
1
kN⋅ m
M2 ( x2 ) = ( A x2 + M)
V 3 ( x3 ) = ( A − F)
1
kN⋅ m
1
kN
M3 ( x3 ) = ⎡⎣A x3 + M − F ( x3 − a − b)⎤⎦
1
kN⋅ m
706
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
6
Engineering Mechanics - Statics
Chapter 7
x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d
V 4 ( x4 ) = w( a + b + c + d − x4 )
M4 ( x4 ) = −w
1
kN
(a + b + c + d − x4)2
1
2
kN⋅ m
60
Force (kN)
V1( x1)
40
V2( x2) 20
V3( x3)
0
V4( x4)
20
40
0
1
2
3
4
5
6
7
8
5
6
7
8
x1 , x2 , x3 , x4
Distance (m)
Moment (kN-m)
0
M1( x1) 20
M2( x2)
M3( x3) 40
M4( x4)
60
80
0
1
2
3
4
x1 , x2 , x3 , x4
Distance (m)
707
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-76
Draw the shear and moment diagrams for
the shaft. The support at A is a thrust
bearing and at B it is a journal bearing.
Units Used:
3
kN = 10 N
Given:
w = 2
kN
m
F = 4 kN
a = 0.8 m
⎛ a2 ⎞
⎟
⎝2⎠
F a + w⎜
⎛ a⎞ = 0
⎟
⎝2⎠
Solution:
b = 0.2 m
B ( a + b ) − F a − w a⎜
B =
A + B − wa − F = 0
A = wa + F − B
x1 = 0 , 0.01a .. a
x2 = a , 1.01a .. a + b
V 1 ( x1 ) = ( A − w x1 )
V 2 ( x2 ) = −B
a+b
B = 3.84 kN
A = 1.76 kN
⎡
⎛ x1 2 ⎞⎤ 1
⎢
⎟⎥
M1 ( x1 ) = A x1 − w⎜
⎣
⎝ 2 ⎠⎦ kN⋅ m
1
kN
M2 ( x2 ) = B ( a + b − x2 )
1
kN
1
kN⋅ m
Force (kN)
5
V1( x1)
V2( x2)
0
5
0
0.2
0.4
0.6
0.8
x1 , x2
Distance (m)
Moment (kN-m)
1
M1( x1) 0.5
M2( x2)
0
0.5
0
0.2
0.4
0.6
0.8
x1 , x2
Distance (m)
708
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-77
Draw the shear and moment
diagrams for the beam.
Given:
w = 20
lb
ft
MB = 160 lb⋅ ft
a = 20 ft
b = 20 ft
Solution:
⎛ a⎞
−w a⎜ ⎟ − MB + Cy( a + b) = 0
⎝ 2⎠
⎛ a2
⎞ 1 ⎞
Cy = ⎜ w
+ MB⎟ ⎛⎜
⎟
⎝ 2
⎠⎝ a + b ⎠
Cy = 104.00 lb
Ay − w a + Cy = 0
Ay = w a − Cy
Ay = 296.00 lb
x1 = 0 , 0.01a .. a
V 1 ( x) = ( A y − w x)
2
⎛
x ⎞ 1
⎜
M1 ( x) = Ay x − w ⎟
2 ⎠ lb⋅ ft
⎝
1
lb
x2 = a , 1.01a .. a + b
V 2 ( x) = −Cy
1
M2 ( x) = Cy( a + b − x)
lb
1
lb⋅ ft
Force (lb)
400
V1( x1) 200
V2( x2)
0
200
0
5
10
15
20
25
30
35
40
x1 x2
,
ft ft
Distance (ft)
709
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment (lb-ft)
3000
M1( x1)2000
M2( x2)
1000
0
0
10
20
30
40
x1 x2
,
ft ft
Distance (ft)
Problem 7-78
The beam will fail when the maximum moment
is Mmax or the maximum shear is Vmax.
Determine the largest distributed load w the
beam will support.
3
kip = 10 lb
Units used:
Given:
Mmax = 30 kip⋅ ft
Solution:
−A b + w
kip
ft
w = 1
Set
V max = 8 kip
w
b
⎞
+ b⎟ + w b = 0
⎜
2⎝3
2
⎠
a
2
b = 6 ft
and then scale the answer at the end
a⎛a
A + B − wb − w
a = 6 ft
A =
b
⎞
+ b⎟ + w
⎜
2⎝3
2
⎠
b
⎛
⎝
B = w⎜ b +
=0
2
a⎛a
a⎞
⎟−A
2⎠
A = 7.00 kip
B = 2.00 kip
Shear limit - check critical points to the left and right of A and at B
⎛
⎝
V big = max ⎜ B , w
⎛ Vmax ⎞
⎟w
⎝ Vbig ⎠
wshear = ⎜
a
2
, w
a
2
⎞
⎠
−A ⎟
wshear = 2.00
V big = 4.00 kip
kip
ft
710
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment limit - check critical points at A and betwen A and B
B
⎛ a⎞⎛ a ⎞
MA = −w⎜ ⎟ ⎜ ⎟ x =
w
⎝ 2 ⎠⎝ 3 ⎠
⎛ x2 ⎞
MAB = B x − w⎜ ⎟
⎝2⎠
Mbig = max ( MA , MAB )
Mbig = 6.00 kip⋅ ft
⎛ Mmax ⎞
⎟w
⎝ Mbig ⎠
⎛ MA ⎞ ⎛ −6.00 ⎞
⎜
⎟=⎜
⎟ kip⋅ ft
⎝ MAB ⎠ ⎝ 2.00 ⎠
kip
ft
wmoment = ⎜
wmoment = 5.00
Choose the critical case
wans = min ( wshear , wmoment)
wans = 2.00
kip
ft
Problem 7-79
The beam consists of two segments
pin connected at B. Draw the shear
and moment diagrams for the beam.
Given:
F = 700 lb
w = 150
lb
ft
Solution:
⎛ c ⎞ − M + Cc = 0
⎟
⎝2⎠
M = 800 lb ft
⎛ c2 ⎞
⎟+M
⎝2⎠
b = 4 ft
c = 6 ft
w⎜
−w c⎜
C =
B + C − wc = 0
B = wc − C
x1 = 0 , 0.01a .. a
V 1 ( x1 ) = ( B + F )
x2 = a , 1.01a .. a + b
a = 8 ft
V 2 ( x2 ) = B
x3 = a + b , 1.01( a + b) .. a + b + c
C = 583.33 lb
c
B = 316.67 lb
M1 ( x1 ) = ⎡⎣−F( a − x1 ) − B ( a + b − x1 )⎤⎦
1
lb
M2 ( x2 ) = −B ( a + b − x2 )
1
lb
V 3 ( x3 ) = ⎡⎣−C + w( a + b + c − x3 )⎤⎦
1
lb⋅ ft
1
lb
2
⎡
⎤ 1
a + b + c − x3 )
(
M3 ( x3 ) = ⎢C( a + b + c − x3 ) − w
− M⎥
2
⎣
⎦ lb⋅ ft
711
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
1
lb⋅ ft
Engineering Mechanics - Statics
Chapter 7
Force (lb)
1500
V1( x1)
1000
V2( x2)
500
V3( x3)
0
500
1000
0
2
4
6
8
10
12
14
16
18
20
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Moment (lb-ft)
5000
M1( x1)
0
M2( x2)
M3( x3)
5000
1 .10
4
0
2
4
6
8
10
12
14
16
18
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Problem 7-80
The beam consists of three segments
pin connected at B and E. Draw the
shear and moment diagrams for the
beam.
3
Units Used:
kN = 10 N
Given:
MA = 8 kN⋅ m
F = 15 kN
w = 3
c = 2m
d = 2m
e = 2m
kN
m
a = 3m
b = 2m
f = 4m
712
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20
Engineering Mechanics - Statics
Guesses
Chapter 7
Ay = 1 N
By = 1 N
Cy = 1 N
Dy = 1 N
Ey = 1 N
Fy = 1 N
Given
Ay + Cy + Dy + Fy − F − w f = 0
F b − MA − Ay( a + b) = 0
−w f⎛⎜
⎟ + Fy f = 0
⎝2⎠
M A + F a + B y ( a + b) = 0
B y + Cy + Dy + E y = 0
−B y c + Dy d + E y( d + e) = 0
⎛⎜ Ay ⎞⎟
⎜ By ⎟
⎜ ⎟
⎜ Cy ⎟ = Find ( A , B , C , D , E , F )
y y y y y y
⎜ Dy ⎟
⎜ ⎟
⎜ Ey ⎟
⎜ Fy ⎟
⎝ ⎠
⎛⎜ Ay ⎞⎟
⎛ 4.40 ⎞
⎜ By ⎟ ⎜ −10.60 ⎟
⎟
⎜ ⎟ ⎜
⎜ Cy ⎟ = ⎜ 15.20 ⎟ kN
⎜ Dy ⎟ ⎜ 1.40 ⎟
⎜ ⎟ ⎜ −6.00 ⎟
⎟
⎜ Ey ⎟ ⎜
⎜ Fy ⎟ ⎝ 6.00 ⎠
⎝ ⎠
f⎞
x1 = 0 , 0.01a .. a
M1 ( x) = ( Ay x + MA)
1
V 1 ( x) = Ay
kN
x2 = a , 1.01a .. a + b
V 2 ( x) = ( A y − F)
1
kN
1
kN⋅ m
M2 ( x) = ⎡⎣Ay x + MA − F ( x − a)⎤⎦
1
kN⋅ m
x3 = a + b , 1.01( a + b) .. a + b + c
V 3 ( x) = B y
1
M3 ( x) = By( x − a − b)
kN
1
kN⋅ m
x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d
V 4 ( x) = ( By + Cy)
1
kN
M4 ( x) = ⎡⎣B y( x − a − b) + Cy( x − a − b − c)⎤⎦
1
kN⋅ m
x5 = a + b + c + d , 1.01( a + b + c + d) .. a + b + c + d + e
V 5 ( x) = −E y
1
kN
M5 ( x) = Ey( a + b + c + d + e − x)
1
kN⋅ m
x6 = a + b + c + d + e , 1.01( a + b + c + d + e) .. a + b + c + d + e + f
713
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Engineering Mechanics - Statics
Chapter 7
V 6 ( x) = ⎡⎣−Fy + w( a + b + c + d + e + f − x)⎤⎦
1
kN
2
⎡
( a + b + c + d + e + f − x) ⎤ 1
⎢
⎥
M6 ( x) = F y( a + b + c + d + e + f − x) − w
2
⎣
⎦ kN⋅ m
10
V1( x1) 5
Force (kN)
V2( x2)
V3( x3) 0
V4( x4)
V5( x5) 5
V6( x6)
10
15
0
2
4
6
8
10
12
14
x1 , x2 , x3 , x4 , x5 , x6
Distance (m)
714
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Engineering Mechanics - Statics
Chapter 7
30
M1( x1) 20
Moment (kN-m)
M2( x2)
M3( x3)
10
M4( x4) 0
M5( x5)
M6( x6)
10
20
30
0
2
4
6
8
10
12
14
x1 , x2 , x3 , x4 , x5 , x6
Distance (m)
Problem 7-81
Draw the shear and moment diagrams for the beam.
Solutions:
Support Reactions:
ΣMx = 0;
⎛ L ⎞ − w0 L ⎛ 4 L ⎞ = 0
⎟
⎜ ⎟
2 ⎝ 3 ⎠
⎝2⎠
B y L − w0 L⎜
By =
7 w0 L
6
715
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Engineering Mechanics - Statics
+
↑
Σ F y = 0;
Chapter 7
⎛ 7w0 L ⎞
⎛ w0 L ⎞
⎟ − w0 L − ⎜
⎟=0
⎝ 6 ⎠
⎝ 2 ⎠
Ay + ⎜
Ay =
w0 L
3
716
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-82
Draw the shear and moment diagrams
for the beam.
Units Used:
3
kip = 10 lb
Given:
F = 2000 lb
a = 9 ft
lb
ft
b = 9 ft
w = 500
Solution:
1
2
⎛ 2 b ⎞ + F b + w a⎛ b + a ⎞ − A ( a + b ) = 0
⎟
⎜ 2⎟
y
⎝3⎠
⎝
⎠
w b⎜
Ay + By − F − w a −
Ay = 5.13 kip
1
2
wb = 0
⎛ w b2 ⎞
⎜
⎟ + F b + w a⎛⎜b + a ⎞⎟
⎝ 3 ⎠
⎝ 2⎠
Ay =
a+b
By = w a +
1
2
w b − Ay + F
B y = 3.63 kip
x1 = 0 , 0.01a .. a
717
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Engineering Mechanics - Statics
V 1 ( x) = ( A y − w x)
Chapter 7
M1 ( x) = ⎡⎢Ay x − w x⎛⎜
1
1
⎟⎥
⎝ 2 ⎠⎦ kip⋅ ft
⎣
kip
x ⎞⎤
x2 = a , 1.01a .. a + b
V 2 ( x) = ⎡⎢−By +
⎣
1
2
w⎛⎜
⎝
a + b − x⎞
b
M2 ( x) = ⎡⎢B y( a + b − x) −
⎣
⎤ 1
⎟ ( a + b − x)⎥
⎠
⎦ kip
1
2
w⎛⎜
⎝
a + b − x⎞
b
⎛ a + b − x ⎞⎤ 1
⎟ ( a + b − x) ⎜ 3 ⎟⎥
⎝
⎠⎦ kip⋅ ft
⎠
Force (kip)
10
V1( x1) 5
V2( x2)
0
5
0
5
10
15
x1 x2
,
ft ft
Distance (ft)
Moment (kip-ft)
30
M1( x1)
20
M2( x2)
10
0
10
0
5
10
15
x1 x2
,
ft ft
Distance (ft)
718
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-83
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kN = 10 N
Given:
w = 3
kN
m
a = 3m
b = 3m
Solution:
wb
⎛ w b ⎞ ⎛ 2b ⎞
− Ay( a + b) + ⎜
w
Ay + By −
2
⎛ w a ⎞⎛
a⎞
2
3
⎟ ⎜ ⎟ + ⎜ ⎟ ⎜b + ⎟ = 0
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠
Ay =
( a + b) = 0
By =
w
⎡
⎣
1
⎛ w a ⎞ ⎛b + a ⎞
⎟⎜
⎟
⎝ 2 ⎠⎝ 3 ⎠
+⎜
a+b
2
( a + b) − A y
x1 = 0 , 0.01a .. a
⎡
⎣
V 1 ( x) = ⎢A y −
⎛ x ⎞ x⎤ 1
⎟⎥
2 ⎝ a ⎠ ⎦ kN
1
w⎜
M1 ( x) = ⎢Ay x −
⎛ x ⎞ x⎛ x ⎞⎤ 1
⎟ ⎜ ⎟⎥
2 ⎝ a ⎠ ⎝ 3 ⎠⎦ kN⋅ m
w⎜
x2 = a , 1.01a .. a + b
⎡
⎣
V 2 ( x) = ⎢−By +
⎛ a + b − x ⎞ ( a + b − x)⎤ 1
⎟
⎥
2 ⎝
b
⎠
⎦ kN
1
w⎜
⎡
⎣
M2 ( x) = ⎢B y( a + b − x) −
⎛ a + b − x ⎞ ( a + b − x) ⎛ a + b − x ⎞⎤ 1
⎟
⎜ 3 ⎟⎥
2 ⎝
b
⎝
⎠⎦ kN⋅ m
⎠
1
w⎜
Force (kN)
5
V1( x1)
V2( x2)
0
5
0
1
2
3
4
5
6
x1 , x2
Distance (m)
719
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment (kN-m)
10
M1( x1) 5
M2( x2)
0
5
0
1
2
3
4
5
6
x1 , x2
Distance (m)
Problem 7-84
Draw the shear and moment
diagrams for the beam.
Units Used:
3
kip = 10 lb
Given:
w = 100
lb
ft
M0 = 9 kip⋅ ft
Guesses:
Ay = 1 lb
Given
−M0 + By( a + b) −
Ay + By −
1
2
⎛ Ay ⎞
⎜ ⎟ = Find ( Ay , By)
⎝ By ⎠
a = 6 ft
b = 6 ft
c = 4 ft
B y = 1 lb
1
2
⎛ 2 a ⎞ − 1 w b⎛ a + b ⎞ = 0
⎟
⎜
⎟
⎝ 3 ⎠ 2 ⎝ 3⎠
w a⎜
w( a + b) = 0
⎛ Ay ⎞ ⎛ −0.45 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ By ⎠ ⎝ 1.05 ⎠
x1 = 0 , 0.01a .. a
⎡
⎣
V 1 ( x) = ⎢A y −
⎛ x ⎞ x⎤ 1
⎟⎥
2 ⎝ a ⎠ ⎦ lb
1
w⎜
⎡
⎣
M1 ( x) = ⎢Ay x −
⎛ x ⎞ x⎛ x ⎞⎤ 1
⎟ ⎜ ⎟⎥
2 ⎝ a ⎠ ⎝ 3 ⎠⎦ kip⋅ ft
1
w⎜
x2 = a , 1.01a .. a + b
720
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
V 2 ( x) = ⎡⎢−By +
⎣
1
2
Chapter 7
w⎛⎜
⎝
a + b − x⎞
b
⎤1
⎟ ( a + b − x)⎥
⎠
⎦ lb
M2 ( x) = ⎡⎢B y( a + b − x) − M0 −
⎣
1
2
w⎛⎜
⎝
a + b − x⎞
⎛ a + b − x ⎞⎤ 1
⎟ ( a + b − x) ⎜ 3 ⎟⎥
⎝
⎠⎦ kip⋅ ft
⎠
b
x3 = a + b , 1.01( a + b) .. a + b + c
V 3 ( x) = 0
M3 ( x) = −M0
1
kip⋅ ft
V1( x1)
V2( x2)
500
V3( x3)
1000
1500
0
2
4
6
8
10
12
14
16
x1 x2 x3
, ,
ft ft ft
Distance (ft)
0
Moment (kip-ft)
Force (lb)
0
M1( x1)
M2( x2)
M3( x3)
5
10
0
5
10
15
x1 x2 x3
, ,
ft ft ft
Distance (ft)
721
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-85
Draw the shear and moment diagrams for the beam.
Solution:
722
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-86
Draw the shear and moment diagrams
for the beam.
Units Used:
3
kN = 10 N
Given:
kN
m
w = 2
a = 3m
b = 3m
Solution:
x1 = 0 , 0.01a .. a
⎡1
⎣2
V 1 ( x) = ⎢ w b +
⎛ a − x ⎞ ( a − x)⎤ 1
⎟
⎥
2 ⎝ a ⎠
⎦ kN
1
w⎜
⎡−1 w b⎛ 2 b + a − x⎞ − 1 w⎛ a − x ⎞ ( a − x) ⎛ a − x ⎞⎤ 1
⎜
⎟ 2 ⎜
⎟
⎜ 3 ⎟⎥
⎠
⎝
⎠⎦ kN⋅ m
⎣2 ⎝3
⎝ a ⎠
M1 ( x) = ⎢
x2 = a , 1.01a .. a + b
⎡1
⎣2
V 2 ( x) = ⎢ w b −
⎛ x − a ⎞ ( x − a)⎤ 1
⎟
⎥
2 ⎝ a ⎠
⎦ kN
1
w⎜
⎡−1 w b⎛a + 2b − x⎞ − 1 w⎛ x − a ⎞ ( x − a) ⎛ x − a ⎞⎤ 1
⎜
⎟ 2 ⎜
⎟
⎜ 3 ⎟⎥
3
⎠
⎝
⎠⎦ kN⋅ m
⎣2 ⎝
⎝ b ⎠
M2 ( x) = ⎢
Force (kN)
10
V1( x1) 5
V2( x2)
0
5
0
1
2
3
4
5
6
x1 , x2
Distance (m)
723
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Moment (kN-m)
0
M1( x1)
M2( x2)
10
20
0
1
2
3
4
5
6
x1 , x2
Distance (m)
Problem 7-87
Draw the shear and moment diagrams for the beam.
Units Used:
3
kip = 10 lb
Given:
w = 5
kip
ft
M1 = 15 kip⋅ ft
M2 = 15 kip⋅ ft
a = 6 ft
b = 10 ft
c = 6 ft
Solution:
⎛ a ⎞ ⎛b + a ⎞ + w b⎛ b ⎞ − w⎛ c ⎞ ⎛ c ⎞ = 0
⎟⎜
⎟
⎜2⎟
⎜ 2 ⎟⎜ 3 ⎟
⎝ 2 ⎠⎝ 3 ⎠
⎝ ⎠
⎝ ⎠⎝ ⎠
M1 − A b − M2 + w⎜
⎛ a + c⎞ = 0
⎟
⎝ 2 ⎠
A + B − w b − w⎜
2
2
⎛ a ⎞ ⎛b + a ⎞ + w⎛⎜ b ⎟⎞ − w⎛⎜ c ⎞⎟
⎟⎜
⎟
⎝ 2 ⎠⎝ 3 ⎠
⎝2⎠
⎝6⎠
M1 − M2 + w⎜
A =
b
⎛
⎝
B = w⎜ b +
a + c⎞
2
x1 = 0 , 0.01a .. a
⎟−A
⎠
⎛ A ⎞ ⎛ 40.00 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ B ⎠ ⎝ 40.00 ⎠
⎛
⎝
V 1 ( x) = ⎜ −w
⎡⎛
⎣⎝
x⎞ x 1
⎟
a ⎠ 2 kip
M1p ( x) = ⎢⎜ −w
x⎞ x x
⎤ 1
⎟ 2 3 − M1⎥
a⎠
⎦ kip⋅ ft
724
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
x2 = a , 1.01a .. a + b
⎡
⎣
V 2 ( x) = ⎢A − w
⎡
⎣
a
2
⎤ 1
⎦ kip
− w( x − a)⎥
M2p ( x) = ⎢−M1 − w
a⎛
2a ⎞
⎛ x − a ⎞⎤ 1
x−
+ A ( x − a) − w( x − a) ⎜
⎜
⎟
⎟⎥
2⎝
3 ⎠
⎝ 2 ⎠⎦ kip⋅ ft
x3 = a + b , 1.01( a + b) .. a + b + c
⎛ a + b + c − x⎞⎛ a + b + c − x⎞ 1
⎟⎜
⎟
2
c
⎠ kip
⎝
⎠⎝
V 3 ( x) = w⎜
⎡ ⎛ a + b + c − x⎞⎛ a + b + c − x⎞⎛ a + b + c − x⎞ − M ⎤ 1
⎟⎜
⎟⎜
⎟
2⎥
2
3
c
⎠⎝
⎠
⎣ ⎝
⎠⎝
⎦ kip⋅ ft
M3p ( x) = ⎢−w⎜
Force (lb)
40
V1( x1)
V2( x2)
V3( x3)
20
0
20
40
0
5
10
15
20
x1 x2 x3
, ,
ft ft ft
Distance (ft)
725
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
20
Moment (kip-ft)
M1p( x1)
M2p( x2)
M3p( x3)
0
20
40
60
0
5
10
15
20
x1 x2 x3
, ,
ft ft ft
Distance (ft)
Problem 7-88
Draw the shear and moment diagrams for
the beam.
Units Used:
3
kip = 10 lb
Given:
w1 = 2
kip
ft
w2 = 1
kip
ft
a = 15 ft
Solution:
x = 0 , 0.01a .. a
⎡
⎣
⎛
⎝
V ( x) = ⎢w2 x − ⎜ w1
x ⎞ x⎤ 1
⎟ ⎥
a ⎠ 2⎦ kip
⎡
⎣
M ( x) = ⎢w2 x
x
2
⎛
⎝
− ⎜ w1
x ⎞ x x⎤ 1
⎟ ⎥
a ⎠ 2 3⎦ kip⋅ ft
726
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Force (kip)
4
V( x ) 2
0
0
2
4
6
8
10
12
14
10
12
14
x
ft
Distance (ft)
Moment (kip-ft)
40
M( x) 20
0
0
2
4
6
8
x
ft
Distance (ft)
Problem 7-89
Determine the force P needed to hold the cable in the position shown, i.e., so segment BC
remains horizontal. Also, compute the sag yB and the maximum tension in the cable.
Units Used:
3
kN = 10 N
727
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
a = 4m
F 1 = 4 kN
b = 6m
F 2 = 6 kN
c = 3m
d = 2m
e = 3m
Solution:
Initial guesses:
yB = 1 m P = 1 kN
TAB = 1 kN TBC = 1 kN TCD = 1 kN TDE = 1 kN
Given
⎛ −a ⎞ T + T = 0
AB
BC
⎜ 2
2⎟
+
y
a
B ⎠
⎝
yB
⎛⎜
⎞⎟
TAB − F1 = 0
⎜ a2 + yB2 ⎟
⎝
⎠
−TBC + ⎡
c
⎤T = 0
CD
⎢ 2
2⎥
+
y
−
e
c
(B )⎦
⎣
yB − e
⎡⎢
⎥⎤ TCD − P = 0
⎢ c2 + ( yB − e) 2⎥
⎣
⎦
−c
⎡
⎤T + ⎛ d ⎞T = 0
CD ⎜
DE
⎢ 2
⎥
2
2
2⎟
+
y
−
e
+
e
c
d
(B )⎦
⎝
⎠
⎣
⎡⎢ −( yB − e) ⎥⎤
e
⎞T − F = 0
TCD + ⎛
⎜
⎟ DE 2
2
2
2
2
⎢ c + ( yB − e) ⎥
+
d
e
⎝
⎠
⎣
⎦
⎛ yB ⎞
⎜
⎟
⎜ P ⎟
⎜ TAB ⎟
⎜
⎟ = Find ( yB , P , TAB , TBC , TCD , TDE)
TBC
⎜
⎟
⎜ TCD ⎟
⎜
⎟
⎝ TDE ⎠
728
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Engineering Mechanics - Statics
Chapter 7
Tmax = max ( TAB , TBC , TCD , TDE)
yB = 3.53 m
P = 800.00 N
Tmax = 8.17 kN
Problem 7-90
Cable ABCD supports the lamp of mass M1 and the lamp of mass M2. Determine the maximum
tension in the cable and the sag of point B.
Given:
M1 = 10 kg
M2 = 15 kg
a = 1m
b = 3m
c = 0.5 m
d = 2m
Solution:
Guesses
Given
yB = 1 m
TAB = 1 N
TBC = 1 N
TCD = 1 N
b
⎛ −a ⎞ T + ⎡
⎤T = 0
AB
BC
⎜ 2
⎢ 2
2⎟
2⎥
+
y
+
y
−
d
a
b
(
)
B
B
⎝
⎠
⎣
⎦
yB
yB − d
⎛⎜
⎡
⎞⎟
⎥⎤ TBC − M1 g = 0
TAB + ⎢
⎜ a2 + yB2 ⎟
⎢ b2 + ( yB − d) 2⎥
⎝
⎠
⎣
⎦
−b
⎡
⎤T + ⎛ c ⎞T = 0
BC ⎜
CD
⎢ 2
2⎥
2
2⎟
+
y
−
d
+
d
b
c
(
)
⎝
⎠
B
⎣
⎦
⎡⎢ −( yB − d) ⎥⎤
d
⎞T − M g = 0
TBC + ⎛
CD
2
⎜
⎟
2
2
⎢ b2 + ( yB − d) 2⎥
+
d
c
⎝
⎠
⎣
⎦
729
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
⎛ yB ⎞
⎜
⎟
⎜ TAB ⎟ = Find ( y , T , T , T )
B AB BC CD
⎜ TBC ⎟
⎜
⎟
⎝ TCD ⎠
⎛ TAB ⎞ ⎛ 100.163 ⎞
⎜
⎟
⎜ TBC ⎟ = ⎜ 38.524 ⎟ N
⎜ T ⎟ ⎜⎝ 157.243 ⎟⎠
⎝ CD ⎠
Tmax = max ( TAB , TBC , TCD)
Tmax = 157.2 N
yB = 2.43 m
Problem 7-91
The cable supports the three loads shown. Determine the sags yB and yD of points B and D.
Given:
a = 4 ft
e = 12 ft
b = 12 ft
f = 14 ft
c = 20 ft
P 1 = 400 lb
d = 15 ft
P 2 = 250 lb
Solution:
Guesses
Given
yB = 1 ft
yD = 1 ft
TAB = 1 lb
TBC = 1 lb
TCD = 1 lb
TDE = 1 lb
c
⎛ −b ⎞ T + ⎡
⎤T = 0
AB ⎢
BC
⎜ 2
⎟
2
2
2⎥
+
y
+
f
−
y
b
c
(
)
B ⎠
B ⎦
⎝
⎣
yB
f − yB
⎛⎜
⎡
⎞⎟
⎤⎥
TAB − ⎢
TBC − P 2 = 0
⎜ b2 + yB2 ⎟
⎢ c2 + ( f − yB) 2⎥
⎝
⎠
⎣
⎦
−c
d
⎡
⎤T + ⎡
⎤T = 0
BC ⎢
CD
⎢ 2
⎥
2
2
2⎥
+
f
−
y
+
f
−
y
c
d
(
)
(
)
B
D
⎣
⎦
⎣
⎦
730
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 7
f − yB
f − yD
⎡⎢
⎥⎤ TBC + ⎡⎢
⎥⎤ TCD − P1 = 0
⎢ c2 + ( f − yB) 2⎥
⎢ d2 + ( f − yD) 2⎥
⎣
⎦
⎣
⎦
−d
e
⎡
⎤T + ⎡
⎤T = 0
CD ⎢
DE
⎢ 2
⎥
⎥
2
2
2
+
f
−
y
+
a
+
y
d
e
( D) ⎦
( D) ⎦
⎣
⎣
a + yD
⎡⎢ −( f − yD) ⎥⎤
⎡
⎥⎤ TDE − P2 = 0
TCD + ⎢
⎢ d2 + ( f − yD) 2⎥
⎢ e2 + ( a + yD) 2⎥
⎣
⎦
⎣
⎦
⎛⎜ TAB ⎞⎟
⎜ TBC ⎟
⎜
⎟
⎜ TCD ⎟ = Find ( T , T , T , T , y , y )
AB BC CD DE B D
⎜ TDE ⎟
⎜
⎟
⎜ yB ⎟
⎜ yD ⎟
⎝
⎠
⎛ TAB ⎞ ⎛ 675.89 ⎞
⎜
⎟ ⎜
⎟
⎜ TBC ⎟ = ⎜ 566.90 ⎟ lb
⎜ TCD ⎟ ⎜ 603.86 ⎟
⎜
⎟ ⎜ 744.44 ⎟
⎠
⎝ TDE ⎠ ⎝
⎛ yB ⎞ ⎛ 8.67 ⎞
⎜ ⎟ =⎜
⎟ ft
⎝ yD ⎠ ⎝ 7.04 ⎠
Problem 7-92
The cable supports the three loads shown. Determine the magnitude of P 1 and find the sag yD for
the given data.
Given:
P 2 = 300 lb
c = 20 ft
yB = 8 ft
d = 15 ft
a = 4 ft
e = 12 ft
b = 12 ft
f = 14 ft
Solution:
Guesses
P 1 = 1 lb
TAB = 1 lb
TBC = 1 lb
TCD = 1 lb
731
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Engineering Mechanics - Statics
TDE = 1 lb
Chapter 7
yD = 1 ft
Given
c
⎛ −b ⎞ T + ⎡
⎤T = 0
AB ⎢
BC
⎜ 2
⎟
2
2
2⎥
+
y
+
f
−
y
b
c
( B) ⎦
B ⎠
⎝
⎣
yB
f − yB
⎛⎜
⎡
⎞⎟
⎤⎥
TAB − ⎢
TBC − P 2 = 0
⎜ b2 + yB2 ⎟
⎢ c2 + ( f − yB) 2⎥
⎝
⎠
⎣
⎦
−c
d
⎡
⎤T + ⎡
⎤T = 0
BC
CD
⎢ 2
⎢ 2
2⎥
2⎥
+
f
−
y
+
f
−
y
c
d
(
)
(
)
B
D
⎣
⎦
⎣
⎦
f − yB
f − yD
⎡⎢
⎥⎤ TBC + ⎡⎢
⎥⎤ TCD − P1 = 0
⎢ c2 + ( f − yB) 2⎥
⎢ d2 + ( f − yD) 2⎥
⎣
⎦
⎣
⎦
−d
e
⎡
⎤T + ⎡
⎤T = 0
CD
DE
⎢ 2
⎢ 2
2⎥
2⎥
+
f
−
y
+
a
+
y
d
e
(
)
(
)
D
D
⎣
⎦
⎣
⎦
a + yD
⎡⎢ −( f − yD) ⎥⎤
⎡
⎥⎤ TDE − P2 = 0
TCD + ⎢
⎢ d2 + ( f − yD) 2⎥
⎢ e2 + ( a + yD) 2⎥
⎣
⎦
⎣
⎦
⎛⎜ TAB ⎞⎟
⎜ TBC ⎟
⎜
⎟
⎜ TCD ⎟ = Find ( T , T , T , T , P , y )
AB BC CD DE 1 D
⎜ TDE ⎟
⎜
⎟
⎜ P1 ⎟
⎜ yD ⎟
⎝
⎠
⎛ TAB ⎞ ⎛ 983.33 ⎞
⎜
⎟ ⎜
⎟
⎜ TBC ⎟ = ⎜ 854.21 ⎟ lb
⎜ TCD ⎟ ⎜ 916.11 ⎟
⎜
⎟ ⎜ 1084.68 ⎟
⎠
⎝ TDE ⎠ ⎝
P 1 = 658 lb
yD = 6.44 ft
Problem 7-93
The cable supports the loading shown. Determine the distance xB the force at point B acts from A.
732
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
P = 40 lb
c = 2 ft
F = 30 lb
d = 3 ft
a = 5 ft
e = 3
b = 8 ft
f = 4
Solution:
The initial guesses:
TAB = 10 lb
TCD = 30 lb
TBC = 20 lb
xB = 5 ft
Given
xB − d
⎛⎜ −xB ⎟⎞
⎡
⎤⎥
TAB − ⎢
TBC + P = 0
⎜ xB2 + a2 ⎟
⎢ ( xB − d) 2 + b2⎥
⎝
⎠
⎣
⎦
a
b
⎛
⎞T − ⎡
⎤T = 0
AB
BC
⎜
⎢
2
2⎟
2
2⎥
+
a
−
d
+
b
x
x
(
)
B
B
⎝
⎠
⎣
⎦
xB − d
⎡⎢
d
⎤⎥
f
⎞T + ⎛
⎞F = 0
TBC − ⎛
CD ⎜
⎜
⎟
⎟
2
2
2
2
⎢ ( xB − d) 2 + b2⎥
⎝ c +d ⎠
⎝ e +f ⎠
⎣
⎦
b
⎡
⎤T − ⎛ c ⎞T − ⎛ e ⎞F = 0
BC ⎜
CD ⎜
⎢
2
2⎥
2
2⎟
2
2⎟
−
d
+
b
+
d
+
f
x
c
e
(
)
⎝
⎠
⎝
⎠
B
⎣
⎦
⎛ TAB ⎞
⎜
⎟
⎜ TCD ⎟ = Find ( T , T , T , x )
AB CD BC B
⎜ TBC ⎟
⎜
⎟
⎝ xB ⎠
⎛ TAB ⎞ ⎛ 50.90 ⎞
⎜
⎟
⎜ TCD ⎟ = ⎜ 36.70 ⎟ lb
⎜ T ⎟ ⎜⎝ 38.91 ⎟⎠
⎝ BC ⎠
xB = 4.36 ft
Problem 7-94
The cable supports the loading shown. Determine the magnitude of the horizontal force P.
733
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Given:
F = 30 lb
c = 2 ft
xB = 6 ft
d = 3 ft
a = 5 ft
e = 3
b = 8 ft
f = 4
Solution:
The initial guesses:
TAB = 10 lb
TCD = 30 lb
TBC = 20 lb
P = 10 lb
Given
xB − d
⎛⎜ −xB ⎟⎞
⎡
⎤⎥
TAB − ⎢
TBC + P = 0
⎜ xB2 + a2 ⎟
⎢ ( xB − d) 2 + b2⎥
⎝
⎠
⎣
⎦
a
b
⎛
⎞T − ⎡
⎤T = 0
AB
BC
⎜ 2
⎢ 2
2⎟
2⎥
+
x
+
x
−
d
a
b
(
)
B
B
⎝
⎠
⎣
⎦
xB − d
⎤⎥
f
⎛ −d ⎞ T + ⎡⎢
⎞F = 0
TBC + ⎛
CD
⎜ 2 2⎟
⎜
⎟
2
2
⎢ b2 + ( xB − d) 2⎥
⎝ c +d ⎠
⎝ e +f ⎠
⎣
⎦
b
⎛ −c ⎞ T + ⎡
⎤T − ⎛ e ⎞F = 0
BC ⎜
⎜ 2 2 ⎟ CD ⎢ 2
2⎥
2
2⎟
+
x
−
d
+
f
b
e
(
)
⎝ c +d ⎠
⎝
⎠
B
⎣
⎦
⎛ TAB ⎞
⎜
⎟
T
BC
⎜
⎟ = Find ( T , T , T , P)
AB BC CD
⎜ TCD ⎟
⎜
⎟
⎝ P ⎠
⎛ TAB ⎞ ⎛ 70.81 ⎞
⎜
⎟
⎜ TBC ⎟ = ⎜ 48.42 ⎟ lb
⎜ T ⎟ ⎜⎝ 49.28 ⎟⎠
⎝ CD ⎠
P = 71.40 lb
734
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-95
Determine the forces P1 and P2
needed to hold the cable in the
position shown, i.e., so segment CD
remains horizontal. Also, compute the
maximum tension in the cable.
Given:
3
kN = 10 N
F = 5 kN
d = 4m
a = 1.5 m
e = 5m
b = 1m
f = 4m
c = 2m
Solution:
Guesses
F AB = 1 kN
F BC = 1 kN
F CD = 1 kN
F DE = 1 kN
P 1 = 1 kN
P 2 = 1 kN
Given
⎛ −c ⎞ F + ⎛ d ⎞ F = 0
⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC
⎝ a +c ⎠
⎝ b +d ⎠
735
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
⎛ a ⎞F − ⎛ b ⎞F − F = 0
⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC
⎝ a +c ⎠
⎝ b +d ⎠
⎛ −d ⎞ F + F = 0
⎜ 2 2 ⎟ BC CD
⎝ b +d ⎠
⎛ b ⎞F − P = 0
⎜ 2 2 ⎟ BC 1
⎝ b +d ⎠
⎡
−F CD + ⎢
f
2
⎤F
2⎥
⎣ f + ( a + b) ⎦
DE = 0
a+b
⎡
⎤F − P = 0
DE
2
⎢ 2
2⎥
+
(
a
+
b
)
f
⎣
⎦
⎛⎜ FAB ⎟⎞
⎜ FBC ⎟
⎜
⎟
⎜ FCD ⎟ = Find ( F , F , F , F , P , P )
AB BC CD DE 1 2
⎜ FDE ⎟
⎜
⎟
⎜ P1 ⎟
⎜ P2 ⎟
⎝
⎠
⎛ FAB ⎞ ⎛ 12.50 ⎞
⎜
⎟ ⎜
⎟
⎜ FBC ⎟ = ⎜ 10.31 ⎟ kN
⎜ FCD ⎟ ⎜ 10.00 ⎟
⎜
⎟ ⎜ 11.79 ⎟
⎠
⎝ FDE ⎠ ⎝
⎛ P1 ⎞ ⎛ 2.50 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ P2 ⎠ ⎝ 6.25 ⎠
Tmax = max ( FAB , F BC , F CD , FDE)
Tmax = 12.50 kN
F max = max ( F AB , FBC , FCD , F DE)
F max = 12.50 kN
736
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-96
The cable supports the loading shown. Determine the distance xB from the wall to point B.
Given:
W1 = 8 lb
W2 = 15 lb
a = 5 ft
b = 8 ft
c = 2 ft
d = 3 ft
Solution:
Guesses
TAB = 1 lb
TBC = 1 lb
TCD = 1 lb
xB = 1 ft
Given
xB − d
⎛⎜ −xB ⎟⎞
⎡
⎤⎥
TAB − ⎢
TBC + W2 = 0
⎜ a2 + xB2 ⎟
⎢ b2 + ( xB − d) 2⎥
⎝
⎠
⎣
⎦
a
b
⎛
⎞T − ⎡
⎤T = 0
AB ⎢
BC
⎜ 2
⎟
2
2
2⎥
+
x
+
x
−
d
a
b
(
)
B ⎠
B
⎝
⎣
⎦
xB − d
⎡⎢
⎤⎥
⎛ d ⎞T = 0
TBC − ⎜
CD
2
2⎟
⎢ b2 + ( xB − d) 2⎥
+
d
c
⎝
⎠
⎣
⎦
b
⎡
⎤T − ⎛ c ⎞T − W = 0
BC ⎜
CD
1
⎢ 2
2⎥
2
2⎟
+
x
−
d
+
d
b
c
(
)
⎝
⎠
B
⎣
⎦
⎛ TAB ⎞
⎜
⎟
TBC
⎜
⎟ = Find ( T , T , T , x )
AB BC CD B
⎜ TCD ⎟
⎜
⎟
⎝ xB ⎠
⎛ TAB ⎞ ⎛ 15.49 ⎞
⎜
⎟
⎜ TBC ⎟ = ⎜ 10.82 ⎟ lb
⎜ T ⎟ ⎜⎝ 4.09 ⎟⎠
⎝ CD ⎠
xB = 5.65 ft
Problem 7-97
Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is
737
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
g
capable of sustaining a maximum tension Tmax before it will break.
pp
Given:
Tmax = 3000 lb
a = 50 ft
b = 6 ft
Solution:
y=
1 ⌠ ⌠
2
wx
⎮ ⎮ w dx dx =
FH ⌡ ⌡
2 FH
⎛ w ⎞ x2
⎟
⎝ 2 FH ⎠
y=⎜
x=
2
a
y=b
2
wa
8b
FH =
⎛ dy ⎞ = tan ( θ ) = w ⎛ a ⎞ = 4b
⎜ ⎟
⎜ ⎟
max
a
FH ⎝ 2 ⎠
⎝ dx ⎠
Tmax =
FH
cos ( θ max)
θ max = atan ⎛⎜
⎟
⎝a⎠
2
=
wa
(
8 b cos θ max
)
4b ⎞
w =
Tmax8 b cos ( θ max)
a
2
θ max = 25.64 deg
w = 51.93
lb
ft
Problem 7-98
The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in
the cable.
Units Used:
3
kip = 10 lb
Given:
w = 250
lb
ft
a = 50 ft
b = 6 ft
738
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
2
y=
⎛ a⎞
b=
⎜ ⎟
2FH ⎝ 2 ⎠
wx
w
2FH
tan ( θ max) =
Tmax =
d
y ⎛⎜ x =
dx ⎝
2
FH =
w ⎛ a⎞
=
⎟
⎜ ⎟
2⎠
FH ⎝ 2 ⎠
a⎞
FH
wa
2
F H = 13021 lb
8b
θ max = atan ⎛⎜
wa ⎞
⎟
⎝ 2 FH ⎠
θ max = 25.64 deg
Tmax = 14.44 kip
cos ( θ max)
The minimum tension occurs at
θ = 0 deg
Tmin = FH
Tmin = 13.0 kip
Problem 7-99
The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine
the equation of the curve y = f(x) which defines the cable shape AB, and the maximum
tension developed in the cable.
Units Used:
kip = 103 lb
Given:
w = 250
lb
ft
a = 20 ft
b = 30 ft
Solution:
⌠ ⌠
1 ⎮ ⎮
y=
FH ⎮ ⎮
⌡ ⌡
y=
wx
b
dx dx
3
⎞
⎜
+ c1 x + c2⎟
FH ⎝ 6 b
⎠
1 ⎛ wx
739
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Engineering Mechanics - Statics
Chapter 7
Apply boundary conditions y = x = 0 and
d
y = 0,x = 0
3
C1 = C2 = 0
3
wx
y=
6FH b
set
y=a
2
wb
FH =
6a
Tmax =
Thus
dx
F H = 1.875 kip
FH
cos ( θ max)
x=b
wb
a=
6FH b
⎛ w b2 ⎞
θ max = atan ⎜
⎟
⎝ 2 FH b ⎠
θ max = 63.43 deg
Tmax = 4.19 kip
Problem 7-100
The cable supports a girder which has weight density γ. Determine the tension in the cable at points
A, B, and C.
Units used:
3
kip = 10 lb
Given:
γ = 850
lb
ft
a = 40 ft
b = 100 ft
c = 20 ft
Solution:
y=
1 ⌠ ⌠
⎮ ⎮ γ dx dx
FH ⌡ ⌡
2
y=
γx
2FH
γx
d
y =
FH
dx
x1 = 1 ft
Guesses
F H = 1 lb
740
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Engineering Mechanics - Statics
c=
Given
tan ( θ A) =
2
2 FH
a=
(x1 − b)
F
γ
cos ( θ A)
⎡ γ x − b⎤
( 1 )⎥
⎣FH
⎦
θ A = −53.79 deg
⎛ γ x⎞
1⎟
⎝ FH ⎠
θ C = 44.00 deg
θ C = atan ⎜
x1
FH
2
b − x1 )
(
2F
θ A = atan ⎢
H
FH
⎛ x1 ⎞
⎜ ⎟ = Find ( x1 , FH)
⎝ FH ⎠
γ
H
γ
tan ( θ C) =
TA =
γ x1
Chapter 7
TB = F H
TC =
FH
cos ( θ C)
F H = 36.46 kip
⎛ TA ⎞ ⎛ 61.71 ⎞
⎜ ⎟
⎜ TB ⎟ = ⎜ 36.46 ⎟ kip
⎜ T ⎟ ⎜⎝ 50.68 ⎟⎠
⎝ C⎠
Problem 7-101
The cable is subjected to the triangular
loading. If the slope of the cable at
point O is zero, determine the equation
of the curve y = f(x) which defines the
cable shape OB, and the maximum
tension developed in the cable.
Units used:
kip = 103 lb
Given:
w = 500
lb
ft
b = 8 ft
a = 15 ft
741
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Engineering Mechanics - Statics
Chapter 7
Solution:
⎛⌠ ⌠
wx
1 ⎜⎮ ⎮
y=
FH ⎜⎮ ⎮
⎝⌡ ⌡
⎞
dx dx⎟
a
⎟
⎠
3
⎤
⎜
⎟ + C1 x + C2⎥
⎢
y=
FH ⎣ a ⎝ 6 ⎠
⎦
1 ⎡w ⎛ x ⎞
2
1 ⎛ w x ⎞ ⎛ C1 ⎞
d
⎜
⎟+⎜ ⎟
y =
FH ⎝ 2a ⎠ ⎝ FH ⎠
dx
At x = 0,
d
dx
y = 0, C1 = 0
At x = 0, y = 0, C2 = 0
3
y=
2
wx
d
y =
2a FH
dx
wx
6a FH
3
At x = a, y = b
b=
wa
6a FH
FH =
1
6
⎛ a2 ⎞
⎟
⎝b⎠
w⎜
F H = 2343.75 lb
742
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 7
⎛ w a2 ⎞
(θ max) = atan ⎜ 2a F ⎟
H⎠
⎝
2
wa
d
y = tan ( θ max) =
2a FH
dx
Tmax =
FH
(θ max) = 57.99 deg
Tmax = 4.42 kip
cos ( θ max)
Problem 7-102
The cable is subjected to the parabolic loading w = w0(1− (2x/a)2). Determine the equation y = f(x)
which defines the cable shape AB and the maximum tension in the cable.
Units Used:
3
kip = 10 lb
Given:
2
⎛ 2x ⎞ ⎥⎤
⎟
⎝a⎠⎦
⎡
w = w0 ⎢1 − ⎜
⎣
a = 100 ft
w0 = 150
lb
ft
b = 20 ft
Solution:
y=
1 ⌠ ⌠
⎮ ⎮ w ( x) dx dx
FH ⌡ ⌡
⌠
y=
y=
⎛⎜ 4 x ⎞⎟
+ C1 dx
⎮ w0 x −
2⎟
FH
⎜
⎮
⎝ 3a ⎠
⌡
3
1 ⎮
⎛
2
1 ⎜ w0 x
FH ⎜ 2
⎝
⎛
⎞
⎟
−
+ C1 x + C2
⎟
2
3a
⎠
4
x w0
3
⎞
4 w0 x
dy
1 ⎜
⎟
=
w0 x −
+ C1
⎟
2
dx
FH ⎜
3a
⎝
⎠
743
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
dy
x=0
At
=0
dx
At x = 0
Chapter 7
C1 = 0
y=0
C2 = 0
x w0 ⎞⎟
dy
⎠
dx
Thus
y=
⎛
2
4
1 ⎜ w0 x
−
FH ⎜ 2
⎝
x=
At
2 ⎟
3a
a
=
⎛
1 ⎜
FH ⎜
⎝
4 w0 x ⎞
⎟
3
w0 x −
3a
2
⎟
⎠
we have
2
2⎞
⎛
b=
⎜ ⎟ − 2 ⎜ 2 ⎟ = 48 ⎜
⎟
FH ⎢ 2 ⎝ 2 ⎠
⎥
⎝ FH ⎠
3a ⎝ ⎠ ⎦
⎣
1 ⎡
⎢w0 ⎛ a ⎞
2
w0 ⎛ a ⎞ 4⎥⎤
5 ⎜ w0 a ⎟
FH =
5 w0 a
2
F H = 7812.50 lb
48b
3
w0 a
a⎞
1 ⎡
a ⎞ 4 w0 ⎛ a ⎞ ⎥⎤
⎛
⎛
⎢
tan ( θ max) = y ⎜ ⎟ =
w0 ⎜ ⎟ −
=
⎜
⎟
FH ⎢ ⎝ 2 ⎠ 3 a2 ⎝ 2 ⎠ ⎥
3 FH
dx ⎝ 2 ⎠
⎣
⎦
d
⎛ w0 a ⎞
⎟
⎝ 3 FH ⎠
θ max = atan ⎜
Tmax =
FH
cos ( θ max)
θ max = 32.62 deg
Tmax = 9.28 kip
Problem 7-103
The cable will break when the maximum tension reaches Tmax. Determine the minimum sag h if
it supports the uniform distributed load w.
Given:
kN = 103 N
Tmax = 10 kN
w = 600
N
m
a = 25 m
Solution:
The equation of the cable:
y=
1 ⌠ ⌠
⎮ ⎮ w dx dx
FH ⌡ ⌡
y=
1 ⎛ wx
2
⎞
⎜
+ C1 x + C2⎟
FH ⎝ 2
⎠
dy
dx
=
1
FH
(w x + C1)
744
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Boundary Conditions:
y = 0 at x = 0, then from Eq.[1]
0 =
d
y = 0 at x = 0, then from Eq.[2]
0 =
w ⎞ 2
y = ⎛⎜
⎟x
⎝ 2 FH ⎠
w
dx
Thus,
tan ( θ max) =
Tmax =
dy
dx
2FH
cos ( θ max)
Guess
h = 1m
Given
Tmax =
FH
FH
1
FH
(C2)
C2 = 0
(C1)
C1 = 0
⎛ a⎞
h=
⎜ ⎟
2FH ⎝ 2 ⎠
w
x
2
4 F H + ( w a)
=
wa
2
2
FH +
( w a)
4
2
=
a
wa
2
2
FH =
w ⎛ a⎞
2
⎜ ⎟
2h ⎝ 2 ⎠
2 FH
cos ( θ max) =
wa
FH
=
1
2
2
2
+1
16h
2
a
16h
2
+1
h = Find ( h)
h = 7.09 m
Problem 7-104
A fiber optic cable is suspended over the poles so that the angle at the supports is θ.
Determine the minimum tension in the cable and the sag. The cable has a mass density ρ and
the supports are at the same elevation.
Given:
θ = 22 deg
ρ = 0.9
kg
m
a = 30 m
g = 9.81
m
2
s
Solution:
θ max = θ
745
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
w0 = ρ g
a ⎞
⎛
⎜
2 ⎟
d
y = tan ( θ ) = sinh ⎜ w0
⎟
dx
⎝ FH ⎠
w0 ⎛⎜
FH =
a⎞
⎟
⎝2⎠
F H = 336 N
asinh ( tan ( θ ) )
Tmax =
FH
Tmax = 363 N
cos ( θ )
⎛
⎛w a ⎞ ⎞
⎜
⎜ 02⎟ ⎟
FH
h =
⎜ cosh ⎜
⎟ − 1⎟
w0 ⎝
⎝ FH ⎠ ⎠
h = 2.99 m
Problem 7-105
A cable has a weight density γ and is supported at points that are a distance d apart and at the
same elevation. If it has a length L, determine the sag.
Given:
γ = 3
lb
d = 500 ft
ft
L = 600 ft
Solution:
Guess
Given
h =
F H = 100 lb
L
2
⎡FH
−⎢
⎣γ
⎡ γ ⎛ d ⎞⎤⎤ = 0
⎜ ⎟⎥⎥
⎣FH ⎝ 2 ⎠⎦⎦
F H = Find ( FH)
sinh ⎢
⎛1 γ ⎞ ⎞
⎜ cosh ⎜ 2 F d⎟ − 1⎟
γ ⎝
⎝ H ⎠ ⎠
FH ⎛
F H = 704.3 lb
h = 146 ft
Problem 7-106
Show that the deflection curve of the cable discussed in Example 7.15 reduces to Eq. (4) in
Example 7.14 when the hyperbolic cosine function is expanded in terms of a series and only the
746
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
p
yp
f
p
y
first two terms are retained. (The answer indicates that the catenary may be replaced by a
parabola in the analysis of problems in which the sag is small. In this case, the cable weight is
assumed to be uniformly distributed along the horizontal.)
Solution:
2
cosh ( x) = 1 +
x
2!
+ ..
Substituting into
2 2
⎞ w0 x2
⎛ w0 ⎞ ⎞ FH ⎛⎜ w0 x
⎟
y=
1+
+ .. − 1 =
⎜cosh ⎜ x⎟ − 1⎟ =
2
w0 ⎝
⎟ 2 FH
⎝ FH ⎠ ⎠ w0 ⎜⎝
2FH
⎠
FH ⎛
Using the boundary conditions y = h at
h=
w0 ⎛ L ⎞ 2
⎜ ⎟
FH =
2FH ⎝ 2 ⎠
w0 L
8h
x=
L
2
2
We get
y=
4h 2
L
2
x
Problem 7-107
A uniform cord is suspended between two points having the same elevation. Determine the
sag-to-span ratio so that the maximum tension in the cord equals the cord's total weight.
Solution:
s=
y=
FH
w0
⎛ w0 ⎞
x⎟
⎝ FH ⎠
sinh ⎜
FH ⎛
⎛ w0 ⎞ ⎞
⎜cosh ⎜ x⎟ − 1⎟
w0 ⎝
⎝ FH ⎠ ⎠
At x =
L
2
⎛ w0 L ⎞
d
y = tan ( θ max) = sinh ⎜
⎟
dx max
⎝ 2FH ⎠
cos ( θ max) =
1
⎛ w0 L ⎞
⎟
⎝ 2 FH ⎠
cosh ⎜
747
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⎛ w0 L ⎞
⎟
⎝ 2 FH ⎠
FH
Tmax =
Chapter 7
w0 2 s = F H cosh ⎜
cos ( θ max)
⎛ w0 L ⎞
⎛ w0 L ⎞
⎟ = FH cosh ⎜
⎟
⎝ 2 FH ⎠
⎝ 2FH ⎠
⎛ w0 L ⎞ 1
⎟=
⎝ 2 FH ⎠ 2
2 F H sinh ⎜
k1 = atanh ( 0.5)
when x =
L
⎛ FH ⎞
⎟
⎝ w0 ⎠
h=
⎛ 2 Fh ⎞
⎟
⎝ w0 ⎠
h = k2 ⎜
w0 L
k1 = 0.55
y=h
2
tanh ⎜
L = k1 ⎜
= k1
2 FH
FH
W0
(cosh (k1) − 1)
ratio =
k2
h
=
2 k1
L
k2 = cosh ( k1 ) − 1
ratio =
k2 = 0.15
k2
2 k1
ratio = 0.14
Problem 7-108
A cable has a weight denisty γ. If it can span a distance L and has a sag h determine the length of
the cable. The ends of the cable are supported from the same elevation.
Given:
γ = 2
lb
ft
L = 100 ft
h = 12 ft
Solution:
From Eq. (5) of Example 7-15 :
⎡⎢⎛ γ L ⎞ 2⎤⎥
⎜
⎟
FH ⎢⎝ 2 FH ⎠ ⎥
h=
⎢ 2 ⎥
γ ⎣
⎦
FH =
⎛ L2 ⎞
⎟
8 ⎝ h ⎠
1
γ⎜
F H = 208.33 lb
From Eq. (3) of Example 7-15:
l
2
⎛ FH ⎞ ⎡ γ ⎛ L ⎞⎤
⎟ sinh ⎢ ⎜ ⎟⎥
⎝ γ ⎠ ⎣FH ⎝ 2 ⎠⎦
=⎜
⎛ FH ⎞ ⎛ 1 L ⎞
⎟ sinh ⎜ γ ⎟
⎝ γ ⎠ ⎝ 2 FH ⎠
l = 2⎜
l = 104 ft
748
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-109
The transmission cable having a weight density γ is strung across the river as shown. Determine the
required force that must be applied to the cable at its points of attachment to the towers at B and C.
Units Used:
3
kip = 10 lb
Given:
lb
ft
b = 75 ft
a = 50 ft
c = 10 ft
γ = 20
Solution:
From Example 7-15,
y=
FH ⎡
⎛ γ ⎞ ⎤
⎢cosh⎜ F x⎟ − 1⎥
γ ⎣
⎝ H ⎠ ⎦
Guess
Given
dy
⎛ γx ⎞
= sinh ⎜
⎟
dx
⎝ FH ⎠
F H = 1000 lb
c=
At B:
⎛ γa ⎞ ⎞
⎜ cosh ⎜ − F ⎟ − 1⎟
γ ⎝
⎝ H⎠ ⎠
FH ⎛
⎛ γa ⎞
⎟
⎝ FH ⎠
θ B = atan ⎜ sinh ⎜
⎛ γb ⎞
⎟
⎝ FH ⎠
θ C = atan ⎜ sinh ⎜
tan ( θ B) = sinh ⎜ −
tan ( θ C) = sinh ⎜
TB =
FH
cos ( θ B)
TC =
FH
cos ( θ C)
F H = Find ( FH)
F H = 2.53 kip
⎛
⎝
⎛ −γ a ⎞ ⎞
⎟⎟
⎝ FH ⎠⎠
θ B = −22.06 deg
⎛
⎝
⎛ γ b ⎞⎞
⎟⎟
⎝ FH ⎠⎠
θ C = 32.11 deg
⎛ TB ⎞ ⎛ 2.73 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ TC ⎠ ⎝ 2.99 ⎠
Problem 7-110
Determine the maximum tension developed in the cable if it is subjected to a uniform load w.
749
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Units Used:
MN = 106 N
Given:
N
w = 600
m
a = 100 m
b = 20 m
θ = 10 deg
Solution:
The Equation of the Cable:
1 ⎛ wx
2
⎞
⎜
y=
⎮ ⎮ w( x) dx dx =
+ C1 x + C2⎟
FH ⌡ ⌡
FH ⎝ 2
⎠
1 ⌠ ⌠
dy
dx
=
1
FH
(w x + C1)
C1 = 1 N
Initial Guesses:
Given
FH = 1 N
Boundary Conditions:
0 =
x=0
1
FH
tan ( θ ) =
C2
FH
(C1)
w ⎞ 2 ⎛ C1 ⎞
⎟a + ⎜ F ⎟a
⎝ 2 FH ⎠
⎝ H⎠
⎛ C1 ⎞
⎜ ⎟
⎜ C2 ⎟ = Find ( C1 , C2 , FH)
⎜F ⎟
⎝ H⎠
tan ( θ max) =
1
b = ⎛⎜
y = b at x = a
Tmax =
C2 = 1 N⋅ m
(w a + C1)
F
1
H
FH
cos ( θ max)
C1 = 0.22 MN
C2 = 0.00 N⋅ m
⎛ w a + C1 ⎞
⎟
⎝ FH ⎠
θ max = atan ⎜
F H = 1.27 MN
θ max = 12.61 deg
Tmax = 1.30 MN
750
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-111
A chain of length L has a total mass M and is suspended between two points a distance d apart.
Determine the maximum tension and the sag in the chain.
Given:
L = 40 m
M = 100 kg
d = 10 m
g = 9.81
m
2
s
Solution:
w = M
g
L
⎛ FH ⎞ ⎛ w ⎞
⎟ sinh ⎜ x⎟
⎝ w ⎠ ⎝ FH ⎠
y=⎜
⎛ FH ⎞ ⎛
w
⎟ ⎜cosh ⎛⎜ x⎟⎞ − 1⎞⎟
⎝ w ⎠⎝
⎝ FH ⎠ ⎠
F H = 10 N
h = 10 m
s=⎜
Guesses
Given
⎛ FH ⎞ ⎛ w d ⎞
⎟ sinh ⎜
⎟
⎝ w ⎠ ⎝ FH 2 ⎠
L
=⎜
2
θ max = atan ⎛⎜ sinh ⎛⎜
⎝
F H = 37.57 N
h=
d
dx
w d⎞
⎞
cosh ⎛⎜
⎜
⎟ − 1⎟
2
w ⎝
F
⎝ H ⎠ ⎠
FH ⎛
w d ⎞⎞
Tmax =
⎟⎟
⎝ FH 2 ⎠⎠
h = 18.53 m
x⎞⎟
F
⎝ H ⎠
y = sinh ⎛⎜
w
⎛ FH ⎞
⎜ ⎟ = Find ( FH , h)
⎝ h ⎠
FH
cos ( θ max)
Tmax = 492 N
Problem 7-112
The cable has a mass density ρ and has length L. Determine the vertical and horizontal components
of force it exerts on the top of the tower.
Given:
ρ = 0.5
kg
m
L = 25 m
θ = 30 deg
d = 15 m
g = 9.81
m
2
s
751
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
⌠
⎮
x=⎮
⎮
⎮
⎮
⎮
⌡
1
1+
⎞
⎛⌠
⎜ ⎮ ρ g ds⎟
2⎜
⎟
FH ⎝⌡
⎠
ds
2
1
Performing the integration yields:
⎞
⎛ ρ g s + C1 ⎞
⎜ asinh ⎜
⎟ + C2⎟
ρg ⎝
⎝ FH ⎠
⎠
FH ⎛
x=
dy
dx
=
1 ⌠
⎮ ρ g ds =
(ρ g s + C1)
FH ⌡
FH
dy
At s = 0;
dy
dx
=
1
dx
ρg s
FH
= tan( θ )
C1 = F H tan ( θ )
Hence
+ tan ( θ )
Applying boundary conditions at x = 0; s = 0 to Eq.[1] and using the result
C1 = F H tan ( θ ) yields C2 = −asinh ( tan ( θ ) ). Hence
Guess
FH = 1 N
Given
⎛ FH ⎞ ⎡
1
⎟ ⎢asinh ⎡⎢⎛⎜ ⎞⎟ ( ρ g L + FH tan ( θ ) )⎥⎤ − ( asinh ( tan ( θ ) ) )⎤⎥
⎝ ρg ⎠⎣
⎣⎝ FH ⎠
⎦
⎦
d=⎜
F H = Find ( FH)
At A
FA =
F H = 73.94 N
tan ( θ A) =
ρg L
FH
F Ax = F A cos ( θ A)
cos ( θ A)
FH
+ tan ( θ )
⎛ ρ g L + tan ( θ )⎞
⎟
⎝ FH
⎠
θ A = atan ⎜
F Ay = F A sin ( θ A)
θ A = 65.90 deg
⎛ FAx ⎞ ⎛ 73.94 ⎞
⎜
⎟=⎜
⎟N
⎝ FAy ⎠ ⎝ 165.31 ⎠
752
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-113
A cable of length L is suspended between two points a distance d apart and at the same elevation.
If the minimum tension in the cable is Tmin, determine the total weight of the cable and the
maximum tension developed in the cable.
Units Used:
kip = 103 lb
Given:
L = 50 ft
d = 15 ft
Tmin = 200 lb
Solution:
Tmin = F H
F H = Tmin
F H = 200 lb
⎛ FH ⎞ ⎛ w0 x ⎞
⎟ sinh ⎜
⎟
⎝ w0 ⎠ ⎝ FH ⎠
s=⎜
From Example 7-15:
w0 = 1
Guess
L
Given
2
lb
ft
⎛ FH ⎞ ⎛ w0 d ⎞
⎟ sinh ⎜
⎟
⎝ w0 ⎠ ⎝ FH 2 ⎠
=⎜
w0 = Find ( w0 )
Totalweight = w0 L
Totalweight = 4.00 kip
w0 L
⎡w ⎛ L ⎞⎤
⎢ 0⎜⎝ 2 ⎟⎠⎥
θ max = atan ⎢
⎥
⎣ FH ⎦
tan ( θ max) =
FH 2
w0 = 79.93
lb
ft
θ max = 84.28 deg
Then,
Tmax =
FH
cos ( θ max)
Tmax = 2.01 kip
753
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-114
The chain of length L is fixed at its ends and
hoisted at its midpoint B using a crane. If the
chain has a weight density w, determine the
minimum height h of the hook in order to lift
the chain completely off the ground. What is
the horizontal force at pin A or C when the
chain is in this position? Hint: When h is a
minimum, the slope at A and C is zero.
Given:
L = 80 ft
d = 60 ft
w = 0.5
lb
ft
Solution:
Guesses
F H = 10 lb
h = 1 ft
Given
h=
FH ⎛
⎛ w d ⎞ − 1⎞
⎟ ⎟
⎝ FH 2 ⎠ ⎠
L
FA = FH
FC = FH
⎜cosh ⎜
w ⎝
⎛ h ⎞
⎜ ⎟ = Find ( h , FH)
⎝ FH ⎠
2
⎛ FH ⎞ ⎛ w d ⎞
⎟ sinh ⎜
⎟
⎝ w ⎠ ⎝ FH 2 ⎠
=⎜
⎛ FA ⎞ ⎛ 11.1 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ FC ⎠ ⎝ 11.1 ⎠
h = 23.5 ft
Problem 7-115
A steel tape used for measurement in surveying has a length L and a total weight W. How
much horizontal tension must be applied to the tape so that the distance marked on the ground
is a? In practice the calculation should also include the effects of elastic stretching and
temperature changes on the tape’s length.
Given:
L = 100 ft
W = 2 lb
a = 99.90 ft
754
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Solution:
w0 =
W
w0 = 0.02
L
lb
ft
F H = 10 lb
Guess
L
Given
2
⎛ FH
−⎜
⎝ w0
⎛ w0 a ⎞⎞
⎟⎟ = 0
⎝ FH 2 ⎠⎠
F H = Find ( FH)
sinh ⎜
F H = 12.9 lb
Problem 7-116
A cable of weight W is attached between two points that are a distance d apart, having equal
elevations. If the maximum tension developed in the cable is Tmax determine the length L of the cable
and the sag h.
Given:
W = 100 lb
d = 50 ft
F H = 20 lb
L = 20 ft
Tmax = 75 lb
Solution:
Guesses
θ max = 20 deg
h = 2 ft
Given
h=
W d⎞
⎞
cosh ⎛⎜
⎜
⎟ − 1⎟
2
W ⎝
L
F
⎝ H ⎠ ⎠
FH L ⎛
Tmax =
FH
cos ( θ max)
⎛ FH ⎞
⎜
⎟
⎜ L ⎟ = Find ( F , L , θ , h)
H
max
⎜ θ max ⎟
⎜
⎟
⎝ h ⎠
tan ( θ max) = sinh ⎛⎜
W
d⎞
⎟
⎝ FH L 2 ⎠
L
2
⎛ FH L ⎞ ⎛ W d ⎞
⎟ sinh ⎜
⎟
⎝ W ⎠ ⎝ FH L 2 ⎠
=⎜
F H = 55.90 lb
⎛ L ⎞ ⎛ 55.57 ⎞
⎜ ⎟=⎜
⎟ ft
⎝ h ⎠ ⎝ 10.61 ⎠
θ max = 41.81 deg
Problem 7-117
Determine the distance a between the supports in terms of the beam's length L so that the moment in
the symmetric beam is zero at the beam's center.
755
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This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics
Chapter 7
Solution:
Support Reactions:
ΣMD = 0;
w
2
( L + a) ⎛⎜
By =
a⎞
⎟ − B y ( a) = 0
⎝ 2⎠
w
4
( L + a)
Internal Forces:
ΣMC = 0;
1 L − a ⎞ ⎛ 2a + L ⎞ w
a
+ w⎛⎜
− ( L + a) ⎛⎜ ⎟⎞ = 0
⎟
⎜
⎟
⎟
⎜
⎟
⎝ 2 ⎠⎝ 4 ⎠ 2 ⎝ 2 ⎠⎝ 6 ⎠ 4
⎝ 2⎠
w⎛⎜
a ⎞⎛ a ⎞
2
2
2a + 2a L − L = 0
b =
−2 +
4
12
b = 0.366
a = bL
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws 756
as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
Problem 7-118
Determine the internal normal force, shear force, and moment at point D.
Given:
w = 150
N
m
a = 4m
b = 4m
c = 3m
Solution:
Ax = 1 N
Guesses
Ay = 1 N
F BC = 1 N
Given
Ax − ⎛
b
⎞
⎜ 2 2 ⎟ FBC = 0
⎝ b +c ⎠
Ay − w( 2 a) + ⎛
c
⎞
⎜ 2 2 ⎟ FBC = 0
⎝ b +c ⎠
w2 a a − Ay( 2 a) = 0
⎛ Ax ⎞
⎜
⎟
⎜ Ay ⎟ = Find ( Ax , Ay , FBC)
⎜F ⎟
⎝ BC ⎠
Guesses
ND = 1 N
Given
Ax + ND = 0
⎛ Ax ⎞ ⎛ 800 ⎞
⎜
⎟
⎜ Ay ⎟ = ⎜ 600 ⎟ N
⎜ F ⎟ ⎜⎝ 1000 ⎟⎠
⎝ BC ⎠
VD = 1 N
MD = 1 N⋅ m
Ay − w a − VD = 0
− Ay a + w a⎛⎜
a⎞
⎟ + MD = 0
⎝2⎠
757
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
⎛ ND ⎞
⎜
⎟
⎜ VD ⎟ = Find ( ND , VD , MD)
⎜M ⎟
⎝ D⎠
⎛ ND ⎞ ⎛ −800.00 ⎞
⎜ ⎟=⎜
⎟N
0.00
V
⎝
⎠
D
⎝ ⎠
MD = 1200 N⋅ m
Problem 7-119
The beam is supported by a pin at C and a rod AB. Determine the internal normal force, shear
force, and moment at point D.
Units Used:
kN = 103 N
Given:
F = 4 kN
a = 6m
b = 5m
c = 3m
d = 6m
θ = 60 deg
Solution:
Guesses
F AB = 1 N
Given
ND = 1 N
VD = 1 N
MD = 1 N⋅ m
−F sin ( θ ) ( b + c + d) + ⎡
a
⎤ F ( b + c) = 0
AB
⎢ 2
2⎥
+
(
b
+
c
)
a
⎣
⎦
b+c
⎤ F + F cos ( θ ) = 0
AB
⎢ 2
2⎥
+
(
b
+
c
)
a
⎣
⎦
−ND − ⎡
VD + ⎡
a
⎤ F − F sin ( θ ) = 0
AB
⎢ 2
2⎥
+
(
b
+
c
)
a
⎣
⎦
a
⎡
⎤ F c − F sin ( θ ) ( c + d) − M = 0
AB
D
⎢ 2
2⎥
⎣ a + ( b + c) ⎦
758
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
⎛ FAB ⎞
⎜
⎟
⎜ ND ⎟ = Find ( F , N , V , M )
AB D D
D
⎜ VD ⎟
⎜
⎟
⎝ MD ⎠
⎛ ND ⎞ ⎛ −6.08 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ VD ⎠ ⎝ −2.6 ⎠
MD = −12.99 kN⋅ m
Problem 7-120
Express the shear and moment acting in the pipe as a function of y, where 0 ≤ y ≤ b ft.
Given:
w = 4
lb
ft
a = 2 ft
b = 4 ft
Solution:
ΣF y = 0;
wb − w y − V = 0
V ( y) = w( b − y)
V ( y) = 4
ΣM = 0;
M + w y⎛⎜
lb
ft
( 4 ft − y)
b
a
+ w b⎛⎜ ⎟⎞ + w a⎛⎜ ⎟⎞ − w b y = 0
⎟
⎝ 2⎠
⎝2⎠
⎝ 2⎠
y⎞
M ( y) = w b y −
1
2
M ( y) = 16 lb y − 2
2
w y −
lb
ft
1
2
2
wb −
1
2
wa
2
2
y − 40 lb⋅ ft
Problem 7-121
Determine the normal force, shear force, and moment at points B and C of the beam.
Given:
kN = 103 N
759
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This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics
a = 5m
Chapter 7
F 2 = 6 kN
b = 5m
w1 = 2
kN
m
w2 = 1
kN
m
c = 1m
d = 3m
F 1 = 7.5 kN M = 40 kN⋅ m
Solution:
Guesses
NC = 1 N
VC = 1 N
MC = 1 N⋅ m
Given
−NC = 0
V C − w2 d − F2 = 0
⎛ d⎞ − F d − M = 0
⎟
2
⎝ 2⎠
−MC − w2 d⎜
⎛ NC ⎞
⎜
⎟
V
⎜ C ⎟ = Find ( NC , VC , MC)
⎜M ⎟
⎝ C⎠
⎛ NC ⎞ ⎛ 0.00 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ VC ⎠ ⎝ 9.00 ⎠
Guesses
NB = 1 N
MC = −62.50 kN⋅ m
VB = 1 N
MB = 1 N⋅ m
Given
−NB = 0
V B − w1 b − w2 ( c + d) − F1 − F2 = 0
⎛ b ⎞ − F b − w ( c + d) ⎛ b + c + d ⎞ − F ( b + c + d) − M = 0
⎟
⎜
⎟
1
2
2
2 ⎠
⎝2⎠
⎝
−MB − w1 b⎜
⎛ NB ⎞
⎜ ⎟
⎜ VB ⎟ = Find ( NB , VB , MB)
⎜M ⎟
⎝ B⎠
⎛ NB ⎞ ⎛ 0.00 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ VB ⎠ ⎝ 27.50 ⎠
MB = −184.50 kN⋅ m
Problem 7-122
The chain is suspended between points A and B. If it has a weight weight density w and the
760
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 7
p
p
sag is h, determine the maximum tension in the chain.
g
g
y
Given:
lb
ft
w = 0.5
L = 60 ft
h = 3 ft
Solution:
Form Example 7-15
y=
FH ⎛
⎛ w x ⎞ − 1⎞
cosh ⎜
⎜
⎟ ⎟
w ⎝
⎝ FH ⎠ ⎠
d
⎛ wx ⎞
y = sinh ⎜
⎟
dx
⎝ FH ⎠
Guess
F H = 1 lb
Given
h=
FH ⎛
⎛ w L ⎞ − 1⎞
cosh ⎜
⎜
⎟ ⎟
w ⎝
⎝ FH 2 ⎠ ⎠
θ max = atan ⎜⎛ sinh ⎛⎜
w L ⎞⎞
⎟⎟
⎝ FH 2 ⎠⎠
⎝
F H = Find ( FH)
Tmax =
FH
cos ( θ max)
F H = 75.2 lb
Tmax = 76.7 lb
Problem 7-123
Draw the shear and moment diagrams for
the beam.
Units Used:
3
kN = 10 N
Given:
w = 2
Solution:
kN
m
a = 5m
Guesses
A = 1N C = 1N
b = 5m
M = 50 kN⋅ m
761
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Given
⎛
⎝
w a⎜ b +
Chapter 7
a⎞
⎟ − A( a + b) − M = 0
2⎠
x1 = 0 , 0.01a .. a
x2 = a , 1.01a .. a + b
A + C − wa = 0
V 1 ( x1 ) = ( A − w x1 )
V 2 ( x2 ) = −C
1
kN
⎛
⎝
⎛A⎞
⎜ ⎟ = Find ( A , C)
⎝C⎠
M1 ( x1 ) = ⎜ A x1 − w x1
x1 ⎞
1
⎟
2 ⎠ kN⋅ m
M2 ( x2 ) = ⎡⎣−M + C( a + b − x2 )⎤⎦
1
kN
1
kN⋅ m
Force (kN)
5
V1( x1) 0
V2( x2)
5
10
0
2
4
6
8
10
6
8
10
x1 , x2
Distance (m)
Moment (kN-m)
20
M1( x1)
0
M2( x2)
20
40
60
0
2
4
x1 , x2
Distance (m)
762
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-1
The horizontal force is P. Determine the normal and frictional forces acting on the crate of weight
W. The friction coefficients are μk and μs.
Given:
W = 300 lb
P = 80 lb
μ s = 0.3
μ k = 0.2
θ = 20 deg
Solution:
Assume no slipping:
ΣF x = 0;
P cos ( θ ) − W sin ( θ ) + F c = 0
F c = −P cos ( θ ) + W sin ( θ )
ΣF y = 0;
Check
F c = 27.4 lb
Nc − W cos ( θ ) − P sin ( θ ) = 0
Nc = W cos ( θ ) + P⋅ sin ( θ )
Nc = 309 lb
F cmax = μ s Nc
F cmax = 92.8 lb
F cmax > Fc
Problem 8-2
Determine the magnitude of force P needed to start towing the crate of mass M. Also determine
the location of the resultant normal force acting on the crate, measured from point A.
Given:
M = 40 kg
c = 200 mm
μ s = 0.3
d = 3
a = 400 mm
e = 4
b = 800 mm
763
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
Initial guesses:
NC = 200 N
P = 50 N
Given
ΣF x = 0;
⎛ d ⎞P − μ N = 0
s C
⎜ 2 2⎟
+
e
d
⎝
⎠
ΣF y = 0;
NC − M g +
eP
2
=0
2
d +e
⎛ NC ⎞
⎜ ⎟ = Find ( NC , P)
⎝P ⎠
NC = 280.2 N
ΣMO = 0;
P = 140 N
⎛ e P ⎞⎛ b ⎞ = 0
⎟ − N1 x + ⎜
⎜ ⎟
2
2 ⎟⎝ 2 ⎠
⎝ 2⎠
+
e
d
⎝
⎠
−μ s NC⎛⎜
x =
a⎞
2
2
−1 μ s NC a
d +e −ePb
2
d +e
Thus, the distance from A is
NC
2
A = x+
x = 123.51 mm
2
b
A = 523.51 mm
2
Problem 8-3
Determine the friction force on the crate of mass M, and the resultant normal force and its
position x, measured from point A, if the force is P.
Given:
M = 40 kg
μ s = 0.5
a = 400 mm
μ k = 0.2
b = 800 mm
d = 3
c = 200 mm
e = 4
P = 300 N
Solution:
Initial guesses:
F C = 25 N
NC = 100 N
764
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given
ΣF x = 0;
ΣF y = 0;
P⎛
d
⎞
⎜ 2 2 ⎟ − FC = 0
⎝ d +e ⎠
NC − Mg + P⎛
e
⎞
⎜ 2 2⎟ = 0
⎝ d +e ⎠
⎛ FC ⎞
⎜ ⎟ = Find ( FC , NC)
⎝ NC ⎠
F Cmax = μ s NC
Since FC = 180.00 N >
F Cmax = 76.13 N
ΣMO = 0;
F C = μ k NC
then the crate slips
⎛ FC ⎞ ⎛ 30.5 ⎞
⎜ ⎟=⎜
⎟N
⎝ NC ⎠ ⎝ 152.3 ⎠
−NC x + P⎛
e
⎛ d ⎞
⎞
⎜ 2 2 ⎟ a − P⎜ 2 2 ⎟ c = 0
⎝ d +e ⎠
⎝ d +e ⎠
x = −P⎛
−e a + d c
⎞
⎜
2
2⎟
⎝ NC d + e ⎠
b
= 0.40 m
2
Then the block does not tip.
Since x = 0.39 m
<
x1 = a + x
x1 = 0.79 m
Problem 8-4
The loose-fitting collar is supported by the pipe for which the
coefficient of static friction at the points of contact A and B is μs.
Determine the smallest dimension d so the rod will not slip when
the load P is applied.
Given:
μ s = 0.2
765
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Engineering Mechanics - Statics
Chapter 8
Solution:
P ⎛⎜ L −
d
ΣMA = 0; NB d − P ⎛⎜ L − ⎟⎞ − μ s NB d = 0
2⎠
⎝
ΣMB = 0;
⎛
⎝
NA d + μ s NA d − P ⎜ L +
d⎞
⎟=0
2⎠
NB =
⎝
⎟
2⎠
(1 − μ s) d
P ⎛⎜ L +
NA =
d⎞
⎝
d⎞
⎟
2⎠
(1 + μ s) d
ΣF y = 0;
μ s ( NA + NB) − P = 0
⎛L+ d L− d⎞
P⎜
2
2⎟
μs ⎜
+
⎟=P
d ⎝ 1 + μs 1 − μs ⎠
Thus,
d = 2μ s L
d = kL
k = 2μ s
Problem 8-5
The spool of wire having a mass M rests on the ground at A and against the wall at B.
Determine the force P required to begin pulling the wire horizontally off the spool. The
coefficient of static friction between the spool and its points of contact is μs.
Units Used:
3
kN = 10 N
Given:
M = 150 kg
μ s = 0.25
a = 0.45 m
b = 0.25 m
Solution:
Initial guesses:
P = 100 N
F A = 10 N
NA = 20 N
NB = 30 N
F B = 10 N
766
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given
ΣF y = 0;
NA + FB − M g = 0
ΣF x = 0;
F A − NB + P = 0
ΣMB = 0;
−P b + M g a − NA a + FA a = 0
F A = μ s NA
F B = μ s NB
⎛P ⎞
⎜ ⎟
⎜ FA ⎟
⎜ FB ⎟ = Find ( P , F , F , N , N )
A B A B
⎜ ⎟
⎜ NA ⎟
⎜N ⎟
⎝ B⎠
⎛ FA ⎞ ⎛ 0.28 ⎞
⎜ ⎟ ⎜
⎟
⎜ NA ⎟ = ⎜ 1.12 ⎟ kN
⎜ FB ⎟ ⎜ 0.36 ⎟
⎜ ⎟ ⎜
⎟
⎝ NB ⎠ ⎝ 1.42 ⎠
P = 1.14 kN
Problem 8-6
The spool of wire having a mass M rests on the ground at A and against the wall at B. Determine
the forces acting on the spool at A and B for the given force P. The coefficient of static friction
between the spool and the ground at point A is μs. The wall at B is smooth.
Units Used:
3
kN = 10 N
Given:
P = 800 N
a = 0.45 m
M = 150 kg
b = 0.25 m
μ s = 0.35
Solution:
Assume no slipping
Initial guesses : F A = 10N
NA = 10N
NB = 10N
F Amax = 10N
767
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Engineering Mechanics - Statics
Chapter 8
Given
ΣF x = 0;
F A − NB + P = 0
ΣF y = 0;
NA − M g = 0
ΣM0 = 0;
−P b + F A a = 0
F Amax = μ s NA
⎛ FA ⎞
⎜
⎟
⎛ FA ⎞ ⎛ 444 ⎞
⎜ FAmax ⎟ = Find ( F , F
,
N
,
N
)
⎜
⎟=⎜
⎟N
A Amax A B
⎜ NA ⎟
⎝ FAmax ⎠ ⎝ 515 ⎠
⎜
⎟
⎝ NB ⎠
If FA = 444 N < F Amax = 515 N
then our no-slip assumption is good.
⎛ NA ⎞ ⎛ 1.47 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ FA ⎠ ⎝ 0.44 ⎠
NB = 1.24 kN
Problem 8-7
The crate has a mass M and is subjected to a towing force P acting at an angle θ1 with the
horizontal. If the coefficient of static friction is μs, determine the magnitude of P to just start the
crate moving down the plane.
Given:
M = 350 kg
θ 1 = 20 deg
θ 2 = 10 deg
μ s = 0.5
g = 9.81
m
2
s
Solution:
Initial guesses:
NC = 10N
P = 20N
768
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given
ΣF x = 0;
P cos ( θ 1 + θ 2 ) − μ s NC + M g sin ( θ 2 ) = 0
ΣF y = 0;
NC − M g cos ( θ 2 ) + P sin ( θ 1 + θ 2 ) = 0
⎛ NC ⎞
⎜ ⎟ = Find ( NC , P)
⎝P ⎠
NC = 2891 N
P = 981 N
Problem 8-8
The winch on the truck is used to hoist the garbage bin onto the bed of the truck. If the loaded bin
has weight W and center of gravity at G, determine the force in the cable needed to begin the lift.
The coefficients of static friction at A and B are μΑ and μB respectively. Neglect the height of the
support at A.
Units Used:
3
kip = 10 lb
Given:
W = 8.5 kip
μ A = 0.3
μ B = 0.2
a = 10 ft
b = 12 ft
θ = 30 deg
Solution:
The initial guesses are
T = 1 lb
NB = 1 lb NA = 1 lb
Given
ΣMB = 0;
W b − NA ( a + b) = 0
769
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Engineering Mechanics - Statics
Chapter 8
+
Σ F x = 0;
→
T cos ( θ ) − μ B NB cos ( θ ) − NB sin ( θ ) − μ A NA = 0
+
NA − W + T sin ( θ ) + NB cos ( θ ) − μ B NB sin ( θ ) = 0
↑Σ Fy = 0;
⎛T ⎞
⎜ ⎟
⎜ NA ⎟ = Find ( T , NA , NB)
⎜ NB ⎟
⎝ ⎠
⎛ NA ⎞ ⎛ 4.64 ⎞
⎜ ⎟=⎜
⎟ kip
⎝ NB ⎠ ⎝ 2.65 ⎠
T = 3.67 kip
Problem 8-9
The motorcyclist travels with constant velocity along a straight, horizontal, banked road. If he
aligns his bike so that the tires are perpendicular to the road at A, determine the frictional force
at A. The man has a mass MC and a mass center at GC, and the motorcycle has a mass Mm and
a mass center at Gm. If the coefficient of static friction at A is μA, will the bike slip?
Given:
MC = 60 kg
Mm = 120 kg
μ A = 0.4
θ = 20 deg
g = 9.81
m
2
s
Solution:
Assume no slipping
ΣF y = 0;
NA − ( Mm + MC) g cos ( θ ) = 0
NA = ( Mm + MC) g cos ( θ )
NA = 1659 N
ΣF x = 0;
F A − ( Mm + MC) g sin ( θ )
F A = ( Mm + MC) g sin ( θ )
F A = 604 N
F Amax = μ A NA
770
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Engineering Mechanics - Statics
Chapter 8
Check: If FA = 604 N < F Amax = 664 N
then our no-slip assumption is good.
Problem 8-10
The block brake is used to stop the wheel from rotating when the wheel is subjected to a
couple moment M0 If the coefficient of static friction between the wheel and the block is μs,
determine the smallest force P that should be applied.
Solution:
ΣMC = 0;
P a − N b + μs N c = 0
N=
ΣMO = 0;
Pa
b − μs c
μ s N r − MO = 0
μs P a r
b − μs c
P=
= MO
MO ( b − μ s c)
μs r a
Problem 8-11
The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple
771
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Engineering Mechanics - Statics
Chapter 8
moment M0 If the coefficient of static friction between the wheel and the block is μs , show that the
b
≤ μs
brake is self locking, i. e., P ≤ 0 , provided
c
Solution:
ΣMC = 0;
P a − N b + μs N c = 0
N=
ΣMO = 0;
Pa
b − μs c
μ s N r − MO = 0
μs P a r
b − μs c
P=
= MO
MO ( b − μ s c)
μs r a
P < 0 if ( b − μ s c) < 0 i.e. if
b
< μs
c
Problem 8-12
The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple
moment M0 If the coefficient of static friction between the wheel and the block is μs , determine the
smallest force P that should be applied if the couple moment MO is applied counterclockwise.
772
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
ΣMC = 0;
P a − N b − μs N c = 0
N=
ΣMO = 0;
Pa
b + μs c
−μ s N r + MO = 0
μs P a r
b + μs c
P=
= MO
MO ( b + μ s c)
μs r a
Problem 8-13
The block brake consists of a pin-connected lever and friction block at B. The coefficient of static
friction between the wheel and the lever is μs and a torque M is applied to the wheel. Determine if
the brake can hold the wheel stationary when the force applied to the lever is (a) P 1 (b) P2.
773
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Engineering Mechanics - Statics
Chapter 8
Given:
μ s = 0.3
M = 5 N⋅ m
a = 50 mm
b = 200 mm
c = 400 mm
r = 150 mm
P 1 = 30 N
P 2 = 70 N
Solution:
To hold lever:
ΣMO = 0;
FB r − M = 0
M
r
FB =
Require
F B = 33.333 N
FB
NB =
NB = 111.1 N
μs
Lever,
ΣMA = 0;
P Reqd ( b + c) − NB b − FB a = 0
P Reqd =
NB b + F B a
b+c
P Reqd = 39.8 N
(a) If P 1 = 30.00 N
> PReqd = 39.81 N
then the break will hold the wheel
(b) If P2 = 70.00 N
> P Reqd = 39.81 N
then the break will hold the wheel
Problem 8-14
The block brake consists of a pin-connected lever and friction block at B. The coefficient of static
friction between the wheel and the lever is μs and a torque M is applied to the wheel. Determine if
the brake can hold the wheel stationary when the force applied to the lever is (a) P1 (b) P2.
Assume that the torque M is applied counter-clockwise.
774
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
μ s = 0.3
M = 5 N⋅ m
a = 50 mm
b = 200 mm
c = 400 mm
r = 150 mm
P 1 = 30 N
P 2 = 70 N
Solution:
To hold lever:
ΣMO = 0;
FB r − M = 0
FB =
Require
NB =
M
r
F B = 33.333 N
FB
NB = 111.1 N
μs
Lever,
ΣMA = 0;
P Reqd ( b + c) − NB b + FB a = 0
P Reqd =
NB b − F B a
b+c
P Reqd = 34.3 N
(a) If P 1 = 30.00 N
> PReqd = 34.26 N
then the break will hold the wheel
(b) If P2 = 70.00 N
> P Reqd = 34.26 N
then the break will hold the wheel
775
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-15
The doorstop of negligible weight is pin connected
at A and the coefficient of static friction at B is μs.
Determine the required distance s from A to the
floor so that the stop will resist opening of the
door for any force P applied to the handle.
Given:
μ s = 0.3
a = 1.5 in
Solution:
ΣF y = 0; NB − ⎛
⎜
⎞F =0
A
⎝ s +a ⎠
s
2⎟
2
ΣF x = 0; μ s NB − ⎛
⎜
⎞F =0
A
⎝ s +a ⎠
a
2
2⎟
⎛⎜ μ s s ⎟⎞
⎛ a ⎞F = 0
FA − ⎜
A
2
2⎟
⎜ s2 + a2 ⎟
⎝
⎠
⎝ s +a ⎠
μs s = a
s =
a
μs
s = 5.00 in
Problem 8-16
The chair has a weight W and center of gravity at G. It is propped against the door as shown.
If the coefficient of static friction at A is μA, determine the smallest force P that must be
applied to the handle to open the door.
776
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Engineering Mechanics - Statics
Chapter 8
Given:
μ A = 0.3
a = 1.20 ft
b = 0.75 ft
c = 3ft
θ = 30 deg
W = 10 lb
Solution:
B y = 1 lb
Guesses
NA = 1 lb
P = 1 lb
Given
ΣF x = 0;
−P + μ A NA = 0
ΣF y = 0;
NA − W − By = 0
ΣΜΒ = 0; μ A NA c cos ( θ ) − NA c sin ( θ ) + W ⎡⎣( c − a)sin ( θ ) + b cos ( θ )⎤⎦ = 0
⎛ By ⎞
⎜ ⎟
⎜ NA ⎟ = Find ( By , NA , P)
⎜P ⎟
⎝ ⎠
B y = 11.5 lb
NA = 21.5 lb
P = 6.45 lb
Problem 8-17
The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is
slowly applied at B. If the hoop begins to slip at A when the angle is θ , determine the
coefficient of static friction between the hoop and the peg.
777
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
θ = 30 deg
Solution:
ΣF x = 0;
μ NA cos ( θ ) + P − NA sin ( θ ) = 0
P = ( μ cos ( θ ) − sin ( θ ) ) NA
ΣF y = 0;
μ NA sin ( θ ) − W + NA cos ( θ ) = 0
W = ( μ sin ( θ ) + cos ( θ ) ) NA
ΣΜΑ = 0; −W r sin ( θ ) + P ( r + r cos ( θ ) ) = 0
W sin ( θ ) = P( 1 + cos ( θ ) )
( μ sin ( θ ) + cos ( θ )) sin ( θ ) = ( sin ( θ ) − μ cos ( θ )) (1 + cos (θ ) )
μ =
sin ( θ )
1 + cos ( θ )
μ = 0.27
Problem 8-18
The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is
slowly applied at B. If the coefficient of static friction between the hoop and peg is μs,
determine if it is possible for the hoop to reach an angle θ before the hoop begins to slip.
778
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Engineering Mechanics - Statics
Chapter 8
Given:
μ s = 0.2
θ = 30 deg
Solution:
ΣF x = 0;
μ NA cos ( θ ) + P − NA sin ( θ ) = 0
P = ( μ cos ( θ ) − sin ( θ ) ) NA
ΣF y = 0;
μ NA sin ( θ ) − W + NA cos ( θ ) = 0
W = ( μ sin ( θ ) + cos ( θ ) ) NA
ΣΜΑ = 0; −W r sin ( θ ) + P ( r + r cos ( θ ) ) = 0
W sin ( θ ) = P ( 1 + cos ( θ ) )
( μ sin ( θ ) + cos ( θ )) sin ( θ ) = ( sin ( θ ) − μ cos ( θ )) (1 + cos (θ ) )
μ =
sin ( θ )
1 + cos ( θ )
μ = 0.27
If μ s = 0.20 < μ = 0.27 then it is not possible to reach θ = 30.00 deg.
Problem 8-19
The coefficient of static friction between the shoes at A and B of the tongs and the pallet is μs1 and
between the pallet and the floor μs2. If a horizontal towing force P is applied to the tongs,
determine the largest mass that can be towed.
779
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Engineering Mechanics - Statics
Chapter 8
Given:
μ s1 = 0.5
a = 75 mm
μ s2 = 0.4
b = 20 mm
P = 300 N
c = 30 mm
m
θ = 60 deg
g = 9.81
2
s
Solution:
Assume that we are on the verge of
slipping at every surface.
Guesses
T = 1N
NA = 1N
F = 1N
Nground = 1N
F A = 1N
mass = 1kg
Given
2 T sin ( θ ) − P = 0
−T sin ( θ ) ( b + c) − T cos ( θ ) a − FA b + NA a = 0
F A = μ s1 NA
2 FA − F = 0
Nground − mass g = 0
F = μ s2 Nground
⎛ T ⎞
⎜
⎟
⎜ NA ⎟
⎜ FA ⎟
⎜
⎟ = Find ( T , NA , FA , F , Nground , mass)
F
⎜
⎟
⎜ Nground ⎟
⎜
⎟
⎝ mass ⎠
780
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ T ⎞ ⎛ 173.21 ⎞
⎜
⎟ ⎜
⎟
N
A
⎜
⎟ ⎜ 215.31 ⎟
⎜ FA ⎟ = ⎜ 107.66 ⎟ N
⎜
⎟ ⎜
⎟
⎜ F ⎟ ⎜ 215.31 ⎟
⎜ Nground ⎟ ⎝ 538.28 ⎠
⎝
⎠
mass = 54.9 kg
Problem *8-20
The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is μs, determine the
smallest dimension b so that any pipe of inner diameter d can be lifted.
Solution:
W − 2 FB = 0
⎛ W⎞ b − N h − F ⎛ d⎞ = 0
⎜ ⎟
B
B⎜ ⎟
⎝2⎠
⎝ 2⎠
Thus
FB =
NB =
W
2
W ( 2 b − d)
4h
Require
F B ≤ μ s NB
W
2
≤
μ s W ( 2b − d)
2 h ≤ μ s ( 2b − d)
4h
b>
h
μs
+
d
2
Problem 8-21
A very thin bookmark having a width a. is in the middle of a dictionary of weight W. If the
pages are b by c, determine the force P needed to start to pull the bookmark out.The coefficient
of static friction between the bookmark and the paper is μs. Assume the pressure on each page
and the bookmark is uniform.
781
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Engineering Mechanics - Statics
Chapter 8
Given:
a = 1 in
W = 10 lb
b = 8 in
c = 10 in
μ s = 0.7
Solution:
Pressure on book mark :
P =
1 W
2 bc
P = 0.06
in
Normal force on bookmark:
F = μs N
ΣF x = 0;
lb
2
N = Pca
F = 0.44 lb
P − 2F = 0
P = 2F
P = 0.88 lb
Problem 8-22
The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction
is μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force F
needed to move the dresser. Also, if the man has a weight Wman,, determine the smallest
coefficient of static friction between his shoes and the floor so that he does not slip.
Given:
W = 90 lb
μ s = 0.25
782
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Wman = 150 lb
θ = 0 deg
Solution:
Dresser:
ND = 1lb
Guesses
F = 1lb
Given
+
↑Σ Fy = 0;
ND − W − F sin ( θ ) = 0
+
Σ F x = 0;
→
F cos ( θ ) − μ s ND = 0
⎛ ND ⎞
⎜ ⎟ = Find ( ND , F)
⎝F ⎠
Man:
F = 22.50 lb
Nm = 1lb
Guesses
μ m = 0.2
Given
+
Nm − Wman + F sin ( θ ) = 0
+
Σ F x = 0;
→
−F cos ( θ ) + μ m Nm = 0
↑Σ Fy = 0;
⎛ Nm ⎞
⎜ ⎟ = Find ( Nm , μ m)
⎝ μm ⎠
μ m = 0.15
Problem 8-23
The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction
is μs. If the man pushes on it in the direction θ, determine the smallest magnitude of force F
needed to move the dresser. Also, if the man has a weight Wman, determine the smallest
coefficient of static friction between his shoes and the floor so that he does not slip.
Given:
W = 90 lb
783
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
μ s = 0.25
Wman = 150 lb
θ = 30 deg
Solution:
Dresser:
ND = 1lb
Guesses
F = 1lb
Given
+
↑Σ Fy = 0;
ND − W − F sin ( θ ) = 0
+
Σ F x = 0;
→
F cos ( θ ) − μ s ND = 0
⎛ ND ⎞
⎜ ⎟ = Find ( ND , F)
⎝F ⎠
Man:
F = 30.36 lb
Nm = 1lb
Guesses
μ m = 0.2
Given
+
Nm − Wman + F sin ( θ ) = 0
+
Σ F x = 0;
→
−F cos ( θ ) + μ m Nm = 0
↑Σ Fy = 0;
⎛ Nm ⎞
⎜ ⎟ = Find ( Nm , μ m)
⎝ μm ⎠
μ m = 0.195
Problem 8-24
The cam is subjected to a couple moment of M. Determine the minimum force P that should be
applied to the follower in order to hold the cam in the position shown.The coefficient of static
friction between the cam and the follower is μs. The guide at A is smooth.
784
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
a = 10 mm
b = 60 mm
M = 5 N⋅ m
μ s = 0.4
Solution:
ΣM0 = 0;
M − μ s NB b − a NB = 0
NB =
M
μs b + a
NB = 147.06 N
Follower:
ΣF y = 0;
NB − P = 0
P = NB
P = 147 N
Problem 8-25
The board can be adjusted vertically by tilting it up and sliding the smooth pin A along the vertical
guide G. When placed horizontally, the bottom C then bears along the edge of the guide, where the
coefficient of friction is μs. Determine the largest dimension d which will support any applied
force F without causing the board to slip downward.
Given:
μ s = 0.4
a = 0.75 in
b = 6 in
785
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
+
↑Σ Fy = 0;
μ s NC − F = 0
ΣMA = 0;
−F b + d NC − μ s NC a = 0
Solving we find
−μ s b + d − μ s a = 0
d = μ s ( a + b)
d = 2.70 in
Problem 8-26
The homogeneous semicylinder has a mass m and mass center at G. Determine the largest angle θ
of the inclined plane upon which it rests so that it does not slip down the plane. The coefficient of
static friction between the plane and the cylinder is μs. Also, what is the angle φ for this case?
Given:
μ s = 0.3
Solution:
The semicylinder is a two-force member:
Since
F=μ N
tan ( θ ) =
μs N
N
= μS
θ = atan ( μ s)
786
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Engineering Mechanics - Statics
Chapter 8
θ = 16.7 deg
Law of sines
4r
r
sin ( 180 deg − φ )
=
3π
sin ( θ )
⎛ 3π sin ( θ )⎞
⎟
⎝4
⎠
φ = asin ⎜
φ = 42.6 deg
Problem 8-27
A chain having a length L and weight W rests on a street for which the coefficient of static
friction is μs. If a crane is used to hoist the chain, determine the force P it applies to the chain
if the length of chain remaining on the ground begins to slip when the horizontal component is
P x. What length of chain remains on the ground?
Given:
L = 20 ft
W = 8
lb
ft
μ s = 0.2
P x = 10 lb
Solution:
ΣF x = 0;
−P x + μ s Nc = 0
Nc =
Px
μs
Nc = 50.00 lb
ΣF y = 0;
P y − W L + Nc = 0
787
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Engineering Mechanics - Statics
Chapter 8
P y = W L − Nc
P y = 110.00 lb
2
P =
Px + Py
2
P = 110 lb
The length on the ground is supported by
L =
Nc = 50.00 lbthus
Nc
W
L = 6.25 ft
Problem 8-28
The fork lift has a weight W1 and center of gravity at G. If the rear wheels are powered,
whereas the front wheels are free to roll, determine the maximum number of crates, each of
weight W2 that the fork lift can push forward. The coefficient of static friction between the
wheels and the ground is μs and between each crate and the ground is μ's.
Given:
W1 = 2400 lb
W2 = 300 lb
μ s = 0.4
μ's = 0.35
a = 2.5 ft
b = 1.25 ft
c = 3.50 ft
Solution:
Fork lift:
ΣMB = 0;
W1 c − NA ( b + c) = 0
NA = W1 ⎛⎜
c
⎞
⎟
⎝ b + c⎠
ΣF x = 0;
NA = 1768.4 lb
μ s NA − P = 0
788
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
P = μ s NA
Chapter 8
P = 707.37 lb
Crate:
Nc − W2 = 0
ΣF y = 0;
Nc = W2
ΣF x = 0;
Nc = 300.00 lb
P' − μ's Nc = 0
P' = μ's Nc
Thus
n =
P
P'
P' = 105.00 lb
n = 6.74
n = floor ( n)
n = 6.00
Problem 8-29
The brake is to be designed to be self locking, that is, it will not rotate when no load P is applied
to it when the disk is subjected to a clockwise couple moment MO. Determine the distance d of
the lever that will allow this to happen. The coefficient of static friction at B is μs.
Given:
a = 1.5 ft
b = 1 ft
μ s = 0.5
Solution:
ΣM0 = 0;
M0 − μ s NB b = 0
NB =
ΣMA = 0;
M0
μs b
P 2 a − NB a + μ s NB d = 0
789
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Engineering Mechanics - Statics
Chapter 8
P=0
d =
a
μs
d = 3.00 ft
Problem 8-30
The concrete pipe of weight W is being lowered from the truck bed when it is in the position
shown. If the coefficient of static friction at the points of support A and B is μs determine
where it begins to slip first: at A or B, or both at A and B.
Given:
W = 800 lb
a = 30 in
μ s = 0.4
b = 18 in
θ = 30 deg
c = 5 in
r = 15 in
Solution:
initial guesses are
NA = 10 lb
Given
NB = 10 lb
F A = 10 lb
F B = 10 lb
Assume slipping at A:
ΣF x = 0;
NA + FB − W sin ( θ ) = 0
ΣF y = 0;
F A + NB − W cos ( θ ) = 0
ΣM0= 0;
FB r − FA r = 0
F A = μ s NA
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( N , N , F , F )
A B A B
⎜ FA ⎟
⎜ ⎟
⎝ FB ⎠
⎛ NA ⎞ ⎛ 285.71 ⎞
⎜ ⎟ ⎜
⎟
⎜ NB ⎟ = ⎜ 578.53 ⎟ lb
⎜ FA ⎟ ⎜ 114.29 ⎟
⎜ ⎟ ⎜
⎟
⎝ FB ⎠ ⎝ 114.29 ⎠
790
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Engineering Mechanics - Statics
Chapter 8
At B,
F Bmax = μ s NB
Since F B = 114.29 lb < F Bmax = 231.41 lb then we conclude that slipping begins at A.
Problem 8-31
A wedge of mass M is placed in the grooved slot of an inclined plane. Determine the maximum
angle θ for the incline without causing the wedge to slip. The coefficient of static friction
between the wedge and the surfaces of contact is μs.
Given:
M = 5 kg
μ s = 0.2
φ = 60 deg
g = 9.81
m
2
s
Solution:
Initial guesses:
NW = 10 N
θ = 10 deg
Given
ΣF x = 0;
M g sin ( θ ) − 2 μ s NW = 0
ΣF z = 0;
2 NW sin ⎜
⎛ φ ⎞ − M g cos ( θ ) = 0
⎟
⎝2⎠
Solving,
⎛ NW ⎞
⎜ ⎟ = Find ( NW , θ )
⎝ θ ⎠
NW = 45.5 N
θ = 21.8 deg
791
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Engineering Mechanics - Statics
Chapter 8
Problem 8-32
A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests
against the wall. If the hanger has a negligible weight and the bearing at O can be considered
frictionless, determine the force P needed to start turning the roll. The coefficient of static
friction between the wall and the paper is μs.
Given:
W = 0.75 lb
θ = 30 deg
φ = 30 deg
μ s = 0.25
a = 3 in
Solution:
Initial guesses:
R = 100 lb
NA = 100 lb
P = 100 lb
Given
ΣF x = 0;
NA − R sin ( φ ) + P sin ( θ ) = 0
ΣF y = 0;
R cos ( φ ) − W − P cos ( θ ) − μ s NA = 0
ΣM0 = 0;
μ s NA a − P a = 0
Solving for P,
⎛⎜ R ⎞⎟
⎜ NA ⎟ = Find ( R , NA , P)
⎜P ⎟
⎝ ⎠
R = 1.14 lb
NA = 0.51 lb
P = 0.13 lb
Problem 8-33
A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests
against the wall. If the hanger has a negligible weight and the bearing at O can be considered
frictionless, determine the minimum force P and the associated angle θ needed to start turning
the roll. The coefficient of static friction between the wall and the paper is μs.
792
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Engineering Mechanics - Statics
Chapter 8
Given:
W = 0.75 lb
φ = 30 deg
μ s = 0.25
r = 3 in
Solution:
ΣF x = 0;
NA − R sin ( φ ) + P sin ( θ ) = 0
ΣF y = 0;
R cos ( φ ) − W − P cos ( θ ) − μ s NA = 0
ΣM0 = 0;
μ s NA r − P r = 0
Solving for P,
P=
μ s W sin ( φ )
cos ( φ ) + μ sin ( θ − φ ) − μ sin ( φ )
For minimum P we must have
dP
dθ
−μ s W sin ( φ ) cos ( θ − φ )
2
=
(cos (φ ) + μ s sin (θ − φ ) − μ s sin (φ ))
One answer is
P =
2
θ = φ + 90 deg
=0
Implies
cos ( θ − φ ) = 0
θ = 120.00 deg
μ s W sin ( φ )
P = 0.0946 lb
cos ( φ ) + μ s sin ( θ − φ ) − μ s sin ( φ )
Problem 8-34
The door brace AB is to be designed to prevent opening the door. If the brace forms a pin
connection under the doorknob and the coefficient of static friction with the floor is μs determine
the largest length L the brace can have to prevent the door from being opened. Neglect the weight
of the brace.
Given:
μ s = 0.5
793
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Engineering Mechanics - Statics
Chapter 8
a = 3 ft
Solution:
The brace is a two-force member.
μs N
2
=
L −a
a
μs a =
L −a
L = a
1 + μs
N
2
2
2
2
L = 3.35 ft
Problem 8-35
The man has a weight W, and the coefficient of static friction between his shoes and the floor is
μs. Determine where he should position his center of gravity G at d in order to exert the maximum
horizontal force on the door. What is this force?
Given:
W = 200 lb
μ s = 0.5
h = 3 ft
Solution:
N−W = 0
N = W
F max = μ s N
+
Σ F x = 0;
→
F max = 100 lb
P − Fmax = 0
P = F max
ΣMO = 0;
N = 200.00 lb
P = 100 lb
W d−P h= 0 d = P
h
W
d = 1.50 ft
794
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-36
In an effort to move the two crates, each of weight W, which are stacked on top of one
another, the man pushes horizontally on them at the bottom of crate A as shown. Determine
the smallest force P that must be applied in order to cause impending motion. Explain what
happens. The coefficient of static friction between the crates is μs and between the bottom
crate and the floor is μs'.
Given:
W = 100 lb
μ s = 0.8
μ's = 0.3
a = 2 ft
b = 3 ft
Solution:
Assume crate A slips:
ΣF y = 0;
NA − W = 0
NA = W
NA = 100.00 lb
ΣF x = 0;
P − μ s NA = 0
P 1 = μ s NA
P 1 = 80.00 lb
Assume crate B slips:
ΣF y = 0;
NB − 2 W = 0
NB = 2 W
NB = 200.00 lb
ΣF x = 0;
P − μ's NB = 0
P 2 = μ's NB
P 2 = 60.00 lb
⎛ a⎞
⎟
⎝ b⎠
P 3 = 66.7 lb
Assume both crates A and B tip:
ΣM = 0;
⎛ a⎞ − P b = 0
⎟
⎝ 2⎠
2W⎜
P = min ( P 1 , P 2 , P 3 )
P3 = W ⎜
P = 60.00 lb
Problem 8-37
The man having a weight of W1 pushes horizontally on the bottom of crate A, which is stacked
on top of crate B. Each crate has a weight W2. If the coefficient of static friction between each
crate is μs and between the bottom crate, his shoes, and the floor is μ's, determine if he can
cause impending motion.
795
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
W1 = 150 lb
W2 = 100 lb
a = 2 ft
b = 3 ft
μ s = 0.8
μ's = 0.3
Assume crate A slips:
ΣF y = 0;
NA − W2 = 0
NA = W2
NA = 100.00 lb
ΣF x = 0;
P − μ s NA = 0
P 1 = μ s NA
P 1 = 80.00 lb
Assume crate B slips:
ΣF y = 0;
NB − 2 W2 = 0
NB = 2 W2
NB = 200.00 lb
ΣF x = 0;
P − μ's NB = 0
P 2 = μ's NB
P 2 = 60.00 lb
a
b
P 3 = 66.7 lb
Assume both crates A and B tip:
ΣM = 0;
⎛ a⎞ − P b = 0
⎟
⎝ 2⎠
2 W2 ⎜
P min = min ( P 1 , P 2 , P 3 )
P 3 = W2
P min = 60.00 lb
Now check to see if he can create this force
ΣF y = 0;
Nm − W1 = 0
Nm = W1
ΣF x = 0;
F m − P min = 0
F m = Pmin
F mmax = μ's Nm
Since Fm = 60.00 lb >
F mmax = 45.00 lb then the man cannot
create the motion.
796
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Engineering Mechanics - Statics
Chapter 8
Problem 8-38
The crate has a weight W and a center of gravity at G. Determine the horizontal force P required
to tow it.Also, determine the location of the resultant normal force measured from A.
Given:
a = 3.5 ft
b = 3 ft
c = 2 ft
W = 200 lb
h = 4 ft
μ s = 0.4
Solution:
ΣF x = 0;
P = FO
ΣF y = 0;
NO = W
NO = 200.00 lb
ΣMo = 0;
−P h + W x = 0
F O = μ s NO
F O = 80.00 lb
P = FO
P = 80.00 lb
x = P
h
W
x = 1.60 ft
The distance of NO from A is
c − x = 0.40 ft
797
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-39
The crate has a weight W and a center of gravity at G. Determine the height h of the tow rope
so that the crate slips and tips at the same time. What horizontal force P is required to do this?
Given:
a = 3.5 ft
b = 3 ft
c = 2 ft
W = 200 lb
h = 4 ft
μ s = 0.4
Solution:
ΣF y = 0;
NA = W
NA = 200.00 lb
ΣF x = 0;
P = FA
F s = μsN;
FA = μ s W
F A = 80.00 lb
ΣMA = 0;
P = 80 lb
−P h + W c = 0
h = W
c
P
h = 5.00 ft
Problem 8-40
Determine the smallest force the man must exert on the rope in order to move the crate of mass M.
Also, what is the angle θ at this moment? The coefficient of static friction between the crate and the
floor is μs.
Given:
M = 80 kg
798
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
μ s = 0.3
α = 30 deg
β = 45 deg
m
g = 9.81
2
s
Solution :
The initial guesses are
T1 = 1 N
NC = 1 N
θ = 30 deg
T = 1N
Given
NC − M g + T1 cos ( θ ) = 0
μ s NC − T1 sin ( θ ) = 0
T cos ( β ) − T cos ( α ) + T1 sin ( θ ) = 0
T sin ( β ) + T sin ( α ) − T1 cos ( θ ) = 0
⎛ T ⎞
⎜T ⎟
⎜ 1 ⎟ = Find ( T , T , N , θ )
1 C
⎜ NC ⎟
⎜ ⎟
⎝ θ ⎠
T = 451.86 N
θ = 7.50 deg
Problem 8-41
The symmetrical crab hook is used to lift packages by means of friction developed between
the shoes Aand B and a package. Determine the smallest coefficient of static friction at the
shoes so that the package of weight W can be lifted.
Given:
a = 1 ft
b = 2 ft
c = 0.8 ft
799
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
d = 1 ft
θ = 45 deg
Solution:
From FBD (a)
ΣF y = 0;
W = 2 F sin ( θ )
W
F=
2 sin ( θ )
From FBD (b)
ΣMD = 0; μ NB d + NB a − F sin ( θ ) c − F cos ( θ ) b = 0
μ NB d + NB a −
NB =
W
2 ( μ d + a)
W
2 sin ( θ )
sin ( θ ) c −
W
2 sin ( θ )
cos ( θ ) b = 0
(c + cot ( θ ) b)
From FBD (c)
ΣF y = 0; 2 μ NB − W = 0
⎡ W
( c + cot (θ ) b)⎥⎤ − W = 0
⎣ 2( μ d + a)
⎦
2μ ⎢
μ =
a
c + b cot ( θ ) − d
μ = 0.56
Problem 8-42
The friction hook is made from a fixed frame which is shown colored and a cylinder of negligible
weight. A piece of paper is placed between the smooth wall and the cylinder. D etermine the
smallest coefficient of static friction μ at all points of contact so that any weight W of paper p
800
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
μ
p
y
g
p p
p
can be held.
Given:
θ = 20 deg
Solution:
Paper:
+
↑ Σ Fy = 0;
F+F−W=0
F=
W
2
F= μ N
N=
W
2μ
Cylinder:
⎛ W⎞ r = 0
⎟
⎝2⎠
F' r − ⎜
⎛ W ⎞ sin ( θ ) − W cos ( θ ) = 0
⎟
2μ
⎝2⎠
N' − ⎜
F' = μ N'
W
2
F' =
W
2
N' =
1
W ⎛
⎞
sin ( θ ) + cos ( θ )⎟
⎜
2 ⎝
μ
⎠
⎛ W ⎞ ⎛ sin ( θ ) + 1 cos ( θ )⎞
⎟⎜
⎟
μ
⎝ 2 ⎠⎝
⎠
=μ ⎜
1 = μ sin ( θ ) + cos ( θ )
μ =
1 − cos ( θ )
sin ( θ )
μ = 0.176
Problem 8-43
The crate has a weight W1 and a center of gravity at G. If the coefficient of static friction
between the crate and the floor is μs, determine if the man of weight W2 can push the crate to
the left. The coefficient of static friction between his shoes and the floor is μ's. Assume the
man exerts only a horizontal force on the crate.
801
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
W1 = 300 lb
W2 = 200 lb
μ s = 0.2
μ's = 0.35
a = 4.5 ft
c = 3 ft
b = 3.5 ft
d = 4.5 ft
Solution:
ΣF y = 0;
NC − W1 = 0
NC = W1
ΣF x = 0;
μ s NC − P = 0
P = μ s NC
ΣΜO = 0;
−W1 x + P d = 0
x =
Pd
W1
Since x = 0.90 ft < a = 4.50 ft
there will not be any tipping.
ΣF y = 0;
Nm − W2 = 0
Nm = W2
Nm = 200.00 lb
ΣF x = 0;
P − Fm = 0
Fm = P
F m = 60.00 lb
F mmax = μ's Nm
F mmax = 70.00 lb
Since F m = 60.00 lb < F mmax = 70.00 lb then the
man can push the crate.
Problem 8-44
The crate has a weight W1 and a center of gravity at G. If the coefficient of static friction
between the crate and the floor is μs, determine the smallest weight of the man so that he can
push the crate to the left. The coefficient of static friction between his shoes and the floor is
μ's. Assume the man exerts only a horizontal force on the crate.
802
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
W1 = 300 lb
W2 = 200 lb
μ s = 0.2
μ's = 0.35
a = 4.5 ft
c = 3 ft
b = 3.5 ft
d = 4.5 ft
Solution:
ΣF y = 0;
NC − W1 = 0
NC = W1
ΣF x = 0;
μ s NC − P = 0
P = μ s NC
ΣΜO = 0;
−W1 x + P d = 0
x =
Pd
W1
Since x = 0.90 ft < a = 4.50 ft
there will not be any tipping.
ΣF x = 0;
ΣF y = 0;
P − Fm = 0
Fm = P
F m = μ's Nm
Nm =
Nm − W2 = 0
W2 = Nm
F m = 60.00 lb
Fm
Nm = 171.4 lb
μ's
W2 = 171.4 lb
Problem 8-45
The wheel has weight WA and rests on a surface for which the coefficient of friction is μB. A
cord wrapped around the wheel is attached to the top of the homogeneous block of weight WC.
803
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
If the coefficient of static friction at D is μD determine the smallest vertical force that can be
applied tangentially to the wheel which will cause motion to impend.
Given:
WA = 20 lb
μ B = 0.2
WC = 30 lb
μ D = 0.3
h = 3 ft
b = 1.5 ft
Solution:
Assume that slipping occurs at B, but that the block does not move.
Guesses
P = 1 lb
NB = 1 lb
F B = 1 lb
T = 1 lb
ND = 1 lb
F D = 1 lb
x = 1 ft
Given
NB − WA − P = 0
T − FB = 0
( P − T − F B) 2 = 0
F B = μ B NB
−T + F D = 0
ND − WC = 0
h
T h − ND x = 0
⎛P ⎞
⎜ ⎟
⎜ NB ⎟
⎜ FB ⎟
⎜ ⎟
⎜ T ⎟ = Find ( P , NB , FB , T , ND , FD , x)
⎜ ND ⎟
⎜ ⎟
⎜ FD ⎟
⎜ x ⎟
⎝ ⎠
Now checke the assumptions
⎛ P ⎞ ⎛ 13.33 ⎞
⎜N ⎟ ⎜
⎟
⎜ B ⎟ ⎜ 33.33 ⎟
⎜ FB ⎟ ⎜ 6.67 ⎟
⎜ ⎟=⎜
⎟ lb
6.67
T
⎜ ⎟ ⎜
⎟
⎜ ND ⎟ ⎜ 30.00 ⎟
⎜ ⎟ ⎜ 6.67 ⎟
⎠
⎝ FD ⎠ ⎝
x = 0.67 ft
F Dmax = μ D ND
Since F D = 6.67 lb < FDmax = 9.00 lb then the block does not slip
804
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Engineering Mechanics - Statics
Since x = 0.67 ft <
Chapter 8
b
= 0.75 ft then the block does not tip.
2
So our original assumption is correct.
P = 13.33 lb
Problem 8-46
Determine the smallest couple moment which can be applied to the wheel of weight W1 that will
cause impending motion. The cord is attached to the block of weight W2, and the coefficients of
static friction are μB and μD.
Given:
W1 = 20 lb
a = 1.5 ft
W2 = 30 lb
b = 3 ft
μ B = 0.2
c = 1.5 ft
μ D = 0.3
Solution:
For the wheel : Assume slipping occurs,
ΣF y = 0;
NB − W1 = 0
NB = W1
NB = 20.00 lb
ΣF x = 0;
T − μ B NB = 0
T = μ B NB
T = 4.00 lb
ΣMB = 0;
M − T2a = 0
M = T2a
M = 12.00 lb⋅ ft
ΣF y = 0;
ND − W2 = 0
ND = W2
ND = 30.00 lb
ΣF x = 0;
FD − T = 0
FD = T
F D = 4.00 lb
ΣMO = 0;
T b − ND x = 0
x = T
For block
b
x = 0.40 ft
ND
F Dmax = μ D ND
F Dmax = 9.00 lb
Since F D = 4.00 lb < FDmax = 9.00 lb then the block doesn't slip
805
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Engineering Mechanics - Statics
Since x = 0.40 ft <
Chapter 8
c
= 0.75 ft then the block doesn't tip.
2
Thus neither slipping nor tipping occurs for the block, and our assumption and answer are
correct.
Problem 8-47
The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading.
It is supported at one end by a pin and at the other end by a post having a mass mp and negligible
thickness. Determine the minimum force P needed to move the post. The coefficients of static
friction at B and C are μB and μC respectively.
Given:
mp = 50 kg
a = 2m
μ B = 0.4
b = 400 mm
μ C = 0.2
c = 300 mm
w = 800
N
g = 9.81
m
m
2
d = 3
e = 4
s
Solution:
Member AB:
ΣMA = 0;
1
2a
−⎛⎜ w a⎟⎞ ⎛⎜ ⎞⎟ + NB a = 0
⎝ 2 ⎠⎝ 3 ⎠
NB =
1
3
wa
NB = 533.33 N
806
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Engineering Mechanics - Statics
Chapter 8
Post:
Assume slipping occurs at C:
F C = μ C NC
The initial guesses are
P = 1N
NC = 1 N
FB = 1 N
Given
−e
2
P c + FB( b + c) = 0
2
d +e
e
2
2
P − FB − μ C NC = 0
d +e
d
2
e +d
2
P + NC − NB − mp g = 0
⎛P ⎞
⎜ ⎟
⎜ NC ⎟ = Find ( P , NC , FB)
⎜ FB ⎟
⎝ ⎠
P = 354.79 N
Now check to see if the post slips at B.
Since F B = 122 N
F Bmax = μ B NB
< F Bmax = 213 N
then our assumptions are correct
P = 355 N
Problem 8-48
The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading.
It is supported at one end by a pin and at the other end by a post having a mass mp and negligible
thickness. Determine the two coefficients of static friction at B and at C so that when the
magnitude of the applied force is increased to P the post slips at both B and C simultaneously.
807
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Engineering Mechanics - Statics
Chapter 8
Given:
mp = 50 kg
P = 150 N
N
w = 800
m
a = 2m
b = 400 mm
c = 300 mm
d = 3
e = 4
Solution:
Member AB:
1
2a
− wa
+ NB a = 0
2
3
NB =
1
3
wa
NB = 533.33 N
Post:
Guesses
NC = 1 N
μ B = 0.2
μ C = 0.2
Given
NC − NB + P⎛
d
⎞
⎜ 2 2 ⎟ − mp g = 0
⎝ d +e ⎠
e
⎛
⎞
⎜ 2 2 ⎟ P − μ C NC − μ B NB = 0
⎝ d +e ⎠
⎛ −e ⎞ P c + μ N ( b + c) = 0
B B
⎜ 2 2⎟
⎝ d +e ⎠
⎛ NC ⎞
⎜ ⎟
⎜ μ B ⎟ = Find ( NC , μ B , μ C)
⎜μ ⎟
⎝ C⎠
⎛ μ B ⎞ ⎛ 0.0964 ⎞
⎜ ⎟=⎜
⎟
⎝ μ C ⎠ ⎝ 0.0734 ⎠
808
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Engineering Mechanics - Statics
Chapter 8
Problem 8-49
The block of weight W is being pulled up the inclined plane of slope α using a force P . If P acts at
the angle φ as shown, show that for slipping to occur, P = W sin( α + θ)/ cos(φ − θ) where θ is the
angle of friction; θ = tan-1 μ..
Solution:
Let
μ = tan ( θ )
Σ F x = 0;
P cos ( φ ) − W sin ( α ) − μ N = 0
Σ F y = 0;
N − W cos ( α ) + P sin ( φ ) = 0
P cos ( φ ) − W sin ( α ) − μ ( W cos ( α ) − P sin ( φ ) = 0
⎛ sin ( α ) + μ cos ( α ) ⎞ = W⎛ sin ( α ) + tan ( θ ) cos ( α ) ⎞
⎟
⎜
⎟
⎝ cos ( φ ) + μ sin ( φ ) ⎠
⎝ cos ( φ ) + tan ( θ ) sin ( φ ) ⎠
P = W⎜
⎛ sin ( α ) cos ( θ ) + sin ( θ ) cos ( α ) ⎞
⎟
⎝ cos ( φ ) cos ( θ ) + sin ( θ ) sin ( φ ) ⎠
P = W⎜
⎛ sin ( α + θ ) ⎞
⎟
⎝ cos ( φ − θ ) ⎠
P = W⎜
( QED)
Problem 8-50
Determine the angle φ at which P should act on the block so that the magnitude of P is as small as
possible to begin pulling the block up the incline. What is the corresponding value of P? The block
has weight W and the slope α is known.
809
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Engineering Mechanics - Statics
Solution:
Let
Chapter 8
μ = tan ( θ )
Σ F x = 0;
P cos ( φ ) − W sin ( α ) − μ N = 0
Σ F y = 0;
N − W cos ( α ) + P sin ( φ ) = 0
P cos ( φ ) − W sin ( α ) − μ ( W cos ( α ) − P sin ( φ ) = 0
⎛ sin ( α ) + μ cos ( α ) ⎞ = W⎛ sin ( α ) + tan ( θ ) cos ( α ) ⎞
⎟
⎜
⎟
⎝ cos ( φ ) + μ sin ( φ ) ⎠
⎝ cos ( φ ) + tan ( θ ) sin ( φ ) ⎠
P = W⎜
⎛ sin ( α ) cos ( θ ) + sin ( θ ) cos ( α ) ⎞ = W sin ( α + θ )
⎟
cos ( φ − θ )
⎝ cos ( φ ) cos ( θ ) + sin ( θ ) sin ( φ ) ⎠
P = W⎜
dP
dφ
⎡sin ( α + θ ) sin ( φ − θ )⎤ = 0
⎥
2
cos ( φ − θ )
⎣
⎦
= W⎢
sin ( α + θ ) sin ( φ − θ ) = 0
sin ( φ − θ ) = 0
φ=θ
P = W sin ( α + φ )
Problem 8-51
Two blocks A and B, each having a mass M, are connected by the linkage shown. If the
coefficient of static friction at the contacting surfaces is μs determine the largest vertical force
P that may be applied to pin C of the linkage without causing the blocks to move. Neglect the
weight of the links.
810
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Engineering Mechanics - Statics
Chapter 8
Given:
M = 6 kg
μ s = 0.5
θ 1 = 30 deg
θ 2 = 30 deg
Solution:
Guesses
P = 1N
NA = 1 N
FA = 1 N
F AC = 1 N
NB = 1 N
FB = 1 N
F BC = 1 N
Assume that A slips first
Given
F AC cos ( θ 2 ) − FBC = 0
F AC sin ( θ 2 ) − P = 0
NA − M g − F AC sin ( θ 2 ) = 0
F A − FAC cos ( θ 2 ) = 0
F BC cos ( θ 1 ) − M g sin ( θ 1 ) − F B = 0
−F BC sin ( θ 1 ) − M g cos ( θ 1 ) + NB = 0
F A = μ s NA
⎛ P1 ⎞
⎜
⎟
⎜ NA ⎟
⎜ FA ⎟
⎜
⎟
N
⎜ B ⎟ = Find ( P , NA , FA , NB , FB , FAC , FBC)
⎜F ⎟
⎜ B ⎟
⎜ FAC ⎟
⎜
⎟
⎝ FBC ⎠
P 1 = 23.9 N
Assume that B slips first
811
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Engineering Mechanics - Statics
Chapter 8
Given
F AC cos ( θ 2 ) − FBC = 0
F AC sin ( θ 2 ) − P = 0
NA − M g − F AC sin ( θ 2 ) = 0
F A − FAC cos ( θ 2 ) = 0
F BC cos ( θ 1 ) − M g sin ( θ 1 ) − F B = 0
−F BC sin ( θ 1 ) − M g cos ( θ 1 ) + NB = 0
F B = μ s NB
⎛ P2 ⎞
⎜
⎟
⎜ NA ⎟
⎜ FA ⎟
⎜
⎟
⎜ NB ⎟ = Find ( P , NA , FA , NB , FB , FAC , FBC)
⎜F ⎟
⎜ B ⎟
⎜ FAC ⎟
⎜
⎟
⎝ FBC ⎠
P = min ( P 1 , P 2 )
P 2 = 51.5 N
P = 23.9 N
Problem 8-52
Block C has a mass mc and is confined between two walls by smooth rollers. If the block rests on top
of the spool of mass ms, determine the minimum cable force P needed to move the spool. The cable is
wrapped around the spool's inner core. The coefficients of static friction at μA and μB.
Given:
g = 9.81
m
2
s
mc = 50 kg
ms = 40 kg
μ A = 0.3
812
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Engineering Mechanics - Statics
μ B = 0.6
Solution:
Chapter 8
r1 = 0.2 m
r2 = 0.4 m
Assume that the spool slips at A but not at B.
The initial guesses are F B = 2 N
Given
P = 3N
NB = 1 N
P − FB − μ A mc g = 0
NB − ms g − mc g = 0
P ( r1 + r2 ) − FB2r2 = 0
⎛ FB ⎞
⎜ ⎟
⎜ P ⎟ = Find ( FB , P , NB)
⎜ NB ⎟
⎝ ⎠
⎛ FB ⎞ ⎛ 441.45 ⎞
⎜ ⎟ ⎜
⎟
⎜ P ⎟ = ⎜ 588.60 ⎟ N
⎜ NB ⎟ ⎝ 882.90 ⎠
⎝ ⎠
Now check the no slip assumption at B
F Bmax = μ B NB
Since F B = 441 N
then our assumptions are correct.
< F Bmax = 530 N
F Bmax = 529.74 N
P = 589 N
Problem 8-53
A board of weight W1 is placed across the channel and a boy of weight W2 attempts to walk
across. If the coefficient of static friction at A and B μs, determine if he can make the crossing;
and if not, how far will he get from A before the board slips?
813
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Engineering Mechanics - Statics
Chapter 8
Given:
W1 = 50 lb
W2 = 100 lb
μ s = 0.4
a = 10 ft
b = 4
c = 3
Solution:
Initial guesses:
NA = 10 lb
NB = 20 lb
d = 3 ft
Given
ΣF x = 0;
μ s NA + μ s NB⎛⎜
c
⎞−N ⎛
⎞
B
⎜ 2 2⎟ = 0
2
2⎟
⎝ b +c ⎠
⎝ b +c ⎠
ΣF y = 0;
NA − W2 − W1 + NB⎜
ΣMB = 0;
W1 ⎜
b
⎛
c
⎞+μ N ⎛
⎞
s
B
⎜ 2 2⎟ = 0
2
2⎟
⎝ b +c ⎠
⎝ b +c ⎠
b
⎛ a ⎞ + W ( a − d) − N a = 0
⎟
2
A
⎝ 2⎠
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( NA , NB , d)
⎜ d ⎟
⎝ ⎠
⎛ NA ⎞ ⎛ 60.34 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ NB ⎠ ⎝ 86.21 ⎠
d = 6.47 ft
Since d = 6.47 ft < a = 10.00 ft then the board will slip
Problem 8-54
Determine the minimum force P needed to push the tube E up the incline. The tube has a
mass of M1 and the roller D has a mass of M2. The force acts parallel to the plane, and the
coefficients of static friction at the contacting surfaces are μA, μB and μC. Each cylinder has a
radius of r.
814
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
M1 = 75 kg
M2 = 100 kg
μ A = 0.3
μ B = 0.25
μ C = 0.4
θ = 30 deg
r = 150 mm
g = 9.81
m
2
s
Solution:
Initial guesses:
NA = 100 N
F A = 10 N
NB = 200 N
F B = 20 N
NC = 300 N
F C = 30 N
P = 100 N
Given
For roller D
ΣF x' = 0;
P − NA − F C − M2 g sin ( θ ) = 0
ΣF y' = 0;
NC + F A − M2 g cos ( θ ) = 0
ΣM0' = 0;
FA r − FC r = 0
For tube E
ΣF x' = 0;
NA − FB − M1 g sin ( θ ) = 0
ΣF y' = 0;
NB − FA − M1 g cos ( θ ) = 0
ΣM0' = 0;
FA r − FB r = 0
Assuming slipping occurs only at A.
F A = μ A NA
815
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Engineering Mechanics - Statics
Chapter 8
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟
⎜ NC ⎟
⎜ ⎟
⎜ FA ⎟ = Find ( NA , NB , NC , FA , FB , FC , P)
⎜F ⎟
⎜ B⎟
⎜ FC ⎟
⎜ ⎟
⎝P ⎠
⎛ NA ⎞
⎜ ⎟ ⎛⎜ 526 ⎞⎟
⎜ NB ⎟ ⎜ 795 ⎟
⎜ NC ⎟ ⎜ 692 ⎟
⎜ ⎟ ⎜
⎟
⎜ FA ⎟ = ⎜ 158 ⎟ N
⎜ F ⎟ ⎜ 158 ⎟
⎜ B⎟ ⎜
⎟
⎜ FC ⎟ ⎜ 158 ⎟
⎜ ⎟ ⎝ 1174 ⎠
⎝P ⎠
F Bmax = μ B NB
⎛ FBmax ⎞ ⎛ 199 ⎞
⎜
⎟=⎜
⎟N
277
F
⎝
⎠
Cmax
⎝
⎠
F Cmax = μ C NC
Since F B = 158 N < FBmax = 199 N and FC = 158 N < F Cmax = 277 N
then our
assumption is correct
P = 1174 N
Problem 8-55
The concrete pipe at A rests on top of B and C. If the coefficient of static friction between the
pipes is μs and at the ground μ's, determine their smallest values so that the pipes will not slip.
Each pipe has a radius r and weight W, and the angle between the centers as indicated is θ.
816
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
Total System :
ΣF y = 0;
2N' − 3W = 0
N' =
3
2
W
Pipe C:
ΣF x = 0;
−F cos ( θ ) + N sin ( θ ) − F' = 0
ΣF y = 0;
N' − W − N cos ( θ ) − F sin ( θ ) = 0
ΣMO = 0;
F r − F' r = 0
Solving,
F = F' =
For Pipe A:
μs N ≥
For Pipe C:
μ's N' ≥
1 ⎛ W sin ( θ ) ⎞
⎜
⎟
2 ⎝ cos ( θ ) + 1 ⎠
1 ⎛ W sin ( θ ) ⎞
⎜
⎟
2 ⎝ cos ( θ ) + 1 ⎠
μs ≥
1 ⎛ W sin ( θ ) ⎞
⎜
N=
⎟
2 ⎝ cos ( θ ) + 1 ⎠
W
2
sin ( θ )
cos ( θ ) + 1
μ's ≥
1
sin ( θ )
3 cos ( θ ) + 1
Problem 8-56
The uniform pole has a weight W and length L. Its end
B is tied to a supporting cord, and end A is placed
against the wall, for which the coefficient of static
friction is μs. Determine the largest angle θ at which
the pole can be placed without slipping.
Solution:
ΣF x = 0;
ΣF y = 0;
ΣMB = 0;
⎛θ⎞ = 0
⎟
⎝2⎠
⎛θ⎞
μ s NA − W + T cos ⎜ ⎟ = 0
⎝2⎠
NA − T sin ⎜
−NA L cos ( θ ) − μ s NA L sin ( θ ) + W ⎛⎜
L⎞
⎟ sin ( θ ) = 0
⎝2⎠
817
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Engineering Mechanics - Statics
Chapter 8
Solving we find
⎛θ⎞
⎟
⎝2⎠
NA = T sin ⎜
⎛θ⎞
⎛θ⎞
μ s T sin ⎜ ⎟ − W + T cos ⎜ ⎟ = 0
⎝2⎠
⎛
⎝
⎛ θ ⎞ + μ sin ⎛ θ ⎞⎞
⎟
s ⎜ ⎟⎟
⎝2⎠
⎝ 2 ⎠⎠
W = T⎜ cos ⎜
⎝2⎠
⎛ θ ⎞ L cos ( θ ) + μ L sin ( θ ) = T⎛ cos ⎛ θ ⎞ + μ sin ⎛ θ ⎞⎞ L sin ( θ )
) ⎜ ⎜ 2 ⎟ s ⎜ 2 ⎟⎟ 2
⎟(
s
⎝2⎠
⎝ ⎝ ⎠
⎝ ⎠⎠
T sin ⎜
⎛ θ ⎞ cos ( θ ) + μ sin ( θ ) = 1 sin ( θ ) ⎛ cos ⎛ θ ⎞ + μ sin ⎛ θ ⎞⎞
) 2
⎟(
⎜ ⎜ ⎟
s
s ⎜ ⎟⎟
⎝2⎠
⎝ ⎝2⎠
⎝ 2 ⎠⎠
sin ⎜
μs
2
⎛ θ ⎞ sin ( θ ) = 1 sin ( θ ) cos ⎛ θ ⎞ − sin ⎛ θ ⎞ cos ( θ )
⎟
⎜ ⎟
⎜ ⎟
2
⎝2⎠
⎝2⎠
⎝2⎠
sin ⎜
⎛ θ ⎞ − 2 sin ⎛ θ ⎞ cos ( θ )
⎟
⎜ ⎟
⎝2⎠
⎝2⎠
⎛θ⎞
sin ⎜ ⎟ sin ( θ )
⎝2⎠
sin ( θ ) cos ⎜
μs =
2
2
⎛ θ ⎞ 2⎛⎜cos ⎛ θ ⎞ − sin ⎛ θ ⎞ ⎟⎞
⎟
⎜ ⎟
⎜ ⎟
⎝2⎠ − ⎝ ⎝2⎠
⎝ 2 ⎠ ⎠ = cot ⎛ θ ⎞ − cos ⎛ θ ⎞ + tan ⎛ θ ⎞
μs =
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
⎛θ⎞
⎛θ⎞ ⎛θ⎞
sin ⎜ ⎟
2 sin ⎜ ⎟ cos ⎜ ⎟
⎝2⎠
⎝2⎠ ⎝2⎠
cos ⎜
⎛θ⎞
μ s = tan ⎜ ⎟
⎝2⎠
θ = 2 atan ( μ )
Problem 8-57
The carpenter slowly pushes the uniform board horizontally over the top of the saw horse. The
board has a uniform weight density γ and the saw horse has a weight W and a center of gravity
at G. Determine if the saw horse will stay in position, slip, or tip if the board is pushed forward
at the given distance d. The coefficients of static friction are shown in the figure.
Given:
γ = 3
lb
ft
818
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Engineering Mechanics - Statics
Chapter 8
L = 18 ft
W = 15 lb
a = 3 ft
b = 1 ft
μ = 0.5
μ' = 0.3
d = 10 ft
Solution:
Board:
1 2⎛ γ ⎞
⎟ + N d = 0 N = L ⎜ ⎟ N = 48.60 lb
2 ⎝ d⎠
⎝2⎠
−L γ ⎛⎜
L⎞
To cause slipping of the board on the saw horse:
P xb = μ N
P xb = 24.30 lb
To cause slipping at the ground:
P xg = μ' ( N + W)
P xg = 19.08 lb
To cause tipping
( N + W)b − P xt a = 0
P xt =
( N + W)b
a
P xt = 21.20 lb
Choose the critical case
P x = min ( P xb , Pxg , P xt)
P x = 19.08 lb
Problem 8-58
The carpenter slowly pushes the uniform board horizontally over the top of the saw horse. The
board has a uniform weight density γ and the saw horse has a weight W and a center of gravity
at G. Determine if the saw horse will stay in position, slip, or tip if the board is pushed forward
at the given distance d. The coefficients of static friction are shown in the figure.
Given:
γ = 3
lb
ft
819
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Engineering Mechanics - Statics
Chapter 8
L = 18 ft
W = 15 lb
a = 3 ft
b = 1 ft
μ = 0.5
μ' = 0.3
d = 14 ft
Solution:
Board:
⎛ L ⎞ + N d = 0 N = 1 L2 ⎛ γ ⎞ N = 34.71 lb
⎟
⎜ ⎟
2
⎝2⎠
⎝ d⎠
−L γ ⎜
To cause slipping of the board on the saw horse:
P xb = μ N
P xb = 17.36 lb
To cause slipping at the ground:
P xg = μ' ( N + W)
P xg = 14.91 lb
To cause tipping
( N + W)b − P xt a = 0
P xt =
( N + W)b
a
P xt = 16.57 lb
Choose the critical case
P x = min ( P xb , Pxg , P xt)
P x = 14.91 lb
Problem 8-59
The disk of mass mo rests on the surface for which the coefficient of static friction is μA
Determine the largest couple moment M that can be applied to the bar without causing motion.
820
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Engineering Mechanics - Statics
Chapter 8
Given:
mo = 45 kg
μ A = 0.2
a = 300 mm
b = 400 mm
r = 125 mm
Solution:
Guesses
M = 1Nm
NA = 1 N
Bx = 1 N
By = 1 N
Given
M − Bx b − By a = 0
B x − μ A NA = 0
NA − mo g − By = 0
B y r − μ A NA r = 0
⎛M⎞
⎜N ⎟
⎜ A ⎟ = Find ( M , N , B , B )
A x y
⎜ Bx ⎟
⎜ ⎟
⎝ By ⎠
⎛ NA ⎞ ⎛ 551.81 ⎞
⎜ ⎟ ⎜
⎟
⎜ Bx ⎟ = ⎜ 110.36 ⎟ N
⎜ B ⎟ ⎝ 110.36 ⎠
⎝ y⎠
M = 77.3 N⋅ m
Problem 8-60
The disk of mass m0 rests on the surface for which the coefficient of static friction is μA
Determine the friction force at A.
821
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Engineering Mechanics - Statics
Chapter 8
Given:
M = 50 N⋅ m
mo = 45 kg
μ A = 0.2
a = 300 mm
b = 400 mm
r = 125 mm
Solution:
Assume no motion
Guesses
B x = 1 N B y = 1 N NA = 1 N
Given
FA = 1 N
M − By a − Bx b = 0
NA − By − mo g = 0
Bx − FA = 0
By r − FA r = 0
⎛ Bx ⎞
⎜ ⎟
⎜ By ⎟ = Find ( B , B , N , F )
x y A A
⎜ NA ⎟
⎜ ⎟
⎝ FA ⎠
⎛ NA ⎞ ⎛ 512.88 ⎞
⎜ ⎟=⎜
⎟N
⎝ FA ⎠ ⎝ 71.43 ⎠
Check assumption:
F Amax = μ A NA
Since F A = 71.4 N
< F Amax = 102.6 N
then our assumption is good.
F A = 71.4 N
Problem 8-61
A block of weight W is attached to a light rod AD that pivots at pin A. If the coefficient of static
friction between the plane and the block is μs, determine the minimum angle θ at which the
block may be placed on the plane without slipping. Neglect the size of the block in the
822
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Engineering Mechanics - Statics
Chapter 8
calculation.
Given:
a = 4 ft
b = 2 ft
c = 3 ft
μ s = 0.4
W = 6 lb
Solution:
Establish a unit vector
perpendicular to the plane.
⎛ −b ⎞
⎜ ⎟
rBC = c
⎜ ⎟
⎝0⎠
⎛ −b ⎞
⎜ ⎟
rBA = 0
⎜ ⎟
⎝a⎠
rBC × rBA
n1 =
rBC × rBA
⎛ 0.77 ⎞
⎜
⎟
n 1 = 0.51
⎜
⎟
⎝ 0.38 ⎠
Establish 2 other unit vectors in this plane, and one in the z direction
⎛0⎞
⎜ ⎟
k = 0
⎜ ⎟
⎝1⎠
⎛0⎞
⎜ ⎟
rCA = −c
⎜ ⎟
⎝a⎠
rCA
n3 = n1 × n2
rCA
θ = 10 deg
Guess
T = 1 lb
Given
Nn 1 + T( cos ( θ ) n 2 − sin ( θ ) n 3) − μ s N( sin ( θ ) n 2 + cos ( θ ) n 3) − Wk = 0
⎛N⎞
⎜ T ⎟ = Find ( N , T , θ )
⎜ ⎟
⎝θ ⎠
N = 1 lb
n2 =
⎛ N ⎞ ⎛ 2.30 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ T ⎠ ⎝ 5.46 ⎠
θ = 20.37 deg
Problem 8-62
Determine the force P needed to lift the load of weight W. Smooth rollers are placed between
823
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Engineering Mechanics - Statics
Chapter 8
g
p
the wedges. The coefficient of static friction between A and C and between B and D is μs.
Neglect the weight of each wedge.
Given:
θ = 10 deg
W = 100 lb
μ s = 0.3
Solution:
Initial guesses:
N' = 10 lb
NA = 15 lb
NB = 20 lb
P = 5 lb
Given
Wedge B:
ΣF x = 0;
N' sin ( θ ) − NB = 0
ΣF y = 0;
N' cos ( θ ) − W − μ s NB = 0
Wedge A:
ΣF x = 0;
P − N' sin ( θ ) − μ s NA = 0
ΣF y = 0;
NA − N' cos ( θ ) = 0
⎛ N' ⎞
⎜ ⎟
⎜ NB ⎟ = Find ( N' , N , N , P)
B A
⎜ NA ⎟
⎜ ⎟
⎝P ⎠
⎛ N' ⎞ ⎛ 107.2 ⎞
⎜ ⎟ ⎜
⎟
⎜ NB ⎟ = ⎜ 18.6 ⎟ lb
⎜ NA ⎟ ⎝ 105.6 ⎠
⎝ ⎠
P = 50.3 lb
Problem 8-63
The wedge is used to level the floor of a building. For the floor loading shown, determine the
horizontal force P that must be applied to move the wedge forward. The coefficient of static
friction between the wedge and the two surfaces of contact is μs. Neglect the size and weight
824
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Engineering Mechanics - Statics
Chapter 8
of the wedge and the thickness of the beam.
Units Used:
3
kN = 10 N
Given:
F 1 = 2 kN
a = 3m
F 2 = 4 kN
b = 2m
F 3 = 4 kN
c = 3m
F 4 = 2 kN
θ = 15 deg
μ s = 0.25
Solution:
Guesses
N1 = 1 kN
N2 = 1 kN
P = 1 kN
Given
(F1 − N1)( a + b + c) + F2( b + c) + F3 c = 0
N2 cos ( θ ) − μ s N2 sin ( θ ) − N1 = 0
μ s N1 + μ s N2 cos ( θ ) + N2 sin ( θ ) − P = 0
⎛ N1 ⎞
⎜ ⎟
⎜ N2 ⎟ = Find ( N1 , N2 , P)
⎜P ⎟
⎝ ⎠
⎛ N1 ⎞ ⎛ 6 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ N2 ⎠ ⎝ 6.66 ⎠
P = 4.83 kN
Problem 8-64
The three stone blocks have weights WA, WB, and WC. Determine the smallest horizontal force P
that must be applied to block C in order to move this block.The coefficient of static friction
between the blocks is μs, and between the floor and each block μ's.
825
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Engineering Mechanics - Statics
Chapter 8
Given:
WA = 600 lb
WB = 150 lb
WC = 500 lb
μ s = 0.3
μ's = 0.5
θ = 45 deg
Solution:
Assume all blocks slip together
N1 = WA + WB + WC
P 1 = μ's N1
P 1 = 625.00 lb
Assume that block A does not move and block
B moves up
Guesses
NC = 1 lb
N' = 1 lb
N'' = 1 lb P = 1 lb
Given
N'' + μ's NC − P = 0
NC − WC − WB − μ s N'' = 0
N' sin ( θ ) − μ s N' cos ( θ ) − μ s N'' − WB = 0
N'' − μ s N' sin ( θ ) − N' cos ( θ ) = 0
⎛ NC ⎞
⎜ ⎟
⎜ N' ⎟ = Find ( N , N' , N'' , P)
C
⎜ N'' ⎟
⎜ ⎟
⎝ P2 ⎠
Choose the critical case
⎛⎜ NC ⎟⎞ ⎛ 838.71 ⎞
⎜
⎟
⎜ N' ⎟ = ⎜ 684.30 ⎟ lb P2 = 1048 lb
⎜ N'' ⎟ ⎝ 629.03 ⎠
⎝ ⎠
P = min ( P 1 , P 2 )
P = 625.00 lb
826
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Engineering Mechanics - Statics
Chapter 8
Problem 8-65
If the spring is compressed a distance δ and the coefficient of static friction between the tapered
stub S and the slider A is μsA, determine the horizontal force P needed to move the slider forward.
The stub is free to move without friction within the fixed collar C. The coefficient of static friction
between A and surface B is μAB. Neglect the weights of the slider and stub.
Given:
δ = 60 mm
μ sA = 0.5
μ AB = 0.4
k = 300
N
m
θ = 30 deg
Solution:
Stub:
+
↑Σ Fy = 0;
NA cos ( θ ) − μ sA NA sin ( θ ) − kδ = 0
NA =
kδ
cos ( θ ) − μ sA sin ( θ )
NA = 29.22 N
Slider:
+
↑Σ Fy = 0;
NB − NA cos ( θ ) + μ sA NA sin ( θ ) = 0
NB = NA cos ( θ ) − μ sA NA sin ( θ )
+
Σ F x = 0;
→
NB = 18 N
P − μ AB NB − NA sin ( θ ) − μ sA NA cos ( θ ) = 0
P = μ AB NB + NA sin ( θ ) + μ sA NA cos ( θ )
P = 34.5 N
827
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Engineering Mechanics - Statics
Chapter 8
Problem 8-66
The coefficient of static friction between wedges B and C is μs1 and between the surfaces of
contact B and A and C and D, μs2. If the spring is compressed to a distance δ when in the position
shown, determine the smallest force P needed to move wedge C to the left. Neglect the weight of
the wedges.
Given:
μ s1 = 0.6
θ = 15 deg
μ s2 = 0.4
k = 500
N
m
δ = 200 mm
Solution:
The initial guesses:
NAB = 10 N
NBC = 20 N
NCD = 30 N
P = 40 N
Given
Wedge B:
NAB − μ s1 NBC cos ( θ ) − NBC sin ( θ ) = 0
NBC cos ( θ ) − μ s1 NBC sin ( θ ) − μ s2 NAB − kδ = 0
Wedge C:
NCD cos ( θ ) − μ s2 NCD sin ( θ ) + μ s1 NBC sin ( θ ) − NBC cos ( θ ) = 0
NCD sin ( θ ) + μ s2 NCD cos ( θ ) + NBC sin ( θ ) + μ s1 NBC cos ( θ ) − P = 0
⎛ NAB ⎞
⎜
⎟
⎜ NBC ⎟ = Find ( N , N , N , P)
AB BC CD
⎜ NCD ⎟
⎜
⎟
⎝ P ⎠
⎛ NAB ⎞ ⎛ 176.39 ⎞
⎜
⎟ ⎜
⎟
⎜ NBC ⎟ = ⎜ 210.40 ⎟ N P = 303.99 N
⎜ N ⎟ ⎝ 197.77 ⎠
⎝ CD ⎠
828
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Engineering Mechanics - Statics
Chapter 8
Problem 8-67
The coefficient of static friction between the wedges B and C is μs1 and between the surfaces of
contact B and A and C and D, μs2. Determine the smalles allowable compression of the spring δ
without causing wedge C to move to the left. Neglect the weight of the wedges.
Given:
μ s1 = 0.6
θ = 15 deg
μ s2 = 0.4
k = 500
N
m
P = 50 N
Solution:
The initial guesses:
NAB = 10 N
NBC = 20 N
NCD = 30 N
δ = 10 mm
Given
Wedge B:
NAB − μ s1 NBC cos ( θ ) − NBC sin ( θ ) = 0
NBC cos ( θ ) − μ s1 NBC sin ( θ ) − μ s2 NAB − kδ = 0
Wedge C:
NCD cos ( θ ) − μ s2 NCD sin ( θ ) + μ s1 NBC sin ( θ ) − NBC cos ( θ ) = 0
NCD sin ( θ ) + μ s2 NCD cos ( θ ) + NBC sin ( θ ) + μ s1 NBC cos ( θ ) − P = 0
⎛ NAB ⎞
⎜
⎟
⎜ NBC ⎟ = Find ( N , N , N , δ )
AB BC CD
⎜ NCD ⎟
⎜
⎟
⎝ δ ⎠
⎛ NAB ⎞ ⎛ 29.01 ⎞
⎜
⎟ ⎜
⎟
N
⎜ BC ⎟ = ⎜ 34.61 ⎟ N
⎜ N ⎟ ⎝ 32.53 ⎠
⎝ CD ⎠
δ = 32.90 mm
Problem 8-68
The wedge blocks are used to hold the specimen in a tension testing machine. Determine the design
angle θ of the wedges so that the specimen will not slip regardless of the applied load. The
coefficients of static friction are μA at A and μB at B. Neglect the weight of the blocks.
829
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Engineering Mechanics - Statics
Chapter 8
Given:
μ A = 0.1
μ B = 0.6
Solution:
Specimen:
+
↑Σ Fy = 0;
2FB − P = 0
FB =
P
2
Wedge:
+
Σ F x = 0;
→
NA cos ( θ ) − μ A NA sin ( θ ) −
+
μ A NA cos ( θ ) + NA sin ( θ ) −
↑
Σ F y = 0;
P
2μ B
P
2
=0
=0
If we eliminate P we have
μ B( NA cos ( θ ) − μ A NA sin ( θ ) ) = μ A NA cos ( θ ) + NA sin ( θ )
(1 + μ Aμ B)sin(θ ) = (μ B − μ A)cos (θ )
⎛ μB − μA ⎞
⎟
⎝ 1 + μ Aμ B ⎠
θ = atan ⎜
θ = 25.3 deg
Problem 8-69
The wedge is used to level the member. Determine the reversed horizontal force - P that must
be applied to pull the wedge out to the left. The coefficient of static friction between the
wedge and the two surfaces of contact is μs. Neglect the weight of the wedge.
830
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Engineering Mechanics - Statics
Chapter 8
Units Used:
3
kN = 10 N
Given:
μ s = 0.15
θ = 5 deg
a = 600 mm
b = 500 mm
c = 250 mm
d = 3
e = 4
F 1 = 8 kN
F 2 = 15 kN
Solution:
Guesses
P = 1N
NA = 1 N
NB = 1 N
Given
−P + μ s NB + μ s NA cos ( θ ) − NA sin ( θ ) = 0
NB − NA cos ( θ ) − μ s NA sin ( θ ) = 0
F1 b + ⎛
e
⎛ d ⎞
⎞
⎜ 2 2 ⎟ F2 a + ⎜ 2 2 ⎟ F2( b + c) − ( NA cos ( θ ) + μ s NA sin ( θ ) ) 2a ... = 0
⎝ e +d ⎠
⎝ e +d ⎠
+ ( μ s NA cos ( θ ) − NA sin ( θ ) ) ( b + c)
⎛P ⎞
⎜ ⎟
⎜ NA ⎟ = Find ( P , NA , NB)
⎜ NB ⎟
⎝ ⎠
⎛ NA ⎞ ⎛ 15.42 ⎞
⎜ ⎟=⎜
⎟ kN
⎝ NB ⎠ ⎝ 15.56 ⎠
P = 3.29 kN
831
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Engineering Mechanics - Statics
Chapter 8
Problem 8-70
If the coefficient of static friction between all the surfaces of contact is μs, determine the
force P that must be applied to the wedge in order to lift the brace that supports the load F.
Solution:
System:
ΣF x = 0;
P − NB − μ s NA = 0
(1)
ΣF y = 0;
NA − μ s NB − F = 0
(2)
Wedge A:
ΣF x = 0;
P − μ s NA − μ s N' cos ( α ) − N' sin ( α ) = 0
(3)
ΣF y = 0;
NA − N' cos ( α ) + μ s N' sin ( α ) = 0
(4)
From Eqs. (3) and (4):
N' =
P − μ s NA
(5)
μ s cos ( α ) + sin ( α )
832
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Engineering Mechanics - Statics
NA =
Chapter 8
P( 1 − μ s tan ( α ) )
(
(6)
)
2μ s + 1 − μ s tan ( α )
2
From Eqs. (1) and (2):
P − μ s NA
NB =
μs
2
(
1 + μ s ) NA − F
P=
μs
Combining Eqs. (5) and (6) :
(
)
⎡ 1 − μ s2 tan ( α ) + 2μ s⎤
⎥
P = F⎢
⎢ 1 − 2μ tan ( α ) − μ 2 ⎥
s
s ⎦
⎣
Problem 8-71
The column is used to support the upper floor. If a force F is applied
perpendicular to the handle to tighten the screw, determine the
compressive force in the column. The square-threaded screw on the jack
has a coefficient of static friction μs, mean diameter d, and a lead h.
Units Used:
3
kN = 10 N
Given:
F = 80 N
μ s = 0.4
h = 3 mm
a = 0.5 m
d = 25 mm
Solution:
φ s = atan ( μ s)
θ p = atan ⎛⎜
φ s = 21.80 deg
h ⎞
⎟
⎝ π d⎠
M = F a = W⎛⎜
θ p = 2.187 deg
d⎞
⎟ tan ( φ s + θ p)
⎝ 2⎠
833
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Engineering Mechanics - Statics
W = 2 F ⎛⎜
a
Chapter 8
⎞
⎟
⎝ d tan ( φ s + θ p ) ⎠
W = 7.19 kN
Problem 8-72
The column is used to support the upper floor. If the force F is removed
from the handle of the jack, determine if the screw is self-locking.. The
square-threaded screw on the jack has a coefficient of static friction μs,
mean diameter d, and a lead h.
Given:
F = 80 N
μ s = 0.4
h = 3 mm
a = 0.5 m
d = 25 mm
Solution:
φ s = atan ( μ s)
θ p = atan ⎛⎜
h ⎞
⎟
⎝ π d⎠
φ s = 21.8 deg
θ p = 2.19 deg
Since φ s = 21.8 deg > θ p = 2.19 deg , screw is self locking.
Problem 8-73
The vise is used to grip the pipe. If a horizontal force F 1 is
applied perpendicular to the end of the handle of length l,
determine the compressive force F developed in the pipe. The
square threads have a mean diameter d and a lead a. How
much force must be applied perpendicular to the handle to
loosen the vise?
Given:
F 1 = 25 lb
d = 1.5 in
μ s = 0.3
L = 10 in
834
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Engineering Mechanics - Statics
Chapter 8
a = 0.2 in
Solution:
r =
d
2
θ = atan ⎛⎜
a ⎞
θ = 2.43 deg
⎟
⎝ 2π r ⎠
φ = atan ( μ s)
φ = 16.70 deg
F 1 L = F r tan ( θ + φ )
F = F 1 ⎛⎜
L
⎞
⎟
⎝ r tan ( θ + φ ) ⎠
F = 961 lb
To loosen screw,
P L = F r tan ( φ − θ )
P = Fr
tan ( φ − θ )
L
P = 18.3 lb
Problem 8-74
Determine the couple forces F that must be applied to the handle of the machinist’s vise in
order to create a compressive force F A in the block. Neglect friction at the bearing A. The
guide at B is smooth so that the axial force on the screw is F A. The single square-threaded
screw has a mean radius b and a lead c, and the coefficient of static friction is μs.
Given:
a = 125 mm
F A = 400 N
b = 6 mm
c = 8 mm
μ s = 0.27
835
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
φ = atan ( μ s)
θ = atan ⎛⎜
c
φ = 15.11 deg
⎞
⎟
θ = 11.98 deg
⎝ 2 π b⎠
F 2 a = FA b tan ( θ + φ )
F = F A ⎛⎜
b⎞
⎟ tan ( θ + φ )
⎝ 2a ⎠
F = 4.91 N
Problem 8-75
If couple forces F are applied to the handle of the machinist’s vise, determine the compressive
force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The
single square-threaded screw has a mean radius of r1 and a lead of r2, and the coefficient of
static friction is μs.
Units Used:
3
kN = 10 N
Given:
F = 35 N
a = 125 mm
r1 = 6 mm
r2 = 8 mm
μ s = 0.27
Solution:
φ = atan ( μ s)
φ = 15.11 deg
⎛ r2 ⎞
⎟
⎝ 2 π r1 ⎠
θ = atan ⎜
θ = 11.98 deg
F 2 a = P r1 tan ( θ + φ )
⎞
⎟
⎝ r1 tan ( θ + φ ) ⎠
P = 2 F ⎛⎜
a
P = 2.85 kN
836
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 8
Problem 8-76
The machine part is held in place using the double-end clamp.The bolt at B has square threads
with a mean radius r and a lead r1, and the coefficient of static friction with the nut is μs. If a
torque M is applied to the nut to tighten it, determine the normal force of the clamp at the
smooth contacts A and C.
Given:
a = 260 mm
b = 90 mm
r = 4 mm
rl = 2 mm
μ s = 0.5
M = 0.4 N⋅ m
Solution:
φ = atan ( μ s)
φ = 26.57 deg
⎛ rl ⎞
⎟
⎝ 2π r ⎠
θ = atan ⎜
θ = 4.55 deg
M = W r tan ( θ + φ )
W =
M
r tan ( θ + φ )
W = 165.67 N
ΣMA = 0;
NC( a + b) − W a = 0
NC = W
ΣF y = 0;
a
a+b
NC = 123 N
NA − W + NC = 0
NA = W − NC
NA = 42.6 N
837
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Engineering Mechanics - Statics
Chapter 8
Problem 8-77
Determine the clamping force on the board A if the screw of the “C” clamp is tightened with a
twist M. The single square-threaded screw has a mean radius r, a lead h, and the coefficient of
static friction is μs.
Units Used:
3
kN = 10 N
Given:
M = 8 N⋅ m
r = 10 mm
h = 3 mm
μ s = 0.35
Solution:
φ s = atan ( μ s)
θ p = atan ⎛⎜
1
φ s = 19.29 deg
h ⎞
⎟
⎝ 2 π r⎠
M = P r tan ( φ s + θ p )
θ p = 2.734 deg
P =
M
r tan ( φ s + θ p )
P = 1.98 kN
Problem 8-78
If the required clamping force at the board A is to be P, determine the
torque M that must be applied to the handle of the “C” clamp to tighten
it down. The single square-threaded screw has a mean radius r, a lead
h, and the coefficient of static friction is μs.
Given:
P = 50 N
r = 10 mm
h = 3 mm
μ s = 0.35
838
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
φ s = atan ( μ s)
θ P = atan ⎛⎜
1
φ s = 19.29 deg
h ⎞
⎟
⎝ 2 π r⎠
M = P r tan ( φ s + θ P)
θ P = 2.73 deg
M = 0.202 N⋅ m
Problem 8-79
Determine the clamping force on the board at A if the screw of the hold-down clamp is
tightened with a twist M. The single square-threaded screw has a mean radius of r and a lead
of rl, and the coefficient of static friction is μs.
Given:
M = 0.2 N m
r = 8 mm
rl = 2 mm
μ s = 0.38
Solution:
φ = atan ( μ s)
⎛ rl ⎞
⎟
⎝ 2π r ⎠
θ = atan ⎜
φ = 20.81 deg
θ = 2.28 deg
M = F r tan ( θ + φ )
F =
M
r tan ( θ + φ )
F = 58.7 N
839
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-80
If the required clamping force at the board A is to be F, determine the torque M that must be
applied to the handle of the hold-down clamp to tighten it down.The single square-threaded
screw has a mean radius r and a lead r1, and the coefficient of static friction is μs.
Given:
F = 70 N
r = 8 mm
rl = 2 mm
μ s = 0.38
Solution:
φ = atan ( μ s)
φ = 20.81 deg
⎛ rl ⎞
⎟
⎝ 2π r ⎠
θ = atan ⎜
θ = 2.2785 deg
M = F r tan ( θ + φ )
M = 0.24 N⋅ m
Problem 8-81
The fixture clamp consist of a square-threaded screw having a coefficient of static friction μs mean
diameter d, and a lead h. The five points indicated are pin connections. Determine the clamping force
at the smooth blocks D and E when a torque M is applied to the handle of the screw.
Given:
μ s = 0.3
d = 3 mm
h = 1 mm
M = 0.08 N⋅ m
a = 30 mm
b = 40 mm
c = 40 mm
β = 45 deg
840
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Frictional Forces on Screw: Here
θ = atan ⎡⎢
⎤
⎥
h
θ = 6.06 deg
d
⎢2 π ⎛⎜ ⎟⎞ ⎥
⎣
⎝ 2 ⎠⎦
φ s = atan ( μ s)
φ s = 16.70 deg
Applying Eq.8-3, we have
M = P ⎛⎜
d⎞
⎟ tan ( θ + φ s)
⎝ 2⎠
P = 2 ⎛⎜
M
⎞
⎟
⎝ d tan ( θ + φ s) ⎠
P = 127.15 N
Note since φ s = 16.70 deg > θ = 6.06 deg , the screw is self-locking. It will not unscrew even if the
moment M is removed.
Equations of Equilibrium and Friction:
ΣMc = 0;
c
⎛
⎞
⎜ 2 2 ⎟ P b − FE cos ( β ) b − FE sin ( β ) a = 0
⎝ b +c ⎠
841
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Engineering Mechanics - Statics
Chapter 8
P cb
FE =
F E = 72.7 N
b + c ( cos ( β ) b + sin ( β ) a)
2
2
The equilibrium of clamped block requires that
FD = FE
F D = 72.7 N
Problem 8-82
The clamp provides pressure from several directions on the edges of the board. If the
square-threaded screw has a lead h, radius r, and the coefficient of static friction is μs, determine
the horizontal force developed on the board at A and the vertical forces developed at B and C if a
torque M is applied to the handle to tighten it further. The blocks at B and C are pin-connected to
the board.
Given:
h = 3 mm
r = 10 mm
μ s = 0.4
M = 1.5 N⋅ m
β = 45 deg
Solution:
φ s = atan ( μ s)
θ = atan ⎛⎜
h ⎞
⎟
⎝ 2π r ⎠
M = Ax r tan ( φ s + θ )
φ s = 21.801 deg
θ = 2.734 deg
Ax =
+
Σ F x = 0;
→
M
r tan ( φ s + θ )
Ax ⎞
⎜
⎟
2 ⎝ cos ( β ) ⎠
1 ⎛
Ax − 2T cos ( β ) = 0
T =
Cy = T sin ( β )
B y = Cy
Ax = 329 N
T = 232.36 N
⎛ By ⎞ ⎛ 164.3 ⎞
⎜ ⎟=⎜
⎟N
⎝ Cy ⎠ ⎝ 164.3 ⎠
842
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This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics
Chapter 8
Problem 8-83
The two blocks under the double wedge are brought together using a left and right
square-threaded screw. If the mean diameter is d, the lead is rl, and the coefficient of static
friction is μs, determine the torque needed to draw the blocks together. The coefficient
of static friction between each block and its surfaces of contact is μ's.
Units Used:
3
kN = 10 N
Given:
F = 5 kN
θ = 20 deg
d = 20 mm
rl = 5 mm
μ s = 0.4
μ's = 0.4
Solution:
Top block:
−F + 2N1 cos ( θ ) − 2μ's N1 sin ( θ ) = 0
N1 =
F
2( cos ( θ ) − μ's sin ( θ ) )
N1 = 3.1138 kN
Bottom block:
N' − N1 cos ( θ ) + μ's N1 sin ( θ ) = 0
N' = N1 cos ( θ ) − μ's N1 sin ( θ )
N' = 2.50 kN
−N1 sin ( θ ) − μ's N1 cos ( θ ) + T − μ's N' = 0
T = N1 sin ( θ ) + μ's N1 cos ( θ ) + μ's N'
T = 3.2354 kN
843
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
φ = atan ( μ s)
φ = 21.80 deg
⎛ rl ⎞
⎟
⎝ πd ⎠
θ = 4.55 deg
θ = atan ⎜
Since there are two blocks,
M = 2T
d
2
tan ( θ + φ )
M = 32 N⋅ m
Problem 8-84
The two blocks under the double wedge are brought together using a left and right
square-threaded screw. If the mean diameter is d, the lead is rl, and the coefficient of static
friction is μs, determine the torque needed to spread the blocks apart. The coefficient
of static friction between each block and its surfaces of contact is μ's.
Units Used:
3
kN = 10 N
Given:
F = 5 kN
θ = 20 deg
d = 20 mm
rl = 5 mm
μ s = 0.4
μ's = 0.4
Solution:
Top block:
−F + 2N1 cos ( θ ) + 2μ's N1 sin ( θ ) = 0
N1 =
F
2( cos ( θ ) + μ's sin ( θ ) )
N1 = 2.3223 kN
844
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Bottom block :
N' − N1 cos ( θ ) − μ's N1 sin ( θ ) = 0
N' = N1 cos ( θ ) + μ's N1 sin ( θ )
N' = 2.50 kN
−N1 sin ( θ ) + μ's N1 cos ( θ ) − T + μ's N' = 0
T = −N1 sin ( θ ) + μ's N1 cos ( θ ) + μ's N'
T = 1.0786 kN
φ = atan ( μ s)
φ = 21.80 deg
⎛ rl ⎞
⎟
⎝ πd⎠
θ = 4.55 deg
θ = atan ⎜
Since there are two blocks,
M = 2T
d
2
tan ( φ − θ )
M = 6.7 N⋅ m
Problem 8-85
The cord supporting the cylinder of mass M passes around three pegs, A, B, C, where the
coefficient of friction is μs. Determine the range of values for the magnitude of the horizontal
force P for which the cylinder will not move up or down.
Given:
M = 6 kg
θ = 45 deg
μ s = 0.2
g = 9.81
m
2
s
Solution:
Total angle
β =
5
2
π − 4θ
β = 270.00 deg
845
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Forces
Chapter 8
− μs β
μ sβ
P max = M g e
P min = M g e
P min = 15.9 N < P < P max = 217.4 N
Answer
Problem 8-86
The truck, which has mass mt , is to be lowered down the slope by a rope that is wrapped around
a tree. If the wheels are free to roll and the man at A can resist a pull P, determine the minimum
number of turns the rope should be wrapped around the tree to lower the truck at a constant
speed. The coefficient of kinetic friction between the tree and rope is μk.
Units Used:
Mg = 1000 kg
Given:
mt = 3.4 Mg
P = 300 N
θ = 20 deg
μ k = 0.3
g = 9.81
m
2
s
Solution:
ΣF x = 0;
T2 − mt g sin ( θ ) = 0
μ kβ
T2 = P e
T2 = mt g sin ( θ )
T2 = 11407.74 N
⎛ T2 ⎞
⎟
⎝P⎠
ln ⎜
β =
β = 694.86 deg
μk
⎛ β ⎞ = 2.00 turns
⎟
⎝ 360 deg ⎠
Use ceil ⎜
846
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This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics
Chapter 8
Problem 8-87
The wheel is subjected to a torque M. If
the coefficient of kinetic friction
between the band brake and the rim of
the wheel is μk, determine the smallest
horizontal force P that must be applied
to the lever to stop the wheel.
Given:
a = 400 mm d = 25 mm
b = 100 mm r = 150 mm
c = 50 mm
M = 50 N⋅ m
μ k = 0.3
Solution:
Initial guesses:
T1 = 5 N
T2 = 10 N
Given
⎛ 3π ⎞
μ k⎜ ⎟
2
T2 = T1 e ⎝ ⎠
Wheel:
ΣM0 = 0;
− T2 r + T1 r + M = 0
⎛ T1 ⎞
⎜ ⎟ = Find ( T1 , T2 )
⎝ T2 ⎠
T1 = 54.66 N
T1 c − F d = 0
F = T1 ⎛⎜
−P a + F b = 0
P = F ⎛⎜
Link:
ΣMB = 0;
c⎞
F = 109.32 N
b⎞
P = 27.3 N
⎟
⎝ d⎠
Lever:
ΣMA = 0;
⎟
⎝ a⎠
Problem 8-88
A cylinder A has a mass M. Determine the smallest force P applied to the handle of the lever
required for equilibrium. The coefficient of static friction between the belt and the wheel is μs.
The drum is pin connected at its center, B.
847
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
M = 75 kg
a = 700 mm
b = 25 mm
c = 300 mm
d = 200 mm
e1 = 60 mm
μ s = 0.3
e = 2.718
Solution:
Initial guesses: T1 = 1 N
T2 = 1 N
P = 1N
Given
Drum:
⎛ 3π ⎞
μ s⎜ ⎟
2
T2 = T1 e ⎝ ⎠
− T2 c + T1 c + M g d = 0
Lever:
−T1 e1 + T2 b − P a = 0
⎛ T1 ⎞
⎜ ⎟
⎜ T2 ⎟ = Find ( T1 , T2 , P)
⎜P⎟
⎝ ⎠
⎛ T1 ⎞ ⎛ 157.7 ⎞
⎜ ⎟=⎜
⎟N
⎝ T2 ⎠ ⎝ 648.2 ⎠
P = 9.63 N
Problem 8-89
Determine the largest mass of cylinder A that can be supported from the drum if a force P is
applied to the handle of the lever. The coefficient of static friction between the belt and the
wheel is μs. The drum is pin supported at its center, B.
Given:
P = 20 N
848
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
a = 700 mm
b = 25 mm
c = 300 mm
d = 200 mm
e1 = 60 mm
μ s = 0.3
e = 2.718
Solution:
Initial guesses: T1 = 1 N
T2 = 1 N
M = 1 kg
Given
Drum:
⎛ 3π ⎞
μ s⎜ ⎟
2
T2 = T1 e ⎝ ⎠
− T2 c + T1 c + M g d = 0
Lever:
−T1 e1 + T2 b − P a = 0
⎛ T1 ⎞
⎜ ⎟
⎜ T2 ⎟ = Find ( T1 , T2 , M)
⎜M ⎟
⎝ ⎠
⎛ T1 ⎞ ⎛ 327.4 ⎞
N
⎜ ⎟=⎜
3⎟
⎝ T2 ⎠ ⎝ 1.3 × 10 ⎠
M = 155.7 kg
Problem 8-90
The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a
fixed peg at D. If the coefficient of static friction between the rope and the peg is μD,
determine the smallest distance x from the end of the bar at which a force F may be placed
and not cause the bar to move.
Given:
F = 20 N
a = 1m
849
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
μ D = 0.3
Solution:
Initial guesses:
TA = 5 N
TB = 10 N x = 10 m
Given
ΣMA = 0;
− F x + TB a = 0
ΣF y = 0;
TA + TB − F = 0
⎛π⎞
μ D⎜ ⎟
2
TA = TB e ⎝ ⎠
⎛ TA ⎞
⎜ ⎟
⎜ TB ⎟ = Find ( TA , TB , x)
⎜ x ⎟
⎝ ⎠
x = 0.38 m
Problem 8-91
Determine the smallest lever force P needed to prevent the wheel from rotating if it is subjected
to a torque M. The coefficient of static friction between the belt and the wheel is μs. The wheel
is pin-connected at its center, B.
Given:
M = 250 N m
μ s = 0.3
r = 400 mm
a = 200 mm
b = 750 mm
Solution:
ΣMA = 0;
− F a + P ( a + b) = 0
850
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Engineering Mechanics - Statics
Chapter 8
⎛ a + b⎞
⎟
⎝ a ⎠
F=P⎜
β =
3π
2
μ sβ
F' = F e
ΣMB = 0;
⎛ a + b ⎞ eμ sβ r + M + P ⎛ a + b ⎞ r = 0
⎟
⎜
⎟
⎝ a ⎠
⎝ a ⎠
−P ⎜
P =
Ma
μ sβ
−1
( a + b)r e
(
P = 42.3 N
)
Problem 8-92
Determine the torque M that can be resisted by the band brake if a force P is applied to the
handle of the lever. The coefficient of static friction between the belt and the wheel is μs. The
wheel is pin-connected at its center, B.
Given:
P = 30 N
μ s = 0.3
r = 400 mm
a = 200 mm
b = 750 mm
Solution:
ΣMA = 0;
− F a + P ( a + b) = 0
⎛ a + b⎞
⎟
⎝ a ⎠
F = P⎜
F = 142.5 N
3π
μs
2
F' = F e
F' = 585.8 N
851
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Engineering Mechanics - Statics
ΣMB = 0;
Chapter 8
−F' r + F r + M = 0
M = F' r − F r
M = 177 N⋅ m
Problem 8-93
Blocks A and B weigh WA and WB and respectively. Using the coefficients of static friction indicated,
determine the greatest weight of block D without causing motion.
Given:
WA = 50 lb
WB = 30 lb
μ = 0.5
μ BA = 0.6
μ AC = 0.4
θ = 20 deg
Assume that B slips on A, but A does not move.
Guesses
NB = 1 lb
Given
WD = 1 lb
TB = 1 lb
NC = 1 lb
F C = 1 lb
π
μ
2
W D = TB e
− TB + F C = 0
NC − WA − WB = 0
−TB + μ BA NB cos ( θ ) − NB sin ( θ ) = 0
NB cos ( θ ) + μ BA NB sin ( θ ) − WB = 0
852
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Engineering Mechanics - Statics
Chapter 8
⎛ WD ⎞
⎜
⎟
⎜ TB ⎟
⎜ NB ⎟ = Find ( W , T , N , N , F )
D B B C C
⎜
⎟
⎜ NC ⎟
⎜F ⎟
⎝ C⎠
⎛ WD ⎞ ⎛ 12.75 ⎞
⎜
⎟ ⎜
⎟
⎜ TB ⎟ ⎜ 5.81 ⎟
⎜ NB ⎟ = ⎜ 26.20 ⎟ lb
⎜
⎟ ⎜
⎟
N
⎜ C ⎟ ⎜ 80.00 ⎟
⎜ F ⎟ ⎝ 5.81 ⎠
⎝ C⎠
Now check the assumption that A does not move
F Cmax = μ AC NC
F Cmax = 32.00 lb
Since F C = 5.81 lb < F Cmax = 32.00 lb then our assumption is good.
WD = 12.75 lb
Problem 8-94
Blocks A and B have weight W, and D weighs WD. Using the coefficients of static
friction indicated, determine the frictional force between blocks A and B and between
block A and the floor C.
Given:
W = 75 lb
μ BA = 0.6
WD = 30 lb
μ AC = 0.4
μ = 0.5
θ = 20 deg
Solution:
π
μ
2
W D = TB e
WD
TB =
1
e
Check
2
μπ
F C = TB
F C = 13.68 lb
NC − 2W = 0
NC = 2W
F Cmax = μ AC NC
F Cmax = 60.00 lb
TB = 13.679 lb
NC = 150.00 lb
Since F C = 13.68 lb < F Cmax = 60.00 lb then the system does not slip at C.
853
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
For block B:
The initial guessess:
NB = 1 lb
F B = 1 lb
Given
+
↑Σ Fy = 0;
+
Σ F x = 0;
→
⎛ NB ⎞
⎜ ⎟ = Find ( NB , FB)
⎝ FB ⎠
NB cos ( θ ) + FB sin ( θ ) − W = 0
F B cos ( θ ) − NB sin ( θ ) − TB = 0
NB = 65.80 lb
F B = 38.51 lb
Check
F Bmax = μ BA NB
F Bmax = 39.48 lb
Since F B = 38.51 lb < FBmax = 39.48 lb then no slipping occurs between the blocks
Problem 8-95
Show that the frictional relationship between the belt tensions, the coefficient of friction μ, and the
angular contacts α and β for the V-belt is T2=T1eμβ/sin(α/2) when the belt is on the verge of slipping.
Solution:
FBD of a section of the belt is shown.
Proceeding in the general manner:
ΣF x = 0;
⎛ dθ ⎞ + T cos ⎛ dθ ⎞ + 2μ dN = 0
⎟
⎜ ⎟
⎝2⎠
⎝2⎠
−( T + dT) cos ⎜
854
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Engineering Mechanics - Statics
Chapter 8
⎛ dθ ⎞ − T sin ⎛ dθ ⎞ + 2dN sin ⎛ α ⎞ = 0
⎟
⎜ ⎟
⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
−( T + dT) sin ⎜
ΣF y = 0;
Since dθ, dN, and dT are small, these become
⎛α⎞
⎟
⎝2⎠
Tdθ = 2dN sin ⎜
dT = 2μ dN
dT
=μ
T
Combine
dθ
⎛α⎞
⎟
⎝2⎠
sin⎜
θ = 0 , T = T1
Integrate from
to
θ = β , T = T2
μβ
We get,
T2 = T1 e
⎛α⎞
⎟
⎝2⎠
sin⎜
Q.E.D
Problem 8-96
A V-fan-belt (V-angle θ) of an automobile engine passes around the hub H of a generator G
and over the housing F to a fan. If the generator locks, and the maximum tension the belt can
sustain is Tmax, determine the maximum possible torque M resisted by the axle as the belt slips
over the hub. Assume that slipping of the belt occurs only at H and that the coefficient of
kinetic friction for the hub is μs.
Given:
θ = 60 deg
b = 2 in
a = 2 in
c = 1.25 ft
Tmax = 175 lb
μ s = 0.25
Solution:
−T1 a + Tmax a − M = 0
⎞
⎟
1
⎜ sin⎛⎜ θ ⎞⎟ ⎟
⎝ ⎝2 ⎠⎠
⎛
− μ s⎜
T1 = Tmax e
π
T1 = 36.4 lb
855
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
M = −T1 a + Tmax a
M = 23.1 lb⋅ ft
Problem 8-97
A cable is attached to the plate B of mass MB, passes over a fixed peg at C, and is attached to
the block at A. Using the coefficients of static friction shown, determine the smallest mass of
block A so that it will prevent sliding motion of B down the plane.
Given:
MB = 20 kg
μ A = 0.2
θ = 30 deg
μ B = 0.3
m
μ C = 0.3
g = 9.81
2
s
Solution:
Iniitial guesses:
T1 = 1 N
T2 = 1 N NA = 1 N
NB = 1 N
MA = 1 kg
Given
Block A:
ΣF x = 0;
T1 − μ A NA − MA g sin ( θ ) = 0
ΣF y = 0;
NA − MA g cos ( θ ) = 0
Plate B:
ΣF x = 0;
ΣF y = 0;
Peg C:
T2 − MB g sin ( θ ) + μ B NB + μ A NA = 0
NB − NA − MB g cos ( θ ) = 0
μ Cπ
T2 = T1 e
856
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ T1 ⎞
⎜ ⎟
⎜ T2 ⎟
⎜ NA ⎟ = Find ( T , T , N , N , M )
1 2 A B
A
⎜ ⎟
⎜ NB ⎟
⎜M ⎟
⎝ A⎠
MA = 2.22 kg
Problem 8-98
The simple band brake is constructed so that the ends of the friction strap are connected to the
pin at A and the lever arm at B. If the wheel is subjected to a torque M, determine the smallest
force P applied to the lever that is required to hold the wheel stationary. The coefficient of static
friction between the strap and wheel is μs.
Given:
M = 80 lb⋅ ft
β = 45 deg
μ s = 0.5
r = 1.25 ft
α = 20 deg
a = 1.5 ft
b = 3 ft
Solution:
The initial guesses: T1 = 10 lb
T2 = 20 lb
P = 30 lb
Given
T1 r + M − T2 r = 0
μ s( π + α + β )
T2 = T1 e
T2 sin ( β ) a − ( a + b)P = 0
857
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ T1 ⎞
⎜ ⎟
⎜ T2 ⎟ = Find ( T1 , T2 , P)
⎜P⎟
⎝ ⎠
⎛ T1 ⎞ ⎛ 8.56 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ T2 ⎠ ⎝ 72.56 ⎠
P = 17.10 lb
Problem 8-99
The uniform beam of weight W1 is supported by the rope which is attached to the end of the
beam, wraps over the rough peg, and is then connected to the block of weight W2. If the
coefficient of static friction between the beam and the block, and between the rope and the
peg, is μs, determine the maximum distance that the block can be placed from A and still
remain in equilibrium. Assume the block will not tip.
Given:
W1 = 50 lb
W2 = 100 lb
μ s = 0.4
a = 1 ft
b = 10 ft
Solution:
Block:
ΣF y = 0; N − W2 = 0
N = W2
N = 100.00 lb
ΣF x = 0; T1 − μ s N = 0
T1 = μ s N
T1 = 40.00 lb
⎛π⎞
μ s⎜ ⎟
2
T2 = T1 e ⎝ ⎠
T2 = 74.97 lb
System:
b⎞
⎟ + T2 b = 0
⎝ 2⎠
ΣΜA = 0; −W2 d − T1 a − W1 ⎛⎜
858
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
T2 b − T1 a − W1 ⎛⎜
b⎞
⎟
⎝ 2⎠
d =
W2
d = 4.6 ft
Problem 8-100
The uniform concrete pipe has weight W and is unloaded slowly from the truck bed using the rope
and skids shown. If the coefficient of kinetic friction between the rope and pipe is μk ,determine
the force the worker must exert on the rope to lower the pipe at constant speed. There is a pulley
at B, and the pipe does not slip on the skids. The lower portion of the rope is parallel to the skids.
Given:
W = 800 lb
μ k = 0.3
α = 15 deg
β = 30 deg
Solution:
−W r sin ( β ) + T2 cos ( α ) ( r cos ( α ) + r cos ( β ) ) + T2 sin ( α ) ( r sin ( α ) + r sin ( β ) ) = 0
T2 =
W sin ( β )
T2 = 203.47 lb
1 + cos ( α ) cos ( β ) + sin ( α ) sin ( β )
− μ k( π + β − α )
T1 = T2 e
T1 = 73.3 lb
Problem 8-101
A cord having a weight density γ and a total length L is suspended over a peg P as shown. If the
coefficient of static friction between the peg and cord is μs ,determine the longest length h which
one side of the suspended cord can have without causing motion. Neglect the size of the peg and
the length of cord draped over it.
859
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
γ = 0.5
lb
ft
L = 10 ft
μ s = 0.5
Solution:
μβ
T2 = T1 e
μ sπ
γ h = γ ( L − h) e
T 1 = γ ( L − h)
T2 = γ h
⎛ eμ sπ ⎞
⎟
h = L⎜
⎜
μ sπ ⎟
⎝1 + e ⎠
h = 8.28 ft
Problem 8-102
Granular material, having a density ρ is transported on a conveyor belt that slides over the fixed
surface, having a coefficient of kinetic friction of μk. Operation of the belt is provided by a
motor that supplies a torque M to wheel A.The wheel at B is free to turn, and the coefficient of
static friction between the wheel at A and the belt is μΑ. If the belt is subjected to a pretension
T when no load is on the belt, determine the greatest volume V of material that is permitted on
the belt at any time without allowing the belt to stop. What is the torque M required to drive the
belt when it is subjected to this maximum load?
Units used:
6
Mg = 10 g
Given:
r = 100 mm
μ A = 0.4
μ k = 0.3
kg
ρ = 1500
m
3
T = 300 N
g = 9.81
m
2
s
860
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Solution:
Chapter 8
(μ A)π
T2 = T e
T2 = 1053.9 N
− M − T r + T2 r = 0
M = − T r + T2 r
T2 − μ k m1 g − T = 0
m1 =
Wheel A :
ΣMA = 0;
Belt
ΣF x = 0;
V =
m1
T2 − T
μk g
M = 75.4 N⋅ m
m1 = 256.2 kg
V = 0.17 m
ρ
3
Problem 8-103
Blocks A and B have a mass MA and MB, respectively. If the coefficient of static friction
between A and B and between B and C is μs and between the ropes and the pegs D and E μ's,
determine the smallest force F needed to cause motion of block B .
Units Used:
3
kN = 10 N
Given:
θ = 45 deg
μ s = 0.25
MA = 100 kg
μ's = 0.5
MB = 150 kg
P = 30 N
g = 9.81
m
2
s
Solution:
Assume no slipping between A & B.
Guesses
F = 1N
NAB = 1 N
F AB = 1 N
NBC = 1 N
F BE = 1 N
F AD = 1 N
Given
⎛π ⎞
μ 's⎜ + θ ⎟
2 ⎠
F = FBE e ⎝
π
μ 's
2
F AD = P e
861
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
F BE cos ( θ ) − μ s NBC − FAB = 0
F BE sin ( θ ) − NAB − MB g + NBC = 0
−F AD + F AB = 0
NAB − MA g = 0
⎛ F ⎞
⎜N ⎟
⎜ AB ⎟
⎜ FAB ⎟
⎜
⎟ = Find ( F , NAB , FAB , NBC , FBE , FAD)
NBC
⎜
⎟
⎜ FBE ⎟
⎟
⎜
⎝ FAD ⎠
F ABmax = μ s NAB
Now check assumption
Since F AB = 65.8 N
< FABmax = 245.3 N
⎛ F ⎞ ⎛ 2.49 ⎞
⎜N ⎟ ⎜
⎟
⎜ AB ⎟ ⎜ 0.98 ⎟
⎜ FAB ⎟ ⎜ 0.07 ⎟
⎜
⎟=⎜
⎟ kN
⎜ NBC ⎟ ⎜ 1.91 ⎟
⎜ FBE ⎟ ⎜ 0.77 ⎟
⎟ ⎜ 0.07 ⎟
⎜
⎠
⎝ FAD ⎠ ⎝
F ABmax = 245.25 N
then our assumption is correct
F = 2.49 kN
Problem 8-104
Blocks A and B weigh W1 and W2, respectively. Using the coefficients of static friction
indicated, determine the greatest weight W of block E without causing motion.
Given:
W1 = 50 lb
W2 = 30 lb
d = 12
μ A = 0.3
a = 1.5 ft
μ B = 0.5
b = 2 ft
μ C = 0.2
c = 5
μ D = 0.3
862
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
Assume that the wedge slips on the ground, but
the block does not slip on the wedge and the
block does not tip.
Guesses
W = 1 lb
TD = 1 lb TC = 1 lb
NA = 1 lb
F A = 1 lb NB = 1 lb
x = 1 ft
F Amax = 1 lb
Given
π
μC
W
2
= TC e
2
⎛
π
μD
W
2
= TD e
2
c
⎞+N ⎛
⎞+T =0
B
D
⎟
⎜
⎟
2
2
2
2
⎝ c +d ⎠
⎝ c +d ⎠
F A − μ B NB⎜
⎛
d
c
⎞+μ N ⎛
⎞−W −N =0
B
B
2
A
⎟
⎜
2
2
2
2⎟
⎝ c +d ⎠
⎝ c +d ⎠
NB⎜
d
TC − F A = 0
NA − W1 = 0
−TC b + NA x = 0
⎛ W ⎞
⎜ T ⎟
⎜ D ⎟
⎜ TC ⎟
⎜
⎟
NA
⎜
⎟ = Find ( W , T , T , N , F , N , x , F
D C A A B
Amax )
⎜ FA ⎟
⎜
⎟
⎜ NB ⎟
⎜ x ⎟
⎜
⎟
F
Amax
⎠
⎝
F Amax = μ A NA
W = 8.15 lb
863
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Check assumptions
Since FA = 2.97 lb < FAmax = 15.00 lb then the block does not slip relative to the wedge.
Since x = 0.12 ft <
a
2
= 0.75 ft then the block does not tip.
Problem 8-105
Block A has mass mA and rests on surface B for which the coefficient of static friction is μsAB. If the
coefficient of static friction between the cord and the fixed peg at C is μsC , determine the greatest
mass mD of the suspended cylinder D without causing motion.
Given:
mA = 50 kg
μ sAB = 0.25
μ sC = 0.3
a = 0.3 m
b = 0.25 m
c = 0.4 m
d = 3
f = 4
g = 9.81
m
2
s
Solution:
Assume block A slips but does not tip.
The initial guesses:
Given
NB = 100 N
T = 50 N
f
β = π − atan ⎛⎜ ⎟⎞
⎝ d⎠
mD = 1 kg
x = 10 mm
μ sCβ
mD g = T e
d
⎛
⎞
⎜ 2 2 ⎟ T − mA g + NB = 0
⎝ f +d ⎠
⎛ − f ⎞T + μ N = 0
sAB B
⎜ 2 2⎟
⎝ f +d ⎠
f
d
⎛
⎛
⎞
⎞ ⎛ b⎞
⎜ 2 2 ⎟ T a − ⎜ 2 2 ⎟ T⎜⎝ 2 ⎟⎠ − NB x = 0
⎝ f +d ⎠
⎝ f +d ⎠
864
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ NB ⎞
⎜ ⎟
⎜ T ⎟ = Find ( N , T , m , x)
B
D
⎜ mD ⎟
⎜ ⎟
⎝ x ⎠
Since x = 51.6 mm <
b
2
⎛ NB ⎞ ⎛ 413.05 ⎞
⎜ ⎟=⎜
⎟N
⎝ T ⎠ ⎝ 129.08 ⎠
mD = 25.6 kg
x = 0.052 m
= 125 mm our assumption is correct
mD = 25.6 kg
Problem 8-106
Block A rests on the surface for which the coefficient of friction is μsAB. If the mass of the
suspended cylinder is mD, determine the smallest mass mA of block A so that it does not slip or tip.
The coefficient of static friction between the cord and the fixed peg at C is μsC.
Units Used:
g = 9.81
m
2
s
Given:
μ sAB = 0.25
mD = 4 kg
μ sC = 0.3
a = 0.3 m
b = 0.25 m
c = 0.4 m
d = 3
f = 4
Solution:
Assume that slipping is the crtitical motion
The initial guesses:
Given
NB = 100 N
μ sCβ
mD g = T e
T = 50 N
f
β = π − atan ⎛⎜ ⎟⎞
⎝ d⎠
mA = 1 kg
x = 10 mm
d
⎛
⎞
⎜ 2 2 ⎟ T − mA g + NB = 0
⎝ f +d ⎠
865
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ − f ⎞T + μ N = 0
sAB B
⎜ 2 2⎟
⎝ f +d ⎠
⎛ NB ⎞
⎜ ⎟
⎜ T ⎟ = Find ( N , T , m , x)
B
A
⎜ mA ⎟
⎜ ⎟
⎝ x ⎠
Since x = 51.6 mm <
b
2
f
d
⎛
⎛
⎞
⎞ b
⎜ 2 2 ⎟ T a − ⎜ 2 2 ⎟ T 2 − NB x = 0
⎝ f +d ⎠
⎝ f +d ⎠
⎛ NB ⎞ ⎛ 64.63 ⎞
⎜ ⎟=⎜
⎟N
⎝ T ⎠ ⎝ 20.20 ⎠
mA = 7.82 kg
x = 0.052 m
= 125 mm our assumption is correct
mA = 7.82 kg
Problem 8-107
The collar bearing uniformly supports an axial force P. If the coefficient of static friction is
μs, determine the torque M required to overcome friction.
Given:
a = 2 in
b = 3 in
P = 800 lb
μ s = 0.3
Solution:
⎡ ⎛ b ⎞ 3 ⎛ a ⎞ 3⎤
⎢⎜ ⎟ − ⎜ ⎟ ⎥
a
⎝ 2⎠ ⎝ 2⎠ ⎥
M = μ s P⎢
2⎥
b
⎢ b 2
a⎞
⎞
⎛
⎛
⎢⎜ ⎟ − ⎜ ⎟ ⎥
⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
⎡1 ⎛ a + a b + b ⎞⎤
a
⎟⎥
μ s P⎢ ⎜
b
⎣2 ⎝ a + b ⎠⎦
2
M =
2
M = 304.00 lb⋅ in
Problem 8-108
The collar bearing uniformly supports an axial force P. If a torque M is applied to the shaft and
causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface
of contact.
866
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
a = 2 in
b = 3 in
P = 500 lb
M = 3 lb ft
Solution:
⎡⎛ b ⎞ 3 ⎛ a ⎞ 3⎤
⎢⎜ ⎟ − ⎜ ⎟ ⎥
3
3
2 ⎠ ⎥ ⎛ aμ k P ⎞ ⎛⎜ b − a ⎟⎞
a
2⎠
⎝
⎝
⎢
M = μk P
=⎜
⎟
2⎥ ⎝ 2b ⎠ ⎜ 2
2⎟
b
⎢ b 2
a
⎝b − a ⎠
⎢⎛⎜ ⎟⎞ − ⎛⎜ ⎟⎞ ⎥
⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
(2
2
aP b − a
3
2M b b − a
μk =
(3
)
)
μ k = 0.0568
Problem 8-109
The double-collar bearing is subjected to an axial force
P . Assuming that collar A supports kP and collar B
supports (1 − k)P, both with a uniform distribution of
pressure, determine the maximum frictional moment M
that may be resisted by the bearing.
Units Used:
3
kN = 10 N
Given:
P = 4 kN
a = 20 mm
b = 10 mm
c = 30 mm
μ s = 0.2
867
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
k = 0.75
Solution:
⎤
2 ⎡⎢ c − b
a −b
μs
kP +
( 1 − k)P⎥
2
2
2
⎥
3 ⎢ 2
a −b
⎣c − b
⎦
3
M =
3
3
3
M = 16.1 N⋅ m
Problem 8-110
The annular ring bearing is subjected to a thrust P. If the coefficient of static friction is μs,
determine the torque M that must be applied to overcome friction.
Given:
P = 800 lb
μ s = 0.35
a = 0.75 in
b = 1 in
c = 2 in
Solution:
⎛ b − c ⎟⎞
2
M = μ s P⎜
⎜ b2 − c2 ⎟
3
⎝
⎠
3
3
M = 36.3 lb⋅ ft
Problem 8-111
The floor-polishing machine rotates at a constant angular velocity. If it has weight W, determine the
couple forces F the operator must apply to the handles to hold the machine stationary. The
coefficient of kinetic friction between the floor and brush is μk. Assume the brush exerts a uniform
pressure on the floor.
Given:
W = 80 lb
μ k = 0.3
868
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
a = 1.5 ft
b = 2 ft
Solution:
Fa =
2
b
μ k W⎛⎜ ⎟⎞
3
⎝ 2⎠
F =
1
b
μ k W ⎛⎜ ⎟⎞
3
⎝ a⎠
F = 10.7 lb
Problem 8-112
The plate clutch consists of a flat plate A that slides over the rotating shaft S. The shaft is fixed to the
driving plate gear B. If the gear C, which is in mesh with B, is subjected to a torque M, determine the
smallest force P, that must be applied via the control arm, to stop the rotation. The coefficient of
static friction between the plates A and D is μs. Assume the bearing pressure between A and D to be
uniform.
Given:
M = 0.8 N⋅ m
μ s = 0.4
a = 150 mm
b = 200 mm
c = 100 mm
d = 125 mm
e = 150 mm
f = 30 mm
Solution:
M
f
F = 26.667 N
M2 = F e
M2 = 4.00 N⋅ m
F =
869
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ d − c ⎟⎞
2
M2 = μ s P' ⎜
3
⎜ d2 − c2 ⎟
⎝
⎠
P' =
P' b − P a = 0
P = P' ⎜
3
3M2 ⎛ d2 − c2 ⎞
3
⎜
⎟
⎝
⎠
P' = 88.5 N
2μ s ⎜ d3 − c3 ⎟
⎛ b⎞
⎟
⎝ a⎠
P = 118 N
Problem 8-113
The shaft of diameter b is held in the hole such that the normal pressure acting around the
shaft varies linearly with its depth as shown. Determine the frictional torque that must be
overcome to rotate the shaft.
Given:
a = 6 in
p0 = 60
lb
in
2
b = 4 in
μ s = 0.2
Solution:
⌠
N = ⎮
⎮
⌡
a
⎛ x ⎞ p 2π ⎛ b ⎞ dx
⎜ ⎟ 0 ⎜ ⎟
⎝ a⎠
⎝ 2⎠
⎛ b⎞
⎟
⎝ 2⎠
T = μ s N⎜
T = 905 lb⋅ in
0
Problem 8-114
Because of wearing at the edges, the pivot bearing is subjected to a conical pressure distribution at its
surface of contact. Determine the torque M required to overcome friction and turn the shaft, which
supports an axial force P. The coefficient of static friction is μs. For the solution, it is necessary to
determine the peak pressure p0 in terms of P and the bearing radius R.
Solution:
⌠
P=⎮
⎮
⌡
2π
0
⌠
⎮
⎮
⌡
R
⎛ p − p r ⎞ r dr dθ = π p0 R
⎜ 0
0 ⎟
R⎠
3
⎝
2
0
p0 =
3P
πR
2
dM = rdF = rμ dN = rμ pdA = rμ p rdθ dr
870
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 8
R
⌠
p0 ⎞ 2 ⌠2 π
⎛
π
π ⎛ 3P ⎞ 3 μ P R
3
⎮
M=
μ ⎜ p0 −
r⎟ r dr ⎮ 1 dθ = μ p0 R = μ ⎜
R =
⎮
2⎟
2
R ⎠
6
6
⌡0
⎝
⌡
⎝ πR ⎠
0
M=
Thus,
μP R
2
Problem 8-115
The conical bearing is subjected to a constant
pressure distribution at its surface of contact. If the
coefficient of static friction is μs, determine the
torque M required to overcome friction if the shaft
supports an axial force P.
Solution:
The differential Area (shaded)
⎛ dr ⎞ = 2π rdr
⎟
⎝ cos ( θ ) ⎠ cos ( θ )
dA = 2πr⎜
⌠
P=⎮
⌡
⌠
p cos ( θ ) d A = ⎮
⎮
⌡
R
⎛ 2π r ⎞ dr = 2π p ⌠
2
⎮ r dr = π p R
⎟
(
)
⌡
⎝ cos θ ⎠
0
p cos ( θ ) ⎜
p=
871
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P
πR
2
Engineering Mechanics - Statics
dN = pdA =
Chapter 8
2P
⎛ 2π rdr ⎞ =
rdr
⎜
⎟
2 ⎝ cos ( θ ) ⎠
2
(
)
πR
R cos θ
P
R
⌠
⌠
⎛ 2μ s P ⎞ ⌠ 2
⎮
M = ⎮ r dF =
rμ s dN = ⎜
r dr
⎟⎮
⎮
2
⌡
⌡
⌡
⎝ R cos ( θ ) ⎠ 0
⎛ 2μ s P ⎞ R3 2μ s P R
⎟ 3 = 3 cos ( θ )
2
(
)
⎝ R cos θ ⎠
M=⎜
Problem 8-116
The tractor is used to push the pipe of weight W. To do this it must overcome the frictional forces
at the ground, caused by sand. Assuming that the sand exerts a pressure on the bottom of the pipe
as shown, and the coefficient of static friction between the pipe and the sand is μs, determine the
force required to push the pipe forward. Also, determine the peak pressure p0.
Given:
W = 1500 lb
μ s = 0.3
L = 12 ft
r = 15 in
872
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
π
⌠2
⎮
2L ⎮ p0 cos ( θ ) r dθ cos ( θ ) − W = 0
⌡
+
↑Σ Fy = 0;
0
π
⌠2
⎮
2
⎛π⎞
W = 2 p0 L r ⎮ cos θ dθ = p0 L r⎜ ⎟
⌡0
⎝ 2⎠
p0 = 2
W
rLπ
+
Σ F x = 0;
→
p0 = 0.442
lb
in
2
π
⌠2
⎮
F =
μ s p0 L r cos ( θ ) dθ
⎮
⌡− π
F = 573 lb
2
Problem 8-117
Assuming that the variation of pressure at the bottom of the pivot bearing is defined as p =
p0(R2/r), determine the torque M needed to overcome friction if the shaft is subjected to an axial
force P. The coefficient of static friction is μs. For the solution, it is necessary to determine p0 in
terms of P and the bearing dimensions R 1 and R2.
Solution:
2π
⌠
P=⎮
⌡0
R2
⌠
⎮
⌡R
p r dr dθ
1
873
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Engineering Mechanics - Statics
⌠
⎮
P=
⎮
⌡
2π
0
R2
⌠
⎮
⎮
⌡R
Chapter 8
⎛ R2 ⎞
⎟ r dr dθ
⎝ r ⎠
p0 ⎜
1
P = 2π p0 R2 ( R 2 − R 1 )
p0 =
P
2π R 2 ( R 2 − R 1 )
2π
⌠
⌠
M = ⎮ r dF = ⎮
⌡
⌡
0
A
⌠
⎮
M=
⎮
⌡
2π
0
⌠
⎮
⎮
⌡
2π
R2
⌠
⎮
⌡R
rμ s p0 r dr dθ
1
(
⎛ R2 ⎞ 2
2
2
⎟ r dr dθ = π μ s p0 R 2 R 2 − R 1
r
⎝ ⎠
μ s p0 ⎜
)
0
(
P
⎤R R 2 − R 2
⎡
M = π μ s⎢
2 2
1
⎥
⎣2π R2( R2 − R1 )⎦
)
M=
μ s P(R2 + R1)
2
Problem 8-118
A disk having an outer diameter a fits loosely over a fixed shaft having a diameter b. If the
coefficient of static friction between the disk and the shaft is μs, determine the smallest
vertical force P , acting on the rim, which must be applied to the disk to cause it to slip over
the shaft. The disk weighs W.
Given:
a = 8 in
b = 3 in
μ s = 0.15
W = 10 lb
Solution:
φ s = atan ( μ s)
874
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Engineering Mechanics - Statics
Chapter 8
⎛ b ⎞ sin ( φ )
⎟
s
⎝ 2⎠
rf = ⎜
⎡⎛ a ⎞ − r ⎤
⎟ f⎥
⎣⎝ 2 ⎠ ⎦
l = ⎢⎜
ΣM0 = 0;
W rf − l P = 0
⎛ rf ⎞
⎟
⎝l⎠
P = W⎜
P = 0.59 lb
Problem 8-119
The pulley has a radius r and fits loosely on the shaft of diameter d. If the loadings acting on the belt
cause the pulley to rotate with constant angular velocity, determine the frictional force between the
shaft and the pulley and compute the coefficient of kinetic friction. The pulley has weight W.
Given:
r = 3 in
d = 0.5 in
W = 18 lb
F 1 = 5 lb
F 2 = 5.5 lb
Solution:
+
↑Σ Fy = 0;
R − W − F1 − F2 = 0
R = W + F1 + F2
ΣMO = 0;
−F 2 r + F1 r + R rf = 0
rf = r⎜
rf =
d
sin ( φ k)
2
R = 28.5 lb
⎛ F2 − F1 ⎞
⎟
⎝ R ⎠
rf = 0.05263 in
φ k = asin ⎜ 2 ⎟
⎛ rf ⎞
⎝ d⎠
φ k = 12.15 deg
μ = tan ( φ k)
μ = 0.215
Also,
ΣMO = 0;
⎛ d⎞ = 0
⎟
⎝ 2⎠
−F 2 r + F1 r + F ⎜
⎛ F2 − F1 ⎞
⎟
⎝ d ⎠
F = 2 r⎜
F = 6 lb
875
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-120
The pulley has a radius r and fits loosely on the shaft of diameter d. If the loadings acting on the belt
cause the pulley to rotate with constant angular velocity, determine the frictional force between the
shaft and the pulley and compute the coefficient of kinetic friction. Neglect the weight of the pulley .
Given:
r = 3 in
d = 0.5 in
W = 0 lb
F 1 = 5 lb
F 2 = 5.5 lb
Solution:
+
↑Σ Fy = 0;
R − W − F1 − F2 = 0
R = W + F1 + F2
ΣMO = 0;
−F 2 r + F1 r + R rf = 0
rf = r⎜
rf =
d
sin ( φ k)
2
R = 10.5 lb
⎛ F2 − F1 ⎞
⎟
⎝ R ⎠
rf = 0.14286 in
φ k = asin ⎜ 2 ⎟
⎛ rf ⎞
⎝ d⎠
φ k = 34.85 deg
μ = tan ( φ k)
μ = 0.696
Also,
ΣMO = 0; −F 2 r + F1 r + F
d
=0
2
⎛ F2 − F1 ⎞
⎟
⎝ d ⎠
F = 2 r⎜
F = 6 lb
Problem 8-121
A pulley of mass M has radius a and the axle has a diameter D. If the coefficient of kinetic
friction between the axle and the pulley is μk determine the vertical force P on the rope required
to lift the block of mass MB at constant velocity.
Given:
a = 120 mm
M = 5 kg
D = 40 mm
876
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
μ k = 0.15
MB = 80 kg
Solution:
φ k = atan ( μ k)
⎛ D ⎞ sin ( φ )
⎟
k
⎝2⎠
rf = ⎜
ΣMp = 0;
MB g( a + rf) + M g rf − P( a − rf) = 0
P =
MB g( a + rf) + M g rf
P = 826 N
a − rf
Problem 8-122
A pulley of mass M has radius a and the axle has a diameter D. If the coefficient of kinetic
friction between the axle and the pulley is μk determine the force P on the rope required to lift
the block of mass MB at constant velocity. Apply the force P horizontally to the right (not as
shown in the figure).
Given:
a = 120 mm
M = 5 kg
D = 40 mm
μ k = 0.15
MB = 80 kg
g = 9.81
Solution:
φ k = atan ( μ k)
rf =
m
2
s
D
sin ( φ k)
2
877
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Guesses
P = 1N
α = 1 deg
R = 1N
Given
R cos ( α ) − MB g − M g = 0
P − R sin ( α ) = 0
MB g a − P a + R rf = 0
⎛P ⎞
⎜ R ⎟ = Find ( P , R , α )
⎜ ⎟
⎝α ⎠
P = 814 N
Problem 8-123
A wheel on a freight car carries a load W. If the axle of the car has a diameter D, determine the
horizontal force P that must be applied to the axle to rotate the wheel. The coefficient of kinetic
friction is μk.
Units Used:
kip = 1000 lb
Given:
W = 20 kip
D = 2 in
μ k = 0.05
r = 16 in
Solution:
ΣF x = 0;
P − R sin ( φ ) = 0
ΣF y = 0;
R cos ( φ ) − W = 0
Thus,
P = W tan ( φ )
φ k = atan ( μ k)
φ k = 2.86 deg
878
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
⎛ D ⎞ sin ( φ )
⎟
k
⎝2⎠
rf = 0.04994 in
⎛ rf ⎞
⎝ r⎠
φ = 0.1788 deg
rf = ⎜
φ = asin ⎜ ⎟
P = W tan ( φ )
P = 62.4 lb
Problem 8-124
The trailer has a total weight W and center of gravity at G which is directly over its axle. If the
axle has a diameter D, the radius of the wheel is r, and the coefficient of kinetic friction at the
bearing is μk, determine the horizontal force P needed to pull the trailer.
Given:
W = 850 lb
r = 1.5 ft
μ k = 0.08
D = 1 in
Solution:
ΣF x = 0;
P − R sin ( φ ) = 0
ΣF y = 0;
R cos ( φ ) − W = 0
Thus,
P = W tan ( φ )
φ k = atan ( μ k)
φ k = 4.57 deg
⎛ D ⎞ sin ( φ )
⎟
k
⎝2⎠
rf = 0.03987 in
φ = asin ⎜ ⎟
⎛ rf ⎞
⎝ r⎠
φ = 0.1269 deg
P = W tan ( φ )
P = 1.88 lb
rf = ⎜
879
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-125
The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic
friction between the shaft and the collar is μk, determine the force P on the horizontal segment
of the belt so that the collar rotates counterclockwise with a constant angular velocity. Assume
that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight
and thickness of the belt and collar. The radius, measured from the center of the collar to the
mean thickness of the belt is R.
Given:
r = 2 in
μ k = 0.3
R = 2.25 in
F = 20 lb
Solution:
φ k = atan ( μ k)
φ k = 16.699 deg
rf = r sin ( φ k)
rf = 0.5747 in
Equilibrium:
+
↑Σ Fy = 0;
Ry − F = 0
Ry = F
+
Σ F x = 0;
→
P − Rx = 0
Rx = P
R=
Guess
P = 1 lb
Given
−
2
2
Rx + Ry =
2
R y = 20.00 lb
P +F
( P2 + F2) rf + F R − P R = 0
2
P = Find ( P)
P = 13.79 lb
Problem 8-126
The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic
friction between the shaft and the collar is μk, determine the force P on the horizontal segment
880
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that
the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and
thickness of the belt and collar. The radius, measured from the center of the collar to the mean
thickness of the belt is R.
Given:
r = 2 in
μ k = 0.3
R = 2.25 in
F = 20 lb
Solution:
φ k = atan ( μ k)
φ k = 16.699 deg
rf = r sin ( φ k)
rf = 0.5747 in
Equilibrium:
+
↑Σ Fy = 0;
+
Σ F x = 0;
→
Ry − F = 0
Ry = F
P − Rx = 0
Rx = P
R=
2
2
Rx + Ry =
Guess
P = 1 lb
Given
( P2 + F2)rf + F R − P R = 0
2
P +F
R y = 20.00 lb
2
P = Find ( P)
P = 29.00 lb
Problem 8-127
The connecting rod is attached to the piston by a pin at B of diameter d1 and to the crank shaft by a
bearing A of diameter d2. If the piston is moving downwards, and the coefficient of static friction at
these points is μs, determine the radius of the friction circle at each connection.
Given:
d1 = 0.75 in
881
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
d2 = 2 in
μ s = 0.2
Solution:
rfA =
1
rfB =
1
2
2
d2 μ s
rfA = 0.2 in
d1 μ s
rfB = 0.075 in
Problem 8-128
The connecting rod is attached to the piston by a pin at B of diameter d1 and to the crank shaft
by a bearing A of diameter d2. If the piston is moving upwards, and the coefficient of static
friction at these points is μs, determine the radius of the friction circle at each connection.
Given:
d1 = 20 mm
d2 = 50 mm
μ s = 0.3
Solution:
rfA =
1
rfB =
1
2
2
d2 μ s
rfA = 7.50 mm
d1 μ s
rfB = 3 mm
882
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-129
The lawn roller has mass M. If the arm BA is held at angle θ from the horizontal and the
coefficient of rolling resistance for the roller is r, determine the force P needed to push the
roller at constant speed. Neglect friction developed at the axle, A, and assume that the
resultant force P acting on the handle is applied along arm BA.
Given:
M = 80 kg
θ = 30 deg
a = 250 mm
r = 25 mm
Solution:
r
θ 1 = asin ⎛⎜ ⎟⎞
⎝ a⎠
θ 1 = 5.74 deg
ΣM0 = 0;
−r M g − P sin ( θ ) r + P cos ( θ ) a cos ( θ 1 ) = 0
P =
rM g
−sin ( θ ) r + cos ( θ ) a cos ( θ 1 )
P = 96.7 N
Problem 8-130
The handcart has wheels with a diameter D. If a crate having a weight W is placed on the cart,
determine the force P that must be applied to the handle to overcome the rolling resistance. The
coefficient of rolling resistance is μ. Neglect the weight of the cart.
883
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
W = 1500 lb
D = 6 in
a = 0.04 in
c = 3
b = 4
Solution:
Guesses
N = 1 lb
P = 1 lb
Given
N − W − P⎛
c
⎞
⎜ 2 2⎟ = 0
⎝ c +b ⎠
⎛ b ⎞ P = N⎛ 2a ⎞
⎜ ⎟
⎜ 2 2⎟
⎝ D⎠
⎝ b +c ⎠
⎛N⎞
⎜ ⎟ = Find ( N , P)
⎝P⎠
N = 1515 lb
P = 25.3 lb
Problem 8-131
The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for
the cylinder's top and bottom surfaces are aA and aB respectively, show that a force having a
magnitude of P = [W(aA + aB)]/2r is required to move the load and thereby roll the cylinder
forward. Neglect the weight of the cylinder.
884
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Solution:
+
Σ F x = 0;
→
R Ax − P = 0
R Ax = P
+
R Ay − W = 0
R Ay = P
↑Σ Fy = 0;
ΣMB = 0;
P ( r cos ( φ A) + r cos ( φ B) ) − W ( aA + aB) = 0
Since φΑ and φB are very small,
cos ( φ A) = cos ( φ B) = 1 Hence from Eq.(1)
P=
W ( aA + aB)
2r
(QED)
Problem 8-132
A steel beam of mass M is moved over a level surface using a series of rollers of diameter D
for which the coefficient of rolling resistance is ag at the ground and as at the bottom surface
of the beam. Determine the horizontal force P needed to push the beam forward at a constant
speed. Hint: Use the result of Prob. 8–131.
Units Used:
Mg = 1000 kg
885
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
M = 1.2 Mg
D = 30 mm
ag = 0.4 mm
as = 0.2 mm
Solution:
P =
M g( ag + as)
D
2⎛⎜ ⎟⎞
⎝2⎠
P = 235 N
Problem 8-133
A machine of mass M is to be moved over a level surface using a series of rollers for which
the coefficient of rolling resistance is ag at the ground and am at the bottom surface of the
machine. Determine the appropriate diameter of the rollers so that the machine can be pushed
forward with a horizontal force P. Hint: Use the result of Prob. 8-131.
Units Used:
Mg = 1000 kg
Given:
M = 1.4 Mg
ag = 0.5 mm
am = 0.2 mm
P = 250 N
Solution:
P=
M g( ag + am)
2r
⎛ ag + am ⎞
⎟
⎝ 2P ⎠
r = M g⎜
r = 19.2 mm
d = 2r
d = 38.5 mm
886
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Problem 8-134
A single force P is applied to the handle of the drawer. If
friction is neglected at the bottom and the coefficient of
static friction along the sides is μs determine the largest
spacing s between the symmetrically placed handles so
that the drawer does not bind at the corners A and B
when the force P is applied to one of the handles.
Given:
μ s = 0.4
a = 0.3 m
b = 1.25 m
Solution:
Equation of Equilibrium and Friction : If
the drawer does not bind at corners A and
B, slipping would have to occur at points
A and B. Hence, FA = μNA and FB = μNB
+
Σ F x = 0;
→
NB − NA = 0
NA = NB = N
+
↑Σ Fy = 0;
ΣMB = 0;
μ s NA + μ s NB − P = 0
P = 2μ s N
⎛ s + b⎞ = 0
⎟
⎝ 2 ⎠
N a + μ s N b − P⎜
⎡a + μ b − 2μ ⎛ s + b ⎞⎤ N = 0
⎢
⎟⎥
s
s⎜
⎣
⎝ 2 ⎠⎦
a + μ s b − μ s ( s + b) = 0
s =
a
μs
s = 0.750 m
Problem 8-135
The truck has mass M and a center of mass at G. Determine the greatest load it can pull if (a) the
truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has
887
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
( )
four-wheel drive. The coefficient of static friction between the wheels and the ground is μst and
between the crate and the ground, it is μsc.
Units Used:
3
kN = 10 N
Mg = 1000 kg
Given:
M = 1.25 Mg
μ st = 0.5
a = 600 mm
b = 1.5 m
μ sc = 0.4
c = 1m
g = 9.81
m
2
d = 800 mm
s
Solution:
Guesses
NA = 1 N
NB = 1 N
T = 1N
NC = 1 N
W = 1N
(a) Rear wheel drive
Given
−T + μ st NA = 0
NA + NB − M g = 0
−M g b + NB( b + c) + T a = 0
T − μ sc NC = 0
NC − W = 0
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟
⎜ T ⎟ = Find ( N , N , T , N , W)
A B
C
⎜ ⎟
⎜ NC ⎟
⎜W⎟
⎝ ⎠
W = 6.97 kN
(b) Four wheel drive
Given
−T + μ st NA + μ st NB = 0
888
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Engineering Mechanics - Statics
Chapter 8
NA + NB − M g = 0
−M g b + NB( b + c) + T a = 0
T − μ sc NC = 0
NC − W = 0
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟
⎜ T ⎟ = Find ( N , N , T , N , W)
A B
C
⎜ ⎟
⎜ NC ⎟
⎜W⎟
⎝ ⎠
W = 15.33 kN
Problem 8-136
The truck has M and a center of mass at G. The truck is traveling up an incline of angle θ.
Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels
are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction between
the wheels and the ground is μst and between the crate and the ground, it is μsc.
Units Used:
3
kN = 10 N
Mg = 1000 kg
Given:
θ = 10 deg
889
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Engineering Mechanics - Statics
Chapter 8
M = 1.25 Mg a = 600 mm
μ st = 0.5
b = 1.5 m
μ sc = 0.4
c = 1m
g = 9.81
m
d = 800 mm
2
s
Solution:
Guesses
NA = 1 N
NB = 1 N
T = 1N
NC = 1 N
W = 1N
(a) Rear wheel drive
Given
−T + μ st NA − M g sin ( θ ) = 0
NA + NB − M g cos ( θ ) = 0
−M g b cos ( θ ) + M g d sin ( θ ) + NB( b + c) + T a = 0
T − μ sc NC − W sin ( θ ) = 0
NC − W cos ( θ ) = 0
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟
⎜ T ⎟ = Find ( N , N , T , N , W)
A B
C
⎜ ⎟
⎜ NC ⎟
⎜W⎟
⎝ ⎠
W = 1.25 kN
(b) Four wheel drive
Given
−T + μ st NA + μ st NB − M g sin ( θ ) = 0
NA + NB − M g cos ( θ ) = 0
−M g b cos ( θ ) + M g d sin ( θ ) + NB( b + c) + T a = 0
T − μ sc NC − W sin ( θ ) = 0
NC − W cos ( θ ) = 0
890
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Engineering Mechanics - Statics
⎛ NA ⎞
⎜ ⎟
⎜ NB ⎟
⎜ T ⎟ = Find ( N , N , T , N , W)
A B
C
⎜ ⎟
⎜ NC ⎟
⎜W⎟
⎝ ⎠
Chapter 8
W = 6.89 kN
Problem 8-137
A roofer, having a mass M, walks slowly in an upright position down along the surface of a
dome that has a radius of curvature ρ. If the coefficient of static friction between his shoes
and the dome is μs determine the angle θ at which he first begins to slip.
Given:
M = 70 kg
ρ = 20 m
μ s = 0.7
Solution:
ΣF y' = 0;
Nm − M g cos ( θ ) = 0
ΣF x' = 0;
M g sin ( θ ) − μ s Nm = 0
μ s = tan ( θ )
θ = atan ( μ s)
θ = 35.0 deg
Problem 8-138
A man attempts to lift the uniform ladder of weight W to an upright position by applying a
force P perpendicular to the ladder at rung R. Determine the coefficient of static friction
between the ladder and the ground at A if the ladder begins to slip on the ground when his
891
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
g
g
p
g
hands reach height c.
Given:
a = 2 ft
b = 14 ft
c = 6 ft
W = 40 lb
Solution:
c
θ = asin ⎛⎜ ⎟⎞
⎝ b⎠
Initial guesses
P = 10 lb
NA = 100 lb
μ A = 100
Given
ΣF x = 0;
μ A NA − P sin ( θ ) = 0
ΣF y = 0;
NA − W + P cos ( θ ) = 0
ΣMA = 0;
−W⎛⎜
b + a⎞
⎟ cos ( θ ) + P b = 0
⎝ 2 ⎠
⎛P ⎞
⎜ ⎟
⎜ NA ⎟ = Find ( P , NA , μ A)
⎜μ ⎟
⎝ A⎠
⎛ P ⎞ ⎛ 20.7 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ NA ⎠ ⎝ 21.3 ⎠
μ A = 0.41
Problem 8-139
Column D is subjected to a vertical load W. It is supported on two identical wedges A and B for
which the coefficient of static friction at the contacting surfaces between A and B and between
B and C is μs. Determine the force P needed to raise the column and the equilibrium force P'
needed to hold wedge A stationary. The contacting surface between A and D is smooth.
892
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Engineering Mechanics - Statics
Chapter 8
Units Used:
3
kip = 10 lb
Given:
W = 8000 lb
θ = 10 deg
μ s = 0.4
Solution:
wedge A:
ΣF y = 0;
N cos ( θ ) − μ s N sin ( θ ) − W = 0
N =
W
cos ( θ ) − μ s sin ( θ )
N = 8739.8 lb
ΣF x = 0;
μ s N cos ( θ ) + N sin ( θ ) − P' = 0
P' = μ s N cos ( θ ) + N sin ( θ )
P' = 4.96 kip
Wedge B:
ΣF y = 0;
NC + μ s N sin ( θ ) − N cos ( θ ) = 0
NC = −μ s N sin ( θ ) + N cos ( θ )
NC = 8000 lb
ΣF x = 0;
P − μ s NC − N sin ( θ ) − μ s N cos ( θ ) = 0
P = μ s NC + N sin ( θ ) + μ s N cos ( θ )
P = 8.16 kip
Problem 8-140
Column D is subjected to a vertical load W. It is supported on two identical wedges A and B
for which the coefficient of static friction at the contacting surfaces between A and B and
between B and C is μs. If the forces P and P' are removed, are the wedges self-locking? The
contacting surface between A and D is smooth.
893
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 8
Given:
W = 8000 lb
θ = 10 deg
μ s = 0.4
Solution:
Wedge A:
ΣF y = 0;
N − W cos ( θ ) = 0
N = W cos ( θ )
N = 7878.5 lb
ΣF x = 0;
W sin ( θ ) − F = 0
F = W sin ( θ )
F = 1389.2 lb
Friction
F max = μ s N
Since F = 1389 lb < Fmax = 3151 lb then the wedges do not slip at the contact surface AB.
Wedge B:
ΣF y = 0;
NC − F sin ( θ ) − N cos ( θ ) = 0
NC = F sin ( θ ) + N cos ( θ )
NC = 8000 lb
ΣF x = 0;
F C + F cos ( θ ) − N sin ( θ ) = 0
F C = −F cos ( θ ) + N sin ( θ )
F C = 0 lb
Friction
F Cmax = μ s NC
Since F C = 0 lb < FCmax = 3200 lb then the wedges do not slip at the contact surface BC.
Therefore the wedges are self-locking.
894
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Engineering Mechanics - Statics
Chapter 9
Problem 9-1
Locate the center of mass of the
homogeneous rod bent in the form of a
parabola.
Given:
a = 1m
b = 2m
Solution:
⎛x⎞
y = b⎜ ⎟
⎝ a⎠
2
d y
2b
=
x
2
dx
a
a
yc =
⌠
2
2
⎮
x⎞
2b ⎞
⎛
⎛
1 + ⎜ x⎟ dx
⎮ b⎜ ⎟
2
⎝ a⎠
⎮
⎝a ⎠
⌡
0
yc = 0.912 m
a
⌠
⎮
⎮
⎮
⌡0
xc = 0 m
2
⎛ 2b ⎞
1 + ⎜ x⎟ dx
2
⎝a ⎠
Problem 9-2
Locate the center of gravity xc of the homogeneous
rod. If the rod has a weight per unit length γ,
determine the vertical reaction at A and the x and y
components of reaction at the pin B.
Given:
γ = 0.5
lb
ft
a = 1 ft
b = 2 ft
895
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⎛x⎞
⎟
⎝ a⎠
2
y = b⎜
d y
2b
=
x
2
dx
a
a
⌠
⎮
L = ⎮
⎮
⌡0
2
⎛ 2 b x⎞ dx
1+⎜
2 ⎟
⎝a ⎠
W = γ L
W = 1.162 lb
a
⌠
2
1 ⎮
⎛ 2 b x⎞ dx
xc =
⎮ x 1+⎜
2 ⎟
L ⎮
a
⎝
⎠
⌡0
Ay = 1lb
Guesses
Given
Bx = 0
B x = 1lb
Ay + By − W = 0
xc = 0.620 ft
B y = 1lb
− Ay a + W ( a − xc) = 0
⎛ Bx ⎞
⎜ ⎟
⎜ By ⎟ = Find ( Bx , By , Ay)
⎜A ⎟
⎝ y⎠
⎛ Bx ⎞ ⎛ 0.000 ⎞
⎜ ⎟ ⎜
⎟
⎜ By ⎟ = ⎜ 0.720 ⎟ lb
⎜ A ⎟ ⎝ 0.442 ⎠
⎝ y⎠
Problem 9-3
Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.
Given:
r = 300 mm
θ = 30 deg
Solution:
yc = 0
Symmetry
896
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Engineering Mechanics - Statics
Chapter 9
π
+θ
⌠2
⎮
r cos ( α ) r dα
⎮
⌡− π − θ
2
xc =
xc = 124.049 mm
π
+θ
⌠2
⎮
r dα
⎮
⌡− π − θ
2
Problem 9-4
Locate the center of gravity xc of the homogeneous rod bent in the form of a semicircular arc.
The rod has a weight per unit length γ. Also, determine the horizontal reaction at the smooth
support B and the x and y components of reaction at the pin A.
Given:
γ = 0.5
lb
ft
r = 2 ft
Solution:
π
⌠2
⎮
r cos ( θ ) r dθ
⎮
−
π
⌡
xc =
2
π
⌠2
⎮
r dθ
⎮
−
π
⌡
xc = 1.273 ft
2
ΣMA = 0;
−π r γ xc + B x ( 2 r) = 0
Bx =
π rγ xc
2r
B x = 1 lb
897
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
+
Σ F x = 0;
→
− Ax + Bx = 0
Ax = Bx
Ax = 1 lb
+
Ay − π rγ = 0
Ay = π rγ
Ay = 3.14 lb
↑Σ Fy = 0;
Problem 9-5
Determine the distance xc to the center of gravity of the homogeneous rod bent into the parabolic
shape. If the rod has a weight per unit length γ determine the reactions at the fixed support O.
Given:
lb
ft
γ = 0.5
a = 1 ft
b = 0.5 ft
Solution:
⎛x⎞
⎟
⎝ a⎠
2
y=b⎜
d y
2bx
=
2
dx
a
a
⌠
⎮
L = ⎮
⎮
⌡0
2
⎛ 2 b x ⎞ dx
1+⎜
2 ⎟
⎝ a ⎠
L = 1.148 ft
a
⌠
1⎮
xc = ⎮ x
L⎮
⌡0
+
Σ F x = 0;
→
2
⎛ 2 b x ⎞ dx
2 ⎟
⎝ a ⎠
1+⎜
Ox = 0
xc = 0.531 ft
Ox = 0 lb
Ox = 0 lb
898
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Engineering Mechanics - Statics
+
↑Σ Fy = 0;
ΣMO = 0;
Chapter 9
Oy − γ L = 0
Oy = γ L
Oy = 0.574 lb
MO − γ L xc = 0
MO = γ L xc
MO = 0.305 lb⋅ ft
Problem 9-6
Determine the distance yc to the center of gravity of the homogeneous rod bent into the parabolic
shape.
Given:
a = 1 ft
b = 0.5 ft
Solution:
⎛x⎞
y=b⎜ ⎟
⎝ a⎠
2
d y
2bx
=
2
dx
a
a
⌠
⎮
L = ⎮
⎮
⌡0
2
⎛ 2 b x ⎞ dx
2 ⎟
⎝ a ⎠
L = 1.148 ft
⎤
⎡⌠a
2
2 ⎥
1 ⎢⎮
x
2
b
x
⎞ dx
⎛ ⎞ 1+⎛
yc = ⎢⎮ b ⎜ ⎟
⎜
⎟ ⎥
2
L ⎮
a
⎢⌡ ⎝ ⎠
⎝ a ⎠ ⎥
⎣0
⎦
yc = 0.183 ft
1+⎜
Problem 9-7
Locate the centroid of the parabolic area.
Solution:
a=
h
2
b
899
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
dA = x dy
xc =
x
2
yc = y
1
h
⌠
A=⎮ b
⎮
⌡
y
⎛ y⎞
dy = h b ⎜ ⎟
h
⎝ h⎠
2
0
⌠
3 ⎮
xc =
2h b ⎮
⌡
h
1⎛
⎜b
2⎝
0
2
y⎞
⎟ dy =
h⎠
3
bh
2
2
8h
3
b
8
xc =
5
h
⌠
3 ⎮
yc =
yb
2h b ⎮
⌡
y
3 ⎛ h⎞
h⎜ ⎟
dy =
h
5 ⎝ h⎠
2
0
yc =
3
h
5
Problem 9-8
Locate the centroid yc of the shaded area.
Given:
a = 100 mm
b = 100 mm
Solution:
⎛x⎞
⎟
⎝ a⎠
2
y = b⎜
900
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 9
y
b
x=a
b
⌠
⎮ y2a
⎮
⌡
y
dy
b
0
yc =
yc = 60 mm
b
⌠
⎮ 2a
⎮
⌡0
y
b
dy
Problem 9-9
Locate the centroid xc of the shaded area.
Solution:
dA = ydx
xc = x
yc =
y
2
b
xc =
yc =
⌠
h 2
⎮ x
x dx
2
⎮
b
⌡
0
b
⌠
2
⎮
x
h
dx
⎮
2
⎮
b
⌡0
⌠
⎮
⎮
⎮
⌡
4
3 b
=
4 3
b
xc =
3
b
4
yc =
3
h
10
b
2
1 ⎛ h 2⎞
x
dx
2 ⎜ b2 ⎟
⎝
⎠
0
⌠
⎮
⎮
⌡
b
h 2
x dx
2
b
2
3 5 h
=
b
5
10
b h
0
901
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 9
Problem 9-10
Determine the location (xc, yc) of the
centroid of the triangular area.
Solution:
a
⌠
1 2
A = ⎮ m x dx =
a m
⌡0
2
a
xc =
2 ⌠
2
⎮ x m x dx =
a
2⌡
3
0
ma
xc =
2
a
3
yc =
m
a
3
a
⌠ 1
1
2
⎮
yc =
( m x) dx = a m
2⎮ 2
3
ma ⌡
2
0
Problem 9-11
Determine the location (xc, yc) of the center of
gravity of the quartercircular plate. Also
determine the force in each of the supporting
wires.The plate has a weight per unit area of γ.
Given:
γ = 5
lb
ft
2
a = 4 ft
902
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
2
2
2
y=
a −x
x +y =a
2
2
2
a
A = π
4
W = Aγ
W = 62.832 lb
1 ⌠
⎮ x a2 − x2 dx
xc =
A ⌡0
xc = 1.698 ft
a
1 ⌠
⎮
yc =
A⎮
⌡
a
(
)
1 2
2
a − x dx
2
yc = 1.698 ft
0
Guesses
TA = 1lb
TB = 1lb
Given
TA + TB − W = 0
⎛ TA ⎞
⎜ ⎟ = Find ( TA , TB)
⎝ TB ⎠
TB a − W xc = 0
⎛ TA ⎞ ⎛ 36.2 ⎞
⎜ ⎟=⎜
⎟ lb
⎝ TB ⎠ ⎝ 26.7 ⎠
Problem *9-12
Locate the centroid of the shaded area.
903
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
dA = ydx
xc = x
yc =
y
2
L
⌠
2
⎛ πx⎞
A = ⎮ a sin ⎜ ⎟ dx = L a
⎮
π
⎝L⎠
⌡
0
xc =
π ⌠
⎮
L
⎛ π x ⎞ dx = 1 L
⎟
2
⎝L⎠
x a sin ⎜
2L a ⎮
⌡
xc =
1
L
2
yc =
1
πa
8
0
⌠
π ⎮
yc =
2L a ⎮
⌡
L
2
1⎛
1
⎛ π x ⎞⎞
⎜ a sin ⎜ ⎟⎟ dx = π a
2⎝
8
⎝ L ⎠⎠
0
Problem 9-13
Locate the center of gravity of the homogeneous
cantilever beam and determine the reactions at the
fixed support.The material has a density of ρ.
3
Mg = 10 kg
Units Used:
Mg
ρ = 8
Given:
m
g = 9.81
3
kN = 10 N
a = 1m
3
b = 4m
m
c = 0.5 m
2
s
Solution:
⌠
⎮
V = ⎮
⌡
0
2
⎛ x ⎞ dx
⎟
⎝ b⎠
ca⎜
−b
⌠
1⎮
xc =
V⎮
⌡
W = ρg V
0
2
−b
⌠
1⎮
yc =
V⎮
⌡
⎛ x ⎞ dx
⎟
⎝ b⎠
xca⎜
0
−b
2
c ⎛x⎞
c a ⎜ ⎟ dx
2 ⎝ b⎠
904
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
⌠
⎮
1⎮
zc =
V⎮
⌡
Chapter 9
0
−b
⎛ xc ⎞ ⎛ −3.00 ⎞
⎜ ⎟ ⎜
⎟
⎜ yc ⎟ = ⎜ 0.25 ⎟ m
⎜ z ⎟ ⎝ −0.30 ⎠
⎝ c⎠
2
−c ⎡ ⎛ x ⎞ ⎤
⎢a ⎜ ⎟ ⎥ dx
2 ⎣ ⎝ b⎠ ⎦
Guesses
Ax = 1 N
Given
Ax = 0
2
Ay = 1 N
Ay = 0
Az = 1 N
Az − W = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( A , A , A , M )
x y z
A
⎜ Az ⎟
⎜ ⎟
⎝ MA ⎠
MA = 1 N⋅ m
MA − W( b + xc) = 0
⎛ Ax ⎞ ⎛ 0.00 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 0.00 ⎟ kN
⎜ A ⎟ ⎝ 52.32 ⎠
⎝ z⎠
MA = 52.32 kN⋅ m
Problem 9-14
Locate the centroid (xc, yc) of the
exparabolic segment of area.
Solution:
0
⌠
1
b 2
A=⎮
x dx =
ab
2
3
⎮
a
⌡− a
0
3 ⌠
−3
b 2
⎮ x
xc =
x dx =
a
2
ab ⎮
4
a
⌡− a
xc =
−3
a
4
yc =
−3
b
10
0
⌠
2
3 ⎮
1 ⎛ b 2⎞
−3
yc =
b
⎮ − ⎜ x ⎟ dx =
ab ⎮
2 2
10
a
⎝
⎠
⌡− a
905
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-15
Locate the centroid of the shaded area.
Solution:
a
xc =
⌠
h n⎞
⎮ x⎛ h −
x dx
⎜
n ⎟
⎮ ⎝
a
⎠
⌡
0
⌠
⎮
⎮
⌡
a
xc =
n+1
a
2( n + 2)
yc =
n
h
2n + 1
⎛h − h xn⎞ dx
⎜
n ⎟
⎝ a ⎠
provided that n ≠ −2
0
yc =
⌠
1⎮
⎮
2⎮
⌡
a
2
⎛ h − h xn⎞ dx
⎜
n ⎟
⎝ a ⎠
0
⌠
⎮
⎮
⌡
a
⎛h − h xn⎞ dx
⎜
n ⎟
⎝ a ⎠
provided that n ≠
−1
2
0
Problem *9-16
Locate the centroid of the shaded area bounded by the parabola and the line y = a.
906
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
3
a
⌠
A=⎮
⌡0
( 2)
2
2
2 a
a y dy =
3
a
3 ⌠
⎮
xc =
2⎮
2a ⌡
2a
A=
3
a
1
3
a y dy =
a
2
8
xc =
3
a
8
yc =
3
a
5
0
5
a
yc =
( )
2
3 ⌠
3
2
⎮ y a y dy =
a
2⌡
4
2a 0
5a
Problem 9-17
Locate the centroid of the quarter elliptical area.
907
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This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics
Chapter 9
Solution:
a
⌠
2
⎮
x⎞
⎛
A = ⎮ b 1 − ⎜ ⎟ dx
⎝ a⎠
⌡
A=
πa b
4
0
a
⌠
2
4 ⎮
⎛ x ⎞ dx = 4 a
xc =
x
b
1
−
⎜ ⎟
πa b ⎮
3π
⎝ a⎠
⌡
xc =
4
3π
a
0
a
⌠
2
⎮
2⎤
4 ⎮ 1⎡
x
⎢b 1 − ⎛⎜ ⎟⎞ ⎥ dx = 4 b
yc =
2⎣
πa b ⎮
3π
⎝ a⎠ ⎦
⌡
yc =
4
3π
b
0
Problem 9-18
Locate the centroid xc of the
triangular area.
Solution:
A=
bh
2
⎡
⎤
⌠
2 ⎢⌠
h
h
⎮
⎮
xc =
x x dx +
x
( b − x) dx⎥
⎮
b h ⎢⎮
a
b−a
⎥
⌡
⌡
a
⎣0
b
a
⎦
xc =
a+b
3
Problem 9-19
Locate the centroid of the shaded area.
908
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 8m
b = 4m
Solution:
⌠
⎮
A = ⎮
⌡
a
⎡
⎡ ⎛ x ⎞ 2⎤⎤
⎢b − b⎢1 − ⎜ ⎟ ⎥⎥ dx
⎣
⎣ ⎝ a ⎠ ⎦⎦
A = 10.667 m
2
0
a
⌠
⎡ ⎛ x ⎞ 2⎤⎤
1 ⎮ ⎡
xc =
x⎢b − b⎢1 − ⎜ ⎟ ⎥⎥ dx
A⎮ ⎣
⎣ ⎝ a ⎠ ⎦⎦
⌡
xc = 6 m
0
⌠
1 ⎮
yc =
A⎮
⌡
a
⎡
⎡
1⎡
x ⎤⎤ ⎡
x ⎤⎤
⎢b + b⎢1 − ⎛⎜ ⎟⎞ ⎥⎥ ⎢b − b⎢1 − ⎛⎜ ⎟⎞ ⎥⎥ dx
2⎣
⎣ ⎝ a ⎠ ⎦⎦ ⎣
⎣ ⎝ a ⎠ ⎦⎦
2
2
yc = 2.8 m
0
Problem 9-20
Locate the centroid xc of the shaded area. Solve the problem by evaluating the integrals using
Simpson's rule.
Given:
a = 2 ft
1
b =
a
2
5
+ 2a
3
Solution:
a
⌠
⎮ ⎛
⎮ ⎜
A = ⎮ ⎝b −
⌡0
5⎞
1
x
2
+ 2x
3⎟
⎠ dx
a
⌠
⎮ ⎛
1 ⎮ ⎜
xc =
⎮ x⎝ b −
A ⌡0
2
2
5⎞
1
x
A = 2.177 ft
+ 2x
3⎟
⎠ dx
xc = 0.649 ft
909
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-21
Locate the centroid yc of the shaded area. Solve the problem by evaluating the integrals using
Simpson's rule.
Given:
a = 2 ft
1
b =
a
5
2
+ 2a
3
Solution:
a
⌠
⎮ ⎛
⎮ ⎜
A = ⎮ ⎝b −
⌡0
⌠
⎮
1 ⎮
yc =
A⎮
⌡
a
5⎞
1
x
2
+ 2x
3⎟
⎠ dx
A = 2.177 ft
1
5 ⎞⎛
1
5⎞
⎛
⎜
⎟
⎜
1
2
3
2
3⎟
⎝b + x + 2x ⎠ ⎝ b − x + 2x ⎠ dx
2
2
yc = 2.04 ft
0
Problem 9-22
The steel plate has thickness t and density ρ. Determine
the location of its center of mass. Also compute the
reactions at the pin and roller support.
Units Used:
3
kN = 10 N
Given:
t = 0.3 m
ρ = 7850
a = 2m
kg
m
3
b = 2m
c = 2m
910
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
g = 9.81
Chapter 9
b
θ = atan ⎛⎜ ⎟⎞
m
⎝c⎠
2
s
Solution:
⌠
A = ⎮
⎮
⌡
c
⎛ x b x⎞
+
⎜a
⎟ dx
⎝ c c⎠
A = 4.667 m
2
0
c
⌠
1 ⎮ ⎛
xc =
x⎜ a
A⎮ ⎝
⌡
x b x⎞
+
⎟ dx
c ⎠
c
xc = 1.257 m
0
c
⌠
1 ⎮ 1⎛
yc =
⎜a
A ⎮ 2⎝
⌡
x b x⎞⎛
+
⎟ ⎜a
c ⎠⎝
c
x b x⎞
−
⎟ dx
c ⎠
c
yc = 0.143 m
0
W = ρAtg
Equilibrium
Guesses
Ax = 1 N
Ay = 1 N
Given
Ay − W + NB cos ( θ ) = 0
⎛ Ax ⎞
⎜ ⎟
⎜ Ay ⎟ = Find ( Ax , Ay , NB)
⎜N ⎟
⎝ B⎠
NB = 1 N
− Ax + NB sin ( θ ) = 0
2
2
NB b + c − W xc = 0
⎛ Ax ⎞ ⎛ 33.9 ⎞
⎜ ⎟ ⎜
⎟
⎜ Ay ⎟ = ⎜ 73.9 ⎟ kN
⎜ N ⎟ ⎝ 47.9 ⎠
⎝ B⎠
Problem 9-23
Locate the centroid xc of the shaded area.
Given:
a = 4 ft
b = 4 ft
911
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⌠
⎮
A = ⎮
⌡
a
2
⎡b x
x ⎤
⎢ − b ⎛⎜ ⎟⎞ ⎥ dx
⎣a
⎝ a⎠ ⎦
0
a
⌠
2
1 ⎮ ⎡b x
x⎞ ⎤
⎛
xc =
x⎢
− b ⎜ ⎟ ⎥ dx
A⎮ ⎣a
⎝ a⎠ ⎦
⌡
xc = 2.00 ft
0
Problem 9-24
Locate the centroid yc of the shaded area.
Given:
a = 4 ft
b = 4 ft
Solution:
⌠
⎮
A = ⎮
⌡
a
2
⎡b x
x⎞ ⎤
⎛
⎢ − b ⎜ ⎟ ⎥ dx
⎣a
⎝ a⎠ ⎦
0
⌠
1 ⎮
yc =
A⎮
⌡
a
1⎡ x
x ⎤ ⎡b x
x ⎤
⎢b + b ⎛⎜ ⎟⎞ ⎥ ⎢ − b ⎛⎜ ⎟⎞ ⎥ dx
2⎣ a
⎝ a ⎠ ⎦⎣ a
⎝ a⎠ ⎦
2
2
yc = 1.60 ft
0
Problem 9-25
Locate the centroid xc of the shaded area.
Given:
a = 4m
912
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 4m
Solution:
⌠
⎮
A = ⎮
⌡
a
2
⎡ x
⎛x⎞ ⎤
⎢b
− b ⎜ ⎟ ⎥ dx
⎣ a
⎝ a⎠ ⎦
0
a
⌠
1 ⎮ ⎡
xc =
x⎢b
A⎮ ⎣
⌡
x
⎛x⎞ ⎤
− b ⎜ ⎟ ⎥ dx
a
⎝ a⎠ ⎦
2
0
xc = 1.80 m
Problem 9-26
Locate the centroid yc of the shaded area.
Given:
a = 4m
b = 4m
Solution:
⌠
⎮
A = ⎮
⌡
a
2
⎡ x
x⎞ ⎤
⎛
⎢b
− b ⎜ ⎟ ⎥ dx
⎣ a
⎝ a⎠ ⎦
0
⌠
1 ⎮
yc =
A⎮
⌡
a
1⎡
⎢b
2⎣
x
⎛ x ⎞ ⎤⎡
+ b ⎜ ⎟ ⎥ ⎢b
a
⎝ a⎠ ⎦⎣
2
x
⎛x⎞ ⎤
− b ⎜ ⎟ ⎥ dx
a
⎝ a⎠ ⎦
2
yc = 1.80 m
0
913
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-27
Locate the centroid xc of the shaded area.
Given:
a = 1 in
b = 3 in
c = 2 in
Solution:
⌠
A = ⎮
⎮
⌡
a+ b
c
x
dx
a+b
a
⌠
1 ⎮
xc =
A⎮
⌡
a+ b
xc
x
dx
a+b
xc = 2.66 in
a
Problem 9-28
Locate the centroid yc of the shaded area.
Given:
a = 1 in
b = 3 in
c = 2 in
914
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⌠
A = ⎮
⎮
⌡
a+ b
x
dx
a+b
c
a
⌠
1 ⎮
yc =
A⎮
⌡
a+ b
1⎛
⎜c
2⎝
2
x ⎞
⎟ dx
a + b⎠
yc = 0.804 in
a
Problem 9-29
Locate the centroid xc of the shaded area.
Given:
a = 4 in
b = 2 in
c = 3 in
Solution:
⌠
A = ⎮
⎮
⌡
a+ b
bc
dy
y
A = 6.592 in
2
b
⌠
1 ⎮
xc =
A⎮
⌡
a+ b
2
1 ⎛ b c⎞
⎜ ⎟ dy
2⎝ y⎠
xc = 0.910 in
b
Problem 9-30
Locate the centroid yc of the shaded area.
915
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 4 in
b = 2 in
c = 3 in
Solution:
⌠
A = ⎮
⎮
⌡
a+ b
bc
dy
y
A = 6.592 in
2
b
1 ⌠
⎮
yc =
A⎮
⌡
a+ b
⎛ b c⎞ dy
⎟
⎝ y⎠
y⎜
yc = 3.64 in
b
Problem 9-31
Determine the location rc of the centroid C of the cardioid, r = a(1 − cosθ).
Solution:
2π
⌠
A=⎮
⌡0
a ( 1 − cos( θ ) )
⌠
⎮
⌡0
r dr dθ =
3 2
a π
2
916
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
2π
⌠
xc =
⎮
2 ⌡
3a π 0
2
Chapter 9
a ( 1 − cos( θ ) )
⌠
⎮
⌡0
r cos ( θ ) r dr dθ =
−5
a
6
rc =
5a
6
yc =
3b
8
Problem 9-32
Locate the centroid of the ellipsoid of revolution.
2
dV = π z dy
Solution:
2
z =a
2⎛
⎜
⎜
⎝
y ⎟⎞
2
1−
2⎟
b ⎠
b
⌠
2
2
2
⎮
y ⎟⎞
1 3b −b
2⎛
2
⎜
V = ⎮ πa 1 −
b
a π
dy =
2
⎜ b2 ⎟
3
⎮
b
⎝
⎠
⌡0
b
⌠
2
3 ⎮
y ⎟⎞
3 2
2⎛
⎜
yc =
b
dy =
⎮ yπ a 1 −
2
⎜ b2 ⎟
8b
2b a π ⎮
⎝
⎠
⌡0
By symmetry
xc = zc = 0
Problem 9-33
Locate the centroid zc of the very thin conical shell. Hint: Use thin ring elements having a center at
(0, 0, z), radius y, and width dL =
2
( dy) + ( dz)
2
917
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
2
2
⎛ dy ⎞
⎛ a⎞
dL = dy + dz = 1 + ⎜ ⎟ dz = 1 + ⎜ ⎟ dz
⎝ dz ⎠
⎝ h⎠
2
h
⌠
⎮
az
A = ⎮ 2π
h
⌡
0
2
az
h
2
⎛ 2 2⎞
⎛ a ⎞ dz = h2 π a ⎜ h + a ⎟
⎟
2 ⎟
h ⎜
⎝ h⎠
h
⎝
⎠
1+⎜
h
⌠
1
⎮
az
zc =
⎮ z2π h
2
2
π a a + h ⌡0
zc =
r=
2
⎛ h2 + a2 ⎞
2
2
⎛ a ⎞ dz =
⎜
⎟
h
π
a
⎟
⎜
2
⎟
h
2
2
⎝ ⎠
h
h +a
⎝
⎠
3π a
1+⎜
(
)
2h
3
Problem 9-34
Locate the centroid zc of the volume.
918
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 2 ft
b = 2 ft
Solution:
b
⌠
2
a z
⎮
V = ⎮ π
dz
b
⌡
V = 12.566 ft
3
0
b
⌠
2
1⎮
a z
zc =
zπ
dz
V⎮
b
⌡
zc = 1.333 ft
0
Problem 9-35
Locate the centroid of the solid.
Solution:
2
2
z =
h
y
a
xc = yc = 0
⎛ z⎞
⎟
⎝ h⎠
2
y = a⎜
By symmetry
h
zc =
⌠
2
⎮
⎡ ⎛ z ⎞ 2⎤
⎮ zπ ⎢a
⎜ ⎟ ⎥ dz
⎮
⎣
⎝ h⎠ ⎦
⌡
0
h
⌠
2
⎮
⎡ ⎛ z ⎞ 2⎤
⎮ π ⎢a ⎜ ⎟ ⎥ dz
⎮
h
⌡0 ⎣ ⎝ ⎠ ⎦
6
=
5 h
6 5
h
zc =
5h
6
919
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-36
Locate the centroid of the quarter-cone.
Solution:
r=
a
( h − z)
h
⌠
⎮
V=⎮
⌡
zc = z
xc = yc =
4r
3π
h
2
π ⎡a
1
2
⎤
ha π
⎢ ( h − z)⎥ dz =
4 ⎣h
12
⎦
0
h
⌠
2
12 ⎮
π ⎡a
1
⎤
zc =
z ⎢ ( h − z)⎥ dz =
h
⎮
2
4 ⎣h
4
⎦
ha π ⌡
0
⎤
⎡⌠h
2 ⎥
12 ⎢⎮ 4 ⎡ a
π ⎡a
a
⎤
⎤
xc =
( h − z)⎥ ⎢ ( h − z)⎥ dz⎥ =
⎢⎮
⎢
2
h
⎦ 4 ⎣h
⎦ ⎥ π
h a π ⎢⌡ 3π ⎣
⎣0
⎦
xc = yc =
a
π
zc =
h
4
920
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-37
Locate the center of mass xc of the hemisphere. The density of the material varies linearly from zero at
the origin O to ρo at the surface. Hint: Choose a hemispherical shell element for integration
Solution:
for a spherical shell
xc =
x
2
x
ρ = ρ 0 ⎛⎜ ⎟⎞
⎝ a⎠
2
dV = 2π x dx
a
xc =
⌠
⎮ ρ ⎛ x ⎞ x 2π x2 dx
0⎜ ⎟
⎮
⎝ a⎠ 2
⌡
0
a
⌠
⎮ ρ ⎛⎜ x ⎟⎞ 2π x2 dx
0
⎮
⎝ a⎠
⌡0
=
2
⋅a
5
xc =
2
a
5
Problem 9-38
Locate the centroid zc of the right-elliptical cone.
Given:
a = 3 ft
b = 4 ft
c = 10 ft
2
2
⎛ x ⎞ + ⎛ y⎞ = 1
⎜ ⎟ ⎜ ⎟
⎝ b⎠ ⎝ a⎠
921
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 9
Solution:
Volume and Moment Arm : From the geometry,
x
b
=
c−z
c
x=
b
( c − z)
c
y
a
=
c−z
c
y=
a
( c − z)
c
The volume of the thin disk differential element is
dV = π
b
a
( c − z) ( c − z) dz
c
c
c
zc =
⌠
⎮ zπ b ( c − z) a ( c − z) dz
⎮
c
c
⌡
0
c
⌠
⎮ π b ( c − z) a ( c − z) dz
⎮
c
c
⌡0
zc = 2.5 ft
Problem 9-39
Locate the center of gravity zc of the
frustum of the paraboloid.The material
is homogeneous.
Given:
a = 1m
b = 0.5 m
c = 0.3 m
Solution
a
⌠
⎡ 2 z 2 2⎤
V = ⎮ π ⎢b − b − c ⎥ dz
⎮
a
⎣
⎦
⌡
(
)
0
922
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
a
Chapter 9
1⌠
⎮ zπ ⎡b2 − z b2 − c2 ⎤ dz
zc =
⎢
⎥
V⎮
a
⎣
⎦
⌡
(
)
0
zc = 0.422 m
Problem 9-40
Locate the center of gravity yc of the volume. The material is homogeneous.
Given:
a = 25 mm
c = 50 mm
d = 50 mm
Solution:
⌠
⎮
V = ⎮
⎮
⌡
c+ d
2
⎡ y 2⎤
π ⎢a ⎛⎜ ⎟⎞ ⎥ d y
⎣ ⎝c⎠ ⎦
c
⌠
⎮
1⎮
yc =
V⎮
⌡
c+ d
2
⎡ ⎛ y ⎞ 2⎤
yπ ⎢a ⎜ ⎟ ⎥ d y
⎣ ⎝c⎠ ⎦
yc = 84.7 mm
c
Problem 9-41
Locate the center of gravity for the homogeneous half-cone.
923
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⌠
⎮
V=⎮
⌡
h
π ⎛ a y⎞
2
1
2
⎜ ⎟ dy = h πa
2⎝ h ⎠
6
0
h
⌠
2
6 ⎮
π ⎛ a y⎞
3
yc =
y ⎜ ⎟ dy = h
⎮
2
2⎝ h ⎠
4
ha π ⌡
yc =
3
h
4
zc =
a
0
h
⌠
2
1
⎮ ⎛ 4a y ⎞ π ⎛ a y ⎞
d
zc =
y
=
a
⎜
⎟
⎜
⎟
2 ⎮ ⎝ 3hπ ⎠ 2 ⎝ h ⎠
π
ha π ⌡
0
6
π
h
⌠
2
6 ⎮
π ⎛ a y⎞
xc =
0 ⎜ ⎟ dy
2 ⎮
2⎝ h ⎠
ha π ⌡
xc = 0
0
Problem 9-42
Locate the centroid zc of the spherical segment.
924
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
a
(
)
⌠
5 3
2
2
V = ⎮ π a − z dz =
a π
24
⎮a
⌡
2
a
(
)
24 ⌠
⎮ zπ a2 − z2 dz = 27 a
zc =
3
40
5π a ⎮
⌡a
zc =
27
a
40
2
Problem 9-43
Determine the location zc of the centroid for the tetrahedron. Suggestion: Use a triangular "plate"
element parallel to the x-y plane and of thickness dz.
925
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
c−z
x
y
=
=
c
a
b
x=
a
( c − z)
c
=
1
c
4
y=
b
( c − z)
c
c
zc =
⌠ ab
2
⎮ z
( c − z) dz
2
⎮
c
⌡
0
c
⌠ ab
2
⎮
( c − z) dz
⎮ c2
⌡
zc =
1
c
4
0
Problem 9-44
Determine the location (x, y) of the particle M1 so that the three particles, which lie in the x–y plane,
have a center of mass located at the origin O.
Given:
M1 = 7 kg
M2 = 3 kg
M3 = 5 kg
a = 2m
b = 3m
c = 4m
Solution:
Guesses
x = 1m
Given
M1 x + M2 b − M3 c = 0
⎛x⎞
⎜ ⎟ = Find ( x , y)
⎝ y⎠
y = 1m
M1 y − M2 a − M3 a = 0
⎛ x ⎞ ⎛ 1.57 ⎞
⎜ ⎟=⎜
⎟m
⎝ y ⎠ ⎝ 2.29 ⎠
926
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-45
Locate the center of gravity (xc, yc, zc) of the
four particles.
Given:
M1 = 2 lb
a = 2 ft
M2 = 3 lb
b = 3 ft
M3 = 1 lb
c = −1 ft
M4 = 1 lb
d = 1 ft
f = 4 ft
e = 4 ft
h = −2 ft
g = 2 ft
i = 2 ft
Solution:
M1 0ft + M2 a + M3 d + M4 g
xc =
xc = 1.29 ft
M1 + M2 + M3 + M4
M1 0ft + M2 b + M3 e + M4 h
yc =
yc = 1.57 ft
M1 + M2 + M3 + M4
zc =
M1 0ft + M2 c + M3 f + M4 i
zc = 0.429 ft
M1 + M2 + M3 + M4
Problem 9-46
A rack is made from roll-formed sheet steel and has the cross section shown. Determine the location
(xc, yc) of the centroid of the cross section. The dimensions are indicated at the center thickness of
each segment.
Given:
a = 15 mm
c = 80 mm
927
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 50 mm
e = 30 mm
Solution:
L = 3a + 2c + e
2a
xc =
2
+ a⎛⎜ e +
⎝
a⎞
⎛ e⎞
⎟ + c( a + e) + e⎜a + ⎟ + ( c − d)a
2⎠
⎝ 2⎠
L
d
yc =
a
d
2
+c
c
+ ( c − d)
2
d+c
2
xc = 24.4 mm
+ ad + ec
yc = 40.6 mm
L
Problem 9-47
The steel and aluminum plate assembly is bolted together and fastened to the wall. Each plate has a
constant width w in the z direction and thickness t. If the density of A and B is ρs, and the density of
C is ρal, determine the location xc, the center of mass. Neglect the size of the bolts.
Units Used:
3
Mg = 10 kg
Given:
w = 200 mm
a = 300 mm
t = 20 mm
b = 100 mm
ρ s = 7.85
Mg
m
ρ al = 2.71
3
c = 200 mm
Mg
m
3
928
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
2( ρ s a t w)
xc =
a
2
+ ⎡⎣ρ al( b + c)t w⎤⎦ ⎛⎜ a − b +
⎝
b + c⎞
2
⎟
⎠
2ρ s a t w + ρ al( b + c)t w
xc = 179 mm
Problem 9-48
The truss is made from five members, each having a length L and a mass density ρ. If the mass of
the gusset plates at the joints and the thickness of the members can be neglected, determine the
distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it
is lifted.
Given:
L = 4m
ρ = 7
kg
m
Solution:
ρ L⎛⎜
d =
L
⎝2
+
L
4
+
3L
4
+L+
5L ⎞
⎟
4⎠
5ρ L
d=3m
Problem 9-49
Locate the center of gravity (xc, yc, zc) of the
homogeneous wire.
929
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 300 mm
b = 400 mm
Solution:
L =
πa
2
2
2
+2 a +b
xc =
1⎡ 2
π a ⎛ 2a ⎞⎤
2a
⎢ a + b + ⎜ ⎟⎥
L⎣
2
2 ⎝ π ⎠⎦
yc =
1⎡ 2
π a ⎛ 2a ⎞⎤
2a
⎢ a + b + ⎜ ⎟⎥
L⎣
2
2 ⎝ π ⎠⎦
zc =
1⎡
2
2 ⎛ b ⎞⎤
⎢2 a + b ⎜ ⎟⎥
L⎣
⎝ 2 ⎠⎦
⎛ xc ⎞ ⎛ 112.2 ⎞
⎜ ⎟ ⎜
⎟
⎜ yc ⎟ = ⎜ 112.2 ⎟ mm
⎜ z ⎟ ⎝ 135.9 ⎠
⎝ c⎠
Problem 9-50
Determine the location (xc, yc) of the center
of gravity of the homogeneous wire bent in
the form of a triangle. Neglect any slight
bends at the corners. If the wire is
suspended using a thread T attached to it at
C, determine the angle of tilt AB makes
with the horizontal when the wire is in
equilibrium.
Given:
a = 5 in
b = 9 in
c = 12 in
930
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
L = a+b+
2
2
2
2
2
2a
a +c +
b +c
xc =
1⎡
a+b
+
⎢( a + b)
L⎣
2
a +c
yc =
1⎛ 2
2c
⎜ a +c +
L⎝
2
b +c
2
2
+
2⎛
2
b + c ⎜a +
⎝
2 c⎞
⎟
b ⎞⎤
⎟⎥
2 ⎠⎦
xc = 6.50 in
yc = 4.00 in
2⎠
⎛ xc − a ⎞
⎟
⎝ c − yc ⎠
θ = atan ⎜
θ = 10.6 deg
Problem 9-51
The three members of the frame each have weight density γ . Locate the position (xc,yc) of the center
of gravity. Neglect the size of the pins at the joints and the thickness of the members. Also, calculate
the reactions at the fixed support A.
Given:
γ = 4
lb
ft
P = 60 lb
a = 4 ft
b = 3 ft
c = 3 ft
d = 3 ft
931
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
W = γ
2
2
2
2
2d
+ γ2 d + c d
γ d + ( b + c)
xc =
2
W
2
W = 88.774 lb
2
xc = 1.6 ft
b + c⎞
2
2⎛
2
2
⎟ + γ d + ( b + c) ⎜a +
⎟ + γ 2 d + c ( a + b + c)
2 ⎠
⎝ 2 ⎠
⎝
γ ( a + b) ⎛⎜
yc =
2
d + ( b + c) + γ 2 d + c + γ ( a + b)
a + b⎞
W
yc = 7.043 ft
Equilibrium
Ax = 0
Ax = 0lb
Ay − W − P = 0
Ax = 0 lb
Ay = W + P
MA − W xc − P2d = 0
Ay = 148.8 lb
MA = W xc + P2d
MA = 502 lb ft
Problem 9-52
Locate the center of gravity G(xc, yc) of the streetlight.
Neglect the thickness of each segment. The mass per
unit length of each segment is given.
Given:
a = 1m
ρ AB = 12
kg
m
b = 3m
ρ BC = 8
kg
m
c = 4m
ρ CD = 5
kg
m
d = 1m
ρ DE = 2
kg
m
e = 1m
f = 1.5 m
932
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⎛
⎝
M = ρ AB c + ρ BC b + ρ CD⎜ a + e +
xc =
yc =
πd⎞
⎟ + ρ DE f
2 ⎠
1⎡
πd ⎛
2d ⎞
f ⎞⎤
⎛ e⎞
⎛
⎢ρ CD ⎜d − ⎟ + ρ CD e⎜d + ⎟ + ρ DE f⎜d + e + ⎟⎥
M⎣
2 ⎝
2 ⎠⎦
π⎠
⎝ 2⎠
⎝
1⎡
πd ⎛
a
2d ⎞
⎡
⎤ ⎤
ρ CD⎢a⎛⎜ c + b + ⎞⎟ +
c+b+a+
+ e( c + b + a + d)⎥ ...⎥
⎜
⎟
⎢
M
2 ⎝
2⎠
π⎠
⎦ ⎥
⎢ ⎣⎝
b
c
⎞
⎢+ ρ DE f( c + b + a + d) + ρ BC b⎛⎜c + ⎟ + ρ AB c
⎥
2
⎣
⎝ 2⎠
⎦
⎛ xc ⎞ ⎛ 0.200 ⎞
⎜ ⎟=⎜
⎟m
⎝ yc ⎠ ⎝ 4.365 ⎠
Problem 9-53
Determine the location yc of the centroid of the
beam's cross-sectional area. Neglect the size of
the corner welds at A and B for the calculation.
Given:
d1 = 50 mm
d2 = 35 mm
h = 110 mm
t = 15 mm
Solution:
2
2
d2 ⎞
⎛ d1 ⎞ d1
h⎞
⎛ d2 ⎞ ⎛
⎛
π⎜ ⎟
+ h t⎜ d1 + ⎟ + π ⎜ ⎟ ⎜ d1 + h +
⎟
2⎠ 2
2⎠ ⎝
2⎠
2⎠
⎝
⎝
⎝
yc =
2
2
⎛ d1 ⎞
⎛ d2 ⎞
π ⎜ ⎟ + ht + π ⎜ ⎟
⎝2⎠
⎝2⎠
yc = 85.9 mm
933
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-54
The gravity wall is made of concrete. Determine the location (xc, yc) of the center of gravity G for
the wall.
Given:
a = 0.6 m
b = 2.4 m
c = 0.6 m
d = 0.4 m
e = 3m
f = 1.2 m
Solution:
A = ( a + b + c)d + ( b + c)e −
xc =
yc =
ce
e
− ( b + c − f)
2
2
A = 6.84 m
2
1⎡
⎛ a + b + c ⎞ + ( b + c)e⎛ a + b + c ⎞ − c e ⎛ a + b + 2c ⎞ ...⎤
( a + b + c)d⎜
⎟
⎜
⎟
⎜
⎟
⎢
A
2 ⎠
2 ⎝
3⎠ ⎥
⎝ 2 ⎠
⎝
⎢
⎥
b + c − f⎞
e
⎢+ −( b + c − f) ⎛⎜ a +
⎥
⎟
3
2⎝
⎣
⎠
⎦
1⎡
d
⎛ e ⎞ c e ⎛ d + e ⎞ ...⎤
( a + b + c)d + ( b + c)e⎜ d + ⎟ −
⎜
⎟
⎢
A
2
2⎠
2 ⎝
3⎠ ⎥
⎝
⎢
⎥
e
2e
⎢+ −( b + c − f) ⎛⎜d + ⎞⎟
⎥
2⎝
3⎠
⎣
⎦
⎛ xc ⎞ ⎛ 2.221 ⎞
⎜ ⎟=⎜
⎟m
⎝ yc ⎠ ⎝ 1.411 ⎠
Problem 9-55
Locate the centroid (xc, yc)of the shaded area.
934
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 1 in
b = 3 in
c = 1 in
d = 1 in
e = 1 in
Solution:
2
A = ( a + b) ( a + e) −
πa
−
4
1
( a + b − d) ( a + e − c)
2
1 ⎡ ( a + b)
π a ⎛ 4a ⎞ 1
a + b − d ⎞⎤
⎛
xc = ⎢
( a + e) −
− ( a + b − d) ( a + e − c) ⎜ a + b −
⎟⎥
⎜
⎟
A⎣
2
4 ⎝ 3π ⎠ 2
3
⎝
⎠⎦
2
yc =
2
1⎡
( a + e)
⎢( a + b)
A⎣
2
2
π a ⎛ 4a ⎞ 1
a + e − c ⎞⎤
⎛
− ( a + b − d) ( a + e − c) ⎜ a + e −
⎟⎥
⎜
⎟
4 ⎝ 3π ⎠ 2
3
⎝
⎠⎦
2
−
⎛ xc ⎞ ⎛ 1.954 ⎞
⎜ ⎟=⎜
⎟ in
⎝ yc ⎠ ⎝ 0.904 ⎠
Problem 9-56
Locate the centroid (xc, yc) of the shaded area.
Given:
a = 1 in
b = 6 in
c = 3 in
d = 3 in
Solution:
A = bd +
πd
4
2
2
−
πa
2
+
1
( d c)
2
935
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
xc =
1 ⎡ b π d ⎛ 4d ⎞ 1 ⎛
c ⎞⎤
⎢b d −
+ d c⎜ b + ⎟⎥
⎜
⎟
A⎣ 2
4 ⎝ 3π ⎠ 2 ⎝
3 ⎠⎦
yc =
1 ⎡ ⎛ d ⎞ π d ⎛ 4d ⎞ π a ⎛ 4a ⎞ 1 ⎛ d ⎞⎤
⎢b d⎜ ⎟ +
⎜ ⎟ − 2 ⎜ ⎟ + 2 d c⎜ 3 ⎟⎥
A ⎣ ⎝ 2⎠
4 ⎝ 3π ⎠
⎝ ⎠⎦
⎝ 3π ⎠
2
2
xc = 2.732 in
2
yc = 1.423 in
Problem 9-57
Determine the location yc of the centroidal axis xcxc of the beam's cross-sectional area. Neglect the
size of the corner welds at A and B for the calculation.
Given:
r = 50 mm
t = 15 mm
a = 150 mm
b = 15 mm
c = 150 mm
Solution:
2
⎛ a⎞
⎟ + a t⎜b + ⎟ + π r ( b + a + r)
2
2
⎝ ⎠
⎝
⎠
b c⎛⎜
yc =
b⎞
yc = 154.443 mm
2
b c + a t + πr
Problem 9-58
Determine the location (xc, yc) of the centroid C of the area.
Given:
a = 6 in
b = 6 in
936
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
c = 3 in
d = 6 in
Solution:
1 ⎛
c⎞ 1
2
⎟ + a c⎜ b + ⎟ + ( b + c)d ( b + c)
3
⎝ 2⎠ 2 ⎝ 3⎠ 2
a b⎛⎜
xc =
b⎞
ab +
2
ca +
1
2
xc = 4.625 in
( b + c)d
1 ⎛ a⎞ 1
⎛ d⎞
⎟ + a c⎜ ⎟ − ( b + c)d⎜ ⎟
⎝ 2⎠ 2 ⎝ 3⎠ 2
⎝ 3⎠
a b⎛⎜
yc =
1
a⎞
ab +
1
2
ca +
1
2
yc = 1 in
( b + c)d
Problem 9-59
Determine the location yc of the centroid C for a beam having the cross-sectional area shown. The
beam is symmetric with respect to the y axis.
Given:
a = 2 in
b = 1 in
c = 2 in
d = 1 in
e = 3 in
f = 1 in
Solution:
A = 2[ ( a + b + c + d) ( e + f) − b f − d e]
A = 40 in
2
937
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Engineering Mechanics - Statics
yc =
Chapter 9
2⎡
( e + f)
⎢( a + b + c + d)
A⎣
2
2
f
⎛ e ⎞⎤
− d e⎜ f + ⎟⎥
2
⎝ 2 ⎠⎦
2
−b
yc = 2.00 in
Problem 9-60
The wooden table is made from a square board having weight W. Each of the legs has wieght Wleg
and length L. Determine how high its center of gravity is from the floor. Also, what is the angle,
measured from the horizontal, through which its top surface can be tilted on two of its legs before it
begins to overturn? Neglect the thickness of each leg.
Given:
W = 15 lb
Wleg = 2 lb
L = 3 ft
a = 4 ft
Solution:
W L + 4Wleg⎛⎜
zc =
L⎞
⎟
⎝2⎠
W + 4Wleg
⎛ a ⎞
⎜ 2 ⎟
θ = atan ⎜ ⎟
⎝ zc ⎠
zc = 2.478 ft
θ = 38.9 deg
Problem 9-61
Locate the centroid yc for the beam’s cross-sectional area.
Given:
a = 120 mm
938
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 240 mm
c = 120 mm
Solution:
A = ( a + b)5c − 3b c
1 ⎡ ( a + b)
⎛ b⎞
⎛ b ⎞⎤
⎢
5c − 2b c⎜ ⎟ − b c⎜ ⎟⎥
A⎣
2
⎝ 2⎠
⎝ 3 ⎠⎦
2
yc =
yc = 229 mm
Problem 9-62
Determine the location xc of the centroid C of the shaded area which is part of a circle having a
radius r.
Solution:
A = α r − r sin ( α ) cos ( α )
2
xc =
2
1 ⎛ 2 2r sin ( α )
2
2
⎞
− r sin ( α ) cos ( α ) r cos ( α )⎟
⎜α r
A⎝
3
3α
⎠
939
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
⎞
sin ( α ) ⎛ 1 − cos ( α )
⎜
⎟
3 ⎝ α − sin ( α ) cos ( α ) ⎠
2
xc = 2r
2r
xc =
3
sin ( α )
α−
3
sin ( 2α )
2
Problem 9-63
Locate the centroid yc for the strut’s
cross-sectional area.
Given:
a = 40 mm
b = 120 mm
c = 60 mm
Solution:
A =
πb
2
2
− 2a c
1 ⎡ π b ⎛ 4b ⎞
⎛ c ⎞⎤
yc = ⎢
− 2a c⎜ ⎟⎥
⎜
⎟
A ⎣ 2 ⎝ 3π ⎠
⎝ 2 ⎠⎦
2
yc = 56.6 mm
Problem 9-64
The “New Jersey” concrete barrier is
commonly used during highway
construction. Determine the location yc
of its centroid.
Given:
a = 4 in
b = 12 in
c = 6 in
940
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 24 in
θ 1 = 75 deg
θ 2 = 55 deg
Solution:
e = b cot ( θ 2 )
f = d − 2e
h =
A = d( c + b + h) − b e − 2h e −
1
( f − a)h
2
1 ⎡ ( c + b + h)
yc = ⎢d
A⎣
2
2
⎛
⎝
− b e⎜ c +
f−a
tan ( θ 1 )
2
2b ⎞
h⎞ 1
2h ⎞⎤
⎛
⎛
⎟ − 2h e⎜ c + b + ⎟ − ( f − a)h⎜c + b + ⎟⎥
3⎠
2⎠ 2
3 ⎠⎦
⎝
⎝
yc = 8.69 in
Problem 9-65
The composite plate is made from both steel (A)
and brass (B) segments. Determine the mass and
location (xc, yc, zc) of its mass center G.
Units Used:
Mg = 1000 kg
Given:
ρ st = 7.85
m
ρ br = 8.74
a = 150 mm
Mg
3
b = 30 mm
Mg
m
3
c = 225 mm
d = 150 mm
Solution:
⎛
⎝
M = ρ st⎜ d b c +
xc =
1
1
⎞
a b c⎟ + ρ br a b c
2
2
⎠
1⎡ ⎡
1
d 1
⎛ a ⎞⎤
⎛ 2a ⎞⎤
⎢ρ st⎢d c b + a b c⎜d + ⎟⎥ + ρ br a b c⎜ d + ⎟⎥
M⎣ ⎣
2
2 2
3 ⎠⎦
⎝ 3 ⎠⎦
⎝
941
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
yc =
−1 ⎡ ⎛
1
b 1
b⎞
b⎤
⎢ρ st⎜ d c b + a b c ⎟ + ρ br a b c ⎥
M⎣ ⎝
2
2 2
2⎠
2⎦
zc =
1⎡ ⎛
1
c 1
2c ⎞
c⎤
⎢ρ st⎜d c b + a b c ⎟ + ρ br a b c ⎥
M⎣ ⎝
2
2 2
3⎠
3⎦
⎛ xc ⎞ ⎛ 152.8 ⎞
⎜ ⎟ ⎜
⎟
⎜ yc ⎟ = ⎜ −15.0 ⎟ mm
⎜ z ⎟ ⎝ 111.5 ⎠
⎝ c⎠
M = 16.347 kg
Problem 9-66
Locate the centroid yc of the concrete beam having the tapered cross section shown.
Given:
a = 100 mm
b = 360 mm
c = 80 mm
d = 300 mm
e = 300 mm
Solution:
1
⎛ b⎞
⎛ b⎞
⎟ + ( d − a)b⎜c + ⎟ + a b⎜ c + ⎟
⎝ 2⎠ 2
⎝ 3⎠
⎝ 2⎠
( d + 2e)c⎛⎜
yc =
c⎞
( d + 2e)c +
1
2
yc = 135 mm
( d − a)b + a b
Problem 9-67
The anatomical center of gravity G of a person can be determined by using a scale and a rigid board
having a uniform weight W1 and length l. With the person’s weight W known, the person lies down on
the board and the scale reading P is recorded. From this show how to calculate the location xc of the
center of mass. Discuss the best place l1 for the smooth support at B in order to improve the accuracy
of this experiment.
942
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 120 mm
b = 240 mm
c = 120 mm
Solution:
ΣMB = 0;
⎛
⎝
W xc − P l1 + W1 ⎜ l1 −
P l1 − W1 ⎛⎜ l1 −
xc =
⎝
l⎞
⎟=0
2⎠
l⎞
⎟
2⎠
W
Put B as close as possible to the center of gravity of the board, i.e., l1 =
l
⎛
⎝
, then W1 ⎜ l1 −
2
the effect of the board's weight will not be a large factor in the measurement.
l⎞
⎟ = 0 and
2⎠
Problem 9-68
The tank and compressor have a mass
MT and mass center at GT and the
motor has a mass MM and a mass
center at GM. Determine the angle of
tilt,θ , of the tank so that the unit will
be on the verge of tipping over.
Given:
a = 300 mm
b = 200 mm
c = 350 mm
d = 275 mm
MT = 15 kg
MM = 70 kg
943
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
xc =
yc =
b MT + ( a + b)MM
xc = 0.4471 m
MT + MM
c MT + ( c + d)MM
yc = 0.57647 m
MT + MM
⎛ xc ⎞
⎟
⎝ yc ⎠
θ = atan ⎜
θ = 37.8 deg
Problem 9-69
Determine the distance h to which a hole of diameter d must be bored into the base of the cone so
that the center of mass of the resulting shape is located at zc. The material has a density ρ.
Given:
d = 100 mm
zc = 115 mm
ρ = 8
mg
m
3
a = 150 mm
b = 500 mm
Solution:
Guess
h = 200 mm
1
Given
zc =
3
2
d
h
2 b
π a b⎛⎜ ⎟⎞ − π ⎛⎜ ⎟⎞ h⎛⎜ ⎟⎞
⎝ 4⎠
⎝ 2⎠ ⎝ 2⎠
2
h = Find ( h)
h = 323 mm
d
1 2
π a b − π ⎛⎜ ⎟⎞ h
3
⎝ 2⎠
944
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-70
Determine the distance to the centroid of the shape which consists of a cone with a hole of height h
bored into its base.
Given:
d = 100 mm
h = 50 mm
ρ = 8
mg
m
3
a = 150 mm
b = 500 mm
Solution:
2 ⎛ b⎞
2
d
h
π a b⎜ ⎟ − π ⎛⎜ ⎟⎞ h⎛⎜ ⎟⎞
3
⎝ 4⎠
⎝ 2⎠ ⎝ 2⎠
zc =
2
d
1 2
π a b − π ⎛⎜ ⎟⎞ h
3
⎝ 2⎠
1
zc = 128.4 mm
Problem 9-71
The sheet metal part has the dimensions shown. Determine the location (xc, yc, zc) of
its centroid.
Given:
a = 3 in
945
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
b = 4 in
c = 6 in
Solution:
−a b⎛⎜
b⎞
⎟
⎝ 2⎠
xc =
ab +
1
2
ac
1 ⎛ 2a ⎞
⎟ + a c⎜ ⎟
⎝ 2⎠ 2 ⎝ 3 ⎠
a b⎛⎜
yc =
a⎞
ab +
−1
zc =
xc = −1.143 in
2
a c⎛⎜
ab +
2
2
yc = 1.714 in
ac
c⎞
⎟
⎝ 3⎠
1
1
zc = −0.857 in
ac
Problem 9-72
The sheet metal part has a weight per unit area of
and is supported by the smooth rod and at C. If the
cord is cut, the part will rotate about the y axis until
it reaches equilibrium. Determine the equilibrium
angle of tilt, measured downward from the negative
x axis, that AD makes with the -x axis.
Given:
a = 3 in
b = 4 in
c = 6 in
946
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
a b⎛⎜
b⎞
ab +
1
⎟
⎝ 2⎠
xc =
1
2
zc =
2
a c⎛⎜
xc = 1.143 in
ac
c⎞
⎟
⎝ 3⎠
ab +
1
2
zc = 0.857 in
ac
⎛ xc ⎞
⎟
⎝ zc ⎠
θ = atan ⎜
θ = 53.13 deg
Problem 9-73
A toy skyrocket consists of a solid conical top of density ρt, a hollow cylinder of density ρc, and a
stick having a circular cross section of density ρs. Determine the length of the stick, x, so that the
center of gravity G of the skyrocket is located along line aa.
Given:
a = 3 mm
ρ t = 600
m
b = 10 mm
ρ c = 400
c = 5 mm
3
kg
m
d = 100 mm
e = 20 mm
kg
ρ s = 300
3
kg
m
3
Solution:
Guess
x = 200 mm
Given
⎛a ⎞
b e
π 2
e
d
x
2
ρ tπ ⎛⎜ ⎟⎞ ⎛⎜ d + ⎞⎟ + ρ c b − c d⎛⎜ ⎟⎞ + ρ sπ ⎜ ⎟ x⎛⎜ d − ⎞⎟ = 0
2
x = Find ( x)
⎝ 2⎠ 3 ⎝
4⎠
(
4
) 2
⎝ ⎠
2
⎝4⎠ ⎝
2⎠
x = 490 mm
947
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-74
Determine the location (xc, yc) of the center of mass of the turbine and compressor assembly. The
mass and the center of mass of each of the various components are indicated below.
Given:
a = 0.75 m
M1 = 25 kg
b = 1.25 m
M2 = 80 kg
c = 0.5 m
M3 = 30 kg
d = 0.75 m
M4 = 105 kg
e = 0.85 m
f = 1.30 m
g = 0.95 m
Solution:
M = M1 + M2 + M3 + M4
xc =
1
⎡M2 a + M3 ( a + b) + M4( a + b + c)⎤⎦
M⎣
xc = 1.594 m
yc =
1
( M 1 d + M 2 e + M 3 f + M 4 g)
M
yc = 0.940 m
Problem 9-75
The solid is formed by boring a conical hole into the hemisphere. Determine the distance zc to the
center of gravity.
948
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
V=
π 3
2 3 π 2
πa − a a = a
3
3
3
zc =
1 ⎡ 5a ⎛ 2 3⎞ 3 ⎛ π 3⎞⎤
⎢ ⎜ π a ⎟ − a ⎜ a ⎟⎥
V⎣ 8 ⎝3
⎠ 4 ⎝ 3 ⎠⎦
zc =
a
2
Problem 9-76
Determine the location xc of the centroid of the solid made from a hemisphere, cylinder, and cone.
Given:
a = 80 mm
b = 60 mm
c = 30 mm
d = 30 mm
Solution:
V =
1 2
2 3
2
πd a + πd b + πd
3
3
xc =
1 ⎡ 1 2 ⎛ 3a ⎞
b ⎞ 2 3⎛
3c ⎞⎤
2 ⎛
⎢ π d a⎜ ⎟ + π d b⎜ a + ⎟ + π d ⎜ a + b + ⎟⎥
V ⎣3
8 ⎠⎦
⎝4⎠
⎝ 2⎠ 3 ⎝
xc = 105.2 mm
949
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-77
The buoy is made from two homogeneous cones each having radius r. Find the distance zc to the
buoy's center of gravity G.
Given:
r = 1.5 ft
h = 1.2 ft
a = 4 ft
Solution:
π 2 ⎛ a⎞
zc =
3
π 2 ⎛ h⎞
⎟ − r h⎜ ⎟
⎝ 4⎠ 3 ⎝ 4⎠
r a⎜
π 2
3
zc = 0.7 ft
r ( a + h)
Problem 9-78
The buoy is made from two homogeneous cones each having radius r. If it is required that the
buoy's center of gravity G be located at zc,determine the height h of the top cone.
Given:
zc = 0.5 ft
r = 1.5 ft
a = 4 ft
Solution:
Guess
h = 1 ft
950
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
π 2 ⎛ a⎞
zc =
Given
3
π 2 ⎛ h⎞
⎟ − r h⎜ ⎟
⎝ 4⎠ 3 ⎝ 4⎠
r a⎜
π 2
3
h = Find ( h)
h = 2 ft
r ( a + h)
Problem 9-79
Locate the center of mass zc of the forked lever, which is made from a homogeneous material and has
the dimensions shown.
Given:
a = 0.5 in
b = 2.5 in
c = 2 in
d = 3 in
e = 0.5 in
Solution:
2
V = b a + 2e a d +
zc =
π⎡
2
2
2
⎣( c + e) − c ⎤⎦ a
1 ⎡ 2⎛ b ⎞
d⎞ ⎛ πa⎞
2⎡
⎛
⎛ c + e ⎞⎤ ...⎤
b a ⎜ ⎟ + 2e a d⎜ b + e + c + ⎟ + ⎜ ⎟ ( c + e) ⎢b + c + e − 4⎜
⎟⎥ ⎥
⎢
V
2⎠ ⎝ 2 ⎠
⎝ 2⎠
⎝
⎣
⎝ 3π ⎠⎦ ⎥
⎢ −π a
⎞ 2⎡
⎛ c ⎞⎤
⎢+ ⎛⎜
⎥
⎟ c ⎢b + c + e − 4⎜ ⎟⎥
⎣ ⎝ 2 ⎠ ⎣
⎝ 3π ⎠⎦
⎦
zc = 4.32 in
951
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-80
A triangular plate made of homogeneous material has a constant thickness which is very small. If it
is folded over as shown, determine the location yc of the plate's center of gravity G.
Given:
a = 6 in
b = 3 in
c = 1 in
d = 3 in
e = 1 in
f = 3 in
Solution:
1
⎛ 2b ⎞ 1
⎛ f⎞
⎟ + ( 2c b) ⎜ ⎟ + ( 2e f) ⎜ ⎟
⎝ 2⎠ 2
⎝3⎠ 2
⎝ 3⎠
2d b⎛⎜
yc =
b⎞
2d b +
1
2
( 2c b) +
1
2
yc = 0.75 in
( 2d) ( a + f)
Problem 9-81
A triangular plate made of homogeneous material has a constant thickness which is very small. If it
is folded over as shown, determine the location zc of the plate's center of gravity G.
Given
a = 6 in
b = 3 in
c = 1 in
952
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 3 in
e = 1 in
f = 3 in
Solution:
1
zc =
2
1
⎛ a⎞
⎟ + 2( d − e)a⎜ ⎟
⎝ 2⎠ 2
⎝ 3⎠
( 2e f a) + 2e a⎛⎜
2d b +
1
2
a⎞
( 2c b) +
1
2
zc = 1.625 in
2d( a + f)
Problem 9-82
Each of the three homogeneous plates welded to the rod has a density ρ and a thickness a. Determine
the length l of plate C and the angle of placement, θ, so that the center of mass of the assembly lies on
the y axis. Plates A and B lie in the x–y and z–y planes, respectively.
Units Used:
Mg = 1000 kg
Given:
a = 10 mm
f = 100 mm
b = 200 mm
g = 150 mm
c = 250 mm
e = 150 mm
ρ = 6
Mg
m
3
Solution:
The thickness and density are uniform
Guesses
θ = 10 deg
l = 10 mm
953
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given
⎛ f ⎞ − g l⎛ g ⎞ cos ( θ ) = 0
⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
b f⎜
⎛l⎞
⎜ ⎟ = Find ( l , θ )
⎝θ⎠
⎛ e ⎞ + g l⎛ g ⎞ sin ( θ ) = 0
⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
−c e⎜
θ = 70.4 deg
l = 265 mm
Problem 9-83
The assembly consists of a wooden dowel rod of length L and a tight-fitting steel collar. Determine
the distance xc to its center of gravity if the specific weights of the materials are γw and γst.The radii
of the dowel and collar are shown.
Given:
L = 20 in
γ w = 150
lb
ft
γ st = 490
3
lb
ft
3
a = 5 in
b = 5 in
r1 = 1 in
r2 = 2 in
Solution:
b⎞
2
2
(
)
⎟
2
⎝ 2⎠
2
2
2
γ wπ r1 L + γ stπ ( r2 − r1 ) b
2 L
γ wπ r1 L
xc =
+ γ stπ r2 − r1 b⎛⎜ a +
xc = 8.225 in
Problem 9-84
Determine the surface area and the volume of the ring formed by rotating the square about the vertical
axis.
954
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
θ = 45 deg
Solution:
⎡ ⎛ a sin ( θ )⎞ a⎤ ...
⎟⎥
⎣ ⎝ 2
⎠⎦
a
⎡
⎛
+ 2⎢2π ⎜ b + sin ( θ )⎞⎟ a⎤⎥
⎣ ⎝ 2
⎠⎦
A = 2⎢2π ⎜ b −
A = 8π b a
V = 2π b a
2
Problem 9-85
The anchor ring is made of steel having specific weight γst. Determine the surface area of the ring.
The cross section is circular as shown.
Given:
γ st = 490
lb
ft
3
a = 4 in
b = 8 in
955
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⎛ a + b − a ⎞ 2π ⎛ b − a ⎞
⎟ ⎜
⎟
4 ⎠ ⎝ 4 ⎠
⎝2
A = 2π ⎜
A = 118 in
2
Problem 9-86
Using integration, determine both the area and the distance yc to the centroid of the shaded area.
Then using the second theorem of PappusGuldinus, determine the volume of the solid generated
by revolving the shaded area about the x axis.
Given:
a = 1 ft
b = 2 ft
c = 2 ft
Solution:
⌠
⎮
A = ⎮
⌡
c
⎡ ⎛ y⎞2 ⎤
⎢a + ⎜ ⎟ b⎥ d y
⎣ ⎝c⎠ ⎦
A = 3.333 ft
2
0
c
⌠
2
1 ⎮ ⎡
⎛ y ⎞ b⎥⎤ d y
⎢
yc =
y
a
+
⎜ ⎟
A⎮ ⎣
⎝c⎠ ⎦
⌡
yc = 1.2 ft
V = 2π yc A
V = 25.1 ft
0
3
956
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-87
The grain bin of the type shown is manufactured by Grain Systems, Inc. Determine the required
square footage of the sheet metal needed to form it, and also the maximum storage capacity (volume)
within it.
Given:
a = 30 ft
b = 20 ft
c = 45 ft
Solution:
A = 2π a c + 2π
a
2
2
a +b
2
3 2
A = 11.9 × 10 ft
⎛ a ⎞ + 2π a ⎛ 1 a b⎞
⎟
⎜
⎟
3⎝2 ⎠
⎝ 2⎠
V = 2π a c⎜
3 3
V = 146 × 10 ft
Problem 9-88
Determine the surface area and the volume of the conical solid.
957
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
A = 2a
3 a
2π
2 2
2
3 πa
A=
⎛ 1 a ⎞ ⎛⎜ 3 a⎟⎞ ⎛⎜ 3 a2π ⎟⎞
⎟
⎝ 2 2 ⎠⎝ 2 ⎠⎝ 6
⎠
V = 2⎜
V=
π 3
4
a
Problem 9-89
Sand is piled between two walls as shown. Assume the pile to be a quarter section of a cone and
that ratio p of this volume is voids (air space). Use the second theorem of Pappus-Guldinus to
determine the volume of sand.
Given:
r = 3m
h = 2m
p = 0.26
Solution:
⎛ π ⎞⎛ r ⎞⎛ h r⎞
⎟⎜ ⎟⎜ ⎟
⎝ 2 ⎠⎝ 3 ⎠⎝ 2 ⎠
V = ( 1 − p) ⎜
V = 3.487 m
3
958
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-90
The rim of a flywheel has the cross section A-A shown. Determine the volume of material needed
for its construction.
Given:
r = 300 mm
a = 20 mm
b = 40 mm
c = 20 mm
d = 60 mm
Solution:
⎛
⎝
V = 2π ⎜ r + b +
6
c⎞
⎛ b⎞
⎟ d c + 2π ⎜ r + ⎟ b a
2⎠
⎝ 2⎠
3
V = 4.25 × 10 mm
Problem 9-91
The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the weight of
the wheel if it is made from steel having a specific weight γ.
Given:
a = 1 in
c = 0.5 in
d = 1 in
e = 1 in
f = 0.25 in
b = 2( c + d + e)
γ = 490
lb
ft
3
959
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
⎡ ⎛
⎣ ⎝
W = γ 2π ⎢d a⎜ c +
d⎞ ⎛
a ⎞⎛ a − f ⎞ ⎤
⎟ + ⎜c + d + ⎟ ⎜
⎟ e⎥
2⎠ ⎝
3 ⎠⎝ 2 ⎠ ⎦
W = 3.01 lb
Problem 9-92
The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the total
surface area of the wheel in order to estimate the amount of paint needed to protect its surface from
rust.
Given:
a = 1 in
c = 0.5 in
d = 1 in
e = 1 in
f = 0.25 in
b = 2( c + d + e)
Solution:
⎡
⎛
⎝
A = 2π ⎢ f( c + d) + a c + 2( d + e) ⎜ c +
⎣
A = 70 in
d + e⎞
e ⎞⎤
2 ⎛ a − f⎞ ⎛
+
2
e
+
c
+
d
+
⎟
⎜
⎟ ⎜
⎟⎥
2 ⎠
2 ⎠⎦
⎝ 2 ⎠ ⎝
2
2
Problem 9-93
Determine the volume of material
needed to make the casting.
Given:
r1 = 4 in
r2 = 6 in
960
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
r3 = r2 − r1
Solution:
4r3 ⎞⎤
⎡ ⎛ π ⎞ 2⎛ 4r2 ⎞
⎛ r2 ⎞ ⎛ π ⎞ 2⎛
⎟⎥
⎟ r2 ⎜ ⎟ + 2r2( 2r3) ⎜ ⎟ − 2⎜ ⎟ r3 ⎜ r2 −
3π ⎠⎦
⎝ 2 ⎠ ⎝ 2⎠ ⎝
⎣ ⎝ 4 ⎠ ⎝ 3π ⎠
V = 2π ⎢2⎜
3
V = 1.40 × 10 in
3
Problem 9-94
A circular sea wall is made of concrete. Determine the total weight of the wall if the concrete has
a specific weight γc.
Given:
γ c = 150
lb
ft
3
a = 60 ft
b = 15 ft
c = 8 ft
d = 30 ft
θ = 50 deg
Solution:
c⎞ ⎤
⎤ 2
⎤ ⎛
⎡ ⎡1
⎡1
W = γ c θ ⎢a⎢ d( b − c)⎥ + ( b − c) ⎢ d( b − c)⎥ + ⎜ a + b − ⎟ d c⎥
2⎠ ⎦
⎣ ⎣2
⎦ 3
⎣2
⎦ ⎝
6
W = 3.12 × 10 lb
Problem 9-95
Determine the surface area of the tank, which consists of a cylinder and hemispherical cap.
961
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 4m
b = 8m
Solution:
⎛
⎝
A = 2π ⎜ a b +
A = 302 m
2a π a ⎞
⎟
π 2 ⎠
2
Problem 9-96
Determine the volume of the tank, which consists of a cylinder and hemispherical cap.
Given:
a = 4m
b = 8m
Solution:
⎡ 4a ⎛ π a2 ⎞ a
⎤
⎜
⎟ + ( b a)⎥
⎣ 3π ⎝ 4 ⎠ 2
⎦
V = 2π ⎢
V = 536 m
3
962
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-97
Determine the surface area of the silo which consists of a cylinder and hemispherical cap. Neglect the
thickness of the plates.
Given:
a = 10 ft
b = 10 ft
c = 80 ft
Solution:
⎡2a ⎛ π a ⎞ + a c⎤
⎜ ⎟
⎥
⎣π ⎝ 2 ⎠
⎦
A = 2π ⎢
3 2
A = 5.65 × 10 ft
Problem 9-98
Determine the volume of the silo which
consists of a cylinder and hemispherical
cap. Neglect the thickness of the plates.
Given:
a = 10 ft
b = 10 ft
c = 80 ft
Solution:
⎡ 4a ⎛ π a2 ⎞
a ⎤
⎜
⎟ + c a⎛⎜ ⎟⎞⎥
⎝ 2 ⎠⎦
⎣ 3π ⎝ 4 ⎠
V = 2π ⎢
3 3
V = 27.2 × 10 ft
963
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-99
The process tank is used to store liquids during manufacturing. Estimate both the volume of the
tank and its surface area. The tank has a flat top and the plates from which the tank is made
have negligible thickness.
Given:
a = 4m
b = 6m
c = 3m
Solution:
⎤
⎡c ⎛ c a ⎞ c
V = 2π ⎢ ⎜ ⎟ + ( c b)⎥
3
2
2
⎣ ⎝ ⎠
⎦
V = 207 m
3
c
⎛c
A = 2π ⎜ c + c b +
2
⎝2
A = 188 m
2
2⎞
a +c ⎟
⎠
2
Problem 9-100
Determine the height h to which liquid should be poured into the cup so that it contacts half the
surface area on the inside of the cup. Neglect the cup's thickness for the calculation.
Given:
a = 30 mm
b = 50 mm
c = 10 mm
Solution:
Total area
⎡ c + a + c b2 + ( a − c) 2⎤
⎥
2
⎣ 2
⎦
Atotal = 2π ⎢c
Guess
h = 1 mm
e = 1 mm
964
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given
a−c
e−c
=
b
h
Atotal
2
⎡ c + e + c h2 + ( e − c) 2⎤
⎥
2
⎣ 2
⎦
= 2π ⎢c
⎛e⎞
⎜ ⎟ = Find ( e , h)
⎝h⎠
e = 21.942 mm
h = 29.9 mm
Problem 9-101
Using integration, compute both the area and the centroidal distance xc of the shaded region. Then,
using the second theorem of Pappus–Guldinus, compute the volume of the solid generated by
revolving the shaded area about the aa axis.
Given:
a = 8 in
b = 8 in
Solution:
a
⌠
2
⎮
⎛x⎞
A = ⎮ b ⎜ ⎟ dx
⎝ a⎠
⌡
0
a
⌠
2
1 ⎮
x⎞
⎛
xc = 2a − ⎮ x b ⎜ ⎟ dx
A
⎝ a⎠
⌡
A = 21.333 in
V = 2π A xc
V = 1.34 × 10 in
2
xc = 10 in
0
3
3
965
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-102
Using integration, determine the area and the centroidal
distance yc of the shaded area. Then, using the second
theorem of Pappus–Guldinus, determine the volume of a
solid formed by revolving the area about the x axis.
Given:
a = 0.5 ft
b = 2 ft
c = 1 ft
Solution:
⌠
⎮
A = ⎮
⌡
b
2
c
dx
x
A = 1.386 ft
2
a
⌠
⎮
1 ⎮
yc =
A⎮
⌡
b
2
1 ⎛c ⎞
⎜ ⎟ dx
2⎝ x⎠
2
yc = 0.541 ft
a
V = 2π A yc
V = 4.71 ft
3
Problem 9-103
Determine the surface area of the roof
of the structure if it is formed by
rotating the parabola about the y axis.
Given:
a = 16 m
b = 16 m
966
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
Centroid : The length of the differential element is
dL =
⎡
2⎤
⎛ dy ⎞ ⎥ dx
⎟
⎝ dx ⎠ ⎦
dx + dy = ⎢ 1 + ⎜
2
2
⎣
and its centroid is
xc = x
dy
bx
= −2
2
dx
a
Here,
Evaluating the integrals, we have
a
⌠
⎮
L = ⎮
⎮
⌡0
2 2
1+
4b x
a
4
dx
L = 23.663 m
a
⌠
2 2
1⎮
4b x
dx
xc = ⎮ x 1 +
4
L⎮
a
⌡0
xc = 9.178 m
3
A = 2π xc L
A = 1.365 × 10 m
2
Problem 9-104
The suspension bunker is made from plates which are curved to the natural shape which a completely
flexible membrane would take if subjected to a full load of coal.This curve may be approximated by a
parabola, y/b = (x/a)2. Determine the weight of coal which the bunker would contain when completely
filled. Coal has a specific weight of γ, and assume there is a fraction loss p in volume due to air voids.
Solve the problem by integration to determine the cross-sectional area of ABC; then use the second
theorem of Pappus–Guldinus to find the volume.
Units Used:
3
kip = 10 lb
Given:
a = 10 ft
b = 20 ft
967
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
lb
γ = 50
ft
3
p = 0.2
Solution:
b
⌠
A = ⎮ a
⎮
⌡
y
dy
b
A = 133.3 ft
2
0
⌠
1 ⎮
xc =
A⎮
⌡
b
1⎛
⎜a
2⎝
2
y⎞
⎟ dy
b⎠
xc = 3.75 ft
0
3 3
V = 2π A xc
V = 3.142 × 10 ft
W = ( 1 − p) γ V
W = 125.7 kip
Problem 9-105
Determine the interior surface area of the
brake piston. It consists of a full circular
part. Its cross section is shown in the figure.
Given:
a = 40 mm
b = 30 mm
c = 20 mm
d = 20 mm
e = 80 mm
968
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
f = 60 mm
g = 40 mm
Solution:
c⎞
3c ⎞
⎤
⎡a ⎛ b ⎞ 2 2 ⎛
⎛
A = 2π ⎢ a + ⎜ a + ⎟ b + e + c⎜ a + b + ⎟ + ( a + b + c) f + ⎜ a + b + ⎟ c ...⎥
2
2
2
2
⎝
⎠
⎝
⎠ ⎥
⎢+ ( a +⎝ b + 2c⎠)g
⎣
⎦
3
2
A = 119 × 10 mm
Problem 9-106
Determine the magnitude of the resultant hydrostatic force acting on the dam and its location H,
measured from the top surface of the water. The width of the dam is w; the mass density is ρw.
Units Used:
3
Mg = 10 kg
6
MN = 10 N
Given:
w = 8m
ρw = 1
Mg
m
3
h = 6m
g = 9.81
m
2
s
Solution:
p = hρ w g
p = 58860
N
m
F =
1
hw p
2
F = 1.41 MN
⎛ 2 ⎞h
⎟
⎝ 3⎠
H=4m
H = ⎜
2
969
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-107
The tank is filled with water to a depth d. Determine the resultant force the water exerts on side A
and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so
that it creates the same resultant forces? The densities are ρ0 and ρw.
Given:
3
kN = 10 N
d = 4m
a = 3m
b = 2m
kg
ρ o = 900
m
ρ w = 1000
3
kg
m
g = 9.81
3
m
2
s
Solution:
For water
At side A:
WA = b ρ w g d
1
WA d
2
F RA =
At side B:
WA = 78480
N
m
F RA = 157 kN
N
m
WB = aρ w g d
WB = 117720
1
WB d
2
F RB = 235 kN
F RB =
For oil
At side A:
F RA =
d1 =
1
bρ o g d1 d1
2
2FRA
d1 = 4.216 m
bρ o g
970
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-108
The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment
about O due to the dam’s weight divided by the overturning moment about O due to the water pressure.
Determine this factor if the concrete has specific weight γconc and water has specific weight γw.
Given:
a = 3 ft
b = 15 ft
c = 9 ft
γ w = 62.4
lb
ft
3
lb
γ conc = 150
ft
3
Solution:
For a 1-ft thick section:
W = γ w b( 1ft)
W = 936
1
Wb
2
F =
lb
ft
F = 7020 lb
W1 = γ conc( 1ft)a b
W1 = 6750 lb
1
W2 = γ conc ( c − a)b( 1ft )
2
W2 = 6750 lb
Moment to overturn:
1
MO = F b
3
MO = 35100 lb ft
Moment to stabilize:
⎡
⎣
MS = W1 ⎢( c − a) +
Fs =
a⎤
⎤
⎡2
⎥ + W2⎢ ( c − a)⎥
2⎦
⎣3
⎦
MS = 77625 lb⋅ ft
MS
F s = 2.21
MO
971
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-109
The concrete "gravity" dam is held in place by its own weight. If the density of concrete is ρc and
water has a density ρw, determine the smallest dimension d that will prevent the dam from
overturning about its end A.
Units Used:
3
Mg = 10 kg
Given:
ρ c = 2.5
Mg
m
3
Mg
ρ w = 1.0
m
3
h = 6m
g = 9.81
m
2
s
Solution:
Consider a dam of width a = 1 m.
w = ρw g h a
W=
w = 58860
N
m
F =
1
wh
2
F = 176580 N
1
ρc g d h a
2
Equilibrium
W
2d
h
−F =0
3
3
1
Fh
2d
ρc g d h a
=
2
3
3
d =
F
ρc g a
d = 2.683 m
Problem 9-110
The concrete dam is designed so that its face AB has a gradual slope into the water as shown.
Because of this, the frictional force at the base BD of the dam is increased due to the hydrostatic
force of the water acting on the dam. Calculate the hydrostatic force acting on the face AB of the
dam. The dam has width w, the water density is γw.
972
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used:
3
kip = 10 lb
Given:
w = 60 ft
lb
γ w = 62.4
ft
3
a = 18 ft
b = 12 ft
Solution:
F AB =
1
2
2
wγ w b a + b
2
F AB = 486 kip
Problem 9-111
The symmetric concrete “gravity” dam is held in place by its own weight. If the density of concrete is
ρc and water has a density ρw, determine the smallest distance d at its base that will prevent the dam
from overturning about its end A.The dam has a width w.
Units Used:
3
Mg = 10 kg
6
MN = 10 N
Given:
a = 1.5 m
ρ c = 2.5
m
b = 9m
w = 8m
Mg
ρ w = 1.0
3
Mg
m
3
Solution:
Guesses
d = 3m
F h = 1 MN
F v = 1 MN
W = 1 MN
973
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given
Fv =
d−a
b wρ w g
4
Fh =
1
ρw g b wb
2
⎡
⎣
⎛ d − a ⎞ b⎤
⎟⎥
⎝ 2 ⎠⎦
W = ρ c g w⎢a b + ⎜
W
d
⎛ d − a⎞ − F b = 0
+ F v⎜ d −
⎟
h
3
2
6 ⎠
⎝
⎛ Fv ⎞
⎜ ⎟
⎜ Fh ⎟ = Find ( F , F , W , d)
v h
⎜W⎟
⎜ ⎟
⎝d ⎠
⎛ Fv ⎞ ⎛ 0.379 ⎞
⎜ ⎟ ⎜
⎟
⎜ Fh ⎟ = ⎜ 3.178 ⎟ MN
⎜ W ⎟ ⎝ 4.545 ⎠
⎝ ⎠
d = 3.65 m
Problem 9-112
The tank is used to store a liquid having a specific weight γ. If it is filled to the top, determine
the magnitude of force the liquid exerts on each of its two sides ABDC and BDFE.
Units used:
3
kip = 10 lb
Given:
γ = 80
lb
ft
3
a = 6 ft
b = 6 ft
c = 12 ft
d = 8 ft
e = 4 ft
974
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
Fluid Pressure: The fluid pressure at points B
and E can be determined using
pB = γ e
pB = 320
lb
ft
pE = γ ( e + d)
pE = 960
2
lb
ft
2
Thus
kip
ft
wB = pB c
wB = 3.84
wE = pE c
wE = 11.52
kip
ft
Resultant Forces: The resultant Force acts on surface ABCD is
F R1 =
1
2
2
wB e + b
2
F R1 = 13.8 kip
and on surface BDFE is
F R2 =
1
(wB + wE)d
2
F R2 = 61.4 kip
Problem 9-113
The rectangular gate of width w is pinned
at its center A and is prevented from
rotating by the block at B. Determine the
reactions at these supports due to
hydrostatic pressure.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
Given:
a = 1.5 m
b = 6m
ρ w = 1.0
Mg
m
3
975
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
w = 2m
g = 9.81
Chapter 9
m
2
s
Solution:
w1 = ρ w g( b − 2a)w
w1 = 59
kN
m
w2 = ρ w g2a w
w2 = 59
kN
m
F1 =
1
2a w1
2
F 2 = w2 2a
F 1 = 88 kN
F 2 = 177 kN
a
− FB a = 0
3
ΣMA = 0;
F1
ΣF x = 0;
F1 + F2 − FB − FA = 0
1
F1
3
F B = 29.4 kN
FA = F1 + F2 − FB
F A = 235 kN
FB =
Problem 9-114
The gate AB has width w. Determine the horizontal and vertical components of force acting on
the pin at B and the vertical reaction at the smooth support A. The density of water is ρw.
Units Used:
3
Mg = 10 kg
3
kN = 10 N
6
MN = 10 N
Given:
w = 8m
ρ w = 1.0
Mg
m
3
a = 5m
b = 4m
976
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
c = 3m
g = 9.81
m
2
s
Solution:
Fluid Pressure: The fluid pressure at points A and B can be determined using Eq. 9-15,
pA = ρ w g( a + b)
pA = 88.29
kN
m
pB = ρ w g a
pB = 49.05
2
kN
m
2
wA = pA w
wA = 706.32
wB = pB w
wB = 392.4
kN
m
kN
m
Equilibrium
2
wB
2
2
2
b +c
1
b +c
+ ( wA − wB)
− Ay c = 0
2
2
3
2
wB
2
b +c
2
Ay =
Ay − wB c −
+
(wA − wB)
2
1
(2
2
2b +c
)
3
Ay = 2.507 MN
c
1
(wA − wB)c − By = 0
2
B y = Ay − wB c −
1
(wA − wB)c
2
−B x + wB b +
1
(wA − wB)b = 0
2
B x = wB b +
1
(wA − wB)b
2
B y = 858.92 kN
B x = 2.197 MN
Problem 9-115
The storage tank contains oil having a specific weight γ. If the tank has width w, calculate the
resultant force acting on the inclined side BC of the tank, caused by the oil, and specify its location
along BC, measured from B. Also compute the total resultant force acting on the bottom of the tank.
977
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used:
3
kip = 10 lb
Given:
lb
γ = 56
ft
3 c = 8 ft
w = 6 ft
d = 4 ft
a = 10 ft
e = 3 ft
b = 2 ft
f = 4 ft
Solution:
wB = wγ b
lb
lb
wC = wγ ( b + c)
wC = 3360
ft
ft
1
1
F h2 = ( wC − wB) c
F v1 = γ w b e
F v2 = γ w c e
2
2
wB = 672
F h1 = wB c
The resultant force
F Rx = F h1 + F h2
F Ry = F v1 + Fv2
The location h measured from point B
F v1
FR =
Guess h = 1 ft
e
2e
c
2c
+ F v2
+ F h1 + Fh2
= F Rx
2
3
2
3
ch
2
2
c +e
2
FRx + FRy
+ F Ry
F bot = γ w f( b + c + d)
F R = 17.225 kip
Given
eh
2
2
c +e
h = Find ( h)
On the bottom of the tank
2
h = 5.221 ft
F bot = 18.816 kip
Problem 9-116
The arched surface AB is shaped in the form of a quarter circle. If it has a length L, determine the
horizontal and vertical components of the resultant force caused by the water acting on the surface.
The density of water is ρw.
978
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Units Used:
3
Mg = 10 kg
3
kN = 10 N
Given:
L = 8m
Mg
ρ w = 1.0
m
3
a = 3m
b = 2m
g = 9.81
m
2
s
Solution:
F3 = ρ w g a b L
F 3 = 470.88 kN
F2 = ρ w g a b L
F 2 = 470.88 kN
b
F1 = ρ w g b L
2
F 1 = 156.96 kN
⎛ 2
πb ⎞
⎝
4 ⎠
W = ⎜b −
2
⎟ Lρ w g
W = 67.368 kN
Fx = F1 + F2
F x = 628 kN
Fy = F2 + W
F y = 538 kN
Problem 9-117
The rectangular bin is filled with coal, which creates a pressure distribution along wall A that
varies as shown, i.e. p = p0(z/b)3. Determine the resultant force created by the coal and specify
its location measured from the top surface of the coal.
Units used:
3
kip = 10 lb
979
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 4 ft
b = 10 ft
p0 = 4000
lb
ft
2
Solution:
Resultant Force and its location:
⌠
⎮
F = ⎮
⌡
b
3
⎛ z ⎞ a dz
⎟
⎝ b⎠
p0 ⎜
F = 40 kip
0
b
⌠
3
1⎮
⎛ z ⎞ a dz
zc =
z
p
0⎜ ⎟
F⎮
⎝ b⎠
⌡
zc = 8 ft
0
Problem 9-118
The semicircular drainage pipe is filled with water. Determine the resultant horizontal and
vertical force components that the water exerts on the side AB of the pipe per foot of pipe
length; water has density γ .
Given:
γ = 62.4
lb
ft
3
r = 2 ft
Solution:
w = γr
w = 124.8
lb
ft
2
Resultant forces (per unit foot):
F Rh =
1
wr
2
F Rh = 124.8
lb
ft
980
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
2
F Rv = γ
πr
F Rv = 196.0
4
lb
ft
Problem 9-119
The load over the plate varies linearly along the sides of the plate such that p = k y (a-x). Determine the
magnitude of the resultant force and the coordinates (xc, yc) of the point where the line of action of the
force intersects the plate.
Given:
a = 2 ft
b = 6 ft
k = 10
lb
ft
4
Solution:
p ( x , y) = k y( a − x)
b
a
⌠ ⌠
F R = ⎮ ⎮ p ( x , y) d y dx
⌡0 ⌡0
a
b
xc =
1 ⌠ ⌠
⎮ ⎮ x p ( x , y) d y dx
FR ⌡0 ⌡0
yc =
1 ⌠ ⌠
⎮ ⎮ y p ( x , y) d y dx
F R ⌡0 ⌡0
a
F R = 360 lb
xc = 0.667 ft
b
yc = 4 ft
Problem 9-120
The drum is filled to its top (y = a) with oil having a density γ. Determine the resultant force of the oil
pressure acting on the flat end of plate A of the drum and specify its location measured from the top of
the drum.
981
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
a = 1.5 ft
lb
γ = 55
ft
3
Solution:
a
⌠
2
2
F R = ⎮ γ 2 a − y ( a − y) d y
⌡− a
F R = 583 lb
a
1 ⌠
⎮ yγ 2 a2 − y2 ( a − y) d y
d = a−
FR ⌡− a
d = 1.875 ft
Problem 9-121
The gasoline tank is constructed with elliptical ends on each side of the tank. Determine the resultant
force and its location on these ends if the tank is half full.
Given:
a = 3 ft
b = 4 ft
lb
γ = 41
ft
3
Solution:
⌠
FR = ⎮
⎮
⌡
0
−a
⎛ b a2 − y2⎞ d y
⎟
⎝a
⎠
−γ y2⎜
F R = 984 lb
982
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
1 ⌠
⎮
yc =
FR ⎮
⌡
0
⎡
⎣
Chapter 9
⎛ b a2 − y2⎞⎤ d y
⎟⎥
⎝a
⎠⎦
y⎢−γ y2⎜
−a
yc = −1.767 ft
xc = 0 ft
Problem 9-122
The loading acting on a square plate is represented by a parabolic pressure distribution. Determine the
magnitude of the resultant force and the coordinates (xc, yc) of the point where the line of action of
the force intersects the plate. Also, what are the reactions at the rollers B and C and the
ball-and-socket joint A? Neglect the weight of the plate.
Units Used:
3
kPa = 10 Pa
3
kN = 10 N
Given:
a = 4m
p0 = 4 kPa
Solution:
Due to symmetry
xc = 0
⌠
FR = ⎮
⎮
⌡
a
p0
y
a dy
a
F R = 42.667 kN
0
a
⌠
1 ⎮
yc =
y p0
FR ⎮
⌡
y
a dy
a
yc = 2.4 m
0
Equilibrium
Given
Guesses
Ay = 1 kN
B y = 1 kN
Cy = 1 kN
Ay + B y + Cy − F R = 0
(By + Cy)a − FR yc = 0
By
a
a
− Cy = 0
2
2
983
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
⎛ Ay ⎞
⎜ ⎟
⎜ By ⎟ = Find ( Ay , By , Cy)
⎜C ⎟
⎝ y⎠
⎛ Ay ⎞ ⎛ 17.067 ⎞
⎜ ⎟ ⎜
⎟
⎜ By ⎟ = ⎜ 12.8 ⎟ kN
⎜ C ⎟ ⎝ 12.8 ⎠
⎝ y⎠
Problem 9-123
The tank is filled with a liquid which has density ρ . Determine the resultant force that it exerts
on the elliptical end plate, and the location of the center of pressure, measured from the x axis.
Units Used:
3
kN = 10 N
Given:
a = 1m
b = 0.5 m
ρ = 900
kg
m
g = 9.81
3
m
2
s
Solution:
⌠
⎮
FR = ⎮
⌡
b
2
y
ρ g2a 1 − ⎛⎜ ⎟⎞ ( b − y) d y
⎝ b⎠
−b
⌠
1 ⎮
yc =
FR ⎮
⌡
F R = 6.934 kN
b
−b
2
⎛ y ⎞ ( b − y) d y
⎟
⎝ b⎠
yρ g2a 1 − ⎜
yc = −0.125 m
Problem 9-124
A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volume
of material required to make the belt.
984
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Given:
r = 600 mm
a = 25 mm
b = 50 mm
c = 75 mm
Solution:
⎡⎛
⎣⎝
V = 2π ⎢⎜ r +
c ⎞ ⎛ 1⎞
⎛ c⎞ ⎤
⎟ 2⎜ ⎟ a c + ⎜ r + ⎟ b c⎥
3⎠ ⎝ 2⎠
⎝ 2⎠ ⎦
V = 22.4 × 10
−3
m
3
Problem 9-125
A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the surface
area of the belt.
Given:
r = 600 mm
a = 25 mm
b = 50 mm
c = 75 mm
Solution:
⎡
⎣
⎛
⎝
A = 2π ⎢r b + 2⎜ r +
c⎞
⎟
2⎠
2
⎤
⎦
2
a + c + ( r + c) ( b + 2a)⎥
A = 1.246 m
2
985
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-126
Locate the center of mass of
the homogeneous rod.
Given:
a = 200 mm
b = 600 mm
c = 100 mm
d = 200 mm
θ = 45 deg
Solution:
L = a+b+c+d
xc =
1⎛ b
⎞
⎜b sin ( θ ) + c b sin ( θ )⎟
L⎝ 2
⎠
xc = 154.3 mm
yc =
1⎛ d
b
⎞
⎜ d + b cos ( θ ) + c b cos ( θ )⎟
L⎝ 2
2
⎠
yc = 172.5 mm
zc =
1⎛ a
c⎞
⎜a + d a − c ⎟
L⎝ 2
2⎠
zc = 50.0 mm
Problem 9-127
Locate the centroid of the solid
986
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Solution:
yc =
⌠
⎮
⎮
⌡
2a
⎛
⎝
zπ a⎜ a −
z⎞
⎟ dz
2⎠
0
2a
⌠
⎮
⎮
⌡0
4
=
3
2a π−a π z
z
π a⎛⎜ a − ⎞⎟ dz
⎝ 2⎠
π a
3
yc =
2
a
3
Problem 9-128
Locate the centroid (xc, yc) of the thin plate.
Given:
a = 6 in
Solution:
2
A = 4a −
2
2
a
πa
−
2
4
A = 97.7 in
1 ⎡ −a ⎛ −2 ⎞ π a ⎛
4a ⎞⎤
⎢ ⎜ a⎟ −
a−
⎜
⎟⎥
A⎣ 2 ⎝ 3 ⎠
4 ⎝
3π ⎠⎦
2
xc =
2
xc = −0.262 in
1 ⎡ −a 2a π a ⎛ 4a
⎞⎤
yc = ⎢
−
− a⎟⎥
⎜
A⎣ 2 3
4 ⎝ 3π
⎠⎦
2
2
2
yc = 0.262 in
987
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-129
Determine the weight and location (xc, yc) of the center of gravity G of the concrete retaining wall.
The wall has a length L, and concrete has a specific gravity of γ.
Units Used:
3
kip = 10 lb
Given:
a = 12 ft
f = 1 ft
b = 9 ft
g = 2 ft
c = 1.5 ft
L = 10 ft
d = 5.5 ft
γ = 150
e = 1.5 ft
lb
ft
3
Solution:
A = bc + a f +
1
a( e − f )
2
W = γAL
W = 42.8 kip
xc =
1⎡ b
f⎞ 1
e − f ⎞⎤
⎛
⎛
⎢b c + a f⎜g + ⎟ + a( e − f) ⎜g + f +
⎟⎥
A⎣ 2
3 ⎠⎦
⎝ 2⎠ 2
⎝
xc = 3.52 ft
yc =
1⎡ c
⎛ a⎞ 1
⎛ a ⎞⎤
⎢b c + a f⎜ c + ⎟ + a( e − f) ⎜c + ⎟⎥
A⎣ 2
⎝ 2⎠ 2
⎝ 3 ⎠⎦
yc = 4.09 ft
Problem 9-130
The hopper is filled to its top with coal. Determine the volume of coal if the voids (air space)
are a fraction p of the volume of the hopper.
Given:
a = 1.5 m
b = 4m
c = 1.2 m
988
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
d = 0.2 m
p = 0.35
Solution:
⎡
⎣
V = ( 1 − p)2π ⎢d c
V = 20.5 m
d 1
⎛ a − d ⎞ + a b a⎤
+ c( a − d) ⎜ d +
⎟
⎥
2 2
3 ⎠
2⎦
⎝
3
Problem 9-131
Locate the centroid (xc, yc) of the shaded area.
Given:
a = 16 ft
b = 4 ft
c = ( a−
b)
2
Solution:
b
⌠
A = ⎮ ( a−
⌡0
x) dx
2
A = 29.3 ft
2
989
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
b
1 ⌠
⎮ x( a −
xc =
A ⌡0
⌠
⎮
1 ⎮
yc =
A⎮
⌡
Chapter 9
x) dx
2
xc = 1.6 ft
b
2
⎡⎣( a − x) 2⎤⎦
dx
2
yc = 4.15 ft
0
Problem 9-132
The rectangular bin is filled with coal, which creates a pressure distribution along wall A that
varies as shown, i.e., p = p0(z/b)1/3. Compute the resultant force created by the coal, and its
location, measured from the top surface of the coal.
Given:
p0 = 8
lb
ft
2
a = 3 ft
b = 8 ft
Solution:
⌠
⎮
⎮
F = ⎮
⎮
⌡
b
1
3
⎛ z⎞
p0 ⎜ ⎟ a dz
⎝ b⎠
F = 144 lb
0
b
⌠
1
⎮
⎮
3
1⎮
⎛ z⎞
zc =
z p0 ⎜ ⎟ a dz
F⎮
⎝ b⎠
⌡
zc = 4.57 ft
0
990
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 9
Problem 9-133
The load over the plate varies linearly along the sides of the plate such that p =
2
x( 4 − y) kPa
3
Determine the resultant force and its position (xc, yc) on the plate.
Solution:
⌠
F = ⎮
⎮
⌡
3
0
⌠
⎮
⎮
⌡
4
2
x( 4 − y) d y dx
3
F = 24 kN
0
1⌠
⎮
xc =
F⎮
⌡
3
0
1⌠
⎮
yc =
F⎮
⌡
3
0
4
⌠
⎮ x 2 x( 4 − y) d y dx
⎮
3
⌡
xc = 2 m
0
4
⌠
⎮ y 2 x( 4 − y) d y dx
⎮
3
⌡
yc = 1.333 m
0
Problem 9-134
The pressure loading on the plate is described by the function p = { -240/(x + 1) + 340 } Pa.
Determine the magnitude of the resultant force and coordinates of the point where the line of action
of the force intersects the plate.
Solution:
Due to symmetry
yc = 3 m
⌠
F = ⎮
⎮
⌡
5
⎛ −240 + 340⎞ 6 dx
⎜
⎟
⎝x + 1
⎠
0
991
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Engineering Mechanics - Statics
Chapter 9
3
F = 7.62 × 10
N
5
1⌠
⎮ x⎛ −240 + 340⎞ 6 dx
xc =
⎜
⎟
F ⎮ ⎝x + 1
⎠
⌡
0
xc = 2.74 m
992
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Engineering Mechanics - Statics
Chapter 10
Problem 10-1
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 2m
b = 4m
b
Solution:
⌠
y
2
Ix = 2 ⎮ y a 1 − d y
⎮
b
⌡
Ix = 39.0 m
4
0
Problem 10-2
Determine the moment of inertia for the shaded area about the y axis.
Given:
a = 2m
b = 4m
a
Solution:
⌠
2
⎮ 2 ⎡
⎛x⎞ ⎤
Iy = 2 ⎮ x b⎢1 − ⎜ ⎟ ⎥ dx
⎣ ⎝ a⎠ ⎦
⌡
Iy = 8.53 m
4
0
993
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Engineering Mechanics - Statics
Chapter 10
Problem 10-3
Determine the moment of inertia for the thin strip
of area about the x axis.The strip is oriented at an
angle θ from the x axis. Assume that t << l.
Solution:
l
⌠
⌠ 2 2
⎮ 2
Ix = ⎮ y d A = ⎮ s sin ( θ ) t ds
⌡0
⌡
A
1 3 2
t l sin ( θ )
3
Ix =
Problem 10-4
Determine the moment for inertia of the
shaded area about the x axis.
Given:
a = 4 in
b = 2 in
Solution:
⌠
⎮
Ix = ⎮
⎮
⌡
a
3
1 ⎡ ⎛x⎞ ⎤
⎢b ⎜ ⎟ ⎥ dx
3 ⎣ ⎝ a⎠ ⎦
3
0
Ix = 1.07 in
4
994
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Engineering Mechanics - Statics
Chapter 10
Problem 10-5
Determine the moment for inertia of the
shaded area about the y axis.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
3
⎮ 2 ⎛x⎞
I y = ⎮ x b ⎜ ⎟ dx
⎝ a⎠
⌡
0
Iy = 21.33 in
4
Problem 10-6
Determine the moment of inertia for the shaded area about the x axis.
Solution:
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
b
3
⎛ x⎞
⎜h ⎟
⎝ b ⎠ dx = 2 b h3
3
15
Ix =
2
3
bh
15
0
995
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Engineering Mechanics - Statics
Chapter 10
Alternatively
h
⌠
2
⎮ 2⎛
y ⎟⎞
2
3
⎜
Ix = ⎮ y b − b
bh
dy =
2⎟
⎜
15
⎮
h ⎠
⎝
⌡0
Ix =
2
3
bh
15
Ix =
ab
3( 1 + 3n)
Problem 10-7
Determine the moment of inertia for the
shaded area about the x axis.
Solution:
b
⌠
1⎤
⎮
⎡
⎢
⎮
n⎥
y⎞ ⎥
2⎢
⎛
⎮
Ix = A y a − a ⎜ ⎟
dy
⎮
b⎠ ⎦
⎣
⎝
⌡
3
0
Problem 10-8
Determine the moment of inertia for
the shaded area about the y axis.
996
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Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
⌠ 2
⎮ 2
I y = ⎮ x d A = ⎮ x y dx
⌡0
⌡
⎡⎢⎛ b ⎞ xn + 3 ⎤⎥
b ⌠ n+ 2
⎮ x
dx = ⎜ ⎟
Iy =
n⌡
⎢ an n + 3⎥
a 0
⎣⎝ ⎠
⎦
a
a
0
3
ba
Iy =
n+3
Problem 10-9
Determine the moment of inertia for
the shaded area about the x axis.
Given:
a = 4 in
b = 2 in
Solution:
b
⌠
2
y⎞ ⎤
⎮ 2⎡
⎛
⎢
Ix = ⎮ y a − a ⎜ ⎟ ⎥ d y
⎣
⎝ b⎠ ⎦
⌡
0
Ix = 4.27 in
4
Problem 10-10
Determine the moment of inertia for the shaded area about the y axis.
997
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
2
Iy = ⎮ x b
⎮
⌡
x
dx
a
0
Iy = 36.6 in
4
Problem 10-11
Determine the moment of inertia for the shaded area about the x axis
Given:
a = 8 in
b = 2 in
Solution:
b
⌠
3
⎮ 2⎛
y ⎞⎟
⎜
Ix = ⎮ y a − a
dy
3⎟
⎜
⎮
b ⎠
⎝
⌡0
Ix = 10.67 in
4
Problem 10-12
Determine the moment of inertia for the shaded area about the x axis
998
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 2m
b = 1m
Solution:
b
⌠
2
⎮
y ⎟⎞
2 ⎛
Ix = ⎮ y a⎜ 1 −
dy
⎜ b2 ⎟
⎮
⎝
⎠
⌡− b
Ix = 0.53 m
4
Problem 10-13
Determine the moment of inertia for the shaded area about the y axis
Given:
a = 2m
b = 1m
Solution:
a
⌠
x
2
Iy = ⎮ x 2b 1 − dx
⎮
a
⌡
Iy = 2.44 m
4
0
Problem 10-14
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 4 in
b = 4 in
999
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Engineering Mechanics - Statics
Chapter 10
Solution:
b
⌠
2
y⎞ ⎤
⎮ 2⎡
⎛
Ix = ⎮ y ⎢a − a ⎜ ⎟ ⎥ d y
⎣
⎝ b⎠ ⎦
⌡
0
Ix = 34.1 in
4
Problem 10-15
Determine the moment of inertia for the
shaded area about the y axis.
Given:
a = 4 in
b = 4 in
Solution:
a
⌠
2
Iy = ⎮ x b
⎮
⌡
x
dx
a
0
Iy = 73.1 in
4
1000
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-16
Determine the moment of inertia of the
shaded area about the x axis.
Given:
a = 2 in
b = 4 in
Solution:
⌠
⎮
Ix = ⎮
⌡
a
3
1⎛
⎛ π x ⎞⎞
⎜ b cos ⎜ ⎟ ⎟ dx
3⎝
⎝ 2a ⎠ ⎠
−a
Ix = 36.2 in
4
Problem 10-17
Determine the moment of inertia for the shaded area
about the y axis.
Given:
a = 2 in
b = 4 in
Solution:
⌠
Iy = ⎮
⎮
⌡
a
2
⎛ π x ⎞ dx
⎟
⎝ 2a ⎠
x b cos ⎜
−a
Iy = 7.72 in
4
Problem 10-18
Determine the moment of inertia for the shaded area about the x axis.
1001
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 4 in
b = 2 in
Solution:
Solutiona
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
3
⎛
⎛ π x ⎞⎞
⎜b cos ⎜ ⎟⎟
⎝
⎝ 2a ⎠⎠ dx
3
−a
Ix = 9.05 in
4
Problem 10-19
Determine the moment of inertia for the shaded area about the y axis.
Given:
a = 4 in
b = 2 in
Solution:
⌠
Iy = ⎮
⎮
⌡
a
2
⎛ π x ⎞ dx
⎟
⎝ 2a ⎠
x b cos ⎜
−a
Iy = 30.9 in
4
Problem 10-20
Determine the moment for inertia of the shaded area about the x axis.
1002
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 2 in
b = 4 in
c =
12 in
Solution:
⌠
⎮
Ix = ⎮
⎮
⌡
a+ b
3
1 ⎛c ⎞
⎜ ⎟ dx
3⎝ x ⎠
2
a
Ix = 64.0 in
4
Problem 10-21
Determine the moment of inertia of
the shaded area about the y axis.
Given:
a = 2 in
b = 4 in
c =
12 in
Solution:
⌠
⎮
Iy = ⎮
⌡
a+ b
2
2⎛ c ⎞
x ⎜
⎟ dx
⎝x⎠
a
1003
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Engineering Mechanics - Statics
Iy = 192.00 in
Chapter 10
4
Problem 10-22
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 2m
b = 2m
Solution:
b
⌠
⎮ 2 ⎛ y2 ⎞
I x = ⎮ y a⎜ ⎟ d y
⎜ b2 ⎟
⎮
⎝ ⎠
⌡0
Ix = 3.20 m
4
Problem 10-23
Determine the moment of inertia for the shaded area about the y axis. Use Simpson's rule to
evaluate the integral.
Given:
a = 1m
b = 1m
1004
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Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
2
⎮
⎛x⎞
⎜a⎟
⎮
2
Iy = ⎮ x b e⎝ ⎠ dx
⌡0
Iy = 0.628 m
4
Problem 10-24
Determine the moment of inertia for the shaded area about the x axis. Use Simpson's rule to
evaluate the integral.
Given:
a = 1m
b = 1m
Solution:
⌠
⎮
⎮
⎮
Iy = ⎮
⎮
⌡
a
3
⎡⎢ ⎛ x ⎞ 2⎥⎤
⎜ ⎟
⎢b e⎝ a ⎠ ⎥
⎣
⎦ dx
3
Iy = 1.41 m
4
0
Problem 10-25
The polar moment of inertia for the area is IC about the z axis passing through the centroid C.
The moment of inertia about the x axis is Ix and the moment of inertia about the y' axis is Iy'.
Determine the area A.
Given:
IC = 28 in
4
1005
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Engineering Mechanics - Statics
Ix = 17 in
Chapter 10
4
Iy' = 56 in
4
a = 3 in
Solution:
IC = Ix + Iy
Iy = IC − Ix
2
Iy' = Iy + A a
A =
Iy' − Iy
a
A = 5.00 in
2
2
Problem 10-26
The polar moment of inertia for the area is Jcc about the z' axis passing through the centroid C. If
the moment of inertia about the y' axis is Iy' and the moment of inertia about the x axis is Ix.
Determine the area A.
Given:
6
4
6
4
6
4
Jcc = 548 × 10 mm
Iy' = 383 × 10 mm
Ix = 856 × 10 mm
h = 250 mm
Solution:
2
Ix' = Ix − A h
Jcc = Ix' + Iy'
2
Jcc = Ix − A h + Iy'
1006
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Engineering Mechanics - Statics
A =
Chapter 10
Ix + Iy' − Jcc
2
h
3
2
A = 11.1 × 10 mm
Problem 10-27
Determine the radius of gyration kx of the column’s
cross-sectional area.
Given:
a = 100 mm
b = 75 mm
c = 90 mm
d = 65 mm
Solution:
Cross-sectional area:
A = ( 2b) ( 2a) − ( 2d) ( 2c)
Moment of inertia about the x axis:
Ix =
1
1
3
3
( 2b) ( 2a) −
( 2d) ( 2c)
12
12
Radius of gyration about the x axis:
kx =
Ix
A
kx = 74.7 mm
Problem 10-28
Determine the radius of gyration ky of the column’s cross-sectional area.
Given:
a = 100 mm
b = 75 mm
1007
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Engineering Mechanics - Statics
Chapter 10
c = 90 mm
d = 65 mm
Solution:
Cross-sectional area:
A = ( 2b) ( 2a) − ( 2d) ( 2c)
Moment of inertia about the y axis:
Iy =
1
12
1
3
( 2a) ( 2b) −
12
( 2c) ( 2d)
3
Radius of gyration about the y axis:
ky =
Iy
A
ky = 59.4 mm
Problem 10-29
Determine the moment of
inertia for the beam's
cross-sectional area with
respect to the x' centroidal axis.
Neglect the size of all the rivet
heads, R, for the calculation.
Handbook values for the area,
moment of inertia, and location
of the centroid C of one of the
angles are listed in the figure.
Solution:
IE =
1
12
⎡
( 6)
( 3)
( 15 mm) ( 275 mm) + 4⎢1.32 10 mm + 1.36 10 mm ⎛⎜
3
⎣
4
2 275 mm
⎝
2
2⎤
⎡1
275 mm
3
+ 2⎢ ( 75 mm) ( 20 mm) + ( 75 mm) ( 20 mm) ⎛⎜
+ 10mm⎞⎟ ⎥
⎣12
⎝ 2
⎠⎦
2⎤
− 28 mm⎟⎞ ⎥ ...
⎠⎦
1008
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
6
Chapter 10
4
IE = 162 × 10 mm
Problem 10-30
Locate the centroid yc of the cross-sectional area for the angle. Then find the moment
of inertia Ix' about the x' centroidal axis.
Given:
a = 2 in
b = 6 in
c = 6 in
d = 2 in
Solution:
⎛ c ⎞ + b d⎛ d ⎞
⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
a c⎜
yc =
Ix' =
yc = 2.00 in
ac + bd
1
12
a c + a c ⎛⎜
3
c
⎝2
2
− yc⎞⎟ +
⎠
1
12
b d + b d ⎛⎜ yc −
3
⎝
d⎞
⎟
2⎠
2
Ix' = 64.00 in
4
Problem 10-31
Locate the centroid xc of the cross-sectional area for the angle. Then find the moment
1009
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
of inertia Iy' about the centroidal y' axis.
Given:
a = 2 in
b = 6 in
c = 6 in
d = 2 in
Solution:
⎛ a ⎞ + b d⎛ a + b ⎞
⎟
⎜
⎟
⎝ 2⎠
⎝ 2⎠
a c⎜
xc =
Iy' =
xc = 3.00 in
ac + bd
1
12
c a + c a ⎛⎜ xc −
3
⎝
a⎞
⎟
2⎠
2
+
1
12
d b + d b ⎛⎜ a +
3
⎝
b
2
− xc⎞⎟
⎠
2
Iy' = 136.00 in
Problem 10-32
Determine the distance xc to the centroid of the beam's cross-sectional area: then find
the moment of inertia about the y' axis.
1010
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4
Engineering Mechanics - Statics
Chapter 10
Given:
a = 40 mm
b = 120 mm
c = 40 mm
d = 40 mm
Solution:
⎛ a + b ⎞ + 2a d a
⎟
2
⎝ 2 ⎠
2( a + b)c⎜
xc =
xc = 68.00 mm
2( a + b)c + 2d a
⎡1
Iy' = 2⎢
⎣ 12
c ( a + b) + c( a + b) ⎛⎜
3
a+b
⎝ 2
6
2⎤
1
⎠⎦
12
− xc⎞⎟ ⎥ +
2d a + 2d a ⎛⎜ xc −
3
⎝
a⎞
2
⎟
2⎠
4
Iy' = 36.9 × 10 mm
Problem 10-33
Determine the moment of inertia of the beam's cross-sectional area about the x' axis.
1011
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Given:
a = 40 mm
b = 120 mm
c = 40 mm
d = 40 mm
Solution:
1
Ix' =
12
3
( a + b) ( 2c + 2d) −
1
12
b ( 2d)
3
6
4
Ix' = 49.5 × 10 mm
Problem 10-34
Determine the moments of inertia for the shaded area about the x and y axes.
Given:
a = 3 in
b = 3 in
c = 6 in
d = 4 in
r = 2 in
Solution:
⎡ 1 3 1 ⎛ 2c ⎞ 2⎤ ⎛ π r4
2 2⎞
Ix = ( a + b) ( c + d) − ⎢ b c + b c ⎜ d + ⎟ ⎥ − ⎜
+ πr d ⎟
3
2 ⎝
3⎠⎦ ⎝ 4
⎣36
⎠
1
3
1012
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ix = 1192 in
Chapter 10
4
1
2b ⎞ ⎤ ⎛ π r
3 ⎡1
3 1 ⎛
2 2⎞
( c + d) ( a + b) − ⎢ c b + b c ⎜ a +
+ πr a ⎟
⎟⎥−⎜
3
2 ⎝
3⎠⎦ ⎝ 4
⎣36
⎠
2
Iy =
Iy = 364.84 in
4
4
Problem 10-35
Determine the location of the centroid y' of the beam constructed from the two channels and the
cover plate. If each channel has a cross-sectional area A c and a moment of inertia about a horizontal
axis passing through its own centroid Cc, of Ix'c , determine the moment of inertia of the beam’s
cross-sectional area about the x' axis.
Given:
a = 18 in
b = 1.5 in
c = 20 in
d = 10 in
Ac = 11.8 in
Ix'c = 349 in
2
4
Solution:
2A c d + a b⎛⎜ c +
yc =
⎝
b⎞
⎟
2⎠
2A c + a b
Ix' = ⎡⎣Ix'c + A c ( yc − d) ⎤⎦ 2 +
2
yc = 15.74 in
1
3
⎞
⎛ b
a b + a b ⎜ c + − yc⎟
12
⎝ 2
⎠
2
Ix' = 2158 in
4
Problem 10-36
Compute the moments of inertia Ix and Iy for the beam's cross-sectional area about
1013
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
the x and y axes.
Given:
a = 30 mm
b = 170 mm
c = 30 mm
d = 140 mm
e = 30 mm
f = 30 mm
g = 70 mm
Solution:
Ix =
1
1
e⎞
3 1 3
3
⎛
a ( c + d + e) + b c +
g e + g e ⎜c + d + ⎟
3
3
12
2⎠
⎝
Iy =
1
3 1 3 1
3
c ( a + b ) + d f + c ( f + g)
3
3
3
2
6
4
6
4
Ix = 154 × 10 mm
Iy = 91.3 × 10 mm
Problem 10-37
Determine the distance yc to the centroid C of the
beam's cross-sectional area and then compute the
moment of inertia Icx' about the x' axis.
Given:
a = 30 mm
e = 30 mm
b = 170 mm
f = 30 mm
c = 30 mm
g = 70 mm
d = 140 mm
1014
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Engineering Mechanics - Statics
Chapter 10
Solution:
( a + b)c⎛⎜
e⎞
⎛ d⎞
⎛
⎟ + d f⎜ c + ⎟ + ( f + g)e⎜ c + d + ⎟
2⎠
⎝ 2⎠
⎝ 2⎠
⎝
yc =
c⎞
( a + b)c + d f + ( f + g)e
yc = 80.7 mm
2
Ix' =
2
1
c⎞
1 3
3
⎞
⎛
⎛ d
( a + b) c + ( a + b)c ⎜ yc − ⎟ +
f d + f d ⎜ c + − yc⎟ ...
12
2⎠
12
⎝
⎝ 2
⎠
e
1
3
⎞
⎛
( f + g) e + ( f + g)e ⎜ c + d + − yc⎟
2
12
⎝
⎠
+
6
2
4
Ix' = 67.6 × 10 mm
Problem 10-38
Determine the distance xc to the centroid C of the beam's cross-sectional area and then compute the
moment of inertia Iy' about the y' axis.
Given:
a = 30 mm
b = 170 mm
c = 30 mm
d = 140 mm
e = 30 mm
f = 30 mm
g = 70 mm
Solution:
b c⎛⎜
xc =
b
⎝2
+ a⎞⎟ + ( c + d) f⎛⎜
f⎞
f+g
⎟ + ( f + g)e
2
⎝ 2⎠
⎠
b c + b c + ( f + g)e
xc = 61.6 mm
1015
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
2
Iy' =
2
1
3
⎛ a + b − x ⎞ + 1 d f 3 + d f ⎛ x − f ⎞ ...
c ( a + b) + c( a + b) ⎜
⎜c
⎟
c⎟
12
2⎠
12
⎝ 2
⎠
⎝
f + g⎞
1
3
⎛
+
e ( f + g) + e( f + g) ⎜ xc −
⎟
2 ⎠
12
⎝
6
2
4
Iy' = 41.2 × 10 mm
Problem 10-39
Determine the location yc of the centroid C of the beam’s cross-sectional area. Then compute
the moment of inertia of the area about the x' axis
Given:
a = 20 mm
b = 125 mm
c = 20 mm
f = 120 mm
g = 20 mm
d =
f−c
2
e =
f−c
2
Solution:
yc =
a + g⎞
b⎞
⎛
⎟ + c b⎜ a + g + ⎟
2⎠
⎝ 2 ⎠
⎝
( a + g) f⎛⎜
( a + g) f + c b
yc = 48.25 mm
2
1
a + g⎞
1 3
3
⎞
⎛
⎛b
Ix' =
f ( a + g) + ( f) ( a + g) ⎜ yc −
⎟ + c b + c b ⎜ + a + g − yc⎟
12
2 ⎠
12
⎝
⎝2
⎠
6
2
4
Ix' = 15.1 × 10 mm
1016
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-40
Determine yc, which locates the centroidal axis x' for the cross-sectional area of the T-beam, and then
find the moments of inertia Ix' and Iy' .
Given:
a = 25 mm
b = 250 mm
c = 50 mm
d = 150 mm
Solutuion:
⎛ b ⎞ b2a + ⎛ b + c ⎞ 2d c
⎜ ⎟
⎜
⎟
2⎠
2⎠
⎝
⎝
yc =
b2a + c2d
yc = 207 mm
2
1
b⎞
1
3
3
⎞
⎛
⎛ c
Ix' =
2a b + 2a b ⎜ yc − ⎟ +
2d c + c2d ⎜ b + − yc⎟
12
2⎠
12
⎝
⎝ 2
⎠
6
2
4
Ix' = 222 × 10 mm
Iy' =
1
1
3
3
b ( 2a) +
c ( 2d)
12
12
6
4
Iy' = 115 × 10 mm
Problem 10-41
Determine the centroid y' for the beam’s cross-sectional area; then find Ix'.
Given:
a = 25 mm
1017
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
b = 100 mm
c = 25 mm
d = 50 mm
e = 75 mm
Solution:
⎛ c ⎞ + 2a b⎛c + b ⎞
⎟
⎜
⎟
⎝ 2⎠
⎝ 2⎠
2( a + e + d)c⎜
yc =
Ix' =
yc = 37.50 mm
2( a + e + d)c + 2a b
2
12
( a + e + d) c + 2( a + e + d)c ⎛⎜ yc −
3
⎡1
+ 2⎢
⎣ 12
a b + a b ⎛⎜ c +
3
⎝
6
b
2
2⎤
⎝
c⎞
2
⎟ ...
2⎠
− yc⎞⎟ ⎥
⎠⎦
4
Ix' = 16.3 × 10 mm
Problem 10-42
Determine the moment of inertia for the beam's cross-sectional area about the y axis.
Given:
a = 25 mm
b = 100 mm
c = 25 mm
d = 50 mm
e = 75 mm
1018
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
⎡1 3
1 3
a ⎤
3
2 ( a + d + e) c + 2⎢ b a + a b ⎛⎜ e + ⎞⎟ ⎥
12
⎣12
⎝ 2⎠ ⎦
2
ly =
6
4
ly = 94.8 × 10 mm
Problem 10-43
Determine the moment for inertia Ix of the shaded area about the x axis.
Given:
a = 6 in
b = 6 in
c = 3 in
d = 6 in
Solution:
3
Ix =
ba
3
+
1
12
3
ca +
1
12
( b + c) d
3
Ix = 648 in
4
Problem 10-44
Determine the moment for inertia Iy of the shaded area about the y axis.
Given:
a = 6 in
b = 6 in
c = 3 in
d = 6 in
1019
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
3
2
c
1
2( b + c) ⎤
3 1
Iy =
+
a c + a c ⎛⎜ b + ⎞⎟ +
d ( b + c) + d( b + c) ⎡⎢
⎥
3
36
2 ⎝
3⎠
36
2
⎣ 3 ⎦
ab
Iy = 1971 in
1
3
1
2
4
Problem 10-45
Locate the centroid yc of the channel's cross-sectional area, and then determine the moment of
inertia with respect to the x' axis passing through the centroid.
Given:
a = 2 in
b = 12 in
c = 2 in
d = 4 in
Solution:
yc =
c
⎛ c + d ⎞ ( c + d)a
b c + 2⎜
⎟
2
⎝ 2 ⎠
b c + 2( c + d)a
yc = 2 in
2
c
2
c+d
3
Ix =
b c + b c ⎛⎜ yc − ⎞⎟ +
a ( c + d) + 2a( c + d) ⎛⎜
− yc⎞⎟
12
2⎠
12
⎝
⎝ 2
⎠
1
3
Ix = 128 in
4
2
Problem 10-46
Determine the moments for inertia Ix and Iy of the shaded area.
Given:
r1 = 2 in
1020
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
r2 = 6 in
Solution:
⎛ π r24 π r14 ⎞
⎟
Ix = ⎜
−
8 ⎠
⎝ 8
Ix = 503 in
4
⎛ π r24 π r14 ⎞
⎟
Iy = ⎜
−
8 ⎠
⎝ 8
Iy = 503 in
4
Problem 10-47
Determine the moment of inertia for the parallelogram about the x' axis, which passes through
the centroid C of the area.
Solution:
h = ( a)sin ( θ )
Ixc =
Ixc =
1
12
3
bh =
1
12
b ⎡⎣( a)sin ( θ )⎤⎦ =
3
1 3
3
a b sin ( θ )
12
1 3
3
a b sin ( θ )
12
Problem 10-48
Determine the moment of inertia for the parallelogram about the y' axis, which passes through
the centroid C of the area.
1021
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
A = b( a) sin ( θ )
xc =
Iy' =
⎡⎡b( a) sin ( θ ) b − 1 ( a) cos ( θ ) ( a) sin ( θ ) ( a)cos ( θ )⎤ ...⎤ = b + ( a)cos ( θ )
⎥ ⎥
⎢⎢
3
2 2
2
b( a) sin ( θ ) ⎣
⎦ ⎥
⎢ 1
(
)
θ
(
a
)cos
⎤
⎢+ ( a) cos ( θ ) ( a) sin ( θ ) ⎡⎢b +
⎥
⎥
3
2
⎣
⎣
⎦
⎦
1
1
2
b
3
( a) sin ( θ ) b + ( a)sin ( θ ) b ⎛⎜ − xc⎟⎞ ...
⎝2
⎠
2
⎡1
( a)cos ( θ )⎤ ⎤
3 1
⎡
+ −⎢ ( a) sin ( θ ) ⎡⎣( a)cos ( θ )⎤⎦ + ( a) sin ( θ ) ( a) cos ( θ ) ⎢xc −
⎥ ⎥ ...
2
3
⎣ 36
⎣
⎦⎦
2
(
)
1
( a)cos θ
3 1
⎤
⎡
(
)
(
)
(
)
(
)
+
( a) sin θ ⎡⎣( a)cos θ ⎤⎦ + ( a) sin θ ( a) cos θ ⎢b +
− xc⎥
36
3
2
⎣
⎦
12
Simplifying we find.
Iy' =
(
)
ab 2
2
2
b + a cos ( θ ) sin ( θ )
12
Problem 10-49
Determine the moments of inertia for the triangular area about the x' and y' axes, which pass
through the centroid C of the area.
1022
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
Ix' =
36
3
bh
2 1
⎛ b − a ⎞ 1 h ( b − a)
a h a + ⎜a +
⎟
3 2
3 ⎠2
⎝
xc =
1
1
h a + h ( b − a)
2
2
Iy' =
Iy' =
1
1
36
1
36
3
ha +
(2
1
2
h a ⎛⎜
b+a
⎝ 3
hb b − ab + a
2
=
b+a
3
2
−
2 ⎞
1
b − a b + a⎞
3 1
a⎟ +
h ( b − a) + h( b − a) ⎛⎜ a +
−
⎟
3 ⎠
36
2
3
3 ⎠
⎝
2
)
Problem 10-50
Determine the moment of inertia for the
beam’s cross-sectional area about the x'
axis passing through the centroid C of
the cross section.
Given:
a = 100 mm
b = 25 mm
c = 200 mm
θ = 45 deg
1023
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
Ix' =
1 ⎡
3
⎣2a ⎡⎣2( c sin ( θ ) + b)⎤⎦ ⎤⎦ ...
12
1
3⎤
⎡1 4⎛
⎞⎤
⎡1
+ 4⎢ ( c cos ( θ ) ) ( c sin ( θ ) ) ⎥ − 2⎢ c ⎜ θ − sin ( 2θ )⎟⎥
2
⎣ 12
⎦ ⎣4 ⎝
⎠⎦
6
4
Ix' = 520 × 10 mm
Problem 10-51
Determine the moment of inertia of
the composite area about the x axis.
Given:
a = 2 in
b = 4 in
c = 1 in
d = 4 in
Solution:
d
⌠
3
⎮
4
2⎤⎤
⎞
⎡
⎛
⎡
1
π
c
1
x
3
2 2
⎢2a⎢1 − ⎛⎜ ⎟⎞ ⎥⎥ dx
Ix = ( a + b) ( 2a) − ⎜
+ πc a ⎟ + ⎮
⎮
4
3
3
⎝
⎠ ⌡ ⎣ ⎣ ⎝ d ⎠ ⎦⎦
0
Ix = 153.7 in
4
1024
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-52
Determine the moment of inertia of
the composite area about the y axis.
Given:
a = 2 in
b = 4 in
c = 1 in
d = 4 in
Solution:
d
⌠
4
2
1
x⎞ ⎤
3 ⎛ πc
2 2⎞ ⎮ 2 ⎡
⎛
⎜
⎟
⎢
Iy = ( 2a) ( a + b) −
+ π c b + ⎮ x 2a 1 − ⎜ ⎟ ⎥ dx
3
⎝ 4
⎠ ⌡
⎣ ⎝ d⎠ ⎦
0
Iy = 271.1 in
4
Problem 10-53
Determine the radius of gyration kx for the column's
cross-sectional area.
Given:
a = 200 mm
b = 100 mm
Solution:
2
⎡1 3
a b⎞ ⎤
⎛
⎢
Ix =
( 2a + b) b + 2
ba + ba⎜ + ⎟ ⎥
12
⎣ 12
⎝ 2 2⎠ ⎦
1
kx =
3
Ix
b( 2a + b) + 2a b
kx = 109 mm
1025
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-54
Determine the product of inertia for the shaded portion of the parabola with respect
to the x and y axes.
Given:
a = 2 in
b = 1 in
a
b
⌠ ⌠
Ixy = ⎮ ⎮
x y d y dx
⎮ ⎮
2
⌡− a ⌡b ⎛ x ⎞
Ixy = 0.00 m
4
⎜a⎟
⎝ ⎠
Also because the area is symmetric about the y axis, the product of inertia must be zero.
Problem 10-55
Determine the product of inertia for the shaded
area with respect to the x and y axes.
Solution:
1
⎛
b ⌠
⎮h ⎜
x⎞
⎟
⌠
⎮ ⎮ ⎝b⎠
Ixy = ⎮ ⎮
⌡ ⌡
0
0
3
x y d y dx =
3
16
2 2
Ixy =
b h
3 2 2
b h
16
1026
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-56
Determine the product of inertia of the shaded area of the ellipse with respect to the x
and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
2
⎮ ⎡⎢
x ⎞ ⎥⎤
⎛
⎮
2
⎢ b 1 − ⎜⎝ a ⎟⎠ ⎥
x⎞
⎛
⎮
Ixy =
x
⎥ b 1 − ⎜ a ⎟ dx
⎮ ⎢⎣
2
⎦
⎝ ⎠
⌡0
Ixy = 8.00 in
4
Problem 10-57
Determine the product of inertia of the parabolic area with respect to the x and y axes.
1027
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
⎮ ⎛⎜ b x ⎟⎞
a
⎮
⎟b
Ixy = ⎮ x⎜
2 ⎠
⎝
⌡0
x
a
dx =
1
6
a
2
3 b
Ixy =
a
1 2 2
a b
6
Problem 10-58
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a = 8 in
b = 2 in
Solution:
a
⌠
1
⎮
3
1
⎮
⎛x⎞
⎮
b⎜ ⎟
3
⎝ a ⎠ b ⎛ x ⎞ dx
Ixy = ⎮ x
⎜ ⎟
⎮
2
⎝ a⎠
⌡0
Ixy = 48.00 in
4
Problem 10-59
Determine the product of inertia for the shaded parabolic area with respect to the x and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
b
Ixy = ⎮ x
⎮
2
⌡0
x
a
b
x
a
dx
1028
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ixy = 10.67 in
Chapter 10
4
Problem 10-60
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a = 2m
b = 1m
Solution:
a
⌠
⎛b
Ixy = ⎮ x⎜
⎮ ⎝2
⌡0
1−
x⎞
x
⎟ b 1 − dx
a⎠
a
Ixy = 0.333 m
4
Problem 10-61
Determine the product of inertia for the shaded area with respect to the x and y axes.
1029
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
h
⌠
2
1⎤
⎮
⎡
⎮
⎢
3⎥
1 ⎛ y⎞
3 2 2
⎮
⎢
Ixy =
y b ⎜ ⎟ ⎥ dy =
b h
⎮
2 ⎣ ⎝ h⎠ ⎦
16
⌡0
Ixy =
3
16
2 2
h b
Problem 10-62
Determine the product of inertia of the shaded area with
respect to the x and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
3
3
⎮ ⎛ b⎞ ⎛ x ⎞ ⎛ x ⎞
Ixy = ⎮ x⎜ ⎟ ⎜ ⎟ b ⎜ ⎟ dx
2 a
a
⌡0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Ixy = 4.00 in
4
1030
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-63
Determine the product of inertia for the shaded area with respect to the x and y axes.
Solution:
a
⌠
⎮ ⎛⎜ b xn ⎞⎟ xn
Ixy = ⎮ x
b
dx
n
n
⎮ ⎜⎝ 2 a ⎟⎠ a
⌡
2 2
Ix =
a b
4( n + 1)
provided n ≠ −1
0
Problem 10-64
Determine the product of inertia for the
shaded area with respect to the x and y axes.
Given:
a = 4 ft
Solution:
a
⌠
( a − x) 2 (
⎮
Ixy = ⎮ x
a−
2
⌡0
Ixy = 0.91 ft
x) dx
2
4
1031
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-65
Determine the product of inertia for the shaded area with respect to the x and y axes. Use Simpson's
rule to evaluate the integral.
Given:
a = 1m
b = 0.8 m
Solution:
a
⌠
2
2
⎮
⎛x⎞ ⎛x⎞
⎮
⎜a⎟
b ⎜a⎟
Ixy = ⎮ x⎛⎜ ⎟⎞ e⎝ ⎠ b e⎝ ⎠ dx
⎮ ⎝ 2⎠
⌡0
Ixy =
Problem 10-66
Determine the product of inertia for the parabolic
area with respect to the x and y axes.
Given:
a = 1 in
b = 2 in
Solution:
Due to symmetry about y axis
Ixy = 0
⌠
Also ⎮
⎮
⎮
Ixy = ⎮
⎮
⌡
a
−a
2
b+b
x
x
2
2
⎜b − b x ⎟⎞ dx
2
⎜
a ⎟⎠
⎝
2⎛
a
Ixy = 0.00 m
4
1032
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-67
Determine the product of inertia for the
cross-sectional area with respect to the x and y axes
that have their origin located at the centroid C.
Given:
a = 20 mm
b = 80 mm
c = 100 mm
Solution:
Ixy = 2b a
a⎞
⎜ − ⎟
2 ⎝ 2 2⎠
c⎛b
4
Ixy = 4800000.00 mm
Problem 10-68
Determine the product of inertia for the beam's cross-sectional area with respect to the x and y axes.
Given:
a = 12 in
b = 8 in
c = 1 in
d = 3 in
Solution:
c ⎞⎛ d ⎞
⎛ a ⎞⎛ c ⎞
⎛
⎟ ⎜ ⎟ c b + ⎜ ⎟ ⎜ ⎟ ( a − 2c)c + d c⎜ a − ⎟ ⎜ ⎟
2 ⎠⎝ 2 ⎠
⎝ 2⎠⎝ 2 ⎠
⎝ 2 ⎠⎝ 2 ⎠
⎝
Ixy = ⎛⎜
c ⎞⎛ b ⎞
Ixy = 97.75 in
4
Problem 10-69
Determine the location (xc, yc) of the centroid C of the angle’s cross-sectional area, and then
1033
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
compute the product of inertia with respect to the x' and y' axes.
Given:
a = 18 mm
b = 150 mm
Solution:
⎛ a ⎞ a b + a( b − a) ⎛ a + b ⎞
⎜ ⎟
⎜
⎟
2⎠
2 ⎠
⎝
⎝
xc =
a b + a( b − a)
xc = 44.1 mm
⎛ b ⎞ a b + ⎛ a ⎞ a( b − a)
⎜ ⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
yc =
a b + a( b − a)
yc = 44.1 mm
⎛
⎝
Ix'y' = a b ⋅ −⎜ xc −
6
a⎞⎛ b
a⎞⎛ b a
⎞
⎞
⎛
⎟ ⎜ − yc⎟ + a( b − a) ⋅ −⎜ yc − ⎟ ⎜ + − xc⎟
2⎠⎝ 2
2
2
2
⎠
⎝
⎠⎝
⎠
4
Ix'y' = −6.26 × 10 mm
1034
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-70
Determine the product of inertia of the beam’s
cross-sectional area with respect to the x and y axes
that have their origin located at the centroid C.
Given:
a = 5 mm
b = 30 mm
c = 50 mm
Solution:
xc =
a + b⎞
⎛ a⎞
⎟ + c a⎜ ⎟
⎝ 2 ⎠
⎝ 2⎠
a( b − a) ⎛⎜
a( b − a) + a c
xc = 7.50 mm
⎛c⎞
⎟ + c a⎜ ⎟
⎝ 2⎠
⎝ 2⎠
a( b − a) ⎛⎜
yc =
a⎞
a( b − a) + c a
yc = 17.50 mm
⎛ a − y ⎞ ⎛ a + b − x ⎞ + a c⎛ a − x ⎞ ⎛ c − y ⎞
⎜
c⎟ ⎜
c⎟
c⎟ ⎜
c⎟
⎝2
⎠⎝ 2
⎠
⎝2
⎠⎝ 2
⎠
Ixy = ( b − a)a⎜
3
4
Ixy = −28.1 × 10 mm
1035
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-71
Determine the product of inertia for the shaded area
with respect to the x and y axes.
Given:
a = 2 in
b = 1 in
c = 2 in
d = 4 in
Solution:
⎛ c + d ⎞ − π b2 a d
⎟
⎝ 2 ⎠
lxy = 2a( c + d)a⎜
lxy = 119 in
4
Problem 10-72
Determine the product of inertia for the beam's cross-sectional area with respect to the x and y
axes that have their origin located at the centroid C.
Given:
a = 1 in
b = 5 in
c = 5 in
Solution:
⎛ a − b ⎞ ⎛c + a ⎞
⎟⎜
⎟
⎝ 2 2⎠⎝ 2 ⎠
Ixy = 2b a⎜
Ixy = −110 in
4
1036
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-73
Determine the product of inertia for the
cross-sec-tional area with respect to the x
and y axes.
Given:
a = 4 in
b = 1 in
c = 6 in
Solution:
⎛ a ⎞ ⎛ c + 3b ⎞ + c b⎛b + c ⎞ ⎛ b ⎞
⎟⎜
⎟
⎜
⎟⎜ ⎟
2⎠
⎝ 2 ⎠⎝
⎝ 2⎠⎝ 2 ⎠
lxy = b a⎜
lxy = 72 in
4
Problem 10-74
Determine the product of inertia for the beam's cross-sectional area with respect to the u and
v axes.
1037
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Given:
a = 150 mm
b = 200 mm
t = 20 mm
θ = 20 deg
Solution:
Moments of inertia Ix and Iy:
Ix =
1
1
3
3
2a ( 2b) −
( 2a − t) ( 2b − 2t)
12
12
Ix = 511.36 × 10 mm
Iy =
2
2
3
3
t ( 2a) +
( b − t) t
12
12
Iy = 90240000.00 mm
6
4
4
The section is symmetric about both x and y axes;
therefore Ixy = 0.
⎛ Ix − Iy ⎞
⎟ sin ( 2θ ) + Ixy cos ( 2θ )
⎝ 2 ⎠
4
Ixy = 0mm
6
Iuv = ⎜
4
Iuv = 135 × 10 mm
Problem 10-75
Determine the moments of inertia Iu and Iv and the product of inertia Iuv for the rectangular area.The
u and v axes pass through the centroid C.
Given:
a = 40 mm
b = 160 mm
θ = 30 deg
1038
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
Ix =
Iu =
1
3
ab
12
Ix + Iy
2
Iy =
1
3
ba
12
4
Ixy = 0 mm
⎛ Ix − Iy ⎞
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠
+⎜
6
4
Iu = 10.5 × 10 mm
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎟−⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iv = ⎜
6
4
Iv = 4.05 × 10 mm
⎛ Ix − Iy ⎞
⎟ sin ( 2θ ) + Ixy cos ( 2θ )
⎝ 2 ⎠
Iuv = ⎜
6
4
Iuv = 5.54 × 10 mm
Problem 10-76
Determine the distance yc to the centroid of the area and then calculate the moments of inertia
Iu and Iv for the channel`s cross-sectional area. The u and v axes have their origin at the
centroid C. For the calculation, assume all corners to be square.
Given:
a = 150 mm
b = 10 mm
c = 50 mm
θ = 20 deg
1039
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This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
2a b
yc =
Ix =
b
⎛ c⎞
+ 2c b⎜ b + ⎟
2
⎝ 2⎠
yc = 12.50 mm
2a b + 2c b
1
12
2a b + 2a b ⎛⎜ yc −
3
⎝
b⎞
⎟
2⎠
2
⎡1
+ 2⎢
⎣12
b c + b c ⎛⎜ b +
3
⎝
c
2
2⎤
− yc⎟⎞ ⎥
⎠⎦
3
4
6
4
6
4
Ix = 908.3 × 10 mm
2
⎡1 3
b⎞ ⎤
⎛
⎢
Iy =
b ( 2a) + 2
c b + c b ⎜a − ⎟ ⎥
12
⎣12
⎝ 2⎠ ⎦
1
3
Iy = 43.53 × 10 mm
4
Ixy = 0 mm
(By symmetry)
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎟+⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iu = 5.89 × 10 mm
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎟−⎜
⎟ cos ( 2θ ) + Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iv = 38.5 × 10 mm
Iu = ⎜
Iv = ⎜
6
4
Problem 10-77
Determine the moments of inertia for the shaded area with respect to the u and v
axes.
Given:
a = 0.5 in
b = 4 in
c = 5 in
θ = 30 deg
1040
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
Moment and Product of Inertia about x and y Axes: Since the
shaded area is symmetrical about the x axis,
1
Ix =
3
12
2a c +
1
12
b ( 2a)
3
3
4
Ix = 10.75 in
4
Iy = 30.75 in
4
2
b
1
3
Iy =
2a b + 2a b ⎛⎜ a + ⎞⎟ +
c ( 2a)
12
12
⎝ 2⎠
1
Ixy = 0 in
Moment of Inertia about the Inclined u and v Axes
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎟+⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iu = 15.75 in
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎟−⎜
⎟ cos ( 2θ ) + Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iv = 25.75 in
Iu = ⎜
Iv = ⎜
4
4
Problem 10-78
Determine the directions of the principal axes with origin
located at point O, and the principal moments of inertia for
the rectangular area about these axes.
Given:
a = 6 in
b = 3 in
Solution:
Ix =
1
Iy =
1
Ixy =
3
Ix = 216 in
3
Iy = 54 in
a b
Ixy = 81 in
3
3
ba
ab
2 2
ab
4
4
4
1041
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
−2Ixy
tan ( 2θ ) =
Imax =
Imin =
Chapter 10
θ =
Ix − Iy
Ix + Iy
2
Ix + Iy
2
Ixy ⎞
1
⎛
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = −22.5 deg
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
2
⎝
⎠
Imax = 250 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 20.4 in
4
Problem 10-79
Determine the moments of inertia Iu , Iv and the product of inertia Iuv for the beam's cross-sectional
area.
Given:
θ = 45 deg
a = 8 in
b = 2 in
c = 2 in
d = 16 in
Solution:
2
1
3
3
⎛ d⎞
I x = ( a + b) c +
2b d + 2b d ⎜ ⎟
3
12
⎝ 2⎠
Iy =
2
1
1
3
3
[ 2( a + b) ] c +
( 2b) d
12
12
Iv =
Ix + Iy
2
Ix + Iy
2
+
−
Ix − Iy
2
Ix − Iy
2
4
3
4
3
4
3
4
Iy = 1.419 × 10 in
Ixy = 0 in
Iu =
3
Ix = 5.515 × 10 in
4
cos ( 2θ ) − Ixy sin ( 2θ )
Iu = 3.47 × 10 in
cos ( 2θ ) + Ixy sin ( 2θ )
Iv = 3.47 × 10 in
1042
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Iuv =
Ix − Iy
Chapter 10
sin ( 2θ ) + Ixy cos ( 2θ )
2
3
Iuv = 2.05 × 10 in
4
Problem 10-80
Determine the directions of the principal axes with
origin located at point O, and the principal moments
of inertia for the area about these axes.
Given:
a = 4 in
b = 2 in
c = 2 in
d = 2 in
r = 1 in
Solution:
Ix =
1
3 ⎛ πr
2 2⎞
( c + d) ( a + b) − ⎜
+ πr a ⎟
3
⎝ 4
⎠
Ix = 236.95 in
4
Iy =
1
3 ⎛ πr
2 2⎞
( a + b) ( c + d) − ⎜
+ πr d ⎟
3
⎝ 4
⎠
Iy = 114.65 in
4
⎛ a + b ⎞ ⎛ d + c ⎞ ( a + b) ( d + c) − d aπ r2
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
Ixy = 118.87 in
4
4
Ixy = ⎜
tan ( 2 θ p ) =
−Ixy
Ix− Iy
θp =
Ixy ⎞
1
⎛
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
4
θ p = −31.39 deg
2
θ p1 = θ p
θ p1 = −31.39 deg
θ p2 = 90 deg + θ p1
θ p2 = 58.61 deg
Imax =
Ix + Iy
2
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
⎝ 2 ⎠
Imax = 309 in
4
1043
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Imin =
Ix + Iy
2
Chapter 10
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
⎝ 2 ⎠
Imin = 42.1 in
4
Problem 10-81
Determine the principal moments of inertia for the
beam's cross-sectional area about the principal
axes that have their origin located at the centroid C.
Use the equations developed in Section 10.7. For
the calculation, assume all corners to be square.
a = 4 in
Given:
b = 4 in
t =
3
in
8
Solution:
⎡1
I x = 2⎢
⎣12
3
Ix = 55.55 in
⎡1
I y = 2⎢
⎛
⎝
a t + a t ⎜b −
1
t⎞ ⎤
3
⎟ ⎥ + t ( 2b − 2t)
2 ⎠ ⎦ 12
2
4
3
2
⎛ a − t + t ⎞ ⎤⎥ + 1 2b t3
⎟
2 ⎠ ⎦ 12
⎝ 2
Iy = 13.89 in
⎡a − t + ⎛ t ⎞⎤ ⎛b − t ⎞ t( a − t)
⎜ ⎟⎥ ⎜
⎟
⎣ 2
⎝ 2 ⎠⎦ ⎝ 2 ⎠
Ixy = −20.73 in
⎣12
Ixy = −2⎢
Imax =
Imin =
4
t ( a − t) + t( a − t) ⎜
Ix + Iy
2
Ix + Iy
2
4
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
2
⎝
⎠
Imax = 64.1 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 5.33 in
4
1044
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-82
Determine the principal moments of inertia for
the angle's cross-sectional area with respect to a
set of principal axes that have their origin located
at the centroid C. Use the equation developed in
Section 10.7. For the calculation, assume all
corners to be square.
Given:
a = 100 mm
b = 100 mm
t = 20 mm
Solution:
tb
xc =
t
2
+ ( a − t)t⎛⎜ t +
⎝
a − t⎞
2
⎟
⎠
xc = 32.22 mm
t b + ( a − t)t
tb
yc =
b
+ ( a − t)t
t
2
2
t b + ( a − t)t
yc = 32.22 mm
2
Ix =
1 3
t⎞
1 3
⎞
⎛
⎛b
t ( a − t) + t( a − t) ⎜ xc − ⎟ +
t b + t b ⎜ − xc⎟
12
2
12
2
⎝
⎠
⎝
⎠
2
2
1 3
t⎞
1
3
⎛
⎛ a−t −x ⎞
Iy =
b t + b t ⎜ xc − ⎟ +
t ( a − t) + t( a − t) ⎜ t +
c⎟
12
2⎠
12
2
⎝
⎝
⎠
⎛
⎝
Ixy = −⎜ xc −
t ⎞⎛ b
⎞
⎛ a − t + t − x ⎞ ⎛ y − t ⎞ ( a − t)t
⎟ ⎜ − yc⎟ b t − ⎜
⎟
c⎟ ⎜ c
2⎠⎝ 2
2⎠
⎠
⎝ 2
⎠⎝
⎛ Ix + Iy Ix − Iy ⎞
−
⎟ − Ixy
2 ⎠
⎝ 2
6
4
6
4
Ix = 3.142 × 10 mm
2
Iy = 3.142 × 10 mm
6
4
6
4
6
4
Ixy = −1.778 × 10 mm
Imax = ⎜
Imax = 4.92 × 10 mm
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
Imin = ⎜
⎟+⎜
⎟ + Ixy
⎝ 2 ⎠ ⎝ 2 ⎠
Imin = 2.22 × 10 mm
1045
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-83
The area of the cross section of an airplane wing has the listed properties about the x and y axes
passing through the centroid C. Determine the orientation of the principal axes and the principal
moments of inertia.
Ix = 450 in
Given:
4
Iy = 1730 in
4
Ixy = 138 in
4
Solution:
tan ( 2θ ) =
Imax =
Imin =
−2Ixy
Ix − Iy
Ix + Iy
2
Ix + Iy
2
θ =
1
⎛ Ixy ⎞
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = 6.08 deg
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
⎝ 2 ⎠
Imax = 1745 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 435 in
4
Problem 10-84
Using Mohr’s circle, determine the principal moments of inertia for the triangular area and the
orientation of the principal axes of inertia having an origin at point O.
Given:
a = 30 mm
b = 40 mm
1046
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
Moment of inertia Ix and Iy:
Ix =
1
3
ba
12
Ix = 90 × 10 mm
Iy =
1
3
ab
12
Iy = 160 × 10 mm
3
4
3
4
Product of inertia Ixy:
⌠
⎮
Ixy = ⎮
⌡
b
2
x⎛
a ⎞
⎜a − x⎟ dx
2⎝
b ⎠
3
4
Ixy = 60 × 10 mm
0
Mohr's circle :
2
OA =
⎞
⎛ Ix + Iy
2
− Ix⎟ + Ixy
⎜
2
⎝
⎠
3
4
OA = 69.462 × 10 mm
⎞
⎛ Ix + Iy
+ OA⎟
⎝ 2
⎠
Imax = ⎜
3
4
Imax = 194.462 × 10 mm
⎞
⎛ Ix + Iy
− OA⎟
⎝ 2
⎠
Imin = ⎜
3
4
Imin = 55.5 × 10 mm
tan ( 2θ ) =
Ixy
Ix + Iy
2
θ =
− Ix
1
⎛ Ixy ⎞
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = 29.9 deg
1047
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics
Chapter 10
Problem 10-85
Determine the directions of the principal axes with origin
located at point O, and the principal moments of inertia
for the rectangular area about these axes.
Solve using Mohr's circle.
Given:
a = 6 in
b = 3 in
Solution:
Ix =
1 3
ba
3
Ix = 216 in
Iy =
1 3
ab
3
Iy = 54 in
Ixy =
a b
ab
2 2
Ixy = 81 in
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
⎣ ⎝ 2 ⎠⎦
4
4
4
2
Imax =
Imin =
Ix + Iy
2
Ix + Iy
2
R = 114.55 in
4
+R
Imax = 250 in
4
−R
Imin = 20.4 in
4
−1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p1 = −22.50 deg
θ p2 = θ p1 + 90 deg
θ p2 = 67.50 deg
θ p1 =
1048
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-86
Determine the principal moments of inertia for the
beam's cross-sectional area about the principal
axes that have their origin located at the centroid
C. For the calculation, assume all corners to be
square. Solve using Mohr's circle.
Given:
a = 4 in
b = 4 in
t =
3
in
8
Solution:
⎡1
I x = 2⎢
1
t⎞ ⎤
3
⎟ ⎥ + t ( 2b − 2t)
2 ⎠ ⎦ 12
2
⎛
⎝
Ix = 55.55 in
4
t ( a − t) + t( a − t) ⎜
Iy = 13.89 in
4
⎡a − t + ⎛ t ⎞⎤ ⎛b − t ⎞ t( a − t)
⎜ ⎟⎥ ⎜
⎟
⎣ 2
⎝ 2 ⎠⎦ ⎝ 2 ⎠
Ixy = −20.73 in
⎣12
⎡1
I y = 2⎢
⎣12
3
a t + a t ⎜b −
2
⎛ a − t + t ⎞ ⎤⎥ + 1 2b t3
⎟
2 ⎠ ⎦ 12
⎝ 2
3
Ixy = −2⎢
4
2
R =
Ix + Iy ⎞
⎛
2
⎜Ix −
⎟ + Ixy
2 ⎠
⎝
Imax =
Imin =
Ix + Iy
2
Ix + Iy
2
R = 29.39 in
4
4
+R
Imax = 64.1 in
−R
Imin = 20.45 in
4
Problem 10-87
Determine the principal moments of inertia for the angle's cross-sectional area with respect to a set
of principal axes that have their origin located at the centroid C. For the calculation, assume all
corners to be square. Solve using Mohr's ciricle.
1049
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
a = 100 mm
Given:
b = 100 mm
t = 20 mm
Solution:
⎛ a − t⎞
⎟ + ( a − t)t⎜ t +
⎟
2 ⎠
⎝ 2⎠
⎝
t b⎛⎜
xc =
xc = 32.22 mm
t b + ( a − t)t
⎛t⎞
⎟ + ( a − t)t⎜ ⎟
⎝ 2⎠
⎝ 2⎠
t b⎛⎜
yc =
t⎞
b⎞
yc = 32.22 mm
t b + ( a − t)t
2
1 3
t⎞
1 3
⎞
⎛
⎛b
Ix =
t ( a − t) + t( a − t) ⎜ xc − ⎟ +
t b + t b ⎜ − xc⎟
12
2⎠
12
⎝
⎝2
⎠
2
2
Iy =
1 3
t⎞
1
3
⎛
⎛ a−t −x ⎞
b t + b t ⎜ xc − ⎟ +
t ( a − t) + t( a − t) ⎜ t +
c⎟
12
2⎠
12
2
⎝
⎝
⎠
6
4
6
4
Ix = 3.142 × 10 mm
2
Iy = 3.142 × 10 mm
⎛
⎝
Ixy = −⎜ xc −
t ⎞⎛ b
⎞
⎛ a − t + t − x ⎞ ⎛ y − t ⎞ ( a − t)t
⎟ ⎜ − yc⎟ b t − ⎜
⎟
c⎟ ⎜ c
2⎠⎝ 2
2⎠
⎠
⎝ 2
⎠⎝
6
2
R =
Ix + Iy ⎞
⎛
2
⎜Ix −
⎟ + Ixy
2 ⎠
⎝
4
Ixy = −1.778 × 10 mm
6
4
R = 1.78 × 10 mm
1050
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ix + Iy
Imax =
Imin =
2
Ix + Iy
2
Chapter 10
6
Imax = 4.92 × 10 mm
−R
Imin = 1364444.44 mm
4
Problem 10-88
Determine the directions of the principal axes with
origin located at point O, and the principal moments
of inertia for the area about these axes. Solve using
Mohr's circle
Given:
a = 4 in
b = 2 in
c = 2 in
d = 2 in
r = 1 in
Solution:
Ix =
1
3 ⎛ πr
2 2⎞
( c + d) ( a + b) − ⎜
+ πr a ⎟
3
⎝ 4
⎠
Ix = 236.95 in
4
Iy =
1
3 ⎛ πr
2 2⎞
( a + b) ( c + d) − ⎜
+ πr d ⎟
3
⎝ 4
⎠
Iy = 114.65 in
4
⎛ a + b ⎞ ⎛ d + c ⎞ ( a + b) ( d + c) − d aπ r2
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
Ixy = 118.87 in
4
4
Ixy = ⎜
4
2
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
⎣ ⎝ 2 ⎠⎦
Imax =
Imin =
4
+R
Ix + Iy
2
Ix + Iy
2
R = 133.67 in
4
+R
Imax = 309 in
4
−R
Imin = 42.1 in
4
1051
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
θ p1 =
Chapter 10
−1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p2 = θ p1 +
θ p1 = −31.39 deg
π
θ p2 = 58.61 deg
2
Problem 10-89
The area of the cross section of an airplane wing has the listed properties about the x and y axes
passing through the centroid C. Determine the orientation of the principal axes and the principal
moments of inertia. Solve using Mohr's circle.
Ix = 450 in
Given:
4
Iy = 1730 in
Ixy = 138 in
4
4
Solution:
2
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
2
⎣ ⎝
⎠⎦
R = 654.71 in
⎞
⎛ Ix + Iy
+ R⎟
⎝ 2
⎠
Imax = 1.74 × 10 in
⎞
⎛ Ix + Iy
− R⎟
⎝ 2
⎠
Imin = 435 in
4
3
Imax = ⎜
Imin = ⎜
4
4
1052
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p1 = 6.08 deg
θ p2 = θ p1 + 90 deg
θ p2 = 96.08 deg
θ p1 =
Problem 10-90
The right circular cone is formed by revolving the shaded area
around the x axis. Determine the moment of inertia lx and
express the result in terms of the total mass m of the cone. The
cone has a constant density ρ.
Solution:
h
⌠
2
rx⎞
1
⎮
2
⎛
m = ⎮ ρ π ⎜ ⎟ dx =
hρ π r
3
⎝h⎠
⌡
0
⌠
3m ⎮
lx =
2⎮
πh r ⌡
h
4
1 ⎛ rx⎞
3
2
π ⎜ ⎟ dx =
mr
2 ⎝h⎠
10
0
lx =
3
2
mr
10
Problem 10-91
Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.
1053
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
m = ρ 2π R
2π
⌠
I=⎮
⎮
⌡0
ρ =
m
2π R
⎛ m ⎞ R 2 R dθ = m R 2
⎜
⎟
⎝ 2π R ⎠
I=mR
2
Problem 10-92
The solid is formed by revolving the shaded area
around the y axis. Determine the radius of gyration
ky. The specific weight of the material is γ.
Given:
a = 3 in
b = 3 in
lb
γ = 380
ft
3
Solution:
b
⌠
2
⎮
⎡ ⎛ y ⎞ 3⎤
m = ⎮ γ π ⎢a ⎜ ⎟ ⎥ d y
⎮
⎣ ⎝ b⎠ ⎦
⌡
m = 2.66 lb
0
b
⌠
2
2
⎮
⎡ ⎛ y ⎞ 3⎤ 1 ⎡ ⎛ y ⎞ 3⎤
⎢a ⎜ ⎟ ⎥ d y
Iy = ⎮ γ π ⎢a ⎜ ⎟ ⎥
⎮
b⎠ ⎦ 2 ⎣ ⎝ b⎠ ⎦
⎣
⎝
⌡
Iy = 6.46 lb⋅ in
2
0
ky =
Iy
ky = 1.56 in
m
1054
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-93
Determine the moment of inertia Ix for the sphere and express the result in terms of
the total mass m of the sphere. The sphere has a constant density ρ.
Solution:
3
4π r
m=ρ
3
ρ =
r
3m
3
4π r
⌠ 1 3m
⎞ π r2 − x2 r2 − x2 dx = 2 m r2
⎛
Ix = ⎮
⎜
⎟
3
5
⎮ 2 4π r
⎠
⌡− r ⎝
(
)(
)
Ix =
2
2
mr
5
Problem 10-94
Determine the radius of gyration kx of the paraboloid. The density of the material is ρ.
Units Used:
Given:
Mg = 1000 kg
ρ = 5
Mg
m
3
a = 200 mm
b = 100 mm
1055
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
⎛ b2 x ⎞
⎮
⎟ dx
mp = ⎮ ρ π ⎜
a ⎠
⎝
⌡
mp = 15.71 kg
0
⌠
⎮
Ix = ⎮
⌡
a
⎛ b x⎞⎛ b x⎞
1
⎟ ⎜ ⎟ dx
ρ π⎜
2
⎝ a ⎠⎝ a ⎠
2
2
Ix = 52.36 × 10
−3
kg⋅ m
2
0
kx =
Ix
mp
kx = 57.7 mm
Problem 10-95
Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and
express the result in terms of the mass m of the semiellipsoid. The material has a
constant density ρ.
Solution:
a
⌠
2
⎮
x ⎟⎞
2
2⎛
2
⎜
m = ⎮ ρ πb 1 −
aρ π b
dx =
2
⎜ a ⎟
3
⎮
⎝
⎠
⌡0
ρ =
3m
2π a b
2
a
⌠
2
2
⎮ 1 ⎛ 3m ⎞ 2 ⎛
x ⎟⎞ 2 ⎛⎜
x ⎟⎞
2
2
⎜
Ix = ⎮
πb 1 −
b 1−
mb
dx =
⎜
⎟
2
2
2
⎜ a ⎟ ⎜ a ⎟
5
⎮ 2 ⎝ 2π a b ⎠
⎝
⎠ ⎝
⎠
⌡0
Ix =
2
2
mb
5
1056
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-96
Determine the radius of gyration kx of the body. The specific weight of the material is γ.
Given:
lb
γ = 380
ft
3
a = 8 in
b = 2 in
Solution:
a
⌠
2
⎮
⎮
3
2⎛x⎞
⎮
mb =
γ π b ⎜ ⎟ dx
⎮
⎝ a⎠
⌡
mb = 13.26 lb
0
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
a
2
1
2 x
γ π b ⎛⎜ ⎟⎞
2
⎝ a⎠
3
2
2⎛x⎞
b ⎜
3
⎟ dx
⎝ a⎠
Ix = 0.59 slug⋅ in
2
0
kx =
Ix
mb
kx = 1.20 in
Problem 10-97
Determine the moment of inertia for the ellipsoid with respect to the x axis and express the result in
terms of the mass m of the ellipsoid. The material has a constant density ρ.
1057
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
2
⎮
x ⎟⎞
4
2⎛
2
m = ⎮ ρ π b ⎜1 −
dx = a ρ π b
2⎟
⎜
3
⎮
⎝ a ⎠
⌡− a
ρ =
3m
4π a b
2
a
⌠
2
2
⎮
1 3m
x ⎞ 2⎛
x ⎞
2
2⎛
2
Ix = ⎮
π b ⎜ 1 − ⎟ b ⎜ 1 − ⎟ dx = m b
2
2⎟ ⎜
2⎟
⎜
2
5
⎮
4π a b
⎝ a ⎠ ⎝ a ⎠
⌡− a
Ix =
2
2
mb
5
Problem 10-98
Determine the moment of inertia of
the homogeneous pyramid of mass m
with respect to the z axis. The density
of the material is ρ. Suggestion: Use a
rectangular plate element having a
volume of dV = (2x)(2y) dz.
Solution:
⌠
⎮
V=⎮
⌡
h
2
z ⎞⎤
1 2
⎡⎛
⎢a⎜1 − ⎟⎥ dz = h a
3
⎣ ⎝ h ⎠⎦
ρ =
0
⌠
3m ⎮
Iz =
2 ⎮
a h⌡
m
3m
=
2
V
a h
h
4
1⎡ ⎛
z ⎞⎤
1
2
⎢a⎜ 1 − ⎟⎥ dz = m a
6⎣ ⎝
h ⎠⎦
10
Iz =
1
2
ma
10
0
1058
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-99
The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of
inertia Iy. The specific weight of concrete is γ.
Given:
γ = 150
lb
ft
3
a = 6 in
b = 4 in
c = 8 in
Solution:
c
⌠
2
2
1
2
2 ⎮ 1 ⎛ a y⎞ a y
⎟
Iy = γ π ( a + b) c ( a + b) − ⎮
γ ⎜π
dy
2
2 ⎝ c ⎠ c
⌡
Iy = 2.25 slug⋅ ft
2
0
Problem 10-100
Determine the moment of inertia of the thin plate about an axis perpendicular to the page and passing
through the pin at O. The plate has a hole in its center. Its thickness is c, and the material has a
density of ρ
Given:
a = 1.40 m
c = 50 mm
b = 150 mm
ρ = 50
kg
m
3
Solution:
IG =
(
)
1
1
2 2
2
2 2
ρa c a + a − ρ πb c b
12
2
IG = 1.60 kg⋅ m
I0 = IG + m d
2
2
1059
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
2
Chapter 10
2
m = ρa c − ρ πb c
m = 4.7233 kg
I0 = IG + m ( a sin ( 45 deg) )
I0 = 6.23 kg⋅ m
2
2
Problem 10-101
Determine the moment of inertia Iz of the frustum of the cone which has a conical depression. The
material has a density ρ.
Given:
kg
ρ = 200
m
3
a = 0.4 m
b = 0.2 m
c = 0.6 m
d = 0.8 m
Solution:
h =
da
a−b
Iz =
3 ⎡ ⎛ 1 2 ⎞⎤ 2
3 ⎡ ⎛ 1 2 ⎞⎤ 2
3 ⎡ ⎡1 2
⎤⎤ 2
⎢ρ ⎜ π a h⎟⎥ a − ⎢ρ ⎜ π a c⎟⎥ a − ⎢ρ ⎢ π b ( h − d)⎥⎥ b
10 ⎣ ⎝ 3
10 ⎣ ⎝ 3
10 ⎣ ⎣ 3
⎠⎦
⎠⎦
⎦⎦
Iz = 1.53 kg⋅ m
2
Problem 10-102
Determine the moment of inertia for the assembly about an axis which is perpendicular to the
page and passes through the center of mass G. The material has a specific weight γ.
Given:
a = 0.5 ft
d = 0.25 ft
1060
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No port