NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 1: Textbook Exercise 3-42
Goal:
Determine the magnitudes of the forces F1, F2 and F3 for
equilibrium of the particle.
Solution:
Strategy:
Write the forces in Cartesian system and then use the equilibrium equations for the particle
in x, y and z axis to find the magnitudes.
Writing Forces in Cartesian form:
K
F1 = F1 {cos 60 D iˆ + sin 60 D kˆ} = {0.5 F1 iˆ + 0.866 F1 kˆ}N
K
3
4
F2 = F2 { iˆ − ˆj} = {0.6 F2 iˆ − 0.8 F2 ˆj}N
5
5
K
F3 = F3 {− cos 30 D iˆ − sin 30 D ˆj} = {−0.866 F3 iˆ − 0.5 F3 ˆj}N
Equilibrium Equations for particle:
For the x-axis:
∑F
x
= 0 ⇒ 0.5 F1 + 0.6 F2 − 0.866 F3 = 0
For the y-axis:
∑F
y
= 0 ⇒ −0.8 F2 − 0.5 F3 + 800 sin 30 D = 0
For the z-axis:
∑F
z
= 0 ⇒ 0.866 F1 − 800 cos 30 D = 0
Solving the above system of equations:
F1 = 800 N ,
F2 = 147 N ,
F3 = 564 N
PAGE
1
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 2: Textbook Exercise 3-51
Cables AB and AC can sustain a maximum
tension of 500 N, and the pole can support a
maximum compression of 300 N.
Goal:
Determine the maximum weight of the lamp
that can be supported in the position shown.
Assumptions:
The force in the pole acts along the axis of
the pole.
Solution:
Strategy:
First write the forces and the weight of the lamp in Cartesian form, then write the
equilibrium equations for point A and solve for the weight.
Writing forces in Cartesian form:
K
2
1.5 ˆ 6 ˆ
FAO = FAO { iˆ −
j+
k }N
6.5
6.5
6.5
K
6
3
6
FAB = FAB {− iˆ + ˆj − kˆ}N
9
9
9
K
2
3
6
FAC = FAC {− iˆ + ˆj − kˆ}N
7
7
7
K
W = {−Wkˆ}N
Equilibrium Equations:
For the x-axis
∑F
x
=0⇒
2
6
2
FAO − FAB − FAC = 0
6.5
9
7
For the y-axis
∑F
x
=0⇒
− 1.5
3
3
FAO + FAB + FAC = 0
6.5
9
7
PAGE
2
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
For the z- axis
∑F
x
=0⇒
6
6
6
FAO − FAB − FAC − W = 0
6.5
9
7
Assuming that one of the members is already under maximum load, we can solve the above
equations to calculate what the load is in the other members.
First Case: FAB = 500 N
2
6
2
FAO − (500) − FAC = 0
6.5
9
7
− 1.5
3
3
FAO + (500) + FAC = 0
6.5
9
7
6
6
6
FAO − (500) − FAC − W = 0
7
6.5
9
⇒ FAC = 388.9 N
, FAO = 1444.46 N > 300 N Not acceptable
Second case: FAC = 500 N
2
6
2
FAO − FAB − (500) = 0
6.5
9
7
− 1.5
3
3
FAO + FAB + (500) = 0
6.5
9
7
6
6
6
FAO − FAB − (500) − W = 0
6.5
9
7
⇒ FAB = 642.85 N > 500 N
, FAO = 1857.14 N > 300 N Not acceptable
Third case: FAO = 300 N
2
6
2
(300) − FAB − FAC = 0
6 .5
9
7
− 1 .5
3
3
(300) + FAB + FAC = 0
6 .5
9
7
6
6
6
(300) − FAB − FAC − W = 0
6 .5
9
7
⇒ FAC = 80.8 N
, FAB = 104 N
, W = 138 N Answer
PAGE
3
12
NAME & ID
Mechatronics Engineering
DATE
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 3: Textbook Exercise 3-65:
Determine the tension developed in
cables OD and OB and the strut OC,
required to support the 50-kg crate. The
spring OA has an un-stretched length of
0.8 m and a stiffness k OA = 1.2 kN / m .
The force in the strut acts along the axis
of the strut.
Solution:
Strategy:
First draw the free body diagram and write the equilibrium equation for point O. Weight and
spring force are known and the other can be found by solving equilibrium equations.
Free Body Diagram:
PAGE
4
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Spring Force:
Fsp = ks = 1.2(1 − 0.8) = 0.24kN = 240 N
Writing forces in Cartesian form:
K
FOB = FOB (
− 2iˆ − 4 ˆj + 4kˆ
1
2
2
) = − FOB iˆ − FOB ˆj + FOB kˆ
3
3
3
(−2) 2 + (−4) 2 + 4 2
K
− 4iˆ + 3kˆ
4
3
FOC = FOC (
) = − FOC iˆ + FOC kˆ
2
2
5
5
(−4) + 3
K
2iˆ + 4 ˆj + 4kˆ
1
2
2
FOD = FOD (
) = FOD iˆ + FOD ˆj + FOD kˆ
3
3
3
22 + 42 + 42
K
Fsp = {−240 ˆj}N
W = {−50(9.81)kˆ}N = {−490.5kˆ}N
Equilibrium equations:
K
K
∑F = 0⇒ F
OB
K
K
K
K
+ FOC + FOD + Fsp + W = 0
1
4
1
2
2
⇒ (− FOB − FOC + FOD )iˆ + (− FOB + FOD − 240) ˆj +
3
5
3
3
3
2
3
2
( FOB + FOC + FOD − 490.5)kˆ = 0
3
5
3
1
4
1
⎧
⎪
(− FOB − FOC + FOD ) = 0
⎪
⎪
3
5
3
⎪
⎪
2
2
⎪
(− FOB + FOD − 240) = 0
⎨
⎪
3
3
⎪
⎪
2
3
2
⎪
( FOB + FOC + FOD − 490.5) = 0
⎪
⎪
5
3
⎪
⎩ 3
Solving the above equations, the forces are:
FOB = 120 N , FOC = 150 N , FOD = 480 N
PAGE
5
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 4: Textbook Exercise 4-14 and 4-15:
4-14 Take FB = 40lb, FC = 50lb
and determine the moment of each
force about the bolt located at A.
4-15 If FB = 30lb and FC = 45lb ,
determine the resultant moment
about the bolt located at A.
Solution:
Strategy:
Find the force component normal to the rod and multiply it by it’s distance from point O.
For problem 4-14
FBNormal = FB cos 25 D = 36.25lb
M B = 36.25(2.5) = 90.6lb. ft
FC Normal = FC cos 30 D = 43.3lb
M B = 43.3(3.25) = 141lb. ft
For problem 4-15
M A = 30 cos 25 D (2.5) + 45 cos 30 D (3.25) = 195lb. ft
PAGE
6
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 5: Textbook Exercise 4-21:
The tool at A is used to hold a power
lawnmower blade stationary while the nut is
being loosened with the wrench at B in the
direction shown.
Goals:
•
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?
Solution:
Strategy:
Use the fact that the summation of moments must be zero
The magnitude of the force normal to the rod exerted by the wrench at B:
FB sin 60D = 50 sin 60D
The magnitude of the moment of FB about C is:
MCB = FB sin 60D (0.3) = 12.99 = 13N.m
The magnitude of the moment exerted by the tool at A about C is:
MCA = FA
( 1213 )( 0.4 ) N.m
Since the two moments balance each other:
12
MCB = MCA : 12.99 = FA ( )(0.4) ⇒ FA = 35.2N
13
PAGE
7
12
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Mechatronics Engineering
Problem 6: Textbook Exercise 4-26:
The towline exerts a force of P=4kN
at the end of the 20m long crane
boom. If θ = 30D
Goal:
•
Determine the placement x
of the hook at A so that this
force creates a maximum
moment about point O.
•
What is this moment?
Solution:
Strategy:
The maximum moment occurs when the force is normal to the crane boom.
Free-Body Diagram:
B
20 m
O
30o
60o
1.5 m
D
A
C
x
By geometry:
LDC = 20 cos 30 D = 17.32m
LBC = 1.5 + 20 sin 30 D = 11.5m
tan 60 D =
LBC
⇒ L AC = 6.63m
L AC
x = L AC + LDC = 23.95 = 24m
The maximum moment of P about O is then:
MOMAX = P ⋅ LOB = 4000(20) = 80kN.m
PAGE
8
12
NAME & ID
Mechatronics Engineering
DATE
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 7: Textbook Exercise 4-65:
If a torque or moment of 80lb.in is required
to looser the bolt at A, determine the force
P that must be applied perpendicular to the
handle of the flex-headed ratchet wrench.
Solution:
Strategy:
Since the force is perpendicular to the handle the resultant moment is the force magnitude
multiplied by the distance from the bolt at A.
M A = P(0.75 + 10 sin 60 D ) = 80
⇒P=
80
= 8.5lb
9.41
PAGE
9
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 8: Textbook Exercise 4-74:
The resultant couple moment created by the
two couples acting on the disk is
K
M R = {10kˆ}kip .in.
Goal:
Determine the magnitude of force T.
Solution:
Let us do:
L1 = 2in.
L2 = ( 4 + 2 + 3 ) in. = 10in.
L1
L2
The resultant couple moment is:
2
MR =
∑ Mi
i =1
M R = 10kip = TL1 + TL2 = T (2) + T (9) = 11T
The magnitude of the force T:
⇒ T = 0.909kip
PAGE
10
12
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Mechatronics Engineering
Problem 9: Textbook Exercise 4-85:
Two couples act on the frame. If d=4 ft, 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)
b) Summing the moments of all the force components
about point B.
Solution:
3 ft
2 ft
F2
B
F1
R1
K
R1 = 4 jˆ
K
4
3
F1 = −80( )iˆ − 80( )jˆ
5
5
K
ˆ
R2 = 3i
K
F2 = −50 sin 30D iˆ − 50 cos 30D jˆ
R2
1 ft
a) Finding the moments of each
couple:
4 ft
K
MC =
2
K
K
∑ (Ri × Fi )
i =1
K
MC =
i
j
k
i
j
k
3
0
0 +
0
4
0 = {126kˆ}lb.ft
−50 sin 30D −50 cos 30D
0
−64 −48 0
b) Find the component of the forces that creates a moment about B and multiply by the
distance from B.
K
4
4
M C = 50 cos 30 D (2) − 50 cos 30 D (5) − (80)(1) + (80)(5) = 126lb. ft
5
5
PAGE
11
12
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 3
SOLUTIONS
Problem 10: Textbook Exercise 4-93:
The gear reducer is subject to the couple
moments shown. Determine the resultant
couple moment and specify its magnitude and
coordinate direction angles.
Solution:
Strategy:
Write the couples in Cartesian system and calculate the sum of them to obtain the resultant
couple moment. Find the couple unit vector to find the coordinate direction angles.
The moments in Cartesian form:
K
M 1 = {60iˆ}lb. ft
K
M 2 = 80(− cos 30D sin 45D iˆ − cos 30D cos 45D ˆj − sin 30D kˆ) = {−48.99iˆ − 48.99 ˆj − 40kˆ}lb. ft
Resultant Moment:
K
K
M R = ∑ M ⇒ M R = {11.01iˆ − 49 ˆj − 40kˆ}lb. ft
Magnitude of the resultant moment:
M R = 11.012 + (−48.99) 2 + (−40) 2 = 64.2lb. ft
Coordinate direction angles: α, β, γ ≤ 120D
11.01
) = 80.1D
64.2
− 48.99
) = 140 D
β = cos −1 (
64.2
− 40
γ = cos −1 (
) = 129 D
64.2
α = cos −1 (
PAGE
12
12