Version 001 – Statics – ramadoss – (171)
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Wheel and Axle in Equilibrium
001 10.0 points
A wheel of diameter 52 cm has an axle of
diameter 11.6 cm. A force 140 N is exerted
along the rim of the wheel.
What force should be applied to the outside
of the axle in order to prevent the wheel from
rotating? Assume the direction of this force
to be tangential to the axle.
1
b
a
m1
m2
Which of the following expresses the condition required for the system to be in static
equilibrium?
1. a m2 = b m1
Correct answer: 627.586 N.
2. a m1 = b m2 correct
Explanation:
3. a2 m1 = b2 m2
4. b2 m1 = a2 m2
Let :
F = 140 N ,
rw = 26 cm , and
ra = 5.8 cm .
5. m1 = m2
Explanation:
In equilibrium, the total torque is zero,
which gives
a m1 = b m2 .
In order to keep the wheel in rotational equilibrium, the torques acting on it must balance
each other, so
X
τ = Fw rw − Fa ra = 0
dw
26 cm
Fa = Fw
= (140 N)
da
5.8 cm
AP M 1993 MC 35 B
003 10.0 points
A rod of negligible mass is pivoted at a point
that is off-center, so that length ℓ1 is different
from length ℓ2 . The figures show two cases in
which masses are suspended from the ends of
the rod. In each case the unknown mass m is
balanced so that the rod remains horizontal.
ℓ1
ℓ2
= 627.586 N .
m
25 kg
ℓ1
AP M 1998 MC 30
002 10.0 points
Consider the wheel-and-axle system shown
below.
22 kg
ℓ2
m
Version 001 – Statics – ramadoss – (171)
What is the value of m ?
M1 M2
2
p
4. m = M1 M2 correct
3. m =
Correct answer: 23.4521.
Explanation:
Let :
M1 + M2
2
Explanation:
X
Applying
τ = 0 to balance the masses
in both cases,
5. m =
M1 = 25 kg and
M2 = 22 kg .
X
Applying
τ = 0 to balance the masses
in both cases,
m ℓ 1 = M1 ℓ 2
M2 ℓ 1 = m ℓ 2 .
2
m ℓ 1 = M1 ℓ 2
M2 ℓ 1 = m ℓ 2 .
and
and
Dividing,
Dividing,
m
M1
=
M2
m
2
m =p
M1 M2
p
m = M1 M2 = (25 kg) (22 kg)
m
M1
=
M2
m
2
m = M1 M2
p
m = M1 M2 .
= 23.4521 kg .
AP M 1993 MC 35 A
004 10.0 points
A rod of negligible mass is pivoted at a point
that is off-center, so that length ℓ1 is different
from length ℓ2 . The figures show two cases in
which masses are suspended from the ends of
the rod. In each case the unknown mass m is
balanced by a known mass M1 or M2 so that
the rod remains horizontal.
ℓ1
ℓ2
m
M1
ℓ1
M2
ℓ2
AP M 1993 MC 29 30
005 (part 1 of 2) 10.0 points
Two objects of masses 18 kg and 45 kg are
connected to the ends of a rigid rod (of negligible mass) that is 70 cm long and has marks
every 10 cm, as shown.
A
BCDE F GH
J
18 kg
10 20 30 40 50 60
Which point represents the center of mass
of the sphere-rod combination?
1. D
m
2. H correct
3. B
What is the value of m in terms of the
known masses?
45 kg
4. G
1. m = M1 M2
5. J
2. m = M1 + M2
6. F
Version 001 – Statics – ramadoss – (171)
7. E
8. C
8. C
9. A correct
Explanation:
The parallel-axis theorem is
9. A
Explanation:
Let : ℓ = 70 cm ,
m1 = 18 kg , and
m2 = 45 kg .
For static equilibrium, τnet = 0.
Denote x the distance from the left end
point of the rod to the center-of-mass point.
X
τ = m1 g x − m2 g (ℓ − x) = 0
(m1 + m2 ) x = m2 ℓ
m2 ℓ
(45 kg) (70 cm)
x=
=
m1 + m2
18 kg + 45 kg
= 50 cm .
Therefore the point should be point H.
006 (part 2 of 2) 10.0 points
The sphere-rod combination can be pivoted
about an axis that is perpendicular to the
plane of the page and that passes through one
of the five lettered points.
Through which point should the axis pass
for the moment of inertia of the sphere-rod
combination about this axis to be greatest?
1. F
2. D
3
I = M h2 + ICM ,
where ICM is the moment of inertia about
an axis through the center of mass. Since
ICM and M = 18 kg + 45 kg = 63 kg are
fixed, maximum h = |50 cm − 0 cm| = 50 cm
will give the maximum moment of inertia.
Therefore, the axis should pass through the
point A.
AP B 1998 MC 7
007 10.0 points
Three forces act on an object.
If the object is in translational equilibrium,
which of the following must be true?
I. The vector sum of the three forces must
equal zero;
II. The magnitude of the three forces must
be equal;
III. The three forces must be parallel.
1. I, II and III
2. I and III only
3. II and III only
4. II only
5. I only correct
Explanation:
If an object is in translational equilibrium,
the vector sum of all forces acting on it must
equal zero.
3. G
4. J
5. E
6. H
7. B
AP B 1998 MC 58
008 10.0 points
A wheel of radius R and negligible mass is
mounted on a horizontal frictionless axle so
that the wheel is in a vertical plane. Three
small objects having masses m, M , and 2 M ,
respectively, are mounted on the rim of the
wheel, as shown.
Version 001 – Statics – ramadoss – (171)
attached if the stick is to remain horizontal
when suspended from the cord?
M
60◦
m
4
R
1. A
2M
2. G correct
3. B
If the system is in static equilibrium, what
is the value of m in terms of M ?
4. C
5. D
1. m = 2 M
6. F
2. m = M
3M
correct
2
M
4. m =
2
5M
5. m =
2
Explanation:
Since the system is in static equilibrium,
the net torque is zero:
3. m =
τ = m g R + M g R cos 60◦ − (2 M ) g R
1
= mgR + M gR − 2M gR
2
3
= mgR − M gR = 0
2
3M
m=
.
2
AP B 1993 MC 57
009 10.0 points
Two objects of masses 10 kg and 25 kg are
hung from the ends of a stick that is 70 cm
long and has marks every 10 cm, as shown.
ABCDE F G
10 20 30 40 50 60
10 kg
25 kg
If the mass of the stick is negligible, at
which of the points indicated should a cord be
7. E
Explanation:
Let : ℓ = 70 cm ,
m1 = 10 kg , and
m2 = 25 kg .
For static equilibrium, τnet = 0.
Let x be the distance from the left end of
the stick to the point of attachment of the
cord:
X
T = m1 g x − m2 g (ℓ − x) = 0
(m1 + m2 ) x = m2 ℓ
m2 ℓ
(25 kg) (70 cm)
x=
=
m1 + m2
10 kg + 25 kg
= 50 cm .
Therefore the point should be point G .
Acrobat Hangs From a Wire
010 10.0 points
An acrobat hangs by his hands from the middle of a tightly stretched horizontal wire so
that the angle between the wire and the horizontal is 11.3◦ .
If the acrobat’s mass is 93.2 kg, what is
the tension in the wire? The acceleration of
gravity is 9.8 m/s2 .
Correct answer: 2330.64 N.
Version 001 – Statics – ramadoss – (171)
5
Explanation:
Let :
θ = 11.3◦ ,
m = 93.2 kg , and
g = 9.8 m/s2 .
Three forces act on the acrobat: gravity
and 2 tensions by the wire. For the acrobat
to remain in static equilibrium, the net force
must equal zero, so in the vertical direction
Let : g = 9.8 m/s2 ,
R = 0.403 m , and
H = 2.21 m .
In the non-inertial frame of the accelerating
dolly, the column is subject to the horizontal
inertial force
~ in = −m~g .
F
Together, the gravity and the inertial force
combine into the apparent weight force
2 T sin θ = m g
mg
(93.2 kg) (9.8 m/s2 )
T =
=
2 sin θ
2 sin 11.3◦
= 2330.64 N .
A Column and a Dolly
011 10.0 points
A cylindrical stone column of diameter 2R =
0.806 m and height H = 2.21 m is transported
in standing position by a dolly.
~ app = m (~g − ~a)
W
in the direction
a
θ = arctan
g
from the vertical.
From the torque point of view, this apparent weight force applies at the center of mass
of the column. The column is stable in the
vertical position when the line of this force
goes through the column’s base
a
CM
Wapp
side view
When the dolly accelerates or decelerates
slowly enough, the column stands upright,
but when the dolly’s acceleration magnitude
exceed a critical value ac , the column topples over. (For a > +ac the column topples
backward; for a < −ac the column toppes
forward.)
Calculate the magnitude of the critical acceleration ac of the dolly. The acceleration of
gravity is 9.8 m/s2 .
but when this line misses the base, the column
topples over
CM
Wapp
For the critical acceleration ac , the line goes
through the edge of the base, hence the direction of the apparent weight force must deviate
from the vertical by the angle θc where
Correct answer: 3.57412 m/s2 .
Explanation:
tan θc =
2R
R
=
.
hcm
H
Version 001 – Statics – ramadoss – (171)
6
Consequently, the critical acceleration of the
dolly is
k
ℓ
2R
H
2
(0.403
m)
= (9.8 m/s2 )
2.21 m
a
ac = g tan θc = g
= 3.57412 m/s2 .
θ
b
m
ℓ
Bars in Equilibrium
012 10.0 points
The bars are in equilibrium, each 5 m long,
and each weighing 87 N. The string pulling
down on the two bars is attached 0.8 m from
the fulcrum on the leftmost bar and 0.7 m
from the left end of the rightmost bar. The
spring (of constant 38 N/cm) is attached at
an angle 35◦ at the left end of the upper bar.
cm
/
N
38
5m
0.8 m
35◦
0.7 m
9 kg
5m
If the suspended mass is 9 kg, by how much
will the spring stretch? The acceleration of
gravity is 9.8 m/s2 .
a
2mg
W
T1
2mg
mg
mg
From rotational equilibrium on the leftmost
lever, the tension 2 m g acts down at a distance ℓ − a from the fulcrum, the tension T1
acts down at a distance a from the fulcrum,
and the weight W of the bar acts down at a
ℓ
distance − a from the fulcrum, so
2
ℓ
T1 a − 2 m g (ℓ − a) − W
−a =0
2
2 T1 a − 4 m g (ℓ − a) − W (ℓ − 2 a) = 0
Correct answer: 45.8315 cm.
Explanation:
Let : W = 87 N ,
m = 9 kg ,
ℓ = 5 m,
a = 0.8 m ,
b = 0.7 m ,
k = 38 N/cm ,
g = 9.8 m/s2 .
Find the tension T1 in the string connecting
the levers by applying rotational equilibrium
on the leftmost bar. The suspended mass
weighs m g and causes a tension of 2 m g on
the string attached to the end of the leftmost
bar.
4 m g (ℓ − a) + W (ℓ − 2 a)
2a
2 (9 kg) (9.8 m/s2 ) (5 m − 0.8 m)
=
0.8 m
(87 N) [5 m − 2 (0.8 m)]
+
2 (0.8 m)
= 1110.98 N .
T1 =
and
Version 001 – Statics – ramadoss – (171)
7
Explanation:
k∆
x
b
Let : m1 = 8 kg ,
x1 = 1 m , and
ℓ = 6 m.
θ
W
T1
ℓ
From rotational equilibrium on the rightmost lever, the weight W acts down at a
ℓ
distance from the fulcrum, the tension T1
2
acts down at a perpendicular distance ℓ − b
from the fulcrum, and the tension k ∆x acts
up at a perpendicular distance ℓ sin θ from the
fulcrum, so
ℓ
=0
k ∆x (ℓ sin θ) − T1 (ℓ − b) − W
2
2 k ∆x ℓ sin θ − 2 T1 (ℓ − b) − W ℓ = 0
2 T1 (ℓ − b) + W ℓ
2 k ℓ sin θ
(1110.98 N) (5 m − 0.7 m)
=
(38 N/cm) (5 m) sin 35◦
(87 N) (5 m)
+
2 (38 N/cm) (5 m) sin 35◦
= 45.8315 cm .
Balancing Rock 02
013 10.0 points
An 8 kg rock is suspended by a massless
string from one end of a 6 m measuring stick.
1
2
3
4
5
xCM =
x1
0
ℓ
2
and
x2 =
ℓ
− 1.
2
4
5
x2
1
2
m1 g
3
m2 g
The static equation for torques gives us
6
Balancing a Truck
014 10.0 points
A truck is balanced on an incline as shown.
The angle of the inclination is 47.2◦ , the outer
part of the large pulley has a radius of 3 r and
the inner part of the large pulley has a radius
of r.
3r
r
8 kg
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s2 .
Correct answer: 39.24 N.
6
m2 g x2 = m1 g x1
m1 g x1
m2 g =
x2
(8 kg) (9.81 m/s2 ) (1 m)
=
2m
= 39.24 N .
∆x =
0
Because the stick is a uniform, symmetric
body, we can consider all of its weight to be
concentrated at the center of mass, so
T
r/2
m
θ
Version 001 – Statics – ramadoss – (171)
16 N
x
µ = 0.12
2.6 m
Find the mass m needed to balance the
2050 kg truck on the incline. The acceleration
of gravity is 9.8 m/s2 . Assume all pulleys are
frictionless and massless and the cords are
massless.
8
T
Correct answer: 250.691 kg.
Explanation:
21 N
µ = 0.23
Let :
θ = 47.2◦ ,
R = 3 r , and
M = 2050 kg .
How far from the 16 N object attachment
should the string be positioned to drag them
evenly (all angles = 90◦ ) and uniformly?
Correct answer: 1.86044 m.
Let T be the tension in the string around the
inner part (radius = r) of the pulley. From
translational equilibrium
Explanation:
ℓ = 2.6 m ,
w1 = 16 N ,
µ1 = 0.12 ,
w2 = 21 N , and
µ2 = 0.23 .
2 T − M g sin θ = 0
1
M g sin θ
2
1
= (2050 kg) 9.8 m/s2 sin 47.2◦
2
= 7370.32 N .
T =
T
1
= M sin θ
3g
6
1
= (2050 kg) sin 47.2◦ = 250.691 kg .
6
m=
Balanced Rod
015 10.0 points
Two objects are to be dragged along a table
by attaching them to the ends of a 2.6 m
rod. One weighs 16 N and has a coefficient of
friction of 0.12. The other weighs 21 N and
has a coefficient of friction of 0.23.
τ=
X
τcw −
X
τccw = 0
Let the fulcrum be at the point of attachment of the pulling string.
W1
x
µ1
ℓ
From rotational equilibrium of the large pulley,
X
τ = m g (3 r) − T r = 0
X
W2
µ2
The weight W1 exerts a frictional force of
µ1 W1 on the bar at a distance x from the
point of attachment. The weight W2 exerts
a frictional force of µ2 W2 on the bar at a
distance ℓ − x from the point of attachment,
so by rotational equilibrium
µ1 W1 x − µ2 W2 (ℓ − x) = 0
(µ1 W1 + µ2 W2 ) x = µ2 W2 ℓ
Version 001 – Statics – ramadoss – (171)
µ2 W2 ℓ
µ1 W1 + µ2 W2
0.23 (21 N) (2.6 m)
=
0.12 (16 N) + 0.23 (21 N)
x=
= 1.86044 m .
Beam in Equilibrium
016 (part 1 of 2) 10.0 points
A uniform horizontal beam of weight 605 N
and length 4.61 m has two weights hanging
from it. One weight of 505 N is located
1.30002 m from the left end; the other weight
of 405 N is located 1.30002 m from the right
end.
What must be the magnitude of the one
additional force on the beam that will produce
equilibrium?
Correct answer: 1515 N.
Explanation:
Let :
W = 605 N ,
W1 = 505 N ,
W2 = 405 N.
W L W1 L1 W2 L2
+
+
2F
F
F
(605 N) (4.61 m)
=
2 (1515 N)
(505 N) (1.30002 m)
+
1515 N
(405 N) (3.30998 m)
+
1515 N
= 2.23866 m .
x=
Beam on a Slanted Wall 01
018 (part 1 of 2) 10.0 points
A solid bar of length L has a mass m1 . The
bar is fastened by a pivot at one end to a
wall which is at an angle θ with respect to
the horizontal. The bar is held horizontally
by a vertical cord that is fastened to the bar
at a distance x from the wall. A mass m2 is
suspended from the free end of the bar.
and
For the beam to preserve translational equilibrium, the net force acting on the beam has
to be zero, so the additional force should be
F = W + W1 + W2
= 605 N + 505 N + 405 N
= 1515 N .
017 (part 2 of 2) 10.0 points
At what distance from the left end must this
force be applied for the beam to be in equilibrium?
Correct answer: 2.23866 m.
Explanation:
The beam also has to preserve its rotational
equilibrium; i.e., the net torque acting on the
beam has to be zero, so
Fx=W
9
L
+ W1 L1 + W2 L2
2
T
θ
m1
x
m2
L
Find the tension in the cord.
1. T = (m1 + m2 ) g sin θ
1
L
2. T =
g correct
m1 + m2
2
x
L
1
g sin θ
m1 + m2
3. T =
2
x
4. T = 0
1
L
5. T = m1 + m2
g sin θ
2
x
1
L
6. T = m1 + m2
g
2
x
7. T = (m1 + m2 ) g cos θ
Version 001 – Statics – ramadoss – (171)
L
g
8. T = (m1 + m2 )
x
2
L
1
g cos θ
m1 + m2
9. T =
2
x
L
1
g cos θ
10. T = m1 + m2
2
x
Explanation:
For equilibrium,
X
F = 0 and
X
τ =0.
Consider the torques around the connection between the bar and the wall, since the
reaction forces will produce no torques around
that point.
L
τ = −L m2 g + x T − m1 g = 0 ,
2
1
Lg
T =
m1 + m2
.
2
x
X
Explanation:
In general, there will be a horizontal reaction force Rx at the connection between an
object and a wall; however, none of the other
forces in this situation act in the horizontal
direction, so
X
Fx = Rx = 0 .
Boom and Cable
020 (part 1 of 3) 10.0 points
A 1040 N uniform boom is supported by a
cable as shown. The boom is pivoted at the
bottom, and a 3000 N object hangs from its
end.
The boom has a length of 25 m and is at an
angle of 48◦ above the horizontal. A support
cable is attached to the boom at a distance
of 0.69 L from the foot of the boom and its
tension is perpendicular to the boom.
019 (part 2 of 2) 10.0 points
Find the the horizontal component of the
force exerted on the bar by the wall. (Let
right be the positive direction.)
1
L
1. Fx =
g
m1 + m2
2
x
L
1
g
2. Fx = m1 + m2
2
x
1
L
3. Fx = m1 + m2
g sin θ
2
x
4. Fx = (m1 + m2 ) g cos θ
L
1
g cos θ
5. Fx = m1 + m2
2
x
1
L
6. Fx =
g cos θ
m1 + m2
2
x
7. Fx = (m1 + m2 ) g sin θ
L
1
g sin θ
m1 + m2
8. Fx =
2
x
9. Fx = 0 correct
L
g
10. Fx = (m1 + m2 )
x
2
10
90◦
0.
69
L
T
◦
48
3000 N
Find the tension in the cable holding up the
boom.
Correct answer: 3413.54 N.
Explanation:
Let : W1 = 1040 N ,
W2 = 3000 N ,
α = 48◦ ,
L = 25 m , and
L1 = 0.69 L = 17.25 m .
Version 001 – Statics – ramadoss – (171)
11
Explanation:
The verticalX
component of the tension is
T cos α. From
Fy = 0 ,
β
T
Fv = W1 + W2 − T cos α
= 1040 N + 3000 N − (3413.54 N) cos 48◦
W1
α
W2
X
For equilibrium
~τ = 0 . Taking torques
about the pivot point,
L
T (0.69 L) = W2 (L cos α) + W1
cos α
2
W2 cos α W1 cos α
+
T =
0.69
2 (0.69)
(3000 N) cos 48◦
=
0.69
(1040 N) cos 48◦
+
2 (0.69)
= 3413.54 N .
021 (part 2 of 3) 10.0 points
Find the horizontal component of the reaction
force on the boom by the floor.
Correct answer: 2536.75 N.
Explanation:
The horizontal
X component of the tension is
T sin α. From
Fx = 0,
Fh = T sin α = (3413.54 N) sin 48◦
= 2536.75 N .
022 (part 3 of 3) 10.0 points
Find the vertical components of the reaction
force on the boom by the floor.
Correct answer: 1755.9 N.
Boom and Weight 01
023 10.0 points
A weight of mass 43 kg is suspended from
a point near the right-hand upper end of a
uniform boom of mass 25 kg . This boom is
supported by a cable that runs from the boom
to a point on the wall (the left-hand vertical
coordinate at a height of 10 m) and by a pivot
(at the origin of the coordinate axes) on the
same wall.
10
9
Vertical Height (m)
0.
69
L
L
= 1755.9 N .
T
8
7
6
5
k
25
4
g
3
2
43 kg
1
0
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
Find the tension in the cable. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 488.311 N.
Explanation:
Let :
mb
mw
ℓx
xb
= 25 kg ,
= 43 kg ,
= 9 m,
= 4.5 m ,
Version 001 – Statics – ramadoss – (171)
x = 8 m,
y = 6.5 m ,
h = 10 m .
and
p
(mb xb + mw x) g (h − y)2 + x2
T =
hx
(25 kg) (4.5 m) + (43 kg) (8 m)
=
(8 m) (10 m)
× (9.8 m/s2 )
q
× (10 m − 6.5 m)2 + (8 m)2
Let the cable make an angle θ with the
horizontal.
(0, h)
10
Vertical Height (m)
9
T
8
Ty
= 488.311 N .
Tx θ
7
(x, y)
Ww
6 (0, y)
5
4
3
2
Wb
1
0
-1
-2
0
1
12
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
Ty
h−y
=
x
Tx
x
Tx =
Ty and
h−y
Boom and Weight 05
024 10.0 points
A weight (with a mass of 48 kg) is suspended
from a point at the right-hand end of a uniform boom with a mass of 9 kg . A horizontal
cable at an elevation of 7.5 m is attached to
the wall and to the boom at this same end
point. The boom is also supported by a pivot
(at the origin of the coordinate axes) on the
same wall.
10
9
Ty
h−y
sin θ =
=p
T
(h − y)2 + x2
T (h − y)
Ty = p
.
(h − y)2 + x2
Applying the static equilibrium condition for
torque (with the origin as the fulcrum),
X
τ = Ty x + Tx y − Wb xb − Ww x = 0
Wb xb + Ww x = Ty x + Tx y
xy
(mb xb + mw x) g = Ty x + Ty
h−y
y
x
= Ty 1 +
h−y
T (h − y)
hx
=p
(h − y)2 + x2 h − y
Vertical Height (m)
tan θ =
T
8
7
6
5
4
9
kg
3
2
48 kg
1
0
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
Figure: Drawn to scale.
Calculate the tension T in the cable. The
acceleration of gravity is 9.8 m/s2 .
Your answer must be within ± 3.0%
Correct answer: 583.1 N.
Explanation:
Version 001 – Statics – ramadoss – (171)
top of the ladder when it slips and falls to the
floor.
What is the coefficient of static friction between the ladder and the floor?
Let : Mb = 9 kg ,
Mw = 48 kg ,
xb = 4.25 m ,
x = 8.5 m , and
y = 7.5 m .
Correct answer: 0.35186.
Explanation:
Let :
10
Vertical Height (m)
9
8 (0, y)
(x, y)
T
7
6
5
4
Wb
3
α
2
Ww
1
0
13
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
The cable is horizontal. Using the pivot
at the origin as the fulcrum and the static
equilibrium condition for torque,
X
τ = Wb xb + Ww x − T y = 0
T y = Wb xb + Ww x
g
y
(9 kg) (4.25 m) + (48 kg) (8.5 m)
=
7.5 m
× (9.8 m/s2 )
T = (Mb xb + Mw x)
= 583.1 N .
Climbing a Ladder
025 10.0 points
A 6 m long, uniform ladder leans against a
frictionless wall and makes an angle of 64.3◦
with the floor. The ladder has a mass 30.4 kg.
A 79.04 kg man climbs 82% of the way to the
Mℓ = 30.4 kg ,
Mm = 79.04 kg ,
R = 0.82 , and
θ = 64.3◦ .
Since the ladder is in static equilibrium the
sum of all forces and torques about any point
must be zero. First, using the sum of the
vertical forces where FN is the normal force
exerted by the floor
X
0=
Fy
0 = FN − Mm g − Mℓ g
FN = (Mm + Mℓ ) g
X
Fx = 0 requires the wall to exert a force
equal in magnitude to the frictional force of
FN µ = (Mm + Mℓ ) g µ exerted by the floor.
Applying torques on the ladder about the
point where it touches the floor,
ℓ (Mℓ + Mm ) g
µ sin θ
ℓ
− Mℓ + R Mm ℓ g cos θ = 0
2
2 (Mℓ + Mm ) µ sin θ = (Mℓ + 2 R Mm ) cos θ
(Mℓ + 2 R Mm ) cos θ
2 (Mℓ + Mm ) sin θ
[30.4 kg + 2(0.82)(79.04 kg)] cos 64.3◦
=
2(30.4 kg + 79.04 kg) sin 64.3◦
µ=
= 0.35186 .
Car in Snow Drift
026 (part 1 of 2) 10.0 points
A student gets her car stuck in a snow drift.
Not at a loss, having studied physics, she
Version 001 – Statics – ramadoss – (171)
attaches one end of a rope to the vehicle and
the other end to the trunk of a nearby tree,
allowing for a small amount of slack. The
student then exerts a force F on the center of
the rope in the direction perpendicular to the
car-tree line as shown. Assume equilibrium
conditions and that the rope is inextensible.
θ
L
d
14
The angle θ is obtained from
2d
d
tan θ = =
L
L
2
2d
θ = arctan
L
2 (3.92 m)
= 27.4386◦ .
= arctan
15.1 m
If T is the tension in the rope, Newton’s
second law in the y direction gives
2 T sin θ = F
F
How does the magnitude of the force exerted by the rope on the car compare to that
of the force exerted by the rope on the tree?
Dishonest Shopkeeper
028 10.0 points
Two pans of a balance are 62.5 cm apart.
The fulcrum of the balance has been shifted
1.58 cm away from the center by a dishonest
shopkeeper.
By what percentage is the true weight of the
goods being marked up by the shopkeeper?
Assume the balance has negligible mass.
1. | Ft |<| Fc |
2. Cannot be determined
3. | Ft |>| Fc |
4. | Ft |=| Fc |= T correct
5. | Ft |= 2 | Fc |
Explanation:
Since the the student pulls on the center of
the rope, the angle between the rope and the
car equals the angle between the rope and the
tree, so
| Ft |=| Fc |= T
027 (part 2 of 2) 10.0 points
What is the magnitude of the force on the car
if L = 15.1 m, d = 3.92 m and F = 585 N?
Correct answer: 634.769 N.
Explanation:
Let :
L = 15.1 m ,
d = 3.92 m ,
F = 585 N .
F
585 N
=
2 sin θ
2 sin 27.4386◦
= 634.769 N .
T =
and
Correct answer: 10.6505 %.
Explanation:
Let :
L = 62.5 cm and
ℓ = 1.58 cm .
Let W be the standard weight and W ′ the
measured weight.
L
L
′
W
−ℓ =W
+ℓ
2
2
W
L+ 2ℓ
=
, so
′
W
L− 2ℓ
L+2ℓ
W − W′
=
− 1 × 100%
W′
L−2ℓ
4ℓ
× 100%
=
L− 2ℓ
4 1.58 cm
=
× 100%
62.5 cm − 2 1.58 cm
= 10.6505 % .
Version 001 – Statics – ramadoss – (171)
Cranes and Booms
029 (part 1 of 2) 10.0 points
A uniform beam of length 5.2 m and mass
19 kg supports a 18 kg mass as shown in the
figure. The cable is at an angle of 58.3◦ with
the wall and the beam is attached to the wall
by a pivoting pin at an angle of 51.1◦ with the
wall.
58.3◦
φ
15
T
α
Ry
L
θ
Mg
mg
Rx
k
19
51.1◦
X
Rotational equilibrium gives us
τ = 0,
so taking the sum of the torques about the
pivot,
L
m g sin θ = 0
L T sin α − L M g sin θ −
2
g
5. 2
m
18 kg
Find the tension in the cable. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 222.362 N.
Explanation:
Let :
θ = 51.1◦ ,
φ = 58.3◦ ,
L = 5.2 m ,
M = 18 kg ,
m = 19 kg .
= 222.362 N .
and
The angle between the tension T and the
beam is
◦
2 T sin α = 2 M g sin θ + m g sin θ
g sin θ
T = (2 M + m)
2 sin α
2 (18 kg) + 19 kg
=
2 sin(70.6◦ )
×(9.8 m/s2 ) sin(51.1◦ )
α = 180 − φ − θ
= 180◦ − 58.3◦ − 51.1◦
= 70.6◦ .
X
The system is in equilibrium, so
F = 0
X
and
τ = 0.
030 (part 2 of 2) 10.0 points
What is the magnitude of the force that the
wall applies to the beam at the pivot?
Correct answer: 310.141 N.
Explanation:
Applying translational equilibrium,
X
Fx = Rx − T sin φ = 0
Rx = T sin φ
= (222.362 N) sin 58.3◦
= 189.188 N , and
Version 001 – Statics – ramadoss – (171)
X
Fy = Ry + T cos φ − (M + m) g = 0
Ry = (M + m) g − T cos φ
= (18 kg + 19 kg) (9.8 m/s2 )
−(222.362 N) cos 58.3◦
= 245.755 N , so
R=
=
q
q
16
Explanation:
At the midpoint of the chain, there is only
a horizontal component of the tension. Since
the chain is in equilibrium, the tension at the
midpoint must equal the horizontal component of the force of a hook.
Tm = Te cos θ = (29.2556 N) cos 40.5◦
= 22.2461 N .
Rx2
+
Ry2
(189.188 N)2 + (245.755 N)2
= 310.141 N .
Hanging Chain
031 (part 1 of 2) 10.0 points
A flexible chain weighing 38 N hangs between
two hooks located at the same height. At each
hook, the tangent to the chain makes an angle
of 40.5◦ with the horizontal.
Hinge and Bar 01
033 10.0 points
A uniform bar of length L and weight W is
attached to a wall with a hinge that exerts
a horizontal force Hx and a vertical force Hy
on the bar. The bar (which makes an angle
θ with respect to wall) is held by a cord that
makes a 90◦ angle with respect to bar.
θ
Find the magnitude of the force each hook
exerts on the chain.
Correct answer: 29.2556 N.
Explanation:
Let : W = 38 N and
θ = 40.5◦ .
X
~ = 0. By symmetry each
In equilibrium
F
hook supports half the weight of the chain, so
Ty = Te sin θ =
Te =
W
θ
L
W
2
38 N
W
=
= 29.2556 N .
2 sin θ
2 sin 40.5◦
032 (part 2 of 2) 10.0 points
Find the tension in the chain at its midpoint.
Consider a free-body diagram for half the
chain.
Correct answer: 22.2461 N.
What is the magnitude of the tension T in
the cord?
1
W sin θ correct
2
1
2. T = W tan θ
2
1
3. T = W sin θ cos θ
2
1. T =
Version 001 – Statics – ramadoss – (171)
1
4. T = W cos2 θ
2
1
5. T = W sin2 θ
2
1
6. T = W cos θ
2
Explanation:
Analyzing the torques on the bar, with the
hinge at the axis of rotation,
17
the symmetry, the tension T1 is the same in
each of these segments:
θ
θ
T1
T1
4W
X
τ = LT −
T =
L
sin θ W = 0
2
1
W sin θ .
2
Applying static equilibrium vertically,
T1 cos θ + T1 cos θ − 4 W = 0
Forces on the Golden Gate
034 (part 1 of 3) 10.0 points
Consider a simplified model of the Golden
Gate bridge, where the bridge is represented
by four equal weights, each weighing 3 N ,
hanging from a wire. The angle between
the hanging wire and the vertical supporting
beam is 31.9◦ and the bridge is symmetric.
T1
T1 =
2W
2 (3 N)
=
= 7.06737 N .
cos θ
cos 31.9◦
035 (part 2 of 3) 10.0 points
Calculate T2 , the tension in the middle segment of the wire.
Correct answer: 3.73467 N.
31.9◦
Explanation:
The wire between the middle two weights
will feel a 2 W weight on either end.
T2
β
θ
T1
T1
3N
3N
3N
3N
T2
θ
Calculate T1 , the tension in the left segment
of the wire.
2W
2W
Correct answer: 7.06737 N.
Explanation:
Let : θ = 31.9◦
W = 3 N.
Considering horizontal equilibrium on the
left knot,
and
The leftmost (or rightmost) segment of wire
“feels” one 4 W weight on its end; because of
T2 = T1 sin θ = (7.06737 N) sin 31.9◦
= 3.73467 N .
Version 001 – Statics – ramadoss – (171)
18
036 (part 3 of 3) 10.0 points
Calculate the angle β in the figure.
Correct answer: 51.2257◦.
φ
Explanation:
Call the tension in the second segment T
and consider the forces on the first knot from
the left.
b
b
ℓ
xcm
T
θ
T1
T
W
β
Horizontal equilibrium for the knot gives
T1 sin θ = T sin β
If the mass of the beam is 8 kg, the tension
4
3
in the wire is 40 N, sin φ = and cos φ = ,
5
5
how far is the center of mass of the beam from
the hinge? The acceleration due to gravity is
10 m/s2 .
1. 0.5 ℓ, of course
2. 0.666667 ℓ
and vertical equilibrium gives
T1 cos θ = T cos β + W .
3. 0.375 ℓ
4. 0.4 ℓ correct
T sin β
T1 sin θ
=
T cos β
T1 cos θ − W
T1 sin θ
tan β =
T1 cos θ − W
T1 sin θ
β = arctan
T cos θ − W
1
(7.06737 N) sin 31.9◦
= arctan
(7.06737 N) cos 31.9◦ − 3 N
= 51.2257◦ .
Hinged Horiz Beam
037 10.0 points
A horizontal, nonuniform beam of mass M
and length ℓ is hinged to a vertical wall at
one side, and attached to a wire on the other
end. The bar is motionless and a wire exerts
a force T at an angle of φ with respect to the
vertical.
5. 0.18 ℓ
6. 0.16 ℓ
7. 0.32 ℓ
8. Not enough information is given.
9. 0.3 ℓ
10. 0.625 ℓ
Explanation:
Consider the free-body diagram, with the
forces due to the wall not shown:
Version 001 – Statics – ramadoss – (171)
19
Since the bat is in static equilibrium, around
any point (in particular around O),
X
φ
b
b
xcm
ℓ
~ =F −W =0
F
F = W = 12.8333 N .
T
m~g
039 (part 2 of 2) 10.0 points
Determine the torque exerted on the bat by
the player about the center of mass of the bat.
Correct answer: 7.95667 N · m.
With the z-axis pointing out of the paper,
the magnitude of the net torque through the
hinge is
τz = T ℓ cos φ − m g xcm = 0
m g xcm = T ℓ cos φ
T cos φ
xcm =
ℓ
mg
4
(40 N)
5
= 0.4 .
=
(8 kg)(10 m/s2 )
Explanation:
The torque exerted by the player is
τ = W d = 7.95667 N · m .
Ladder 14
040 (part 1 of 2) 10.0 points
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.633 .
Fw
Holding a Baseball Bat
038 (part 1 of 2) 10.0 points
A baseball player holds a 46.2 oz bat
(12.8333 N) with one hand at the point O.
The bat is in equilibrium. The weight of the
bat acts along a line 62 cm to the right of O.
ℓ
h
θ
N
O
b
µ = 0.633
w
Determine the force exerted on the bat by
the player.
Correct answer: 12.8333 N.
W
f
b
Consider the following expressions:
A1: f = Fw
A2: f = Fw sin θ
W
B1: N =
2
B2: N = W
Explanation:
Let : W = 12.8333 N .
C1: ℓ Fw sin θ = 2 Fw cos θ
C2: ℓ Fw sin θ = ℓ W cos θ
Version 001 – Statics – ramadoss – (171)
1
C3: ℓ Fw sin θ = ℓ W cos θ ,
2
where
f : force of friction between the ladder and
the ground,
Fw : normal force on the ladder due to the
wall,
θ: angle between the ladder and the ground,
N : normal force on the ladder due to the
ground,
W: weight of the ladder, and
ℓ: length of the ladder.
Identify the set of equations which is correct.
20
So the correct
A1 , B2 , C3 .
set
of
A2, B1, C3
2.
A1, B2, C3 correct
3.
A1, B2, C2
is
041 (part 2 of 2) 10.0 points
Determine the smallest angle θ for which the
ladder remains stationary.
Correct answer: 38.3048◦ .
Explanation:
f = µs N = Fw ,
1.
equations
so
1
W cos θ
2
1
µs W sin θ = W cos θ
2
1
µs =
2 tan θ
1
tan θ =
2 µs
Fw sin θ =
4.
A1, B1, C2
5.
A2, B1, C2
6.
A1, B2, C1
7.
A1, B1, C3
8.
A2, B2, C1
9.
A2, B1, C1
1
θ = tan
2 µs
1
−1
= tan
2 (0.633)
−1
10.
A1, B1, C1
Explanation:
From translational equilibrium,
X
Fx = f − Fw = 0
f = Fw
X
= 38.3048◦ .
and
Fy = N − W = 0
N =W.
Applying rotational equilibrium about the
foot of the ladder,
X
ℓ
τ = W cos θ − Fw ℓ sin θ = 0
2
1
ℓ Fw sin θ = ℓ W cos θ
2
Ladder 11
042 10.0 points
A 19.8 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.2 . The angle between the horizontal and
the ladder is 52◦ .
Version 001 – Statics – ramadoss – (171)
21
X
τ◦ = m g d cos θ − Nw L sin θ = 0
4m
µ=0
Nw L sin θ = m g d cos θ
f L sin θ = m g d cos θ
mgd
f=
L tan θ
52◦
d
The ladder may slip when f = fmax = Nw ,
19.8 kg
b
so
µ = 0.2
When the person attempts to climb the
ladder, how far up the ladder will the person
reach before the ladder slips (kaboom)? The
acceleration of gravity is 9.8 m/s2 .
mgd
L tan θ
d ≤ µ L tan θ
≤ (0.2) (4 m) tan 52◦
fmax ≡ µ m g ≥
≤ 1.02395 m .
Correct answer: 1.02395 m.
Explanation:
Let : g = 9.8 m/s2 ,
θ = 52◦ ,
L = 4 m,
m = 19.8 kg , and
µ = 0.2 .
Nw
Leaning Bar 02
043 10.0 points
A uniform bar of mass M and length ℓ is
propped against a very slick vertical wall as
shown. The angle between the wall and the
upper end of the bar is θ. The force of static
friction between the upper end of the bar and
the wall is negligible, but the bar remains at
rest (in equilibrium).
~w
F
θ
~n
θ
f
mg
b
P ivot
Nf
X
Fx = f − Nw = 0
X
Fy = Nf − m g = 0
f = Nw
~fs
If we take the pivot at the point where the
barX
touches
the floor, which expression below
is
~τ , where x is along the floor and y
z
is along the wall?
Mg
1. ℓ
cos θ − Fw sin θ = 0
2
Mg
2. ℓ Fw cos θ −
sin θ = 0 correct
2
3. ℓ (−fs sin θ − n cos θ) = 0
Version 001 – Statics – ramadoss – (171)
Mg
4. ℓ
− Fw = 0
2
Mg
sin θ = 0
5. ℓ Fw cos θ +
2
Mg
cos θ = 0
6. ℓ Fw sin θ +
2
Mg
7. ℓ
+ Fw = 0
2
Correct answer: 382.993 N.
Explanation:
Let :
Explanation:
Taking the pivot at the point where the bar
touches the floor, the torque into the paper is
M g ℓ sin θ
and the torque out of the paper is
2
Fw ℓ cos θ, so
X
Mg
sin θ = 0 .
~τz = ℓ Fw cos θ −
2
Levers in Equilibrium
044 (part 1 of 2) 10.0 points
The levers are in equilibrium, and are considered massless. The supporting string on the
upper bar is attached 1.17 m from the wall,
and 0.305 m from the other end. It acts at
an angle 39 ◦ with the bar. The string between the levers is vertical. The fulcrum on
the lower bar is 1.43 m from the left end, and
3.26 m units from the right end. The spring
is pulled up vertically.
c
T1 c − k x d = 0 ,
(24 N/cm)(7 cm)(3.26 m)
kxd
=
c
1.43 m
= 382.993 N .
045 (part 2 of 2) 10.0 points
What is the tension in the slanted string?
Correct answer: 767.229 N.
Explanation:
Let T be the tension in the slanted string.
b
3.26 m
If the spring of constant 24 N/cm is
stretched 7 cm, what is the tension in the
string between the bars?
T
T
a
b
θ
b
b
1.17 m
0.305 m
1.43 m
so
T1 =
θ
39
d
From rotational equilibrium,
X
X
X
τ=
τCW −
τCCW = 0
7 cm
b
c = 1.43 m and
d = 3.26 m .
Let T1 be the tension in the vertical string
between the bars. The spring exerts a force
Fs = k x straight up at a distance d from the
fulcrum, and the string T1 acts straight up at
a distance c from the fulcrum:
T1
kx
8. ℓ (fs sin θ − n cos θ) = 0
−M g
9. ℓ
cos θ − Fw sin θ = 0
2
Mg
cos θ = 0
10. ℓ Fw sin θ −
2
◦
22
a
From rotational equilibrium on the upper
lever, the tension T1 acts down at a distance
a + b from the fulcrum, and the tension T
acts up at a distance a sin θ from the fulcrum.
Thus
T1 (a + b) − T a sin θ = 0
Version 001 – Statics – ramadoss – (171)
T =
(0.1 kg) (50 m)
1.00512 kg
(0.701 kg)(5.71 cm)
−
1.00512 kg
T1 (a + b)
382.993 N(1.17 m + 0.305 m)
=
a sin θ
(1.17 m) sin 39◦
−
= 767.229 N .
Meter Stick in Equilibrium
046 10.0 points
A 0.1 kg meter stick is supported at its 40 cm
mark by a string attached to the ceiling. A
0.701 kg mass hangs vertically at the 5.71 cm
mark. A second mass is attached at another
mark to keep it horizontal and in rotational
and translational equilibrium.
If the tension in the string attached to the
ceiling is 17.7 N , determine the mark at which
the second mass is attached. The acceleration
of gravity is 9.8 m/s2 .
Correct answer: 62.9199 cm.
23
= 62.9199 cm .
Moving a Block
047 (part 1 of 2) 10.0 points
~ acts on a rectangular block to slide
A force F
the block with constant speed, as seen in the
figure below.
N
247
=
F
39◦
50 cm
1.4 m
N
0.499 m
Explanation:
b
Let :
m = 0.1 kg ,
m1 = 0.701 kg ,
T = 17.7 N ,
ℓ = 40 cm ,
ℓ1 = 5.71 cm , and
g = 9.8 m/s2 .
405 N
x
Find the coefficient of friction.
Correct answer: 0.769181.
Explanation:
From translational equilibrium,
Let :
T = m1 g + m g + m2 g
T
m2 = − m1 − m
g
17.7 N
− 0.701 kg − 0.1 kg
=
9.8 m/s2
= 1.00512 kg .
F = 247 N ,
W = 405 N ,
θ = 39◦ .
B
and
F
θ
From rotational equilibrium, for a fulcrum at
the 0 cm mark,
T ℓ = m2 g ℓ2 + m g (50 m) + m1 g ℓ1
m2 g ℓ2 = T ℓ − m g (50 m) − m1 g ℓ1
Tℓ
m (50 m) m1 ℓ1
ℓ2 =
−
−
m2 g
m2
m2
(17.7 N)(40 cm)
=
(1.00512 kg)(9.8 m/s2 )
A
N
h
b
W
x
Version 001 – Statics – ramadoss – (171)
Vertically,
X
Fy = F sin θ + N − W = 0
N = W − F sin θ
= 405 N − (247 N) sin 39◦
= 249.558 N
and horizontally,
X
Fx = F cos θ − µ N = 0
F cos θ
(247 N) cos 39◦
=
N
249.558 N
= 0.769181 .
µ=
24
Old MacDonald Had a Farm
049 (part 1 of 5) 10.0 points
Old MacDonald had a farm, e i , e i , oh ! And
on that farm he had a gate, e i , e i , oh ! And
a squeak, squeak, here. A squeak, squeak,
there. And a squeak, squeak, everywhere . . .
The gate is 3.37 m wide and 2.8 m tall with
hinges attached to the top and bottom. The
guide wire makes an angle of 14◦ with the top
of the gate and has a tension of 215 N. The
mass of the gate is 39.9 kg.
l
048 (part 2 of 2) 10.0 points
How far from the lower left corner does the
normal force act?
α
Mac Donald's
Ranch
h
Correct answer: 47.8104 cm.
Explanation:
Let :
a = 1.4 m ,
b = 50 cm = 0.5 m , and
h = 0.499 m = 49.9 cm .
The torque about the right-hand edge must
be 0. This reduces the normal force to a single
point on the bottom of the block acting at a
distance x from the lower left corner. Consid~
ering the point of attachment of the force F
to be the pivot point, rotational equilibrium
gives us
X
b
− N (b − x) − µ N h = 0
2
W b − 2N b +2N x− 2µN h = 0
τ =W
2µN h −W b+ 2N b
x=
2N
Wb
+b
= µh−
2N
= 0.769181 (49.9 cm)
(405 N) (50 cm)
−
+ 50 cm
2 (249.558 N)
= 47.8104 cm .
Determine the magnitude of the horizontal
force exerted on the gate by the bottom hinge.
The acceleration of gravity is 9.8 m/s2 .
Correct answer: 172.709 N.
Explanation:
Let : m = 39.9 kg ,
g = 9.8 m/s2 ,
ℓ = 3.37 m ,
T = 215 N ,
α = 14◦ , and
h = 2.8 m .
For equilibrium
X
~ = 0 and
F
T
X
FBy
~τ = 0 .
T
Tx
FBx
FAy
FAx
mg
Ty
Version 001 – Statics – ramadoss – (171)
Applying torques about the top hinge,
T sin α ℓ − m g
FAx
ℓ
+ FAx h = 0
2
ℓ
m g − T sin α ℓ
2
3.37 m
1 h
(39.9 kg) (9.8 m/s2 )
=
2.8 m
2
i
◦
−(215 N) sin 14 (3.37 m)
1
=
h
= 172.709 N ,
with a magnitude of 172.709 N .
050 (part 2 of 5) 10.0 points
Determine the magnitude of the horizontal
force exerted on the gate by the upper hinge.
Correct answer: 35.9049 N.
Explanation:
X
Fx :
25
with a magnitude of 208.614 N .
052 (part 4 of 5) 10.0 points
Determine the magnitude of the total vertical
force exerted on the gate by the two hinges.
Correct answer: 339.007 N.
Explanation:
X
Fy :
FAy + FBy − m g + T sin α = 0
FAy + FBy = m g − T sin α
= (39.9 kg) (9.8 m/s2 )
− (215 N) sin 14◦ ,
with a magnitude of 339.007 N .
053 (part 5 of 5) 10.0 points
What must be the tension in the guy wire so
that the horizontal force exerted by the upper
hinge is zero?
FAx + FBx − T cos α = 0
FBx = T cos α − FAx
= (215 N) cos 14◦ − 172.709 N
= 35.9049 N ,
Correct answer: 186.537 N.
Explanation:
Applying torques about bottom hinge,
−m g
ℓ
+ T ℓ sin α + T h cos α = 0
2
with a magnitude of 35.9049 N .
051 (part 3 of 5) 10.0 points
Determine magnitude of the total horizontal
force exerted on the gate by the two hinges.
ℓ
2
T =
ℓ sin α + h cos α
mg
Correct answer: 208.614 N.
3.37 m
2
=
(3.37 m) sin 14◦ + (2.8 m) cos 14◦
Explanation:
= 186.537 N .
FAx + FBx = FAx + FBx
= 172.709 N + 35.9049 N
= 208.614 N .
Alternate Solution:
FAx + FBx = T cos α
= (215 N) cos 14◦ ,
(39.9 kg) (9.8 m/s2 )
Painter on a Ladder 10 MC
054 10.0 points
A stepladder has negligible weight and is
constructed as shown. The horizontal bar
connecting the two legs of the ladder is pivoted half way up the ladder. A 68 kg painter
stands on the ladder a distance 4 m from the
Version 001 – Statics – ramadoss – (171)
Let N1 be normal force on the right-hand
leg of the ladder, N2 the normal force on the
left-hand leg of the ladder, and Px and Py the
components of the action/reaction pair that
one half of the ladder exerts on the other half.
Consider the forces acting on the each half of
the ladder.
4m
11
m
foot of the ladder. The length of the ladder is
11 m, and the legs of the ladder are 4 m apart
on the floor.
26
Py
Px
θ
4m
Find the normal force on the left-hand leg
of the ladder. The acceleration of gravity
is 9.8 m/s2 . Assume the floor is frictionless
and that the ladder is symmetric and forms
an isosceles triangle. Treat each half of the
ladder separately.
Py
T
N2
Px
θ
T
mg
N1
X
~ =0
F
Applying translational equilibrium
vertically to the right half of the ladder,
Py + N1 − m g = 0
1. 175.7 N
and to the left half
2. 127.2 N
N2 − Py = 0 .
3. 163.6 N
Adding,
N1 + N2 = m g .
4. 157.5 N
X
Applying rotational equilibrium
~τ = 0
to the left half about the top of the ladder,
5. 121.2 N correct
T
6. 151.5 N
L
cos θ − N2 L sin θ = 0
2
and to the right half,
7. 139.3 N
L
cos θ
2
−N1 L sin θ = 0 .
m g (L − ℓ) sin θ + T
8. 133.3 N
9. 145.4 N
Adding,
m g (L − ℓ) sin θ + T L cos θ
− (N1 + N2 ) L sin θ = 0
10. 169.6 N
Explanation:
Let : ℓ = 4 m ,
L = 11 m ,
w = 4 m.
and
T L cos θ = (N1 + N2 ) L sin θ
− m g (L − ℓ) sin θ
= m g L sin θ − m g (L − ℓ) sin θ
= m g ℓ sin θ .
Version 001 – Statics – ramadoss – (171)
From the torque equation,
T L cos θ = 2 N2 L sin θ
m g ℓ sin θ = 2 N2 L sin θ
mgℓ
N2 =
2L
(68 kg) (9.8 m/s2 ) (4 m)
=
2 (11 m)
27
left-hand leg of the ladder, and Px and Py the
components of the action/reaction pair that
one half of the ladder exerts on the other half.
Consider the forces acting on the each half of
the ladder.
Py
Px
θ
= 121.2 N .
Painter on a Ladder 10
055 10.0 points
4m
7. 3
m
A stepladder has negligible weight and is
constructed as shown. The horizontal bar
connecting the two legs of the ladder is pivoted half way up the ladder. A 79 kg painter
stands on the ladder a distance 4 m from the
foot of the ladder. The length of the ladder
is 7.3 m, and the legs of the ladder are 8 m
apart on the floor.
8m
θ
T
mg
N1
X
~ =0
Applying translational equilibrium
F
vertically to the right half of the ladder,
Py + N1 − m g = 0
and to the left half
N2 − Py = 0 .
Adding,
N1 + N2 = m g .
X
Applying rotational equilibrium
~τ = 0
to the left half about the top of the ladder,
L
T
cos θ − N2 L sin θ = 0
2
and to the right half,
L
m g (L − ℓ) sin θ + T
cos θ
2
−N1 L sin θ = 0 .
Adding,
Find the normal force on the left-hand leg
of the ladder. The acceleration of gravity
is 9.8 m/s2 . Assume the floor is frictionless
and that the ladder is symmetric and forms
an isosceles triangle. Treat each half of the
ladder separately.
Correct answer: 212.11 N.
m g (L − ℓ) sin θ + T L cos θ
− (N1 + N2 ) L sin θ = 0
T L cos θ = (N1 + N2 ) L sin θ
− m g (L − ℓ) sin θ
= m g L sin θ − m g (L − ℓ) sin θ
= m g ℓ sin θ .
From the torque equation,
Explanation:
Let : ℓ = 4 m ,
L = 7.3 m ,
w = 8 m.
Py
T
N2
Px
and
Let N1 be normal force on the right-hand
leg of the ladder, N2 the normal force on the
T L cos θ = 2 N2 L sin θ
m g ℓ sin θ = 2 N2 L sin θ
mgℓ
N2 =
2L
(79 kg) (9.8 m/s2 ) (4 m)
=
2 (7.3 m)
= 212.11 N .
Version 001 – Statics – ramadoss – (171)
Perpendicular Ropes
056 10.0 points
Two ropes support a load of 275 kg. The two
ropes are perpendicular to each other, and the
tension in the first rope is 2.05 times that of
the second rope.
Find the tension in the second rope. The
acceleration of gravity is 9.8 m/s2 .
Pole Vaulter
057 (part 1 of 2) 10.0 points
A vaulter holds a 27.1 N pole in equilibrium
by exerting an upward force U with his leading hand, and a downward force D with his
trailing hand, as shown in the figure. Let
ℓ1 = 0.75 m, ℓ2 = 1.6 m, and ℓ3 = 2.35 m.
U
l1
Correct answer: 1181.55 N.
Explanation:
28
l2
l3
C
A
B
Let :
m = 275 kg ,
f = 2.05 , and
g = 9.8 m/s2 .
T1 = f T2 .
The load is not moving, so the net force exerted on it must be zero. Let θ be the angle
between the first rope and the vertical direction; when the net force on the load is zero, its
horizontal component and its vertical component should have a sum of zero. For the
horizontal direction,
T1 sin θ − T2 cos θ = 0
f T2 sin θ − T2 cos θ = 0
f sin θ = cos θ
1
tan θ =
f
1
1
−1
= tan
θ = tan
f
2.05
◦
= 26.0034 .
The vertical components of the net force
should also have a sum of zero:
T1 cos θ + T2 sin θ = M g
f T2 cos θ + T2 sin θ = M g
T2 (f cos θ + sin θ) = M g
−1
Mg
f cos θ + sin θ
(275 kg) (9.8 m/s2 )
=
2.05 cos 26.0034◦ + sin 26.0034◦
T2 =
= 1181.55 N .
W
D
What is the magnitude of the upward force
U?
Correct answer: 84.9133 N.
Explanation:
Let :
ℓ1
ℓ2
ℓ3
W
= 0.75 m ,
= 1.6 m ,
= 2.35 m ,
= 27.1 N .
and
For equilibrium,
X
~ = 0 and
F
X
~τ = 0
To find U , measure the distances and forces
from point A. Then, balancing the torques,
(0.75 m) U = (27.1 N)(0.75 m + 1.6 m)
0.75 m + 1.6 m
U = (27.1 N)
0.75 m
= 84.9133 N .
058 (part 2 of 2) 10.0 points
What is the magnitude of D?
Version 001 – Statics – ramadoss – (171)
29
Correct answer: 57.8133 N.
Let :
Explanation:
From Newton’s second law,
X
Fy = U − D − W = 0
D = U − W = 84.9133 N − 27.1 N
= 57.8133 N .
Plank on a Scaffold MC
059 10.0 points
A uniform plank of length 6.4 m and mass
19 kg rests horizontally on a scaffold, with
1.5 m of the plank hanging over one end of
the scaffold.
L
l
x
How far can a painter of mass 52 kg walk on
the overhanging part of the plank x before it
tips? The acceleration of gravity is 9.8 m/s2 .
L = 6.4 m ,
ℓ = 1.5 m ,
m = 19 kg ,
M = 52 kg , and
g = 9.8 m/s2 .
For equilibrium,
X
~ = 0 and
F
X
~τ = 0 .
Considering the edge of the scaffold to be
the pivot point, then
X
L
−ℓ −M gx = 0
τ0 = m g
2
m L
−ℓ
x=
M 2
19 kg 6.4 m
=
− 1.5 m
52 kg
2
= 0.6212 m .
1. 0.6212 m correct
2. 0.4659 m
3. 0.5590 m
4. 0.6522 m
5. 0.4969 m
Seesaw Tied Down
060 (part 1 of 5) 10.0 points
A child weighing 38 N climbs onto the right
end of a see-saw, little knowing that it is tied
down with a rope at its left end. The see-saw
is 3 m long, weighs 11 N and is tilted at an
angle of 22◦ from the horizontal. The center
of mass of the see-saw is half way along its
length, and lies right above the pivot.
6. 0.7143 m
l
θ
7. 0.5280 m
8. 0.7764 m
9. 0.5901 m
What torque does the weight of the child
exert about the pivot point? Take counterclockwise to be positive.
10. 0.6833 m
Correct answer: −52.8495 N · m.
Explanation:
Explanation:
Version 001 – Statics – ramadoss – (171)
ℓ
τC = MC g
cos θ
2
3m
cos 22◦
= (38 N)
2
= −52.8495 N · m .
061 (part 2 of 5) 10.0 points
What torque does the weight of the see-saw
exert about the pivot point? Take counterclockwise to be positive.
Correct answer: 0 N · m.
Explanation:
This must be zero since the center of mass
of the see-saw is just at the center, namely at
the pivot.
062 (part 3 of 5) 10.0 points
What torque does the rope exert about the
pivot point? Take counter-clockwise to be
positive.
30
Correct answer: 87 N.
Explanation:
Besides the balance of the total torque, the
total force must be zero also since the see-saw
actually doesn’t move. The total downward
forces, including those from the child, the rope
and the see-saw itself can be easily added to
be 87 N, so that the total upward from the
pivot should be 87 N also.
Serway CP 08 25
065 10.0 points
A 7.7 m, 270 kg uniform ladder rests against a
smooth wall. The coefficient of static friction
between the ladder and the ground is 0.65 ,
and the ladder makes a 44.5◦ angle with the
ground.
How far up the ladder can a 1080 kg person
climb before the ladder begins to slip? The
acceleration of gravity is 9.8 m/s2 .
Correct answer: 5.1855 m.
Explanation:
Correct answer: 52.8495 N · m.
Explanation:
Since the see-saw doesn’t rotate, the total
torque should be zero. The torque which
balances the torque of the child is from the
rope. They have opposite directions but the
same magnitude.
063 (part 4 of 5) 10.0 points
What force does the rope exert downwards on
the see-saw? Give a positive answer.
Correct answer: 38 N.
Explanation:
As mentioned above, the torques from the
rope is equal to that from the child, since they
have the same force arm, the force should be
the same also, namely the force from the rope
on the see-saw is 38 N .
064 (part 5 of 5) 10.0 points
What is the total positive force exerted upwards by the pivot?
Let : g = 9.8 m/s2 ,
m = 270 kg ,
α = 44.5◦ ,
µs = 0.65 ,
L = 7.7 m , and
M = 1080 kg .
Let x be the distance up the ladder. Applying
rotational equilibrium with the pivot at the
point of contact with the ground,
X
X
τCW =
τCCW
mgL
cos α + M g x cos α = fmax L sin α
2
m g L cos α + 2 M g x cos α = 2 fmax L sin α
2 M g x cos α = 2 fmax L sin α − m g L cos α
x=
fmax L tan α m L
−
Mg
2M
Version 001 – Statics – ramadoss – (171)
=
(8599.5 N) tan 44.5◦ (7.7 m)
(1080 kg) (9.8 m/s2 )
(270 kg) (7.7 m)
−
2 (1080 kg)
= 5.1855 m .
Sphere on Ramp
066 10.0 points
A sphere of radius R and mass M is held
on a ramp that makes an angle θ with the
horizontal by a string that pulls with force F
directed up the ramp and running over the
top of the sphere.
~F
R
Mg
fs
n
θ
What is the minimum value of µs , the coefficient of static friction between sphere and
ramp, required to hold the sphere at rest?
1.
cos θ
tan θ
2. Zero; no static friction is needed in this
case.
3.
1
2
4. 1.0 precisely
5. tan 2 θ
6. tan θ
2
tan θ
tan θ
8.
correct
2
1
9.
2 tan θ
31
sin θ
tan θ
Explanation:
Let x be parallel to the ramp and y perpendicular to it.
10.
X
Fx = F + fs − M g sin θ = 0 and
X
Fy = n − M g cos θ = 0 .
Taking torques about the point where the
sphere touches the ramp,
X
τz = (M g sin θ) R − 2 F R = 0
F =
M g sin θ
2
to keep the sphere from rotating.
In order for F to be a minimum, fs must be
a maximum, so
fsmax = µs n = µs M g cos θ
and
F min = M g (sin θ − µs cos θ)
M g sin θ
= M g (sin θ − µs cos θ)
2
tan θ
.
µs =
2
Sphere in Smooth Wedge
067 10.0 points
Consider a solid sphere of radius R and mass
m placed in a wedge, where one wall is vertical
and the other wall has an angle θ with respect
to the vertical wall.
7.
A
mg
θ
B
Version 001 – Statics – ramadoss – (171)
Assuming that the walls are smooth, which
expression is appropriate if you consider a
free body diagram for the ball which involves
FA and FB , the forces acting on the contact
points A and B, respectively, and the weight
mg?
32
F
L
1
FA tan θ
2
1
2. m g = FA cos θ
2
1. m g =
h
45
N
mg
b
3. m g = FA cos θ
f
µs = 0.4
b
4. m g = 2 FA tan θ
The ladder will be
5. m g = FA tan θ correct
1. at the critical point of slipping.
6. m g = 2 FA cos θ
2. unstable. correct
1
7. m g = FA sin θ
2
3. stable.
8. m g = 2 FA sin θ
Explanation:
Stability requires
9. m g = FA sin θ
mg
θ
FA
FB
θ
FA
The three forces form a right triangle with
FB as the hypotenuse, so
mg
FA
m g = FA tan θ .
tan θ =
Stability or Not
068 10.0 points
A ladder is leaning against a smooth wall.
There is friction between the ladder and the
floor, which may hold the ladder in place; the
1
ladder is stable when µs ≥
. Otherwise
2 tan θ
the net torque on it is not zero.
Tipping a Block 01
069 (part 1 of 3) 10.0 points
A force acts on a rectangular block as shown.
0.8 m
0N
16
◦
N
38
1m
FB
1
= 0.5 .
2 tan 45◦
Since µs = 0.4 < 0.5 , the ladder is unstable.
In other words, there is a net torque on it.
µs ≥
Explanation:
The free body diagram is
mg
◦
0.45 m
380 N
x
If the block slides with constant speed, find
the position x of the resultant normal force N
to the left of the leading (right-hand) edge.
Correct answer: 33.842 cm.
Explanation:
Version 001 – Statics – ramadoss – (171)
F = 160 N ,
W = 380 N ,
h = 0.45 m ,
H = 1 m,
w = 0.8 m , and
α = 38◦ .
w
F
N
H
X
Fx = F cos α − µN = 0
F cos α
(160 N) cos 38◦
=
N
281.494 N
= 0.447902 .
µ=
Correct answer: 0.535808 m.
X
~τ = 0 .
Locate the origin at the bottom left corner
of the block and let x be the distance between
the resultant normal force and the front of the
block; then
X
Fy = F sin α + N − W = 0
N = W − F sin α
= 380 N − (160 N) sin 38◦
= 281.494 N and
X
1
τ = N (w − x) − W
w
2
+(F sin α) w − (F cos α) h = 0
1
W w + F w sin α − F h cos α
2
x=
N
W w F w sin α − F h cos α
+
=w−
2N
N
(380 N) (0.8 m)
= 0.8 m −
2 (281.494 N)
(160 N) (0.8 m) sin 38◦
+
281.494 N
(160 N) (0.45 m) cos 38◦
−
281.494 N
= 0.33842 m = 33.842 cm .
Nw−
Explanation:
071 (part 3 of 3) 10.0 points
If the force acting on the block becomes 360 N
find the value of h for which the block just
begins to tip over in the forward direction.
x
For equilibrium,
X
~ = 0 and
F
Correct answer: 0.447902.
α
h
W
070 (part 2 of 3) 10.0 points
Find the coefficient of sliding friction.
Explanation:
Locate the origin at the bottom right corner
of the block. Since the block is about to tip,
X
1
τ = W w − F h cos α = 0
2
1
1
Ww
(380 N) (0.8 m)
= 2
h= 2
F cos α
(360 N) cos 38◦
= 0.535808 m .
Tipping a Block 02
072 10.0 points
A string provides a horizontal force which acts
on a 500 N rectangular block at top right-hand
corner as shown.
F
0.8 m
1m
Let :
33
500 N
Version 001 – Statics – ramadoss – (171)
34
If the block slides with constant speed, find
the tension in the string required to start to
tip the block over.
A
Correct answer: 200 N.
42 m
Explanation:
Let :
W = 500 N ,
h = 1 m , and
w = 0.8 m .
W
For equilibrium,
and
X
~τ = 0 .
The block is trying to tip over its lower
right-hand corner (the fulcrum). The string
tension T pulls to the right (clockwise) at a
distance h from that corner and the weight W
acts down (counter-clockwise) at a distance
w
from the corner, so
2
X
τ =W
Correct answer: 1.00343 × 106 N.
Let : L = 69 m ,
M = 1.5 × 105 kg ,
m = 45000 kg , and
x = 42 m .
h
~ =0
F
69 m
Calculate the normal force on the bridge at
point B (the right support). The acceleration
of gravity is 9.8 m/s2 .
Explanation:
F
w
X
B
1
w −F h=0
2
Ww
2
Ww
(500 N) (0.8 m)
F =
=
2h
2 (1 m)
Fh=
= 200 N .
Truck on a Bridge 02
073 (part 1 of 2) 10.0 points
A bridge of length 69 m and mass 1.5 × 105 kg
is supported at each end (points A and B on
the picture below). A truck of mass 45000 kg
is located 42 m from the left support A.
For equilibrium
X
~ =0
F
and
X
~τ = 0.
The truck’s weight m~g , the bridge’s own
~ A on the bridge
weight M ~g , the normal force N
~ B on the
at point A, and the normal force N
bridge at point B are all vertical. Applying
rotational equilibrium about A,
X
L
τz,A = x (m g) + (M g)
2
+ 0 (NA ) − L (NB ) = 0
L
L NB = x m g + M g
2
x
1
NB = m g + M g
L
2
42 m
(45000 kg) (9.8 m/s2 )
=
69 m
1
+ (1.5 × 105 kg) (9.8 m/s2 )
2
= 1.00343 × 106 N .
074 (part 2 of 2) 10.0 points
Calculate the normal force on the bridge at
point A (the left support).
Correct answer: 9.07565 × 105 N.
Explanation:
Version 001 – Statics – ramadoss – (171)
Applying translational equilibrium vertically,
X
Fy = NA + NB − m g − M g = 0
NA = m g + M g − NB
= (45000 kg) (9.8 m/s2 )
+ (1.5 × 105 kg) (9.8 m/s2 )
− 1.00343 × 106 N
= 9.07565 × 105 N .
Winning Without Tipping
075 (part 1 of 2) 10.0 points
Bill Nomoore has entered a race to push the
11.7 kg cart loaded with the 18.5 kg trash can
of height 1.49 m and width 0.44 m as shown
in the figure.
ω
mt h
mc
What is the minimum possible time for the
can to finish the required 12 m distance without tipping? The acceleration of gravity is
9.8 m/s2 . Assume the frictional coefficient
between the cart and the trash can is large
enough so that the trash can does not slip.
Correct answer: 2.87978 s.
Explanation:
35
The trash can will just start to tip
(metastable equilibrium) when
a
w
=
g
h
wg
a=
.
h
From kinematics,
1
1
s = so + vo t + a t2 = a t2
2
s2
r
2s
2 (12 m)
t=
=
a
2.89396 m/s2
= 2.87978 s .
076 (part 2 of 2) 10.0 points
Brakes are put on the cart so that the wheels
are in a locked position.
If there is a frictional coefficient of 0.157
between cart’s wheels and the surface, what
force must Bill maintain to accomplish his
goal in this minimum time?
Correct answer: 133.863 N.
Explanation:
Let : mc = 11.7 kg ,
mt = 18.5 kg ,
µ = 0.157 , and
g = 9.8 m/s2 .
a
Let : s = 12 m and
a = 2.89396 m/s2 .
The acceleration is defined by the dimensions of the trash can:
ω
a
h
g
F
m c+m t
µN
Bill exerts a force F to the right on the
system, friction acts to the left, and his net
acceleration is to the right, so from dynamics
Fnet = mtotal a = F − µ N 6= 0
(mc + mt ) a = F − µ (mc + mt ) g
Version 001 – Statics – ramadoss – (171)
F = (mc + mt ) a + µ (mc + mt ) g
= (mc + mt ) (a + µ g)
= (11.7 kg + 18.5 kg)
× 2.89396 m/s2 + 0.157 (9.8 m/s2 )
h
= 133.863 N .
α
y1
Drop the Drawbridge
077 10.0 points
Sir Lost-a-Lot dons his armor and bursts out
of the castle on his trusty steed Tripper in
his quest to rescue fair damsels from dragons. Unfortunately his valiant aide Doubtless
lowered the drawbridge too far and finally
stopped it 28 ◦ below the horizontal. Losta-Lot screeches to a stop when his and his
steed’s combined mass (420 kg) is 0.84 m
from the end of the bridge, which is 9.1 m
long with a mass of 2800 kg. The lift cable is
attached to the bridge 4.1 m from the hinge
and to the parapets 10 m above the bridge.
α
l
L
x
Find the tension in the cable. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 48.2835 kN.
Explanation:
θ
l
β
α
x1
h + y1
h + ℓ sin α
=
x1
ℓ cos α
−1 h + ℓ sin α
θ = tan
ℓ cos α
◦
−1 10 m + (4.1 m) sin 28
= tan
(4.1 m) cos 28◦
= 73.113◦ .
tan θ =
Taking the sum of the torques about the
hinge,
X
h
36
L
cos α + T ℓ sin(θ − α)
2
− m g (L − x) cos α
= 0.
τ = −M g
2 T ℓ sin(θ − α)
= [M L + 2 m (L − x)] g cos α
g cos α [M L + 2 m (L − x)]
2 ℓ sin(θ − α)
(9.8 m/s2 ) cos 28◦
=
2 (4.1 m) sin(73.113◦ − 28◦ )
h
× (2800 kg) (9.1 m)
T =
+ 2 (420 kg) (9.1 m − 0.84 m)
Let :
L = 9.1 m ,
g = 9.8 m/s2 ,
α = 28◦ ,
x = 0.84 m ,
ℓ = 4.1 m ,
M = 2800 kg ,
h = 10 m , and
m = 420 kg .
×
i
1 kN
1000 N
= 48.2835 kN .
Knot in Equilibrium
078 10.0 points
The system shown in the figure is in equilibrium. A 15 kg mass is on the table. A string
Version 001 – Statics – ramadoss – (171)
attached to the knot and the ceiling makes an
angle of 64◦ with the horizontal. The coefficient of the static friction between the 15 kg
mass and the surface on which it rests is 0.43.
Alternate Solution: Equations 1 and 2
come directly from the free-body diagram for
the knot. Dividing Eq. 2 by Eq. 1,
mmax
µM
= µ M tan θ = (0.43) (15 kg) tan 64◦
= 13.2245 kg .
tan θ =
mmax
◦
64
15 kg
m
What is the largest mass m can have and
still preserve the equilibrium? The acceleration of gravity is 9.8 m/s2 .
Correct answer: 13.2245 kg.
Explanation:
Let : M = 15 kg ,
m = 13.2245 kg ,
θ = 64◦ .
and
For the system to remain in equilibrium,
the net forces on both M and m should be
zero, so the tension in the rope has an upper
bound value Tmax , where
Tmax cos θ = µ M g
(1)
µM g
Tmax =
cos θ
(0.43) (15 kg) (9.8 m/s2 )
=
cos 64◦
= 144.193 N .
For m to remain in equilibrium
Tmax sin θ = mmax g
(2)
(144.193 N) sin 64◦
Tmax sin θ
=
mmax =
g
9.8 m/s2
= 13.2245 kg .
37