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Holt Physics
Problem 8B
ROTATIONAL EQUILIBRIUM
PROBLEM
In 1960, a polar bear with a mass of 9.00 × 102 kg was discovered in
Alaska. Suppose this bear crosses a 12.0 m long horizontal bridge that
spans a gully. The bridge consists of a wide board that has a uniform
mass of 2.50 × 102 kg and whose ends are loosely set on either side of the
gully. When the bear is two-thirds of the way across the bridge, what is
the normal force acting on the board at the end farthest from the bear?
SOLUTION
Given:
mb = mass of bridge = 2.50 × 102 kg
mp = mass of polar bear = 9.00 × 102 kg
l = length of bridge = 12.0 m
g = 9.81 m/s2
Unknown:
Fn,1 = ?
Diagram:
12.0 m
Fn,1
mb
2. PLAN
mp
Fn,2
Choose the equation(s) or situation:
Apply the first condition of equilibrium: The unknowns in this problem are the
normal forces exerted upward by the ground on either end of the board. The known
quantities are the weights of the bridge and the polar bear. All of the forces are in the
vertical (y) direction.
Fy = Fn,1 + Fn,2 − mb g − mp g = 0
Because there are two unknowns and only one equation, the solution cannot be
obtained from the first condition of equilibrium alone.
Choose a point for calculating net torque: Choose the end of the bridge farthest
from the bear as the pivot point. The torque produced by Fn,1 will be zero.
Apply the second condition of equilibrium: The torques produced by the bridge’s
and polar bear’s weights are clockwise and therefore negative. The normal force on
the end of the bridge opposite the axis of rotation exerts a counterclockwise (positive)
torque.
tnet = −(mb g)db − (mp g)dp + Fn,2 d2 = 0
88
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. DEFINE
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The lever arm for the bridge’s weight (db ) is the distance from the bridge’s
center of mass to the pivot point, or half the bridge’s length. The lever arm for the
polar bear is two-thirds the bridge’s length. The lever arm for the normal force
farthest from the pivot equals the entire length of the bridge.
1
db = 2l ,
2
dp = 3l ,
d2 = l
The torque equation thus takes the following form:
mb gl 2mp gl
tnet = − − + Fn,2 l = 0
2
3
Rearrange the equation(s) to isolate the unknowns:
mb gl 2mp gl m 2 m
Fn,2 = + = b + p g
2l
3l
2
3
Fn,1 = mb g + mp g − Fn,2
3. CALCULATE
Substitute the values into the equation(s) and solve:
2.50 × 102 kg (2)(9.00 × 102 kg)
Fn,2 = + (9.81 m/s2)
2
3
Fn,2 = (125 kg + 6.00 × 102 kg)(9.81 m/s2)
Fn,2 = (725 kg)(9.81 m/s2)
Fn,2 = 7.11 × 103 N
Fn,1 = (2.50 × 102 kg)(9.81 m/s2) + (9.00 × 102 kg)(9.81 m/s2)
− 7.11 × 103 N
Fn,1 = 2.45 × 103 N + 8.83 × 103 N − 7.11 × 103 N
Fn,1 = 4.17 × 103 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. EVALULATE
The sum of the upward normal forces exerted on the ends of the bridge must
equal the weight of the polar bear and the bridge. (The individual normal forces
change as the polar bear moves across the bridge.)
(4.17 kN + 7.11 kN) = (2.50 × 102 kg + 9.00 × 102 kg)(9.81 m/s2)
11.28 kN = 11.28 × 103 N
ADDITIONAL PRACTICE
1. The heaviest sea sponge ever collected had a mass of 40.0 kg, but after
drying out, its mass decreased to 5.4 kg. Suppose two loads equal to the
wet and dry masses of this giant sponge hang from the opposite ends of a
horizontal meterstick of negligible mass and that a fulcrum is placed
70.0 cm from the larger of the two masses. How much extra force must
be applied to the end of the meterstick with the smaller mass in order to
provide equilibrium?
2. A Saguaro cactus with a height of 24 m and an estimated age of 150 years
was discovered in 1978 in Arizona. Unfortunately, a storm toppled it in
1986. Suppose the storm produced a torque of 2.00 × 105 N • m that
acted on the cactus. If the cactus could withstand a torque of only
Problem 8B
89
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1.2 × 105 N • m, what minimum force could have been applied to the cactus keep it standing? At what point and in what direction should this
force have been applied? Assume that the cactus itself was very strong
and that the roots were just pulled out of the ground.
3. In 1994, John Evans set a record for brick balancing by holding a load of
bricks with a mass of 134 kg on his head for 10 s. Another, less extreme,
method of balancing this load would be to use a lever. Suppose a board
with a length of 7.00 m is placed on a fulcrum and the bricks are set on
one end of the board at a distance of 2.00 m from the fulcrum. If a force
is applied at a right angle to the other end of the board and the force has
a direction that is 60.0° below the horizontal and away from the bricks,
how great must this force be to keep the load in equilibrium? Assume the
board has negligible mass.
4. In 1994, a vanilla ice lollipop with a mass of 8.8 × 103 kg was made in
Poland. Suppose this ice lollipop was placed on the end of a lever 15 m in
length. A fulcrum was placed 3.0 m from the lollipop so that the lever
made an angle of 20.0° with the ground. If the force was applied perpendicular to the lever, what was the smallest magnitude this force could
have and still lift the lollipop? Neglect the mass of the lever.
6. Goliath, a giant Galápagos tortoise living in Florida, has a mass of 3.6 ×
102 kg. Suppose Goliath walks along a heavy board above a swimming
pool. The board has a mass of 6.0 × 102 kg and a length of 15 m, and it is
placed horizontally on the edge of the pool so that only 5.0 m of it extends over the water. How far out along this 5.0 m extension of the board
can Goliath walk before he falls into the pool?
7. The largest pumpkin ever grown had a mass of 449 kg. Suppose this
pumpkin was placed on a platform that was supported by two bases
5.0 m apart. If the left base exerted a normal force of 2.70 × 103 N on the
platform, how far must the pumpkin have been from the platform’s left
edge? The platform had negligible mass.
8. In 1991, a giant stick of Brighton rock (a type of rock candy) was made
in England. The candy had a mass of 414 kg and a length of 5.00 m.
Imagine that the candy was balanced horizontally on a fulcrum. A child
with a mass of 40.0 kg sat on one end of the stick. How far must the fulcrum have been from the child in order to maintain equilibrium?
90
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. The Galápagos fur seals are very small. An average adult male has a mass
of 64 kg, and a female has a mass of only 27 kg. Suppose one average
adult male seal and one average adult female seal sit on opposite ends of
a light board that has a length of 3.0 m. How far from the male seal
should the board be pivoted in order for equilibrium to be maintained?
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Additional Practice 8B
Givens
Solutions
1. t1 = 2.00 × 105 N • m
t2 = 1.20 × 105 N • m
h = 24 m
Apply the second condition of equilibrium, choosing the base of the cactus as the
pivot point.
tnet = t1 − t2 − Fd(sin q) = 0
Fd(sin q) = t1 − t2
For F to be minimum, d and sin q must be maximum. This occurs when the force is
perpendicular to the cactus (q = 90°) and is applied to the top of the cactus (d = h =
24 m).
2.00 × 105 N • m − 1.20 × 105 N • m
t1 − t2
=
Fmin =
24 m
h
8.0 × 104 N•m
Fmin = = 3.3 × 103 N applied to the top of the cactus
24 m
2. m1 = 40.0 kg
Apply the first condition of equilibrium.
m2 = 5.4 kg
Fn − m1g − m2g − Fapplied = 0
d1 = 70.0 cm
Fn = m1g + m2g + Fapplied = (40.0 kg)(9.81 m/s2) + (5.4 kg)(9.81 m/s2) + Fapplied
d2 = 100.0 cm − 70.0 cm
= 30.0 cm
Fn = 392 N + 53 N + Fapplied = 455 N + Fapplied
g = 9.81 m/s2
II
Apply the second condition of equilibrium, using the fulcrum as the location for the
axis of rotation.
Fappliedd2 + m2gd2 − m1gd1 = 0
2
2
m1gd1 − m2gd2 (40.0 kg)(9.81 m/s )(0.700 m) − (5.4 kg)(9.81 m/s )(0.300 m)
=
Fapplied =
0.300 m
d2
275 N•m − 16 N•m
259 N•m
Fapplied = =
0.300 m
0.300 m
Fapplied = 863 N
Substitute the value for Fapplied into the first-condition equation to solve for Fn.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fn = 455 N + 863 N = 1318 N
3. m = 134 kg
Apply the first condition of equilibrium in the x and y directions.
d1 = 2.00 m
Fx = Fapplied (cos q) − Ff = 0
d2 = 7.00 m − 2.00 m = 5.00 m
Fy = Fn − Fapplied (sin q) − mg = 0
q = 60.0°
To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum as
the pivot point.
2
g = 9.81 m/s
Fapplied (sin q)d2 − mgd1 = 0
(134 kg)(9.81 m/s2)(2.00 m)
mgd1
Fapplied =
=
(5.00 m)(sin 60.0°)
d2(sin q)
Fapplied = 607 N
Substitute the value for Fapplied into the first-condition equations to solve for Fn and
Ff .
Fn = Fapplied(sin q) + mg = (607 N)(sin 60.0°) + (134 kg)(9.81 m/s2)
Fn = 526 N + 1.31 × 103 N = 1.84 × 103 N
Ff = Fapplied(cos q) = (607 N)(cos 60.0°) = 304 N
Section Two—Problem Workbook Solutions
II Ch. 8–3
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Givens
Solutions
4. m = 8.8 × 103 kg
Apply the first condition of equilibrium in the x and y directions.
d1 = 3.0 m
Fx = Ffulcrum,x − F (sin q) = 0
d2 = 15 m − 3.0 m = 12 m
Fy Ffulcrum,y − F (cos q) − mg = 0
q = 20.0°
To solve for F, apply the second condition of equilibrium •, using the fulcrum as the
pivot point.
g = 9.81 m/s2
Fd2 − mg d1 (cos q) = 0
(8.8 × 103 kg)(9.81 m/s2)(3.0 m)(cos 20.0°)
mg d1 (cos q)
=
F=
12 m
d2
F = 2.0 × 104 N
Substitute the value for F into the first-condition equations to solve for the components of Ffulcrum.
Ffulcrum,x = F (sin q) = (2.0 × 104 N)(sin 20.0°)
Ffulcrum,x = 6.8 × 103 N
Ffulcrum,y = F (cos q) + mg = (2.0 × 104 N)(cos 20.0°) + (8.8 × 104 kg)(9.81 m/s2)
Ffulcrum,y = 1.9 × 104 N + 8.6 × 105 N = 8.8 × 105 N
II
5. m1 = 64 kg
Apply the first condition of equilibrium to solve for Fapplied.
m2 = 27 kg
Fn − m1 g − m2 g − Fapplied = 0
3.00 m
d1 = d2 = = 1.50 m
2
Fapplied = Fn − m1 g − m2 g = 1.50 × 103 N − (64 kg)(9.81 m/s2) − (27 kg)(9.81 m/s2)
Fn = 1.50 × 103 N
g = 9.81 m/s2
Fapplied = 1.50 × 103 N − 6.3 × 102 N − 2.6 × 103 N = 6.1 × 102 N
To solve for the lever arm for Fapplied, apply the second condition of equilibrium,
using the fulcrum as the pivot point.
kg)(9.81 m/s2)(1.50 m) − (27 kg)(9.81 m/s2)(1.50 m)
m1 g d1 − m2 g d2 (64
=
d=
6.1 × 102 N
Fapplied
9.4 × 102 N • m − 4.0 × 102 N • m
5.4 × 102 N • m
d =
=
2
6.1 × 10 N
6.1 × 102 N
d = 0.89 m from the fulcrum, on the same side as the less massive seal
6. m1 = 3.6 × 102 kg
Apply the second condition of equilibrium, using the pool’s edge as the pivot point.
m2 = 6.0 × 102 kg
Assume the total mass of the board is concentrated at its center.
l = 15 m
l 1 = 5.0 m
m1 g d − m2 g
g = 9.81 m/s2
2 − l = 0
l
l
m g − l
2 m 2 − l 1
1
2
d=
l
m1 g
1
2
=
m1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fapplied d + m2 g d2 − m1 g d1 = 0
15m
(6.0 × 102 kg) − 5.0 m
(6.0 × 102 kg)(7.5 m − 5.0 m)
(6.0 × 102 kg)(2.5 m)
2
d =
=
=
2
3.6 × 10 kg
3.6 × 102 kg
3.6 × 102 kg
d = 4.2 m from the pool’s edge
II Ch. 8–4
Holt Physics Solution Manual
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Givens
Solutions
7. m = 449 kg
l
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − mg = 0
= 5.0 m
3
F1 = 2.70 × 10 N
2
g = 9.81 m/s
F2 = mg − F1
F2 = (449 kg)(9.81 m/s2) − 2.70 × 103 N = 4.40 × 103 N − 2.70 × 103 N = 1.70 × 103 N
Apply the second condition of equilibrium, using the left end of the platform as the
pivot point.
F2 l − m g d = 0
F2 l (1.70 × 103 N)(5.0 m)
d = =
m g (449 kg)(9.81 m/s2)
d = 1.9 m from the platform’s left end
8. m1 = 414 kg
l
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − m1 g − m2 g = 0
= 5.00 m
m2 = 40.0 kg
F2 = m1 g + m2 g − F1 = (m1 + m2) g − F1
F1 = 50.0 N
F2 = (414 kg + 40.0 kg)(9.81 m/s2) − 50.0 N = (454 kg)(9.81 m/s2) − 50.0 N = 4.45 ×
103 N − 50.0 N
2
g = 9.81 m/s
II
F2 = 4.40 × 103 N
Apply the second condition of equilibrium, using the supported end (F1) of the stick
as the rotation axis.
2 − m g l = 0
m
414 kg
+ m g l
+ 40.0 kg (9.81 m/s )(5.0 m)
2
2
d =
=
F2 d − m1 g
l
1
2
2
2
4.40 × 103 N
F2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(207 kg + 40.0 kg)(9.81 m/s2)(5.00 m)
(247 kg)(9.81 m/s2)(5.00 m)
d =
=
3
4.40 × 10 N
4.40 × 103 N
d = 2.75 m from the supported end
Additional Practice 8C
M = 1.20 × 106 kg
1.0 × 109 N • m
t
t
a = = 2 =
(1.20 × 106 kg)(50.02)
I MR
t = 1.0 × 109 N • m
a = 0.33 rad/s2
1. R = 50.0 m
2. M = 22 kg
R = 0.36 m
5.7 N•m
t
t
a = = 2 = 2
(22 kg)(0.36 m)
I MR
t = 5.7 N • m
a = 2.0 rad/s2
Section Two—Problem Workbook Solutions
II Ch. 8–5
