Problem 8B

NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 8B
ROTATIONAL EQUILIBRIUM
PROBLEM
SOLUTION
1. DEFINE
In 1960, a polar bear with a mass of 9.00 × 10 2 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 × 10 2 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
Given:
Unknown: F n,1 =
Diagram:
m b = mass of bridge = 2.50 × 10 2 kg
m p = mass of polar bear = 9.00 × 10 2 kg
l = length of bridge = 12.0 m
g = 9.81 m/s 2
12.0 m
F n,1
mb
m p
F n,2
2. PLAN
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.
F y = F n,1 + F n,2 − m b g − m p 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 F n,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.
t net =−(m b g)d b − (m p g)d p + F n,2 d 2 = 0
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88
Holt Physics Problem Workbook

NAME ______________________________________ DATE _______________ CLASS ____________________
The lever arm for the bridge’s weight (d b ) 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.
d b = ⎯ 1 2 ⎯ l , d p = ⎯ 2 3 ⎯ l , d 2 = l
The torque equation thus takes the following form:
m b gl 2m p gl
t net =−⎯ − ⎯ +F n,2l = 0
2 3
Rearrange the equation(s) to isolate the unknowns:
m b gl 2m p gl
F n,2 = + =
⎯ m b
⎯ + ⎯ 2 m
⎯ ⎯ p
2 3
g
2l
⎯ 3l
F n,1 = m b g + m p g − F n,2
3. CALCULATE
Substitute the values into the equation(s) and solve:
F n,2 =
⎯ 2.50 × 10 2 kg
⎯ + ⎯ (2)(9.00 × 10 2 kg)
⎯
2
3
(9.81 m/s 2 )
F n,2 = (125 kg + 6.00 × 10 2 kg)(9.81 m/s 2 )
F n,2 = (725 kg)(9.81 m/s 2 )
F n,2 = 7.11 × 10 3 N
F n,1 = (2.50 × 10 2 kg)(9.81 m/s 2 ) + (9.00 × 10 2 kg)(9.81 m/s 2 )
− 7.11 × 10 3 N
F n,1 = 2.45 × 10 3 N + 8.83 × 10 3 N − 7.11 × 10 3 N
F n,1 =
4.17 × 10 3 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. EVALULATE
ADDITIONAL PRACTICE
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 × 10 2 kg + 9.00 × 10 2 kg)(9.81 m/s 2 )
11.28 kN = 11.28 × 10 3 N
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 × 10 5 N •m that
acted on the cactus. If the cactus could withstand a torque of only
Problem 8B 89

NAME ______________________________________ DATE _______________ CLASS ____________________
1.2 × 10 5 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 × 10 3 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.
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
6. Goliath, a giant Galápagos tortoise living in Florida, has a mass of 3.6 ×
10 2 kg. Suppose Goliath walks along a heavy board above a swimming
pool. The board has a mass of 6.0 × 10 2 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 × 10 3 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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
90
Holt Physics Problem Workbook

Additional Practice 8B
Givens
1. t 1 = 2.00 × 10 5 N •m
t 2 = 1.20 × 10 5 N •m
h = 24 m
Solutions
Apply the second condition of equilibrium, choosing the base of the cactus as the
pivot point.
t net = t 1 − t 2 − Fd(sin q) = 0
Fd(sin q) = t 1 − t 2
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).
F min = ⎯ t 1 − t 2 2.00 × 10 5 N •m − 1.20 × 10 5 N •m
⎯ = ⎯⎯⎯⎯
h
24 m
8.0 × 10 4 N•m
F min = ⎯⎯ = 3.3 × 10 3 N applied to the top of the cactus
24 m
2. m 1 = 40.0 kg
m 2 = 5.4 kg
d 1 = 70.0 cm
d 2 = 100.0 cm − 70.0 cm
= 30.0 cm
g = 9.81 m/s 2
Apply the first condition of equilibrium.
F n − m 1 g − m 2 g − F applied = 0
F n = m 1 g + m 2 g + F applied = (40.0 kg)(9.81 m/s 2 ) + (5.4 kg)(9.81 m/s 2 ) + F applied
F n = 392 N + 53 N + F applied = 455 N + F applied
Apply the second condition of equilibrium, using the fulcrum as the location for the
axis of rotation.
F applied d 2 + m 2 gd 2 − m 1 gd 1 = 0
F applied = ⎯ m 1gd 1 − m 2 gd 2 (40.0 kg)(9.81 m/s 2 )(0.700 m) − (5.4 kg)(9.81 m/s 2 )(0.300 m)
⎯ = ⎯⎯⎯⎯⎯⎯⎯
d2
0.300 m
275 N•m − 16 N•m 259 N•m
F applied = ⎯⎯ = ⎯ 0.300 m 0.300 m
F applied = 863 N
Substitute the value for F applied into the first-condition equation to solve for F n .
F n = 455 N + 863 N = 1318 N
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. m = 134 kg
d 1 = 2.00 m
d 2 = 7.00 m − 2.00 m = 5.00 m
q = 60.0°
g = 9.81 m/s 2
Apply the first condition of equilibrium in the x and y directions.
F x = F applied (cos q) − F f = 0
F y = F n − F applied (sin q) − mg = 0
To solve for F applied , apply the second condition of equilibrium, using the fulcrum as
the pivot point.
F applied (sin q)d 2 − mgd 1 = 0
mgd F applied = ⎯ 1 (134 kg)(9.81 m/s 2 )(2.00 m)
⎯ = ⎯⎯⎯
d2 (sin
q) (5.00 m)(sin 60.0°)
F applied = 607 N
Substitute the value for F applied into the first-condition equations to solve for F n and
F f .
F n = F applied (sin q) + mg = (607 N)(sin 60.0°) + (134 kg)(9.81 m/s 2 )
F n = 526 N + 1.31 × 10 3 N = 1.84 × 10 3 N
F f = F applied (cos q) = (607 N)(cos 60.0°) =
304 N
Section Two—Problem Workbook Solutions II Ch. 8–3

Givens
Solutions
II
4. m = 8.8 × 10 3 kg
d 1 = 3.0 m
d 2 = 15 m − 3.0 m = 12 m
q = 20.0°
g = 9.81 m/s 2
6. m 1 = 3.6 × 10 2 kg
m 2 = 6.0 × 10 2 kg
l = 15 m
l 1 = 5.0 m
g = 9.81 m/s 2
Apply the first condition of equilibrium in the x and y directions.
F x = F fulcrum,x − F (sin q) = 0
F y F fulcrum,y − F (cos q) − mg = 0
To solve for F, apply the second condition of equilibrium •, using the fulcrum as the
pivot point.
Fd 2 − mg d 1 (cos q) = 0
F = ⎯ mg d 1 ( cos q) (8.8 × 10 3 kg)(9.81 m/s 2 )(3.0 m)(cos 20.0°)
⎯ = ⎯⎯⎯⎯⎯
d
12 m
2
F = 2.0 × 10 4 N
Substitute the value for F into the first-condition equations to solve for the components
of F fulcrum .
F fulcrum,x = F (sin q) = (2.0 × 10 4 N)(sin 20.0°)
F fulcrum,x = 6.8 × 10 3 N
F fulcrum,y = F (cos q) + mg = (2.0 × 10 4 N)(cos 20.0°) + (8.8 × 10 4 kg)(9.81 m/s 2 )
F fulcrum,y = 1.9 × 10 4 N + 8.6 × 10 5 N = 8.8 × 10 5 N
5. m 1 = 64 kg
Apply the first condition of equilibrium to solve for F applied .
m 2 = 27 kg
F n − m 1 g − m 2 g − F applied = 0
d 1 = d 2 = ⎯ 3.0 0m ⎯ = 1.50 m
F applied = F n − m 1 g − m 2 g = 1.50 × 10 3 N − (64 kg)(9.81 m/s 2 ) − (27 kg)(9.81 m/s 2 )
2
F n = 1.50 × 10 3 F applied = 1.50 × 10 3 N − 6.3 × 10 2 N − 2.6 × 10 3 N = 6.1 × 10 2 N
N
g = 9.81 m/s 2 To solve for the lever arm for F applied , apply the second condition of equilibrium,
using the fulcrum as the pivot point.
F applied d + m 2 g d 2 − m 1 g d 1 = 0
d = ⎯ m 1 gd1
− m2 gd
⎯ 2
=
Fapplied
9.4 × 10 2 N •m − 4.0 × 10 2 N •m
d = ⎯⎯⎯⎯ =
6.1 × 10 2 N
Apply the second condition of equilibrium, using the pool’s edge as the pivot point.
Assume the total mass of the board is concentrated at its center.
m 1 g d − m 2 g
− l 1 = 0
m 2 g
⎯ l ⎯
2
l
⎯
− l 1
m 2
⎯l ⎯ − l 1
2
d = ⎯ = ⎯
m1 g
m1
2
(64 kg)(9.81 m/s 2 )(1.50 m) − (27 kg)(9.81 m/s 2 )(1.50 m)
⎯⎯⎯⎯⎯⎯
6.1 × 10 2 N
5.4 × 10 2 N •m
⎯⎯
6.1 × 10 2 N
d = 0.89 m from the fulcrum, on the same side as the less massive seal
(6.0 × 10 2 kg)
(6.0 × 10 2 kg)(7.5 m − 5.0 m)
d =
⎯ 15 m
⎯ − 5.0 m
2
⎯⎯⎯
= ⎯⎯⎯ =
3.6 × 10 2 kg
3.6 × 10 2 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(6.0 × 10 2 kg)(2.5 m)
⎯⎯⎯
3.6 × 10 2 kg
d = 4.2 m from the pool’s edge
II Ch. 8–4
Holt Physics Solution Manual

Givens
Solutions
7. m = 449 kg
Apply the first condition of equilibrium to solve for F 2 .
l = 5.0 m
F 1 + F 2 − mg = 0
F 1 = 2.70 × 10 3 N
F 2 = mg − F 1
g = 9.81 m/s 2 F 2 = (449 kg)(9.81 m/s 2 ) − 2.70 × 10 3 N = 4.40 × 10 3 N − 2.70 × 10 3 N = 1.70 × 10 3 N
Apply the second condition of equilibrium, using the left end of the platform as the
pivot point.
F 2 l − m g d = 0
F (1.70 × 10 3 2 l
N)(5.0 m)
d = ⎯ = ⎯⎯⎯
mg (449 kg)(9.81 m/s 2 )
d =
1.9 m from the platform’s left end
8. m 1 = 414 kg
Apply the first condition of equilibrium to solve for F 2 .
l = 5.00 m
F 1 + F 2 − m 1 g − m 2 g = 0
m 2 = 40.0 kg
F 2 = m 1 g + m 2 g − F 1 = (m 1 + m 2 ) g − F 1
F 1 = 50.0 N
F 2 = (414 kg + 40.0 kg)(9.81 m/s 2 ) − 50.0 N = (454 kg)(9.81 m/s 2 ) − 50.0 N = 4.45 ×
g = 9.81 m/s 2 10 3 N − 50.0 N
F 2 = 4.40 × 10 3 N
Apply the second condition of equilibrium, using the supported end (F 1 ) of the stick
as the rotation axis.
F 2 d − m 1 g
− m 2 g l = 0
l
⎯2
⎯m 1
⎯ + m 2
2 g l
d = ⎯⎯ = F2
⎯414 kg
⎯ + 40.0 kg
2
(9.81 m/s 2 )(5.0 m)
⎯⎯⎯⎯
4.40 × 10 3 N
(207 kg + 40.0 kg)(9.81 m/s 2 )(5.00 m)
d = ⎯⎯⎯⎯ =
4.40 × 10 3 N
(247 kg)(9.81 m/s 2 )(5.00 m)
⎯⎯⎯
4.40 × 10 3 N
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice 8C
1. R = 50.0 m
M = 1.20 × 10 6 kg
t = 1.0 × 10 9 N •m
2. M = 22 kg
R = 0.36 m
t = 5.7 N •m
d = 2.75 m from the supported end
a = ⎯ t I ⎯ = ⎯ t 1.0 × 10 9 N •m
⎯
MR 2 = ⎯⎯⎯
(1.20 × 10 6 kg)(50.0 2 )
a = 0.33 rad/s 2
a = ⎯ t I ⎯ = ⎯ t 5.7 N•m
⎯
MR 2 = ⎯⎯
(22 kg)(0.36 m)
2
a = 2.0 rad/s 2
Section Two—Problem Workbook Solutions II Ch. 8–5