TORQUE

A beam that is hinged near one end can be lowered to stop traffic at a rail-road crossing or border checkpoint. Consider a beam with a mass of12.0 kg that is partially balanced by a 20.0 kg counterweight. The counter-weight is located 0.750 m from the beam’s fulcrum. A downward force of1.60 � 102 N applied over the counterweight causes the beam to move up-ward. If the net torque on the beam is 29.0 N•m when the beam makes anangle of 25.0° with respect to the ground, how long is the beam’s longersection? Assume that the portion of the beam between the counterweightand fulcrum has no mass.

S O L U T I O NGiven: mb = 12.0 kg

mc = 20.0 kg

dc = 0.750 m

Fapplied = 1.60 × 102 N

tnet = 29.0 N•m

q = 90.0° − 25.0° = 65.0°

g = 9.81 m/s2

Unknown: l = ?

Diagram:

Choose the equation(s) or situation: Apply the definition of torque to each

force and add up the individual torques.

t = F d (sin q)

tnet = ta + tb + tc

where ta = counterclockwise torque produced by applied force = Fapplied dc (sin q)

tb = clockwise torque produced by weight of beam

= −mb g� � (sin q)

tc = counterclockwise torque produced by counterweight

= mc g dc (sin q)

tnet = Fapplied dc (sin q) − mb g� � (sin q) + mc g dc (sin q)

Note that the clockwise torque is negative, while the counterclockwise torques are

positive.

l2

l2

mb g

dc

θ

�

1. DEFINE

2. PLAN

Rearrange the equation(s) to isolate the unknown(s):

mb g� � = (Fapplied + mc g) dc − �stinne

qt�

l =

Substitute the values into the equation(s) and solve:

l =

l =

l =

l =

l =

l =

For a constant applied force, the net torque is greatest when q is 90.0° and de-

creases as the beam rises. Therefore, the beam rises fastest initially.

3.99 m

(2)(235 N•m)(12.0 kg)(9.81 m/s2)

(2)(2.67 × 102 N•m − 32.0 N•m]

(12.00 kg)(9.81 m/s2)

(2) [356 N)(0.750 m) − 32.0 N•m]

(12.0 kg)(9.81 m/s2)

(2) [(1.60 × 102 N + 196 N)(0.750 m) − 32.0 N•m]

(12.0 kg)(9.81 m/s2)

(2)([1.60 × 102 N + (20.0 kg)(9.81 m/s2)] (0.750 m) − �2s

9

i

.

n

0

6

N

5.

•

0

m

°�

(12.0 kg)(9.81 m/s2)

2�(Fapplied + mc g)dc − �stinne

qt��

mb g

l2

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The nests built by the mallee fowl of Australia can have masses as large as

3.00 × 105 kg. Suppose a nest with this mass is being lifted by a crane.

The boom of the crane makes an angle of 45.0° with the ground. If the

axis of rotation is the lower end of the boom at point A, the torque pro-

duced by the nest has a magnitude of 3.20 × 107 N•m. Treat the boom’s

mass as negligible, and calculate the length of the boom.

2. The pterosaur was the most massive flying dinosaur. The average mass

for a pterosaur has been estimated from skeletons to have been between

80.0 and 120.0 kg. The wingspan of a pterosaur was greater than 10.0 m.

Suppose two pterosaurs with masses of 80.0 kg and 120.0 kg sat on the

middle and the far end, respectively, of a light horizontal tree branch.

The pterosaurs produced a net counterclockwise torque of 9.4 kN •m

about the end of the branch that was attached to the tree. What was the

length of the branch?

Problem 8A 87

3. A meterstick of negligible mass is fixed horizontally at its 100.0 cm mark.

Imagine this meterstick used as a display for some fruits and vegetables

with record-breaking masses. A lemon with a mass of 3.9 kg hangs from

the 70.0 cm mark, and a cucumber with a mass of 9.1 kg hangs from the

x cm mark. What is the value of x if the net torque acting on the meter-

stick is 56.0 N•m in the counterclockwise direction?

4. In 1943, there was a gorilla named N’gagi at the San Diego Zoo. Suppose

N’gagi were to hang from a bar. If N’gagi produced a torque of –1.3 ×104 N•m about point A, what was his weight? Assume the bar has negli-

gible mass.

5. The first—and, in terms of the number of passengers it could carry, the

largest—Ferris wheel ever constructed had a diameter of 76 m and held

36 cars, each carrying 60 passengers. Suppose the magnitude of the

torque, produced by a ferris wheel car and acting about the center of the

wheel, is –1.45 × 106 N•m. What is the car’s weight?

6. In 1897, a pair of huge elephant tusks were obtained in Kenya. One tusk

had a mass of 102 kg, and the other tusk’s mass was 109 kg. Suppose both

tusks hang from a light horizontal bar with a length of 3.00 m. The first

tusk is placed 0.80 m away from the end of the bar, and the second, more

massive tusk is placed 1.80 m away from the end. What is the net torque

produced by the tusks if the axis of rotation is at the center of the bar?

Neglect the bar’s mass.

7. A catapult, a device used to hurl heavy objects such as large stones, con-

sists of a long wooden beam that is mounted so that one end of it pivots

freely in a vertical arc. The other end of the beam consists of a large hol-

lowed bowl in which projectiles are placed. Suppose a catapult provides

an angular acceleration of 50.0 rad/s2 to a 5.00 × 102 kg boulder. This can

be achieved if the net torque acting on the catapult beam, which is 5.00 m

long, is 6.25 × 105 N•m.

a. If the catapult is pulled back so that the beam makes an angle of

10.0° with the horizontal, what is the magnitude of the torque pro-

duced by the 5.00 × 102 kg boulder?

b. If the force that accelerates the beam and boulder acts perpendicu-

larly on the beam 4.00 m from the pivot, how large must that force

be to produce a net torque of 6.25 × 105 N•m?

II

1. m = 3.00 × 105 kg

q = 90.0° − 45.0° = 45.0°

t = 3.20 × 107 N •m

g = 9.81 m/s2

t = Fd(sin q) = mgl (sin q)

l = mg(s

tin q)

l =

l = 15.4 m

3.20 × 107 N •m(3.00 × 105 kg)(9.81 m/s2)(sin 45.0°)

Additional Practice 8A

Givens Solutions

2. tnet = 9.4 kN •m

m1 = 80.0 kg

m2 = 120.0 kg

g = 9.81 m/s2

tnet = t1 + t2 = F1d1(sin q1) + F2d2(sin q2)

q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1g � � + m2gl

l =

l =9.4 × 103 N •m

+ (120.0 kg)(9.81 m/s2)

=

l = = 6.0 m9.4 × 103 N •m1.57 × 103 N

9.4 × 103 N•m392 N + 1.18 × 103 N(80.0 kg)(9.81 m/s2)

2

tnetm

21g + m2g

l2

3. tnet = 56.0 N •m

m1 = 3.9 kg

m2 = 9.1 kg

d1 = 1.000 m − 0.700 m =0.300 m

g = 9.81 m/s2

tnet = t1 + t2 = F1d1(sin q1) + F2d2(sin q2)

q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1gd1 + m2g(1.000 m − x)

x = 1.000 m − tnet

m

−

2

m

g1gd1

x = 1.000 m −

x = = = 1.000 m − 0.50 m

x = 0.50 m = 5.0 × 101 cm

45 N•m(9.1 kg)(9.81 m/s2)

56.0 N•m − 11 N•m(9.1 kg)(9.81 m/s2)

56.0 N •m − (3.9 kg)(9.81 m/s2)(0.300 m)

(9.1 kg)(9.81 m/s2)

Rotational Equilibrium and Dynamics

II

4. t = −1.3 × 104 N •m

l = 6.0 m

d = 1.0 m

q = 90.0° − 30.0° = 60.0°

t = Fd(sin q) = −Fg(l − d)(sin q)

Fg = = =

Fg = 3.0 × 103 N

1.3 × 104 N •m(5.0 m)(sin 60.0°)

−(−1.3 × 104 N •m)(6.0 m − 1.0 m)(sin 60.0°)

−t(l − d)(sin q)

Givens Solutions

5. R = 76

2

m = 38 m

q = 60.0°

t = −1.45 × 106 N •m

t = Fd(sin q) = −FgR(sin q)

Fg = R(s

−in

tq)

=

Fg = 4.4 × 104 N

−(−1.45 × 106 N •m)

(38 m)(sin 60.0°)

6. m1 = 102 kg

m2 = 109 kg

l = 3.00 m

l 1 = 0.80 m

l 2 = 1.80 m

g = 9.81 m/s2

tnet = t1 + t2 = F1d1(sin q1) + F2d2(sin q2)

q1 = q2 = 90°, so

tnet = F1d1 + F2d2 = m1g� − l 1� + m2g� − l 2�tnet = (102 kg)(9.81 m/s2)�3.0

2

0 m − 0.80 m� + (109 kg)(9.81 m/s2)�3.0

2

0 m − 1.80 m�

tnet = (102 kg)(9.81 m/s2)(1.50 m − 0.80 m) + (109 kg)(9.81 m/s2)(1.50 m − 1.80 m)

tnet = (102 kg)(9.81 m/s2)(0.70 m) + (109 kg)(9.81 m/s2)(−0.30 m)

tnet = 7.0 × 102 N •m − 3.2 × 102 N •m

tnet = 3.8 × 102 N •m

l2

l2

7. m = 5.00 × 102 kg

d1 = 5.00 m

t = 6.25 × 105 N •m

g = 9.81 m/s2

q1 = 90.0° − 10.0° = 80.0°

d2 = 4.00 m

q2 = 90°

a. t � = Fd(sin q) = mgd1(sin q1)

t � = (5.00 × 102 kg)(9.81 m/s2)(5.00 m)(sin 80.0°)

t � =

b. tnet = Fd2(sin q2) − t � = Fd2(sin q2) − mgd1(sin q1)

F =

F = =

F = 1.62 × 105 N

6.49 × 105 N •m

4.00 m

6.25 × 105 N •m + 2.42 × 104 N •m

4.00 m (sin 90°)

tnet + mgd1(sin q1)

d2(sin q2)

2.42 × 104 N •m