Assignment 12:
Equilibrium and Torque
Name ………………….
1. Define the word torque. Turning effect of a force
2. Write the formula and units for torque.   F  d
( Nm)
3. What is meant by equilibrium? No acceleration, balanced force and torque
4. What are the two conditions for something to be in equilibrium?
F  0
  0
5. Can you tell where the baby’s centre of mass is?
Above the person’s hand
6. Where is the center of mass of the bottle and
chain?
Above the support
7. Why must you bend forward when carrying a heavy load on your back?
When you have a heavy load on your back, your centre of gravity is behind your feet. You therefore
need to lean forward to bring your centre of gravity over your feet, to prevent toppling over
backwards.
8. Jenny is holding the torque feeler. Explain what happens to the torque she exerts if:
(a) The mass was doubled.
Force is doubled so torque is doubled   F  d
( Nm)
(b) The mass stays the same but the distance from the pivot doubles.
torque is doubled   F  d
( Nm)
(c) She hangs half the mass at twice the distance from the pivot
Torque is the same   1 F  2d
2
9. What direction is the torque due to the weight?
clockwise
10. What direction is the torque she exerts?
anticlockwise
Why is it easier to carry the same amount of water in two buckets, one in each hand, than in a
single bucket?
Two buckets are easier because the total torque is zero. One bucket creates a torque and you have
to lean against it, (Or carry a single bucket on your head!)
11. Explain clearly why this balances.
Clockwise torque = (Fx1 + Fx3)
anticlockwise torque = (2Fx2 )
torques are equal and opposite…. Total torque = 0
12. Jess is using the winch on her parents’ boat. Would a
longer handle make it easier or harder to pull the rope
tight. Explain
Easier, to produce the same torque, a smaller force is
needed if the distance to the pivot is greater
13. Nobody at the playground wants to play with the
obnoxious boy, so he makes a seesaw as shown so he
can play by himself. Explain how this works.
The weight of the boy is counterbalanced by the weight of
the board, which can be considered to be concentrated at
its centre of mass on the opposite side of the fulcrum. He is in balance when his weight multiplied by
his distance from the pivot is equal to the weight of the board multiplied by the distance between
the pivot and the midpoint (COM) of the board.
The boy is 50 kg. He is 1.0m from the pivot. The board is 5.0m long.
Calculate the mass of the board.
50 x 1 = m x d
50 x 1 = m x 1.5
m = 33kg
nb we are using mass instead of force because “g” cancels out.
14. Draw arrows showing the wind force and the weight force on the
windsurfer. Explain how these keep her in equilibrium? (Her feet is the pivot)
Wind applies a clockwise torque.
Her weight applies an anticlockwise torque is equal (but opp)
that is How does the size of the wind force compare with the size of her weight force
Total torque = 0
15. A meter stick (negligible mass) has 2 weights hanging from each end – 1 kg on the left and 3 kg on
the right. Where is the centre of mass of this system? How does this relate to torque?
Centre of mass is where the fulcrum would balance both masses i.e. torques are balanced. If distance
from 1 kg mass to fulcrum is x then balancing torques gives
1
x
100-x
3
1 x X = 3(100-x)
x = 300 – 3x
4x = 300 x = 75 cm
i.e. centre of mass is below the 75 cm mark.
Clockwise torque about c. of m = antilockwise torque about c. of m
16. The board is 6.0 m long. The pivot
is in the middle. Assume the mass
is at the end of the board.
Calculate the boy’s mass.
1.5 m
20 kg
M x 1.5 = 20 x 3
M= 40 kg
17. A 10 000 kg truck is on a 10 m long bridge. It is 7 m from A. The mass of the bridge is 20 000 kg.
(a) Draw all the force acting on the bridge.
(b) Calculate the support force at B (hint consider A to be the pivot)
A
B
A
B
 bridge   truck   sup portB
200, 000  5  100, 000  7  Fbridge 10
Fbridge  170, 000 N
18. A 10 000 kg truck is stalled one
quarter of the way across a 10 m
long bridge. The mass of the bridge
is 20 000 kg.
Calculate the extra support force supplied at the supports on both ends of the bridge due to
the truck
consider torques about A with the truck:
truck is a distance 2.5m from A
Support B is 10m from A
Clockwise torque = Anticlockwise torque
100 000 x 2.5 = FB x 10
FB = 25 000 N
i.e. the additional reaction force at B = 25000 N (quarter weight of the car)
A must support the rest of the weight of the truck so the additional reaction force at A is 75 000 N
19. Lydia is 50 kg. She abseiling as shown. The diagram show
the forces on her.
Calculate the tension in the rope.
cos 30 
1.0 m
o
30
0.90 m
cos 30 
Fup
T
Fup  T cos 30
 clock   anticlock
(50  9.8)  0.90  (T cos 30) 1.0
T  520 N
Fup
T