# Physics Torque Question?

Can someone give me a question about physics torque that can be answered with yes or no that could be tested? A simple torque question dealing with a fulcrum and a meter stick.

### 2 Answers

- Steve4PhysicsLv 76 years agoFavorite Answer
A meter rule is balanced in the middle (50cm)

A 20g mass is placed at 10cm.

A 40g mass is placed at 65cm.

Does the heavier side go down?

The answer is no.

The 20g mass is 40cm from the middle.

The 40g mass is 15cm from the middle.

20x40 is bigger than 40x15, so the lighter mass has the bigger moment (torque) and goes down.

If you haven't got masses, use 1 coin and 2 coins (all 3 coins of the same type).

- SpacemanLv 76 years ago
T1 = torque applied to the input end of the lever

T2 = torque applied to the output end of the lever

F1 = force on the input end of the lever

F2 = force on the output end of the lever

d1 = distance from the fulcrum to the input end of the lever

d2 = distance from the fulcrum to the output end of the lever

MA = mechanical advantage

The mechanical advantage of a lever can be determined by considering the balance of moments

or torque, T, about the fulcrum:

T1 = (F1 x d1) = (F2 x d2) = T2

The mechanical advantage of the lever is the ratio of output force to input force:

MA = F2/F1 = d1/d2

A load of 4,613 N is to be placed on the right (output) end of a lever that is 36.75 m long. The

lever is to have a mechanical advantage of 23.5. The distance from the output load to the fulcrum

of the lever is 1.5. (a) Determine the magnitude of the necessary input force. (b) Calculate the

distance from the fulcrum to the point of application of the input force. (c) What is the applied

input torque?

(a) MA = F2/F1

23.5 = 4,613 N/F1

23.5F1 = 4,613 N

F1 = 4,613 N/23.5 = 196.3 N ANSWER

(b) MA = d1/d2

23.5 = d1/1.5

d1 = 1.5(23.5) = 35.25 m ANSWER

(c) T1 = d1 x F1

T1 = 35.25 m x 196.3 N

T1 = 6,919.575 N-m ≈ 6,920 N-m ANSWER