Question

A wheel is free to rotate about a horizontal axis through \( O \). A force of 200 N is applied at a point \( P_2 \) from the center \( O \). \( OP \) makes an angle of 55° with the \( x \)-axis and the force is in the plane of the wheel making an angle of 25° with the horizontal axis. What is the torque?

• A. 4 N m
• B. 3.2 N m
• C. 2 N m
• D. 3.4 N m

Hint:

To calculate torque, use the formula \( \tau = F \times r \times \sin(\theta) \), where \( F \) is the applied force, \( r \) is the distance from the axis of rotation, and \( \theta \) is the angle between the force and the lever arm.

Answer 1

The Correct Option is C

The torque \( \tau \) is given by the formula: \[ \tau = F \times r \times \sin(\theta) \] where: - \( F = 200 \, \text{N} \) is the applied force, - \( r = 0.02 \, \text{m} \) is the distance from the axis of rotation (since \( 2 \, \text{cm} = 0.02 \, \text{m} \)), - \( \theta = 25^\circ \) is the angle between the force and the horizontal axis (because the force is in the plane of the wheel and the angle given is with respect to the horizontal axis).
Now, substitute the values into the torque formula: \[ \tau = 200 \times 0.02 \times \sin(25^\circ) \] Using \( \sin(25^\circ) \approx 0.4226 \), we get: \[ \tau = 200 \times 0.02 \times 0.4226 = 2 \, \text{N m} \] Thus, the torque is \( 2 \, \text{N m} \).