Question

A meter stick is pivoted at the 0.50 m line. A 3.0 kg object is hung from the 0.15 m line. Where should a 5.0 kg object be hung to achieve equilibrium (the meter stick oriented horizontal and motionless)?

a. 0.56-m line
b. 0.71-m line
c. 0.24-m line
d. 0.06-m line
e. a 5.0-kg object cannot be placed anywhere on the meter stick to result in equilibrium in this system.

Answer 1

To achieve equilibrium on a meter stick pivoted at the 0.50 m line with a 3.0 kg mass at the 0.15 m line, a 5.0 kg mass should be placed at the 0.71 m line.

option b is the correct

Step-by-step explanation:

The question asks where to place a 5.0 kg object on a meter stick pivoted at the 0.50 m line to achieve equilibrium, given a 3.0 kg object is hung from the 0.15 m line. To solve this, we must use the principle of torque balance. Torque (τ) is the product of the force applied and the distance from the pivot point (lever arm), and is given by the formula τ = r × F, where r is the distance from the pivot point and F is the force due to the object's weight (mass × gravity).

For equilibrium, the sum of the torques around the pivot must equal zero; hence, the torque caused by the 3.0 kg object must be equal and opposite to the torque caused by the 5.0 kg object. The 3.0 kg object exerts torque of 3.0 kg × 9.81 m/s2 × (0.50 m - 0.15 m). To find the place (lever arm) for the 5.0 kg object, we set up the equation: 3.0 × 9.81 × (0.50 - 0.15) = 5.0 × 9.81 × (0.50 - x), where x is the position on the meter stick where the 5.0 kg object should be hung. Solving for x, we find that x = 0.71 m, which corresponds to option (b).