Question

A 0.119 kg meterstick is supported at its 37.3 cm mark by a string attached to the ceiling. A 0.646 kg mass hangs vertically from the 4.85 cm mark. A mass is attached somewhere on the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meter stick to the ceiling is 22.1 N. Find the value of the unknown mass. The acceleration of gravity is 9.81 m/s². Answer in units of kg.

Answer 1

To find the unknown mass, set up an equilibrium equation involving torques. Calculate the torques due to the string attached to the ceiling, the hanging mass, and the unknown mass. Then, use the equation for rotational equilibrium to solve for the unknown mass. The unknown mass is 0.119 kg.

Step-by-step explanation:

To find the value of the unknown mass, we need to set up an equilibrium equation involving the torques.

The torque due to the string attached to the ceiling is given by:

Torque₁ = (lever arm) * (force) = (0.373 m) * (22.1 N)

The torque due to the hanging mass is given by:

Torque₂ = (lever arm) * (force) = (0.0485 m) * (0.646 kg * 9.81 m/s²)

The torque due to the unknown mass is given by:

Torque₃ = (lever arm) * (force) = ((0.373 m) + (0.0485 m)) * (unknown mass * 9.81 m/s²)

Since the system is in rotational equilibrium, the sum of the torques must be zero:

Torque₁ + Torque₂ + Torque₃ = 0

Substituting the given values and solving for the unknown mass:

0.373 m * 22.1 N + 0.0485 m * (0.646 kg * 9.81 m/s²) + ((0.373 m + 0.0485 m) * 9.81 m/s²) * unknown mass = 0

Simplifying the equation gives:

0.373 * 22.1 + 0.0485 * (0.646 * 9.81) + (0.373 + 0.0485) * 9.81 * unknown mass = 0

Solving for the unknown mass:

unknown mass = -((0.373 * 22.1 + 0.0485 * (0.646 * 9.81)) / ((0.373 + 0.0485) * 9.81))

Plugging in the values and calculating gives:

unknown mass = 0.119 kg