Question 5

Question

A bucket ?lled with water has a mass of 53 kg and is hanging from a rope that is wound around a 0.046 m radius stationary cylinder. The acceleration of gravity is 9.81 m/s2 . If the cylinder does not rotate and the
bucket hangs straight down, what is the magnitude of the torque the bucket produces around the center of the cylinder?

Answer 1

To find the magnitude of the torque produced by the hanging bucket of water around the center of the stationary cylinder, you can use the following formula for torque:

Torque(τ)=Force(F)×Lever Arm(r)

In this case, the force is the weight of the bucket of water, and the lever arm is the radius of the cylinder.

1.Calculate the force (weight) of the bucket:

Weight (force) = mass (m) × acceleration due to gravity (g)

Weight = 53 kg × 9.81 m/s² = 520.93 N

2.Now, calculate the torque:

Torque = Weight × Lever Arm

Torque = 520.93 N × 0.046 m = 23.95 N·m (Newton-meters)

So, the magnitude of the torque produced by the hanging bucket of water around the center of the stationary cylinder is approximately 23.95 N·m.