A clock has a second hand which rubs against the inside of the glass cover. If the frictional force between the glass cover and the tip of the hand is 0.0020 N and the length of the hand is 8 cm, what minimum torque must be supplied to the hand by the clock mechanism if the hand is not to stop?

Question

A clock has a second hand which rubs against the inside of the glass cover. If the frictional force between the glass cover and the tip of the hand is 0.0020 N and the length of the hand is 8 cm, what minimum torque must be supplied to the hand by the clock mechanism if the hand is not to stop?

Help please! Equation then plug the numbers into that equation

Answer 1

To determine the minimum torque required to prevent the clock's second hand from stopping, we can use the equation for torque:

Torque = Force x Distance

In this case, the force is the frictional force between the glass cover and the tip of the hand, given as 0.0020 N. The distance is the length of the hand, which is 8 cm (or 0.08 m) in this case.

Plugging in the values into the equation:
Torque = 0.0020 N x 0.08 m

Now, let's calculate the torque:
Torque = 0.0020 N x 0.08 m
Torque = 0.00016 Nm

Therefore, the minimum torque the clock mechanism must supply to the hand is 0.00016 Nm, in order to prevent the hand from stopping due to the frictional force between the glass cover and the tip of the hand.