Aclock has a second hand whose tip rubs against the inside of the glass cover. if the frictional force between the glass cover and the tip of the second hand is 0.0020 and the length of the hand is 8.0 cm, what is the minimum torque that must be applied to the second hand if the clock is not to be stopped? what is the direction of the torque from the friction?

1.60×10^{-4} N-m

opposite to each other

Explanation:

Given the frictional force F = 0.0020 N

length of hand clock L= 8.0 cm

therefore the minimum torque required to the second hand if the clock is not to be stopped, \tau= F×L

= 0.0020×8×10^{-2}

=1.60×10^{-4} N-m

And the direction of the torque is always opposite to the direction of frictional force.

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