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A 55 g mouse runs out to the end of the 17 cm-long minute hand of a grandfather clock when the clock reads 10 minutes past the hour. What torque does the mouse's weight exert about the rotation axis of the clock hand?

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meerkat18
The equation for torque is written below:

τ = rFsinθ
where
τ is torque 
r is the length of minute hand
F is the force exerted by the mouse
θ is the angle

Every number interval in the clock has an angle of 30°. Because 10 minutes past the hour coincides with the number 10, θ = 30°*2 = 60°. For the force, it is simply the weight of the mouse which is
F = mg = (55g)*(1 kg/1000g)*(9.81 m/s²) = 0.53955 N

Substituting the rest of the values,

τ = (17cm)*(1 m/100 cm)*(0.53955 N)*(sin60°)
τ = 0.079 N·m

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