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?
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
