I am pushing with a contact force of 40 N on the handle of the winch shown in th

Question

I am pushing with a contact force of 40 N on the handle of the winch shown in the diagram.

The handle of the winch is L = 0.750 m long, the radius of the winch R = 0.300 m. The moment of inertia (rotational inertia) of the winch is 24 kg m2 and friction in the bearings exerts a torque of 14.0 N m, which resists the winch turning.

What is the torque I exert with my push (2 s.f.)?

[When you use this answer in the later parts, make sure to keep 3 significant figures during your working, so you can give your answers to 2. This applies all the way through.]

Explanation / Answer

Torque = r X F

Torque = rF sin theta

here r = L = 0.75

F = 40

theta = 60 degree

therefore

T = 40x0.75x sin60

T = 25.98 Nm

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