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.

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

b) What are the simplifications and assumptions?

c) Define the axis system.

d) What is the angular accleration of the winch?

Explanation / Answer

A) torque exerted = F cos 30 ×L

=40 × 0.866×0.750 = 26 Nm

B) The assumption is mass of handle is very small compared to mass of winch

C) the axis of rotation is passing through center of the winch and into the page.

d) I alpha = torque exerted by handle - torque resisted by bearings

I alpha = 26 - 14 = 12 Nm

Angular acceleration=alpha = 12/24 = 0.5 /s2

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