Question

The mechanism shown in the figure (Figure 1)is used to raise a crate of supplies from a ship's hold. The crate has total mass 43 kg . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.34 m and a moment of inertia I = 2.9 kg⋅m2 about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.12 m , the cylinder turns, and the crate is raised.

What magnitude of the force F⃗  applied tangentially to the rotating crank is required to raise the crate with an acceleration of 1.40 m/s2? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

Answer 1

the torque required to raise the load, τ = T*r ; where T is the tension

the torque required to accelerate the drum, τ = Iα

To raise the load with the given acceleration,

T = m(g + a) = 43kg(9.8 + 1.4)m/s² = 481.6 N

For a non-slipping rope, α = a/r, so the combined applied torque

τ = T*r + Ia/r = 481.6N * 0.34m + (2.9kg·m² * 1.4m/s² ) / 0.34m

τ = 175.69 N·m

To find the applied force, use the crank's radius and τ = F*r:

175.69 N·m = F * 0.12m

F = 1464.1 N

so F = 1464 N