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The mechanism shown in the figure (Figure 1) is used to raise a crate of supplie

ID: 1548019 • Letter: T

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 39 kg . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.25 m and a moment of inertia I = 2.7 kgm2about 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.

Part A

What magnitude of the force F  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.)

Express your answer using two significant figures.

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Figure 1 of 1

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 39 kg . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.25 m and a moment of inertia I = 2.7 kgm2about 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.

Part A

What magnitude of the force F  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.)

Express your answer using two significant figures.

F =   kN  

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

12m (0.12 m to

Explanation / Answer

This problem is all about the two expressions for torque:
i) the torque required to raise the load, = T*r where T is the tension
ii) the torque required to accelerate the drum, = I

To raise the load with the given acceleration,
T = m(g + a) = 39kg*(9.8 + 1.4)m/s² = 436.8 N

For a non-slipping rope, = a/r, so the combined applied torque
= T*r + Ia/r = 436.8 N * 0.25m + 2.7kg·m² * 1.4m/s² / 0.25m
= 124.32 N·m

To find the applied force, use the crank's radius and = F*r:
124.32 N·m = F * 0.12m
F = 1036 N

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