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## Homework Statement

A colliery lift cage can be considered as a simple hoist system. An investigation found that when raised from the bottom of the mineshaft, the cage accelerates uniformly for 10 seconds, travels for 70 seconds at constant speed of 3 m/s and just before reaching the pit head, decelerates uniformly in 4 seconds.

Frictional torque at the drum shaft is constant at 1500 Nm.

Mass of drum = 1.5 Tonne

Dia. = 3m

k = 1.4m

Mass of Cage = 0.5 Tonne

For the period during which the cage accelerates, determine:

(i) The tension in the cable with a labelled free body diagram (assume

that the mass of the cable is negligible.)

(ii) The torque required at the drum shaft.

## Homework Equations

a = Δv/t

∑F = ma

∑τ = Iα

I = mk^2

## The Attempt at a Solution

Acceleration calculated to be 0.3m/s^2.

Tension in cable:

∑F = ma

∴ T = m(a + g)

= 5055N (correct according to answer sheet)

∑τ = Iα

I = mk^2 = 2940kgm^2

α = a/r = 0.2m/s^2

∴τ = Iα + τƒ + Tr, where τƒ = frictional torque of 1500Nm and Tr = 7582.5Nm

= 9671Nm

The answer given is 8171Nm -- which is simply the above calculation without consideration to the constant frictional torque of 1500Nm.

Am I missing something or is the answer given simply incorrect?

Thanks for any help!

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