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MechEngJordan
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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 TonneFor 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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