As Simple Hoist System Problem - Given Answer Wrong?

  • #1

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!
 
Last edited:

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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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?
The sources of torque on the drum are the shaft (positive) and cable tension (negative). It is a poorly worded question, but you are asked to find the torque applied by the shaft to the drum. You have found the torque of the motor that is turning the shaft (τmotor - τf = τshaft).

Other than that, your approach is perfectly correct.

AM
 
  • #3
The sources of torque on the drum are the shaft (positive) and cable tension (negative). It is a poorly worded question, but you are asked to find the torque applied by the shaft to the drum. You have found the torque of the motor that is turning the shaft (τmotor - τf = τshaft).

Other than that, your approach is perfectly correct.

AM

Ah! I see that now. I need to be a bit more careful when reading the question!

Thanks a lot.
 

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