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Physics/Torques, a lift, and a dose of confusion

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Question
I received a question about torques and lifts and I gave it a crack but I don't think my logic is quite sound.

---Question---

A lift operates in a vertical mine-shaft to a depth of 600 m below ground level. It has a
cage of mass 300 kg, which can carry a maximum payload of 500 kg. The cage is
suspended on a steel wire cable 20 mm in diameter. The cable has an effective steel
cross-sectional area of 180 mm2
and a mass of 1.64 kg/m. The lift is operated by a
winder mounted on a framework 10 m above ground level.
A differential band brake is to be mounted on the same shaft as the winding drum, to
slow the cage's descent and to stop it at the 600 m level. The rate of descent is to be
restricted to 4.5 m/s, and the cage is to be brought to rest at the 600 m level in a distance
of 6.0 m. The effective radius of the cable on the winding drum is 1.5 m. The mass of
the rotating parts in the last stage of descent is estimated to be around 1100 kg, with an
effective radius of gyration of 1.2 m. Hydraulic actuation of the brake lever-arm is
proposed.
You are required to estimate the required braking torque for the lift.  The torque is to be calculated for the steady descent and stopping conditions.

---Attempt at calculations---

Assumptions: cable is taut, friction is ignored.

I had assumed forces going upwards is positive and downwards is negative.

Total mass = mass of lift(300kg) + mass of max payload(500kg) + mass of cable (1000.4kg as 1.64kg/m * 610m) = 1800.4kg

For steady descent:

Tension in cable(T) - total mass of cage and all(m)*gravity(9.81m/s^2) = 0 (I think if speed is constant net force is 0); T-mg=0

T = 1800.4 * 9.81 = 17661.924N

Torque(Y) = T * radius of drum(r) = 17611.924N * 1.5 = 26492.886Nm

For deceleration:

v^2-u^2 =2*a*s where is final velocity, u is initial velocity, s is stopping distance (6m)

( (0)^2 - (4.5)^2 )/ (2 * 6) = a = -1.6875m/s^2 (This is the acceleration need to stop the elevator in given conditions)

T - mg = ma (I think the acceleration  of the elevator is it slows is the tangential acceleration)(ma is positive as force is applied upwards on elevator to slow it down?)

T = mg + ma = 1800.4 * 9.81 + 1800.4 * (-1.6875) = 14623.749

Y1 = T * r = 14623.749 * 1.5 = 21935.623Nm

I think I also need to account for the torque required to accelerate the drum.

Y2 = moment of inertia(I) * angular acceleration(&)

I = mass of drum(M) * radius of gyration(k) = 1100kg * (1.2)^2 = 1584kgm

& = tangential acceleration(a) / radius of drum(r) = -1.6875/1.5 =  -1.125 rad/s^2

Y2 = I& = 1584 * -1.125 = -1782Nm

Total torque = Y1 + Y2 = 21935.623 + -1782 = 20153.623Nm ???

It seems strange that the torque required to keep the lift in a steady state is greater that to decelerate it. I think my mistake may involve a misapplication of positive and negative signs in my equations. And I don't even know if I have the right concept of solving the problem at all?  I would much appreciate it if you could offer any help.

Answer
Hello Peter,

I appreciate that you sent a complete description of what you have done. I compliment you on your analysis of what needs to be done. I believe you're correct in your suspicion that polarity is the key to the reason the torque required to stop the cage is less than that required to maintain constant velocity. I will make comments on your work from the top down in your write-up.

I agree with your value of the torque required to maintain constant velocity when the cage is at the bottom. But, the cable is the largest contributor to the hanging mass at the bottom of the shaft and when higher in the shaft, consider that cable that is still wrapped around the drum is not hanging mass. We don't know the dimensions of the cage, so we must assume that 10 m of cable has unwound from the drum when the miners are getting in the cage. That is consistent with your use of 610 m to calculate the mass of the hanging cable when it has all unwound from the drum. When at the top of the shaft, the torque required to maintain 4.5 m/s would be considerably less than 26492.886Nm. I suggest that you calculate the torque required when at the top and then indicate that it would increase to the higher value in a linear fashion. Incidentally, the constant velocity phase of the descent ends when 6 m from the bottom, so the cable unwound at that point is 604 m.

I found 2 places in your work on the deceleration phase of the descent that polarity needs to be rethought.  
Regarding your lines:
v^2-u^2 =2*a*s where is final velocity, u is initial velocity, s is stopping distance (6m)
( (0)^2 - (4.5)^2 )/ (2 * 6) = a = -1.6875m/s^2
Notice that the value of s should be negative. The cage is moving downward. That will give you a positive acceleration - as it should be when decreasing a negative velocity. And that will give you a more positive value for T.

Regarding your line:
I = mass of drum(M) * radius of gyration(k) = 1100kg * (1.2)^2 = 1584kgm
radius of gyration(k) should be to the power 2 and the units of I should be kg.m^2
(You may consider this quibbling.) This obviously was a simple oversight since you did square the radius when doing the math, but some teachers/professors are fussy about those things.

Regarding your line:
& = tangential acceleration(a) / radius of drum(r) = -1.6875/1.5 =  -1.125 rad/s^2
When calculating torque, consider the rotation direction. The torque in this case is the same direction as would be used to raise the cage, so you should just use the magnitude of the linear acceleration and then assign the polarity according to your understanding of the affect of the rotation direction.

I hope this helps,
Steve

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

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