# Elevator Mechanics Problem

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Thread starter 10 months ago
#1
Hi guys, where am I going wrong in this question ?

A miners’ cage of mass 420 kg contains 3 miners of total mass 280 kg. The cage is lowered from rest by a cable. For the first 10 seconds the cage accelerates uniformly and descends a distance of 75 m. What is the force in the cable during the first 10 seconds?

Here are my workings :

Thanks !!
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Thread starter 10 months ago
#2

0
10 months ago
#3
I don't see why you've introduced S; it looks like you start by treating the lift as a single object of 700kg, in which case you dont need to (and shouldn't) consider the miners separately.

(I think).
1
10 months ago
#4
(Original post by seals2001)
...
Acceleration is correct.

You seem to have an extra force "S".

It looks as if you're mixing up treating the whole mass (cage plus miners) as one thing, and treating the miners and cage separately.

In your first diagram you have 700kg, so you've treated the miners and cage together, which is what I would do. But there is no extra force "S". That is an internal force to the overall mass - the cage exerts an upward force on the miners and the miners exert an equal and opposite force on the cage; resultant is 0.

So, you can remove "S" from your diagram. And then calculate just using "T", to get the desired result.
1
Thread starter 10 months ago
#5
(Original post by ghostwalker)
Acceleration is correct.

You seem to have an extra force "S".

It looks as if you're mixing up treating the whole mass (cage plus miners) as one thing, and treating the miners and cage separately.

In your first diagram you have 700kg, so you've treated the miners and cage together, which is what I would do. But there is no extra force "S". That is an internal force to the overall mass - the cage exerts an upward force on the miners and the miners exert an equal and opposite force on the cage; resultant is 0.

So, you can remove "S" from your diagram. And then calculate just using "T", to get the desired result.
S is the normal force contact force exerted on the floor by the miners. When looking at the forces acting on the lift in isolation to the forces acting on the miners , you have to consider the weight of the lift , the weight of the miners and the normal contact force exerted on the lift by the miners ("S"). Why can we just ignore "S" ?

Thanks.
0
10 months ago
#6
(Original post by seals2001)
S is the normal force contact force exerted on the floor by the miners. When looking at the forces acting on the lift in isolation to the forces acting on the miners , you have to consider the weight of the lift , the weight of the miners and the normal contact force exerted on the lift by the miners ("S"). Why can we just ignore "S" ?

Thanks.
If you want to treat the cage in isolation, then the weight of the cage is 420g, not 700g, and you'd have the tension in the cable, T, and the normal contact force "S", as you had.

Edit: With that correction, your working should produce the desired result - not checked it in detail. BUT, it involves more work than treating the cage and miners as one thing.
Last edited by ghostwalker; 10 months ago
0
10 months ago
#7
(Original post by seals2001)
Hi guys, where am I going wrong in this question ?

A miners’ cage of mass 420 kg contains 3 miners of total mass 280 kg. The cage is lowered from rest by a cable. For the first 10 seconds the cage accelerates uniformly and descends a distance of 75 m. What is the force in the cable during the first 10 seconds?

Here are my workings :

Thanks !!
(Original post by DFranklin)
I don't see why you've introduced S; it looks like you start by treating the lift as a single object of 700kg, in which case you dont need to (and shouldn't) consider the miners separately.

(I think).
Yeah exactly that. In your second equation, just count the whole object as being one mass, which is the lift and the miners together, then it should be correct. The S, I believe, should be written as T. S is usually written only for displacement.
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Thread starter 10 months ago
#8
(Original post by ghostwalker)
If you want to treat the cage in isolation, then the weight of the lift is 420g, not 700g, and you'd have the tension in the cable, T, and the normal contact force "S", as you had.

Edit: With that correction, your working should produce the desired result - not checked it in detail. BUT, it involves more work than treating the cage and miners as one thing.
I see, that makes a lot more sense now. Thank you for this.
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