Elevator Mechanics Problem

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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.

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 !!

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).

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.