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A and B are in a lift. The mass of the lift is 250kg. As it moves upward, the floor of the lift exerts a force of 678N and 452N on A and B, respectively. The tension in the cable of the lift is 3955N.

So I drew a diagram of the lift with the tension, T, of 3955N and the exerting force of the floor (678+452=1130N) acting upwards, and the 250gN force acting downwards. I had to find the acceleration (which is upwards), so therefore did the following calculation:

$F=ma$

$=> (3955+1130)-250g=250a$ (I added those two together because [I thought] they both exert upward forces.

However, I apparently took the wrong approach, and the statement that the "floor of the lift exerts a force on A and B of 1130N" doesn't mean that it's an upward force, but is instead a downward force, making the following the correct approach:

$=> 3955-(250g+1130)=250a$

How can this be? Surely if something (the floor) is exerting a force on something above it (A and B) then the force exerted is an upwards one.

How can the floor exert an downward force on something above it? Or is the question worded incorrectly?

Please note that the reason for the simplicity is that it's only AS Level maths.

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Hi Olly Price. I just added the "homework" tag because your question suites the tag definition :-) –  Waffle's Crazy Peanut Jan 12 '13 at 15:34

1 Answer 1

up vote 3 down vote accepted

Newton's 3rd law is what you are missing: the floor of the lift exerts an upward force on A and B, so the equal, opposite forces that A and B exert on the lift are downward. And you are after all writing Newton's 2nd law for the lift, so it is forces acting on it that you need.

Free body diagrams usually make this much easier to figure out.

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Thanks, that seems right. Yeah I tend to use free body diagrams but I have no idea how to draw them on here, is there any way you could explain to me how? –  Olly Price Jan 12 '13 at 15:43
    
Like, what software and stuff do I use and how do I get the diagrams onto this website? –  Olly Price Jan 13 '13 at 0:32

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