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Mechanics question, answered on forum, but need explanation watch

1. Hello,

The question is:

2) Albert and Bella are both standing in a lift. The mass of the lift is 250kg. As the lift moves upward with constant acceleration, the floor of the lift exerts forces of magnitude 678N and 452N respectively on Albert and Bella. The tension in the cable which is pulling the lift upwards is 3955N.

a) Find the acceleration of the lift
b) Find the mass of the Albert
c) Find the mass of Bella
It has been answered here: http://www.thestudentroom.co.uk/show....php?t=1030138 by insane2642.

However I don't understand why the forces acting on Albert and Bella are subtracted from the tension in the cable. In fact I don't even know what direction the forces (678N and 452N) are acting in.

Are they acting upwards in the direction of the cable tension? Or are they acting downwards in the direction of gravity?

Thanks,
yacc
2. Have you got the correct answer so i can compare to check my method. Then ill explain if correct
3. The correct answer provided by cravendale is in the link in my post:

Question 2 part a:

3955-678-452=2825

a=F/m= 2825/250
=11.3
4. However I don't understand why the forces acting on Albert and Bella are subtracted from the tension in the cable. In fact I don't even know what direction the forces (678N and 452N) are acting in.
The force of the floor on the people is directed upwards, and they push on the floor of the lift with an equal/opposite force.

Basically, imagine the lift, being pulled up by the string and pushed down by the passangers. The Tension in the string is pulling up both the lift and the passangers, so you subtract the force nessecary to move the passangers to obtain the force needed just to move the lift with that accelleration.

So, you have a constant mass and a constant force, and you're trying to find the acceleration. F = ma.

Once you have the acceleration, you know the passangers travel with the same accelleration and you have the force required to accellerate the passangers, so their mass is trivial.
5. Why are we finding the force to just move the lift? the lift contains Albert and Bella..it says so in the question "Albert and Bella are both standing in a lift." Part a) then asks "Find the acceleration of the lift".
6. Why are we finding the force to just move the lift? the lift contains Albert and Bella..it says so in the
question "Albert and Bella are both standing in a lift." Part a) then asks "Find the acceleration of the lift".
Because you have the mass of the lift, you don't have the mass of the lift and the passangers. With the resultant force on the lift, and the mass of the lift, you can work out the accelleration of the lift.

The tension in the rope is the resultant force for all 3 objects. The accelleration of all 3 objects (which is the same as the accelleration for any one object) is the force moving all 3 divided by the total mass. This is the same as the resultant force on any one object divided by that object's mass; as you know they all must have the same accelleration, so given the mass of just the lift you must find the force on the lift
7. OK. Lets see if I have understood:

Forces up: 3955N (pulling up lift + Albert + Bella)

Forces down: Lift (250gN) + Albert (unknown) + Bella (unknown)

Problem: Cannot calculate all the downward forces because mass of Albert and Bella are unknown.

Floor exerts force on Albert of 678N, so Albert must be exerting the same force on the lift floor (equal and opposite reaction, Newton's 3rd law).
Floor exerts force on Bella of 452N, so Bella must be exerting the same force on the lift floor (equal and opposite reaction, Newton's 3rd law).

Subtracting these forces from the 3955N force required to move all 3, gives force required to move just lift: 3955- (678 + 452) = 2825N.

Now it doesn't matter that the downward force of Albet and Bella cannot be calculated, because the upward force moving them is no longer in the picture...so to speak.

Now use F = ma to find acceleration of lift:

2825 - 250g = 250a
375 = 250a
a = (375/250)
a = 1.5ms^2
8. Can I then assume that the force required to lift Albert and Bella without the lift is 3995 - 2825 = 1130N?

Not sure that's going to help me with parts b) and c) though.

b) I know acceleration for Albert is 1.5ms^2, and that he and the floor exert a force of 678N on each other. Also, Albert's downward force is still unknown (? x g)

c) I know acceleration for Bella is 1.5ms^2, and that she and the floor exert a force of 452N on each other. Also, Bella's downward force is still unknown (? x g).

...but then I can't see how to proceed.
9. As me teacher always says. Draw a BIG picture!!
10. I have drawn plenty of diagrams, but I'm confused about how to use F=ma.

I don't know the force required to lift Albert and Bella individually, only combined (1130N).
The acceleration is 1.5ms^2, which means using F=ma I get

1130 = 1.5m
m = (1130/1.5)
Which means the mass of both Albert and Bella is 753kg.

this is obviously wrong...

Then I try with F=ma

678 = 1.5m
m = 678/1.5 = 452

...which is also clearly wrong.

I think my confusion lies around the two forces being exerted by Albert:

There is the gravitational force (Albert's mass*g)
But then there is also the force he's exerting on the lift floor (678N).

I don't really understand how these two relate.
11. Think I have it:

Albert:
Force upwards: 678N
Mass: unknown kg (call it 'm')
Acceleration: 1.5ms^2

Resultant Force = 678 - mg.
So F = 678 - mg

F=ma

678 - mg = 1.5m
678 = 1.5m + mg
678 = m(1.5 + g)
678 / m = 1.5 + g
678 / m = 1.5 + 9.8
678 / m = 11.3
m = 678 / 11.3

m = 60 kg

But this seems pretty messy / complex.

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