# Forces involved in an elevator

1. Jul 2, 2007

### danielle36

Safety engineers estimate that an elevator can hold 20 persons of 75kg average mass. The elevator itself has a mass of 500kg. Tensile strength tests show that the cable supporting the elevator can tolerate a maximum force of 29600 N. What is the greatest acceleration that the elevator's motor can produce without breaking the cable?

The answer to the question is listed as 4.8 m/s/s.

Here's what I'm comming up with (and i can't get superscripts working which should explain the odd formatting):

m(elevator) = 500kg
m(people)= 75kg x 20 = 1500 kg
m(total)=500kg + 1500kg = 2000kg

F = m/a so a=F/m
29600N/2000kg = a
14.8 m/s/s = a

I'm really not sure where I'm going wrong here, it seems simple enough...

2. Jul 2, 2007

### cristo

Staff Emeritus
Good.

You're using an incorrect value for the force here. You are told that the wire can support 29600N, however this is not the force that causes the acceleration. There is another force present in this scenario-- the combined weight of the lift.

To answer the question you need to look at the resultant force of the system.

3. Jul 2, 2007

### danielle36

Oh thats right, thanks!

Well I made some adjustments and I'm still not getting 4.8 but I'm closer now - I'm getting 5 m/s/s.

W=mg
=2000kg(9.80 m/s/s)
=19600N

F(net) = W + F(max)
F(net) = -19600N + 29600N
F(net) = 10000N

F/m=a
10000N(2000kg) = a
5m/s/s = a

Something still feels wrong about the way I'm doing this, I think it must be an issue with the net force...?

4. Jul 2, 2007

### cristo

Staff Emeritus
I think your book must be using g=10m/s2

Think of it this way: the wire can provide a force of 29600N. The combined mass of the lift produces a downward force (a weight) of 20000N, so maximum force that the wire can provide is a force of 9600N. This is the force that drives the acceleration. So, the maximum acceleration occurs when this force is maximum; namely when this force is 9600N.

As an aside, if you plan on sticking around, there is an easy way to create formulae on posts here: at the top of the reply box is a button labelled $\Sigma$. Click on this for various mathematical typesetting options.

Last edited: Jul 2, 2007
5. Jul 2, 2007

### danielle36

Well thanks so much for helping me with both those issues. I'm doing all this through correspondance so its a big help to have been able to discuss this with someone. I think I've got it down now so I can finally go on to write the final test for the chapter. Thanx again!

6. Jul 2, 2007

### cristo

Staff Emeritus
You're welcome! Feel free to ask any future questions you may have.

7. Jul 2, 2007

### nrqed

You have already solved it with Cristo's help, but I thought I would make a comment about how to approach these problems in general. First, it is key to your understanding to realize that the F that appears in f=ma is the NET force on the system. It is actually the sum (and I mean by sum, th e*vector* sum) of all the forces. Th ereal equation si really

$$\sum \vec{F} = m \vec{a}$$
(well, there is an even more general expression with the derivative of the momentum but no need to mention this here)

If there is only one force acting on the system, then it' sthat force that enters the equation. If there are more forces, however, you must add them (as vectors) before setting the sum equal to ma.

A second, related, point is that it is crucial in those problems, even in simpel ones, to draw a free body diagram showing all teh forces involved before starting to write equations. In this example, you may treat the elevator an dthe people inside as one, single, object. The forces acting on this object are: a tension due to the cable acting upward and the weight (the force of gravity) acting downward with a magnitude equal to $m_{total} g$. So the free body diagram will show the object with two forces acting on it, a tension acting up and the weight acting down. The eqation reduces to
[tex] T - mg = m a_y [/itex]
where T is the tension in the cable.

You then plug in the numbers and solve.