# Elevator Problem(dynamics)

1. Oct 13, 2009

### ur5pointos2sl

Inside the elevator, masses m1 = 2.00kg and m2 = 4.00 kg are conncted by a cable A. Mass m1 is connected to the roof of the elevator by a cable B. The elevator has a downward acceleration of g/2. Calculate the tensions in cables A and B.

m1 = 2.00 kg
m2 = 4.00 kg
a = 9.8/2

I set up two free body diagrams.. One for m1 and one for m2

So for m1 FBD

SUMFy= 0 = Fb - 2(9.8/2) - Fa
Fb = 9.8 N

for m2 FBD

SUMFy = 0 = Fa - mg
Fa = 4(9.8/2) = 19.6 N NOW PLUGGING BACK IN

2. Oct 13, 2009

### kuruman

There are no answers to check. When you post them, we will check them. Nevertheless, I can see mistakes already. You say

SUMFy = 0 = Fa - mg

Why is the sum of all the forces in the y-direction zero? Are the masses not accelerating?

Also, when you say "mg" which mass are you talking about, 1 or 2?

3. Oct 13, 2009

### ur5pointos2sl

m1 = 2.00kg m2 = 4.00 kg

ok let me try this over..

FBD of m1....so here the mg will be 2kg (9.8) or would it be 2(9.8/2)?

so SUMFy = Fb - Fa - mg = ma
= Fb - Fa -(2)(9.8)=2(-9.8/2)
Fb = -9.8 N which couldnt be possible since its in tension

FBD of m2...so here mg will be 4kg (-9.8/2)???

so SUMFy= Fa - 4(9.8)= 4(-9.8/2)
Fa = 19.6 N

again I am getting the same answers as before..

4. Oct 13, 2009

### kuruman

How did you get -9.8 N? I don't get that when I put in the numbers. Check your algebra.

This looks correct. Do you think it is not?

5. Oct 13, 2009

### ur5pointos2sl

The instructor said 19.6 N was not the correct answer. I am not really sure of the actual correct answer.

I checked my algebra again for Fb and ended up getting 49N. I really am not sure what to think now.

6. Oct 13, 2009

### kuruman

The Fa = 19.6 N is correct. I agree with your instructor. Now show me exactly what you do after you write down

Fb - Fa -(2)(9.8)=2(-9.8/2)

7. Oct 13, 2009

### ur5pointos2sl

Fb - 19.6 -(2)(9.8)=2(-9.8/2)

Fb = 19.6 + 98/5 - 49/5
= 29.4 N

8. Oct 13, 2009

### kuruman

Looks good. I think you are done.