# Acceleration of elevator and force in cable

1. Dec 17, 2015

### Wilson123

1. The problem statement, all variables and given/known data

Mass of elevator = 500kg
Counterweight = 134kg
Force in cable C1 is 1.2kn

Determine the acceleration of the elevator and the force in cable C2

2. Relevant equations

3. The attempt at a solution

a = f / m

a = (C2 - C1) / m2

a = ((m2 * g) - C1) / m2

a = ((134 * 9.81) - 1200) / 134 = 0.86m/s2

Not sure if I need to be taking the mass of the elevator into account somewhere?

2. Dec 17, 2015

### Staff: Mentor

Ignoring the elevator and motor for the moment, draw a free body diagram for the 134 kg counterweight alone. What forces are acting on it? Which ones do you know? What can you conclude about the motion of the counterweight?

3. Dec 17, 2015

### Wilson123

The counterweight has a force of 1.314kN acting downwards and 1.2 kN acting upwards. Therefore the counterweight is going down and pulling the elevator upwards?

4. Dec 17, 2015

### jbriggs444

That appears to be what OP has done already. That takes care of the motion of the counterweight.

The next step would be to see what that implies about the motion of the elevator.

5. Dec 17, 2015

### Staff: Mentor

I saw the C2 in the acceleration equation (second line) and had my doubts as to what the OP had in mind.

6. Dec 17, 2015

### Staff: Mentor

Yes. And what's more, you can find the acceleration of the counterweight, which you actually did when you replaced the "C2" term in your equation with the weight of the counterweight. I couldn't tell if this was a fortuitous mistake or intentional... ... which is why I asked you to concentrate on the counterweight alone.

7. Dec 17, 2015

### Wilson123

I am confused now as to whether or not I have used C2 in the right application?

8. Dec 17, 2015

### Staff: Mentor

If you concentrate on the counterweight alone, C2 is nowhere in the picture. What appears in the FBD for the counterweight?

We'll get to C2 later when we look at the elevator.

9. Dec 17, 2015

### Wilson123

Disregarding C2. The acceleration on the counterweight alone is:

F=ma
a = F/ m
a = (1314N - 1200N) / 134
a = 0.85 ??

10. Dec 17, 2015

### Staff: Mentor

That's right. Be sure to include units on results, and indicate the direction of the acceleration.

Now, what does that tell you about the direction of motion of the elevator car and its acceleration?

11. Dec 17, 2015

### Wilson123

Units are in m/s^2 for acceleration? The elevator car will be moving upwards as the counterweight is moving down, I'm not sure what this suggests about the acceleration..

12. Dec 17, 2015

### Staff: Mentor

Yes, units of acceleration are m/s2.

How are the counterweight and elevator car connected? Can they have different velocities or accelerations?

13. Dec 17, 2015

### Wilson123

They are connected through the same pulley therefore will have the same velocity and acceleration. Which would mean the acceleration of the elevator is the same as the counter block... Is that the logic you were trying to make me realise?

14. Dec 17, 2015

### Staff: Mentor

It is indeed

So with that fact in mind, can you use the known forces operating on the elevator and its acceleration to find the "missing" force C2? Think Newton's 2nd law. An FBD for the isolated elevator car is in order.

15. Dec 17, 2015

### Wilson123

Is it as simple as doing F=ma, f=500*0.85, giving me an answer of 0.425kN. Or do I need to take the 1.2 kN and weight of the elevator into account as well?

16. Dec 17, 2015

### Staff: Mentor

Yup. Hence the FBD for the elevator car. F = MA relates the mass, acceleration, and NET force acting.

17. Dec 17, 2015

### Wilson123

(m1*a) + (m1*g) - C1 = C2
(500 * 0.85) + (500 * 9.81) - 1200 = 4.13kN

Is this correct taking all variables into account?

18. Dec 17, 2015

### Staff: Mentor

Looks good to me!

19. Dec 17, 2015

### Wilson123

Thanks for your help! Got there in the end...

20. Dec 17, 2015

### Saints-94

So far i believe this is correct, however I'm
Well Done!