# Homework Help: Force- Major Help needed

1. Oct 12, 2006

### parwana

Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 4.00 kg and m2 = 8.00 kg, what value of f will reduce the acceleration of the system by 50%?

Now I got the equations for these two as

for M1= T-4g= 4a
for M2= 8g-T= 8a

I got a as 3.266

I got T as 52.264

2. Oct 12, 2006

### Staff: Mentor

Now you have an additional force acting on m1 and a given acceleration. Rewrite your equations accordingly and solve for that force.

3. Oct 12, 2006

### parwana

4. Oct 12, 2006

### Staff: Mentor

How did you get those first equations? It's the same analysis, only now there is an additional force acting on m1.

5. Oct 12, 2006

### parwana

So is it going to be

T-(4+x)g= (4+x)a

6. Oct 12, 2006

### parwana

7. Oct 12, 2006

### OlderDan

No. What you have written assumes mass is being added to m1. No mass is being added. There is an additional force acting, but no additional mass. Also, you have been given the acceleration as half what it was without the additional force, so (a) is now known.

8. Oct 12, 2006

### parwana

I dont understand still, can you write it in a equation like I did?

9. Oct 12, 2006

### parwana

So will it be

T-4g-F= 4(.5a)

In that case I got F= 19.596

10. Oct 12, 2006

### OlderDan

Did you use the old T, or caclulate the new T? The old T will not work.

11. Oct 13, 2006

### Staff: Mentor

Try this. Take your original equations:
Now add the new force F on M1. The equations become:
(for M1) T -4g -F = 4a
(for M2) 8g - T = 8a

You have two unknowns: T and F. Solve for F. (This time around "a" is not an unknown--it's given as 3.266/2 m/s^2.)

12. Oct 13, 2006

### parwana

thank u so much