How to Find the Force Needed to Reduce Acceleration by 50% in Atwood Setup

[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

But I can't get the answer to the asked question. Please HELP me solve this.

 

[Doc Al]

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

 

[parwana]

I don't know how to do that, please help

 

[Doc Al]

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

 

[parwana]

So is it going to be

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

Please help me, I am so fed up of trying to get this right.

 

[parwana]

please help, I am so fed up of this problem.

 

[OlderDan]

So is it going to be

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

Please help me, I am so fed up of trying to get this right.

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.

 

[parwana]

I don't understand still, can you write it in a equation like I did?

 

[parwana]

So will it be

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

In that case I got F= 19.596

 

[OlderDan]

So will it be

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

In that case I got F= 19.596

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

 

[Doc Al]

Try this. Take your original equations:

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

I got a as 3.266

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.)

 

[parwana]

thank u so much