# Find acceleration? Are two solution possible?

1. Jul 5, 2011

### CICCI_2011

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

Find acceleration-x ̈ of the load. Force F=10G, friction μ=1/3, mass of he body is G.

I tried two ways of solving this problem but not sure if any is good. I think second is good. But wondering if first is also good?

2. Relevant equations

Ek=1/2 G/g x ̇^2

Fμ=G μ

3. The attempt at a solution

Please look at the picture.There you can see my effort in solving this problem. Thanks for any help.

2. Jul 5, 2011

### HallsofIvy

I don't recognize this formula. Is "g" the acceleration due to gravity? This problem does not involve falling under gravity. You have an object moving across a surface with friction, under a force. And, although you don't say it, I suspect $\mu$ is the coefficient of friction, not the friction force itself.

3. Jul 5, 2011

### vela

Staff Emeritus
The way you're using it, G is the object's weight, not its mass.

Both approaches will work.

In the first approach, the initial equation, though correct, seems to come out of nowhere. It would be clearer if you used A.L. Bruce's suggestion and started with Newton's second law.

In the second approach, you didn't calculate the work (A) correctly.

4. Jul 5, 2011

### CICCI_2011

Yes μ is the coefficient of friction. "g" is the acceleration due to gravity. Ek-would be kinetics energy of the system. G/g*X-Inertial force of the body.Fμ-friction force.G is the object's weight. Is any of the solution correct? If not how it should be done?

Why work is not correctly calculated?

Work shouldn't have coefficient of friction wright?

I was thinking that pulling force is acting on one side and friction force+ inertial force on the other. That is how I come up with initial equation of the first approach. I don't see how to get same equation with Newton's second law?

Thanks for all for your fast response

Last edited: Jul 5, 2011
5. Jul 5, 2011

### CICCI_2011

Work would be dA=Fdx-Gμdx

dA=10Gdx-G/3dx
dA=(10-1/3)Gdx
dA=29/3 Gdx

from which acceleration is x=29/3 g

I still don't see how to get G μ+G/g x ̈ =F from Newton's second law?

Last edited: Jul 6, 2011
6. Jul 5, 2011

### Staff: Mentor

Fnet = M*A.

M is the mass of the object. What is M in this case? What do you get for Fnet?

7. Jul 6, 2011

### CICCI_2011

Mass of the object G/g. For Fnet=F-Gμ. Is this correct ?

G1 is weight on first load and 1 is just a subscript. That also stand for G2,G3. (Can't type subscript don't know why)

Can the same principle be used for multiple loads. If there were 3 loads tied one to another with rope(light unstretchable) G1,G2,G3 than would it be correct to write (G1+G2+G3)μ+(G1+G2+G3)/g *x ̈ =F ?

S- force in rope
S1-force in first rope etc. Again 1 is just a subscript That also stand for S2,S3.

And the force in the ropes S3 (G3 is the last in line)

S3=G3 μ+G3/g x ̈

S2=S3+G2 μ+G2/g x ̈

S1=F-S2+G3 μ+G3/g x ̈

Is this correct? If not any suggestion?

Last edited: Jul 6, 2011
8. Jul 6, 2011

### Staff: Mentor

Yes. So you can plug these values into the Fnet = M*A expression and solve for A.