# Draw a free-body diagram

1. Oct 30, 2009

### swimchica93

What formula do I use to find the force that is used to pull on a box, going a constant speed.

I have mu-k, the degree of the ramp, and the weight of the box.

Last edited: Oct 30, 2009
2. Oct 30, 2009

### ehild

Re: Formula?

First draw a free-body diagram and than take into account that he resultant force is 0.

ehild

3. Oct 30, 2009

### swimchica93

Re: Formula?

I drew a free body diagram, now Ijust don't know what to do with it.

4. Oct 30, 2009

### ehild

Re: Formula?

You have got four forces:

gravity (G=mg) pointing vertically downward;
normal force Fn, normal to the ramp;
friction, Ff=mu *Fn parallel to the ramp and opposite to the velocity of the box
pull Fp, parallel with the slope.

Decompose each forces into components parallel and perpendicular to the ramp. Both the parallel and normal components have to cancel. The normal component of G is opposite to Fn, G-Fn=0, from that you get Fn, and the force of friction. Do you pull the box up or down the ramp?

ehild

5. Oct 30, 2009

### swimchica93

Re: Formula?

The box goes up the ramp.

So, there isn't a formula to use? I know that the box weights 500kg, and the angle of the ramp is 50 degrees and the mu-k is .10. Then it has to be at a constant speed.

I tried:

F-mg(sin(theta))=0 is that kind of what you are talking about? The problem was I couldn't add friction.

6. Oct 30, 2009

### milagro

Re: Formula?

the pulling force along the surface will be

F=coeff of friction*mg*sin(angle of surface with horizontal)

7. Oct 30, 2009

### ehild

Re: Formula?

Well, the normal component of G is mgcos(theta). Do you know why?
So the normal force is Fn=mgcos(theta).
The magnitude of friction is mu*Fn. As the box moves uphill, the friction points downhill.

F(pull)-mgsin(theta)-mu*mgcos(theta)=0. Calculate F(pull).

ehild