# Finding Acceleration from Friction Coefficient and Force Applied

1. Oct 13, 2008

### ChrisMihm31

1. The problem statement, all variables and given/known data
To move a large crate across a rough floor, you push on it with a force F at an angle of 21°, below the horizontal, as shown in Figure 6-21. Find the acceleration of the crate, given that the mass of the crate is m = 32 kg, the applied force is 344 N and the coefficient of kinetic friction between the crate and the floor is 0.54.

http://www.webassign.net/walker/06-16alt.gif

2. Relevant equations

a = Fnet/m
Ffriction = Fnormal * muk

3. The attempt at a solution

First I found the frictional force by (.54)(32 kg)(9.81 m/s2) = 169.5168
Now this is where I am stuck. Do you subtract that from the horizontal force? The horizontal force I got was 344Cos(21) = 321.15. Then I do 321.15 - 169.5168 = 151.63 and then divide that by the mass(32kg) and get 4.74

This is obviouley what you aren't supposed to do, so HELP!

Last edited: Oct 13, 2008
2. Oct 13, 2008

### Rake-MC

Sorry but your frictional force is wrong.

Force due to friction is the coefficient of friction multiplied by the normal reaction force.

First you must find the normal reaction force.

3. Oct 13, 2008

### ChrisMihm31

Isn't the normal force just mass x gravity?

4. Oct 13, 2008

### Rake-MC

No, because there is also a vertical component of the force applied by the worker.

Sum up the forces in the y plane to find the normal reaction force.

5. Oct 13, 2008

### ChrisMihm31

so would the normal reaction force be: mg + Fsin(21) ???
probably wrong, but its my best guess

6. Oct 13, 2008

### Rake-MC

That's correct.

7. Oct 13, 2008

### ChrisMihm31

Thanks for your help, I plugged everything in and got the correct answer: 2.66 m/s2.