# Finding acceleration of an object (with friction)

1. Oct 15, 2011

### Joe26

1. The problem statement, all variables and given/known data
A 1,300-N crate is being pulled across a level floor at a constant speed by a force of 310 N at an angle of 20.0° above the horizontal. If coefficient of kinetic friction is 0.207, find the acceleration of the crate.

2. Relevant equations
Fk= (uk)(N)
F=ma

3. The attempt at a solution
Fx= 310cos20
Fy= 310sin20

N=1300-310sin20

Fk=(uk)(N) = (0.207)(1300-310sin20) = 247.15 N

sumFx= Fx-Fk =310cos20-247.15 = 44.15 N

F=ma=44.15
a=44.15/1300 =0.0340 m/s^2

Can someone point out what I did wrong? Thanks

Last edited: Oct 16, 2011
2. Oct 15, 2011

### SammyS

Staff Emeritus
Doesn't gravity exert a force in the y direction, contributing to Fy ?

The Fy you have is downward. Do you want that to be negative ?

3. Oct 16, 2011

### Joe26

im not sure if I quite follow. Yes, force of gravity (1300N) is exerted in the y-direction:

F1y= 310sin20
N=1300-310sin20
sumFy=N+F1y-1300N=0

I'm not sure what is wrong. Can someone help me out?

4. Oct 16, 2011

### Staff: Mentor

The applied force is downward:

5. Oct 16, 2011

### Joe26

sorry that should be above the horizontal

6. Oct 16, 2011

### Staff: Mentor

1300 N is the weight of the crate, not its mass.

7. Oct 16, 2011

### Joe26

so, how would i be able to find the acceleration?

8. Oct 16, 2011

### Staff: Mentor

Just like you tried to do, only first find the mass of the crate. How do weight and mass relate to each other?

9. Oct 16, 2011

### Joe26

Got it, thanks. W=mg...find m using weight and use f=ma to find acceleration.