# Simple friction problem?

1. Oct 18, 2009

### zhangster11

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

a box having mass 100kg is dragged accross a level floor by means of a rope that makes a 30.0 degree angle with the floor. the coefficient of friction between the box and floor is .330. what force f just be applied to hte rope to move the box along at a constant velocity?
meow

2. Relevant equations

joooooooooooooooooooooooooooooooooonwooooooooooooooooooooooooooo nhoooooooooooooooooooo

3. The attempt at a solution

i drew a body diagram
x=zcos30-f
y=n-980 +zsin30

z is the force needed to pull the rope

Last edited: Oct 18, 2009
2. Oct 18, 2009

### A(s)

The box moves at a constant velocity when the force of kinetic friction equals the force of horizontal pull (net horizontal force = 0). You know for this problem that your net normal force is 980 - F*sin(30).

so your force of friction in the negative horizontal direction, plus your Force of pull in the positive horizontal direction must be add to be zero.

[980 - F*sin(30)]*(.330) + F*Cos(30) = 0

Edit* where F = your force of pull

3. Oct 18, 2009

### A(s)

4. Oct 18, 2009

### zhangster11

johnwoooooooooooooooo

Last edited: Oct 18, 2009
5. Oct 18, 2009

### A(s)

Also, i think i missed a negative sign

[980 - F*sin(30)]*(.330) - F*Cos(30) = 0

6. Oct 18, 2009

### A(s)

stupid typos heh

7. Oct 18, 2009

### zhangster11

yea i got it. my answer is 314nt...is that wat you got joonwoo??

Last edited: Oct 18, 2009
8. Oct 18, 2009

### A(s)

Ya, that is the answer I get as well. The trick to the problem was definitely remembering that Fnet = 0 at constant velocity. That would have probably given you the extra equation you would have needed to not have too many variables

9. Oct 18, 2009

### zhangster11

yea thanks for your help jooonwoo !~!@\$!#%!

Last edited: Oct 18, 2009
10. Oct 18, 2009

### A(s)

No problem : )