How to Calculate Friction Force for a Box Being Dragged on a Level Floor?

[zhangster11]

Homework Statement

a box having mass 100kg is dragged across 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

Homework Equations

joooooooooooooooooooooooooooooooooonwooooooooooooooooooooooooooo nhoooooooooooooooooooo

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

 

[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

 

[A(s)]

Do you follow the reasoning?

 

[zhangster11]

johnwoooooooooooooooo

 

[A(s)]

Also, i think i missed a negative sign


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

 

[A(s)]

stupid typos heh

 

[zhangster11]

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

 

[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

 

[zhangster11]

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

 

[A(s)]

No problem : )