# Friction question (introductory level)

1. Jul 17, 2010

### Fallinleave

Hello everyone, I am new here. If there is anything I did not explain well in stating the question, please point it out. thanks.

A block weighing 45N is initially at rest on a plane inclined at angle of 15 degree above the horizontal, and force "P" is acting on the block with direction down the plane. The static and kinetic coefficients are 0.50 and 0.34. If the force "P" is 5.0 N(direction is down the plane), what is the frictional force acting on the block?

Plans of attack:
1. find whether the force "P" can overcome the frictional force.
2. find the friction.

attempt solution:
1. Free body diagram is drawn and positive x-axis is chosen to be up the plane.

Horizontal forces:
Ff -- Fp -- mgsin15 = unknown force = max
Vertical forces:
Fn -- mgcos15 = 0(no vertical acceleration), which simplify as Fn = mgcos15

normal force multiply static coefficient = max static friction, I then combine my two equations and found that the friction is bigger.

2.The applied force "P" could not make the block slide down the plane, so the magnitude of the friction is the same as net force down the plane.
Ff = fp + mgsin15 = 16.647N with direction up the plane.

Is my solution correct? thanks,

2. Jul 17, 2010

### kuruman

Hi Fallinleave and welcome to PF. Your analysis and solution are correct. For future reference, please use the homework help template when you submit posts.

3. Jul 17, 2010

thanks