# Force Friction at an angle

1. Oct 20, 2008

### halo9909

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

What force would you have to exert on a 326-N trunk up a 21.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.322.)

2. Relevant equations

not sure what to use by
Ff=coefficient friction *Fn

3. The attempt at a solution
Ihave tried drawing a picture of this
but what i done is
figure that the Fw=326N
from there I am not sure where to use it or what equations to

2. Oct 21, 2008

### alphysicist

Hi halo9909,

From your picture, what force components are parallel to the incline? what force components are perpendicular to the incline? Once you have those, you can write two force equations. What do you get?

3. Oct 21, 2008

### halo9909

well since 326N is the Fw
so i would use
326sin21=115.82N=Fay
326cos21=304.34=Fax
so Fnormal=326+115.82=442.8
Ff=.322*442.8=142.59 since Ff=mu*Fnormal

then
Fnet= Fa-Ff
Fnet=304.34-142.6
=161.74
From here I am unsure what to do,

4. Oct 21, 2008

### alphysicist

Didn't you get 116.828N for this?

These are fine; these are the components of the weight force parallel (x) and perpendicular (y) to the incline.

Now you can use this in the force equations. I would start with the perpendicular (y) direction. What forces have components in the y direction? And what is the acceleration in the y direction? Once you have those you use them in:

(sum of all y-components of all forces ) = (mass) (acceleration in y direction)

which is what Fnet,y=may means.

So you should be able to find the normal force from that equation. What do you get for that? (Remember that the components can be positive or negative.)

After you have the normal force, repeat the process for the parallel (x) direction. Using the normal force and the equation for kinetic friction in it, you should be able to find F. What do you get?

5. Oct 21, 2008

### halo9909

so the Fnormal for the vertical force would be 442.82=326+116.82
and the horizontal would still be 161.74=394.34-142.6

6. Oct 21, 2008