Frictional force acting on mass

[demonslayer42]

Homework Statement

A 40 kg block slides down arough incline of 35 degrees with a constant velocity. Find the frictional force acting on the block.

Homework Equations

Fgx = Fg sin theta
Fgy = Fg cos theta
Fg = mg

The Attempt at a Solution

Well I drew a diagram and solved for my Fgx component and my Fgy component:

Fgx = 40(9.8)sin35 = 224.8 N
Fgy = 40(9.8)cos35 = 321.1 N

Now what do I do? Take the sum of Fx ? Won't that just be 224.8 N ? That doesn't sound correct I'm confused. Am I doing something wrong?

 

[demonslayer42]

But I can't take the sum of Fx because I don't have an acceleration. I'm stuck :(

 

[PhanthomJay]

There is no acceleration along the incline, the acceleration is zero along the incline when the block is moving at constant velocity along the incline. Looks like Newton 1 applies.

 

[demonslayer42]

That's what I was thinking, but if the a = 0 because it's at a constant velocity wouldn't that just mean the sum of my Fx would be 224.8 N ? So in this case Fx = fs ?Is this correct?

 

[PhanthomJay]

That's what I was thinking, but if the a = 0 because it's at a constant velocity wouldn't that just mean the sum of my Fx would be 224.8 N ?

the sum of your Forces in x direction (parallel to incline) would be 0 (no net force per Newton 1)!

So in this case Fx = fs ?Is this correct?

It's sum of forces (also called F_net) along incline = 0. Thus

mgsin theta (which is the component of the gravity force acting down the plane) - F_k (which is the kinetic friction force acting up the plane) = 0.

224.8 - F_k =0

F_k = 224.8 N acting up the incline and parallel to it.

Which i think it what you meant.

 

[demonslayer42]

Yes, you are right that's what I meant. O.k., I think I'm starting to get the hang of this now thank you.