Frictional force acting on mass

AI Thread Summary
A 40 kg block sliding down a rough incline at 35 degrees with constant velocity experiences a frictional force that balances the gravitational component acting down the slope. The gravitational force components were calculated as Fgx = 224.8 N and Fgy = 321.1 N. Since the block moves at constant velocity, the net force along the incline is zero, leading to the equation mgsin(theta) - F_k = 0. This results in the frictional force F_k equaling 224.8 N, acting up the incline. Understanding that the forces balance due to zero acceleration clarifies the problem.
demonslayer42
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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?
 
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But I can't take the sum of Fx because I don't have an acceleration. I'm stuck :(
 
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.
 
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?
 
demonslayer42 said:
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.
 
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.
 
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