Coefficient of Friction on an Inclined Plane

[Bamshakalaka]

The problem states: An 800N box rest on a surface inclined 30 degrees. If a force 200N parallel to the surface is applied, the box will not slide down the incline. Whats the greatest force that can be applied to the box before it starts sliding up the incline??

I know Fnorm=mg*cos theta Fnorm=692.8N
Is the mu=0.298 because Ffriction=mu*Fnorm ---> 200N=mu*692.8N
I also know Fparallel=mg*sin theta Fparallel= 400N
How do I find Fnet?
I don't know what to do from there! Can anybody help??! Please and thank you!

 

[PhanthomJay]

Bamshakalaka, welcome to PF!

The problem states: An 800N box rest on a surface inclined 30 degrees. If a force 200N parallel to the surface is applied, the box will not slide down the incline. Whats the greatest force that can be applied to the box before it starts sliding up the incline??

I know Fnorm=mg*cos theta Fnorm=692.8N
Is the mu=0.298 because Ffriction=mu*Fnorm ---> 200N=mu*692.8N

are you assuming that the 200N applied force is the friction force?

I also know Fparallel=mg*sin theta Fparallel= 400N

yes, correct.

How do I find Fnet?

If you look at all the forces acting parallel to the incline, you know that the 400N gravity force acts down the plane, and it is given that there is a 200N force applied up the plane, parallel to the incline. You are also given that the box does not slide; thus Newton 1 applies. So you have 400 down, 200 up, PLUS the max friction force, all three of which forces act parallel to the incline. So what's the friction force? Then determine what the applied max force must be to prevent motion up the plane, using that same value of friction force. Watch your plus and minus signs; the friction force opposes the direction of pending motion .

 

[Bamshakalaka]

Thank you for the warm welcome! =)

What I'm confused about is the max friction force. How do i know what the max friction force is?

I kind of understand what you are saying so let me take a stab at it. If I use the Fnet=Fapp-(Fparallel+Ffriction) formula then Fnet would equal 0 because the box is not in motion yet. Fparallel is the 400N and Ffriction would be the 200N. Therefore 0=Fapp-(400N+200N) Fapp=600N.

So 600N is the greatest force that can be applied to the box just before it starts sliding up? Is that correct?

Thank you in advance for helping me! I really appreciate it!

 

[PhanthomJay]

Thank you for the warm welcome! =)

What I'm confused about is the max friction force. How do i know what the max friction force is?

I kind of understand what you are saying so let me take a stab at it. If I use the Fnet=Fapp-(Fparallel+Ffriction) formula then Fnet would equal 0 because the box is not in motion yet. Fparallel is the 400N and Ffriction would be the 200N. Therefore 0=Fapp-(400N+200N) Fapp=600N.

So 600N is the greatest force that can be applied to the box just before it starts sliding up? Is that correct?

Thank you in advance for helping me! I really appreciate it!

Yes, you are correct. Remember that Ffriction (static) is less than or equal to (mu)N. It becomes equal to (mu)N (its max value) only when it is just about to start sliding. 200N is the max available static friction force here. For example, if the block were on a level table and you applied a 100N force, it wouldn't move, and the friction force would just be 100N for equilibrium.

 

[Bamshakalaka]

Ohh i see. Thank you sooo much you have been such a great help! ^___^