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Forces and Friction

  • Thread starter JasonAdams
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1. A 3 kg block slides down a 30 degree incllined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2m.
a) What is the velocity of the block at the bottom?
b) What is the magnitude of the normal force?
c) What is the coefficient of friction between the plane and the block?


I solved the questions, I'm just not sure if I got the answers right, so I wanted to double check with somebody here. For a) I got 2m/s (I forget how I got it, I didn't write down my work). For b), I did Fn = mg, and I got 29.4 N. For c), I did a diagram and got 13.3 N as the applied force. For Fnet, I did Fnet = ma, and got 29.4 N. Fnet = F applied - F opposing or friction, and I got -16.1 N as the force of friction (I'm not sure if having a negative force is possible, so if I'm wrong, just stay with me here). µ = Fs/Fn = -16.1N / 29.4N = -0.547. If any of this is wrong, please tell me how I was supposed to do it.
 

Answers and Replies

cristo
Staff Emeritus
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1. A 3 kg block slides down a 30 degree incllined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2m.
a) What is the velocity of the block at the bottom?
b) What is the magnitude of the normal force?
c) What is the coefficient of friction between the plane and the block?


I solved the questions, I'm just not sure if I got the answers right, so I wanted to double check with somebody here. For a) I got 2m/s (I forget how I got it, I didn't write down my work).
Well, I can't check that if I can't see how you've done it-- and besides, do you not need to show your working for the homework?
For b), I did Fn = mg, and I got 29.4 N.
This isn't right. The normal force is not equal to the weight, since the block is on an inclined plane, it is equal to the component of the weight which is perpendicular to the plane. Draw a diagram-- it will help you!
 
To find Fn, what equation do I use? I can't seem to figure it out.
 
957
0
diagram will help. Think about it this way, if the block is "sliding" down a vertical face what would be Fn? If its horizontal what is Fn? At intermediate angles?
 
Fn = m a cosΘ

Fn = mass x acceleration (the accel. of the object, not of gravity)

Are either of these right? I'm just not getting it! The diagram isn't helping me as much as I thought it would.
 
957
0
the normal force is the force pushing back against the block right? It would be zero in the case of sliding along a vertical cliff, and the full weight, mg, when it is horizontally supported.

Maybe this will help more, with inclines we need to break down the gravitaional force into two components: those perpendicular to the block which does not tend to accelerate the object but causes static friction and the component along the surface of the block which accelerates the block.

In effect it is like we are rotating the coordinate system.

Still murky?
John
 

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