If a 12.0 kg object is on a surface that is inclined 30 degrees and the coefficient of static friction is 0.65, what is the force of static friction? Show your work.

Question

vincent-lyon.fr · Answer

Can you draw a sketch of the situation, so that we can discuss it?

anonymous · Answer

I believe it's like this:
|dw:1388162874029:dw|

vincent-lyon.fr · Answer

Add the forces acting on the block and give the relation showing that the block is at rest.

anonymous · Answer

|dw:1388163365828:dw|
Something like this?

vincent-lyon.fr · Answer

Great! Now, now what can you say since this block is at rest?

anonymous · Answer

That the forces are all equal?

vincent-lyon.fr · Answer

Maybe your thinking is ok, but the way you put it into words is wrong.

vincent-lyon.fr · Answer

What about "net force" ?

anonymous · Answer

Ohh that the net force is 0.

vincent-lyon.fr · Answer

Ok, so now it is easy to find the force of static friction.

anonymous · Answer

I'm not quite sure how to do that, would you mind walking me through it?

vincent-lyon.fr · Answer

|dw:1388163957036:dw|
Since net force = 0, the two highlighted forces must compensate.
So:  friction = // component of weight = mg ......... (fill in that part)