Homework hellp. coefficent of friction.

  • #1

Homework Statement


An 18.0 kg box is released on a 37.0 degree incline and accelerates down the incline at 0.270 m/s/s. Find the friction force impeding it's motion. How large is the coefficient of friction?


Homework Equations


Ff=uFn


The Attempt at a Solution


I have no idea how to do this.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
961
0
Can you do the problem if there were no friction, that is just asking for acceleration down an inclined plane?
 
  • #3
ohh, maybe how can i do it without friction though?
 
  • #4
961
0
Well you can't but the principles used apply:

In other words the acceleration from gravity along the surface of the ramp is sin(theta) * g. It is not 9.8. So if there were no friction a=9.8 sin(37). What is this come out to be?
 
  • #6
961
0
OK. Now we are told that the acceleration is not 5.9 but a measly 0.27.

So lets write an eqn using Newtons laws.

ma=sum of all forces ? is the frictional force

18.0*0.27=18*5.9-? Lets not worry about the components that make up ? yet. Can you compute ?
 
  • #7
4.86=106.2-? so 101.34? or is there more than one other force?, Thank you for helping I'm sorry I'm really bad at physics.
 
  • #8
961
0
Should be 18(5.9-0.27)

Ok now the tricky part. This frictional force is equal to the normal force * u where u is the coefficient of friction.

On level ground normal force is just opposite the weight. But on a plane it is
weight* cos(37)

So normal force is 18*9.8 cos (37). Lets call this N.

Then we have N*u=the same ? as you calculated before. can you get u?
 
  • #9
176.4 cos(37)=140.9
N*u=101.34
140.9u=101.34
so u=.719
 
  • #10
961
0
By george I think you got it. It makes more sense with a free body diagram.
 
  • #11
Thank you so much!
 

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