Finding coefficient of kinetic friction

1. Apr 19, 2014

chococho

1. The problem statement, all variables and given/known data

A 100 N force directed 37° below the horizontal is applied to a 30 kg object on a horizontal surface. If the magnitude of the acceleration of the object is 1.3 m/s2, what is the coefficient of kinetic friction (μk) between the object and the surface?

2. Relevant equations

F = ma

3. The attempt at a solution

I drew the body diagram and tried calculating the net force in the x direction.
This is what I got:

F = Fpx (the 100 N force) - Ffr
Fpx is 100cos37 and Ffr is μmg, so
ma = 79.86 - 294μ

I plugged in 1.3 for a and got 0.24 for the answer.
Not sure where I went wrong, any help would be appreciated. Thanks!

2. Apr 19, 2014

Simon Bridge

... this is not correct.
Did you remember to sum the forces in the y direction?

3. Apr 19, 2014

chococho

It said friction force equals u*normal force.. isn't normal force mg?

4. Apr 19, 2014

Simon Bridge

No.

You have a free body diagram - add up the forces in the y direction.

5. Apr 19, 2014

chococho

ok.
In the diagram I drew I just have the 100N force, normal force, friction force and mg. Is that correct?
So the forces in the y direction would be just "100sin37 + Fn - mg" ?
But can I still set it equal to m*a with a being 1.3?

6. Apr 19, 2014

Simon Bridge

What direction does the magnitude 100sin37 force act in?
What do these forces add up to? (hint: does the object move in the y direction?)
Solve for Fn.

7. Apr 20, 2014

chococho

So it's 100sin37 - Fn - mg = 0 ?
When I solve for Fn I get 233.81. I solved for the forces in the x direction (ma = 100cos37 - Ffr) and I got Ffr to be 40.86.
So do I just divide that by Fn to get the coefficient?
It's still wrong, I'm confused

8. Apr 20, 2014

haruspex

No, that's still wrong. The normal force has to balance the sum of the other vertical forces.

9. Apr 20, 2014

chococho

OHHHH I got one of the signs wrong... Just making sure, so it's Fn = 100sin37 + mg? Because then I get 0.115 which should be right

10. Apr 20, 2014

Simon Bridge

Well done.

Use your free body diagram to get the signs right.
You should draw the diagonal forces as diagonal arrows - then divide into components on the diagram.
Thus the 100N force would be diagonally downwards and you cannot get the component directions wrong.

Note: The forces all add head-to-tail to make a resultant force pointing horizontally.
If you fiddle with the order of the addition, you should get an easy triangle.