Finding coefficient of kinetic friction

[chococho]

Homework Statement

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?

Homework Equations

F = ma

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 = F_px (the 100 N force) - F_fr
F_px is 100cos37 and F_fr is μmg, so
ma = 79.86 - 294μ

I plugged in 1.3 for a and got 0.24 for the answer.
The correct answer is 0.12.
Not sure where I went wrong, any help would be appreciated. Thanks!

 

[Simon Bridge]

F_fr is μmg

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

 

[chococho]

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

 

[Simon Bridge]

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

No.

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

 

[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?

 

[Simon Bridge]

100sin37 + Fn - mg

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.

 

[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

 

[haruspex]

So it's 100sin37 - Fn - mg = 0 ?

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

 

[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

 

[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.