How Much Force to Slide a 10 kg Box with Friction?

[lxared]

Homework Statement

A 10 kg box rests on a horizontal surface. The coefficient of static friction is 0.85. If
a force is applied to the object at 30° above the horizontal, what is the minimum
magnitude this force must have in order for the box to slide?

Homework Equations

Fnet = ma
Fs = usFn

The Attempt at a Solution

m=10kg
us=0.85
Theta=30

Okay so Fnet = ma and since the box is at rest, Fnet = 0
The net force is the force we're trying to find minus the friction force making F - Fs = 0, F = Fs
Fs is us*Fn which is 0.85*mg which is 0.85*10kg*9.8 which calculates to 83.3N

Now here is where I'm stuck, the equation now looks like F = 83.3N, which should be the answer but it isn't since the answer is 64.5N and I have no clue how to get it.
Dividing 83.3N by cos30 still doesn't get me 64.5N so I think I might have done something wrong in setting it up?

 

[hage567]

Fn which is 0.85*mg

This is incorrect. Fn will not be just mg. There is also the vertical component of the applied force to consider. You must sum up all the forces in each direction. You can't ignore any!

 

[lxared]

Oh I see now, thank you! Fn is actually mg - F*sin 30, making F = us*mg/(cos 30 + us*sin 30)