Physics
posted by Randy .
To move a large crate across a rough floor, you push on it with a force at an angle of 21 below the horizontal, as shown in the figure.
Find the acceleration of the crate if the applied force is 400 , the mass of the crate is 32 and the coefficient of kinetic friction is 0.31.

I assume you are pushing downward, the angle is confusing.
If you are pushing downward, some of that is pressing downward increasing fn
fn=mg+Fsin21
fhorizontal=Fcos21
horizontal forcefriction= ma
F*cos21mu*(mg+Fsin21)=m(400)
solve for F. You did notput units in the statement, you need to check them 
Fc = mg = 32kg * 9.8Nkg = 313.6N = Force of crate.
Fp = mg*sin(0) = 313.6sin(0) = 0 N =
Force parallel to the plane = Fh.
Fv = mg*cos(0) + Fap*sin21,
Fv = 313.6cos(0) + 400sin21 = 456.9N =
Force perpendicular to plane. Fap = Applied force.
Ff = u*Fv = 0.31 * 456.9 = 141.7N =
Force of friction.
Fn = Fap*cos21  Fp  Ff,
Fn = 400cos21  0  141.7 = 231.7N =
Net force.
a = Fn / m = 231.7 / 32 = 7.2m/s^2.
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