Question

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.

Answer 1

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 force-friction= ma

F*cos21-mu*(mg+Fsin21)=m(400)

solve for F. You did notput units in the statement, you need to check them

Answer 2

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.