Can someone please help w/ Physics?

Question

A 1100 N crate is being pushed across a level floor at a constant speed by a force of 300 N at an angle of 20.0° below the horizontal, as shown in Figure P4.37a.
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 300 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in Figure P4.37b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).

Answer 1

a) Calculate the horizontal component of the applied force. Trig req'd here.

Since the speed is constant, the net force on the box is zero. That means the horizontal component of the applied force is equal and opposite the force of friction.
Ff = mu*N
where N is the Normal force pressing the crate down on the floor. All of the weight is pressing down on the floor, since the floor is level. But added to that is the vertical component of the applied force. So
N = W + (the 300 N)*a trig function

b) Calculate the horizontal component of the applied force.

You now know mu. Calculate the force of friction.
Ff = mu*N
where N is the Normal force pressing the crate down on the floor. All of the weight is pressing down on the floor, since the floor is level. But subtracted from that is the vertical component of the applied force. So
N = W - (the 300 N)*a trig function

Subtract Ff from the horizontal component of the applied force. The acceleration will come from use of
F = m*a