Question

A box is pushed down at an angle of 32 degrees on a rough surface. The box moves to the right. What equation should be used to find the net force in the y-direction?

Answer 1

The net vertical forces must cancel since it does not accelerate vertically.  The 3 forces are gravity (down), the vertical component of the pushing force (down) and the normal force from the ground to the box (up).  That is:
$F_{y}=-mg- F_{push}sin(32)+ F_{n}=0$
Where I have assumed the 32 degrees was measured down from the horizontal so the sine would give the vertical component.

Answer 2

Answer:

$F_{net}=\mu mg cos 32^{\circ} sin 32^{\circ} +mg cos^2 32^{\circ} - mg - F sin 32^{\circ}$

Explanation:

We have four forces acting on the box. Taking the upward direction as positive direction for the vertical axis:

- The push, which pushes down along the ramp inclined at 32 degrees. Its vertical component will be $-F sin 32^{\circ}$, where F is the magnitude of the push

- The weight of the box, of magnitude $-mg$, with m being the mass of the box and g being the acceleration of gravity (9.8 m/s^2). This force is already directed vertically downward, so we don't need to resolve it into the vertical direction

- The frictional force, of magnitude $\mu mg cos 32^{\circ}$, where $\mu$ is the coefficient of friction, directed upward along the inclined plane. Its vertical projection will be $\mu mg cos 32^{\circ} sin 32^{\circ}$

- The normal reaction, whose magnitude is equal to the component of the weight perpendicular to the inclined plane, but with opposite direction (upward): $N=mg cos \theta$. If we resolve it along the vertical direction, we get $N_y = mg cos 32^{\circ} cos 32^{\circ} = mg cos^2 32^{\circ}$

So, the net force along the y-direction will be

$F_{net}=\mu mg cos 32^{\circ} sin 32^{\circ} + mg cos^2 32^{\circ} - mg - F sin 32^{\circ}$