Question

A cube of mass m = 6.75 kg is placed on an inclined plane of mass M = 14.27 kg, which makes an angle ? = 48.50

Answer 1

Answer 2

let's start with mass m. get the weight of the cube. 6.75*9.81 = 66.2175 N pointing downwards,
the normal force between the inclined plane and the cube is perpendicular to the surface of the inclined plane with angle 41.50 from the horizontal pointing NW

Solving for the N between cube and inclined plane: 0 = 66.2175cos(48.50) + N... N= 64.2175cos(48.50) or 42.55N

Now for the inclined plane, forces acting are the normal force between cube and inclined plane, normal force between inclined plane and floor, and weight of the inclined plane.

since the direction of the normal force between inclined plane and floor is just upwards, simply break the stated forces above in their vertical components.

0 = 42.55N + 64.2175cos(48.50)sin(41.50) - N
N = 42.55N + 28.19N
N = 70.74 N