Abasket has a mass of 5.5 kg. find the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal at:
a) 0 degrees (horizontal surface)
b) 12 degrees
c) 25 degrees
d) 45 degrees

The free body diagram consists of:

1) Weight force, which is a vertical arrow downward

2) A Normal force that is perpendicular to the ramp.

The Normal force can be decomposed in a vertical component and an horizontal component.

The vertical component, Ny,  must equal the weight force, due to equilibrium (rest) condiition =>

Ny = weight = m*g = 5.5kg*9.8m/s^2 = 53.9 N

By geometry the vertical component of the Normal force, Ny is such that

cos (angle) = N / Ny => N = Ny * cos(angle)

=> N = 53.9N * cos(angle)

Now use that equation for each option:

a) 0 ° => N = 53.9N * cos(0) = 53.9N

b) 12° => N = 53.9 * cos(12) = 52.7N

c) 25° => N = 53.9 * cos(25) = 48.9N

d) 45° => N = 53.9 * cos(45) = 38.1N

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