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A 250 N crate is sitting on a 30 degree inclined plane. The frictional force is larger than the component of the crate's weight parallel to the inclined plane, so that the crate does not slide down on its own. What is the normal force on the crate? What is the minimum value of the frictional force?
W = 250 Newtons
θ = 30°
N = ??
Ff > Wx = ??
For 99% of inclined plane problems it is best to choose your x-axis parallel to the inclined plane and the y-axis perpendicular to the incline as shown. The weight is perpendicular to the base of the inclined plane and must be resolved into x and y components. The angle between the weight and the y component is equal to the angle of the inclined plane. (If this interests you it can be proven geometrically, but you can just take my word for it.)
For the crate to be sitting still (at equilibrium) and not moving down the inclined plane or jumping up off the inclined plane, we know that the forces in the x-direction must be balanced and the forces in the y-direction must also be balanced.
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