1. Resolving forces on slopes

Hi, could anyone help me with understanding this please?

"A crate of mass 30kg is at rest on a slope which is at 18 degrees to the horizontal. Find the frictional force."

The frictional force = 30gcos(72 degrees)

I don't understand this because cos(72 degrees)= adjacent/hypotenuse, so cos(72 degrees)=30g/frictional force
which means that the frictional force=30g/cos(72 degrees)

2. Originally Posted by michaeljoannou1990
Hi, could anyone help me with understanding this please?

"A crate of mass 30kg is at rest on a slope which is at 18 degrees to the horizontal. Find the frictional force."

The frictional force = 30gcos(72 degrees)

I don't understand this because cos(72 degrees)= adjacent/hypotenuse, so cos(72 degrees)=30g/frictional force
You are looking at the wrong triangle. You need to look at the force triangle, not the triangle formed by the slope. If you draw a picture, representing the slope as a right triangle with angle 18 and 72 degrees, the 72 degree angle being at the top of the triangle. But you divide the force into components, not the slope. The gravitational force, 30g, is downward and that forms the hypotenuse of the force triangle, while the component along the slope, the force friction must be equal to in order that the crate not slide, is the near side in this triangle: friction force/30g = cos(72), not the other way around.