Force Required at 40 Degrees on an Inclined Ramp: Vector Diagram Included

[madeeeeee]

I have not been able to solve this question:

A crate with a weight of 45.5 N sits on a ramp inclined at 20 degrees to the horizontal. There is a frictional force of 4.5 N exerted on the crate directed up the ramp. Determine the force that must be applied at an angle of 40 degrees to the ramp to keep the crate at rest. Include a labeled vector diagram with your solution. [6 marks]

What is the force that must be applied at an angle of 40 degree to the ramp, how does that look like on a diagram.

 

[HallsofIvy]

You are told that the force vector is "at 40 degrees to the ramp" but it is not clear if that is 40 degrees up from the ramp or 40 degrees down toward the ramp.
(Wait- those two different forces will change how hard the crate presses against the ramp but does not change the force parallel to the ramp- so it doesn't matter which you take!)

In either case, you need to divide the forces into components parallel to and perpendicular to the ramp. The "weight vector" for the crate is downward, perpendicuar to the ramp. If you draw components parallel to and perpendicular to the ramp, you get a right triangle that has angles of 20 degrees and 90- 20= 70 degrees and hypotenuse of "length" 45.5 N. The component parallel to the ramp is 45.5 sin(20) N.

If the force applied to the crate, at 40 degrees to the ramp makes a right angle with angle 40 degrees, hypotenuse of "length" F (the unknown force to be applied) and so the force component parallel to the ramp is F cos(40) N.

In order that the crate not move along the ramp, those must be equal (and in opposite directions): 45.5 sin(20)= F cos(40).

 

[madeeeeee]

Ok i see, that makes sense ! thank you so much :) I wording kind of threw me off but thank you.