at what slope angle are these two components equal?

Question

The weight of a ball rolling down an inclined plane can be broken into two vector components: one acting parallel to the plane, and the other acting perpendicular to the plane.
a) at what slope angle are these two components equal?
b)at what slope angle is the component parallel to the plane equal to zero.
c) at what slope is the component parallel to the plan equal to the weight?


When you answer the question, please also explain WHY, because I don’t just want the answer, I want to learn also =)

Answer 1

Define slope angle = theta
F(parallel) = mg*sin(theta)
F(normal) = mg*cos(theta)
a) When F(parallel) = F(normal), sin(theta) = cos(theta)
Solving for theta:
sin^2(theta) + cos^2(theta) = 1
2sin^2(theta) = 1
sin^2(theta) = 0.5
sin(theta) = sqrt(0.5)
theta = 45 deg
I think the best way to visualize the "why" is to construct a parallelogram of forces. Draw a line of length L from the origin at 45 deg (the parallel force) and one at 135 deg (the normal force), forming two adjacent lines of the parallelogram. Now add the two lines that complete the parallelogram. They converge on the y axis at a value which we can call mg, a vertical force. The result is that the two equal forces sum to the resultant mg, and considered in reverse, mg resolves to the two equal forces. Symmetry and perpendicularity are required for mg to be vertical, and 45 deg is the only angle that can satisfy that.
b) F(parallel) = mg*sin(theta) = 0
sin(theta) = 0
theta = 0
The ramp is horizontal so all the weight is normal to the ramp, none is parallel to it.
c) F(parallel) = mg*sin(theta) = mg
sin(theta) = 1
theta = 90 deg
The ramp is vertical so all the weight is parallel to the ramp.