斜面的問題(下滑力和下壓力)

Let r1 and r2 be the forces acting on the two plates by the balls
By the law of action and reaction, the same forces r1 and r2 will be acting on the balls (only in opposite direction).
The force r1 is acting on the ball horizontally , and r2 is acting upward on the ball parallel to the plane.

Let R1 and R2 be the normal reaction on balls 1 and 2 respectively given by the plane.

(1) For ball 1, because the ball is in equilibrium
(R1).cos(theta) = mg -------------- (1)
where m is the mass of the ball, and g is the acceleration due to gravity
and (R1).sin(theta) = r1 ----------------- (2)

For ball2, it is in equilibrium, hence
r2 = mg.sin(theta) ---------------- (3)
R2 = mg.cos(theta) ---------------- (4)

Therefore, (2)/(3): r1/r2 = (R1).sin(theta)/mg.sin(theta) = R1/mg
But from (1) R1 = mg/cos(theta)
r1/r2 = [mg/cos(theta)]/mg = 1/cos(theta)

(2) R1/R2 = [mg/cos(theta)]/mg.cos(theta) = 1/[cos(theta)]^2