Translational Equilibrium

Question

A 0.12 kg ball is placed between two inclined frictionless surfaces. One makes an angle 20° with respect to the ground and the other is angled 40°. What is the magnitude of the force acting on the ball exerted by each surface?

Answer 1

Gnarls,

Since the ball is sitting between the two inclines and we are asked for the magnitude of the force ... "exerted by each surface" separately, we can work on each incline separately.

Suppose we use A as the angle of an incline with respect to the ground. We will use A = 20 deg and then A = 40 deg.

The force acting on the ball by a surface is the normal force Fn, that is, it's normal or perpendicular to the surface, and it's equal in magnitude and opposite to the weight of the ball pushing directly into the surface (the incline).

The weight of the ball is the gravitational force Fg = mg that pushes vertically down. We can decompose Fg into a component into the incline Fy = Fg cos A (the angle A is adjacent to this component) and a component down the incline Fx = Fg sin A (this component is opposite the angle A). I will let you find a diagram for this decomposition of forces on a ramp (incline) in your physics book.

The normal force is equal in magnitude and opposite in direction (negative) to the component of the gravitational force into the incline Fy = Fg cos A, so Fn = -Fy = -mg cos A for each incline.

For each incline, plug in m = 0.12 kg, g = 9.8 m/s^2, and its value for the angle A.

I'll let you do the numbers.

Good Luck!

Michael.