piston ramp spring A piston is placed on top of a spring inside a wedge-shaped block fitted with small rollers so that it can roll down an inclined plane, as shown. When the block is at rest the spring is compressed 1.6 inches from it's equilibrium position, but when  the block is rolling down the plane the spring is compressed only 1.2 inches. What is the inclination angle of the plane? (Neglect friction.)

Solution by Toby Crisford

By Hooke's law, the force exerted by a spring is proportional to the displacement from its equilibrium position. Calling this constant of proportionality k, we know that:

mg = 1.6k,

where m is the mass of the piston, and g is acceleration due to gravity, since the force exerted by the spring exactly balances the weight of the piston when they are at rest. From this,

k = mg/1.6.

Once the block is rolling down the ramp, its acceleration along the ramp will be given by

g sin θ,

where θ is the angle of inclination, since the component of the block's weight perpendicular to the ramp will be balanced by a reaction force, and so only the component acting along the ramp is relevant. In turn, the component of this acceleration in the vertical direction will be

g sin2θ.

Because the piston accelerates with the block, the resultant vertical force on the piston must be mg sin2θ, and since its weight provides a downward vertical force of mg, the upward force from the spring must make up the difference. We are given that the compression of the spring is 1.2, thus

mg - mg sin2θ = 1.2k,

but k = mg/1.6, so

mg - mg sin2θ  = mg (1.2/1.6).

Hence,

1 - sin2θ = 3/4,

sin2θ = 1/4,

sin θ = ±1/2,

θ = 30°.