Question

A mass of M-kg rests on a frictionless ramp inclined at 30°. A string with a linear mass density of μ=0.025" kg/m" is attached to the M-kg mass. The string passes over a frictionless pulley of negligible mass and is attached to a hanging mass (m). The system is in static equilibrium. A wave is induced on the string and travels up the ramp.
(a) Show that the mass of M is equal to twice the mass of the hanging mass: M = 2m.
(b) and at what wave speed does the wave travel up the string, if m = 5 kg?

Answer 1

Answer:

44.3 m/s

Explanation:

a) Draw a free body diagram of the mass M.  There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m.  There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N.  Therefore:

v = √(49 N / 0.025 kg/m)

v = 44.3 m/s