Question

a 269-kg block is on the horizontal surface of a single-axis vibration table that is oscillating horizontally with simple harmonic motion of frequency 2.10 hz.the coefficient of static friction between the surface and the block is 0.550. what is the maximum amplitude of the shm tolerable for the block to not slip along the surface?

Answer 1

Fm = μ M g maximum force that can be applied to object with object not slipping

F = M * ω^2 A since ω^2 A is the maximum acceleration of SHM

μ M g = M * ω^2 A equating forces

A = μ g / ω^2 = μ g / (2 π f)^2

A = .55 * 9.80 m/s^2/ (6.28 * 2.10 / s)^2 = .031 m