Question

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.20 Hz. On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is 5.50 10-2 m.

Answer 1

Coefficient of friction will be 0.9738

Step-by-step explanation:

We have given frequency of the motion f = 2.20 Hz

So angular frequency
`\omega =2\pi f=2* 3.14* 2.20=13.816rad/sec`

Amplitude of the motion
`A=5.50* 10^(-2)m`

Maximum acceleration is given by
`a=\omega ^2A=13.816^2* 5.50* 10^(-2)=9.544rad/sec^2`

Acceleration due to gravity
`g=9.8m/sec^2`

We know that acceleration is given by
`a=\mu g`

`\mu =(a)/(g)=(9.544)/(9.8)=0.9738`

So coefficient of friction will be 0.9738