Question

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 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 3.13 cm.

Answer 1

Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

$a=\mu g$

and maximum acceleration during SHM is

$a=A\omega ^2$

$a=A(2\pi f)^2$

$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

$a=5.296\ m/s^2$

$\mu =\frac{a}{g}$

$\mu =\frac{5.296}{9.8}=0.54$

Answer 2

Answer:

The coefficient of static friction between the tray and the cup is 0.53.

Explanation:

Given that,

Frequency f= 2.07 Hz

Amplitude x= 3.13 cm

We need to calculate the acceleration

Using formula of acceleration

$a=\omega^2\times x$

$a=(2\pi\times f)^2\times x$

Put the value into the formula

$a=(2\pi\times2.07)^2\times3.13\times10^{-2}$

$a=5.29\ m/s^2$

We need to calculate the coefficient of static friction between the tray and the cup

Using formula of normal force

$F = ma$

For slipping just start,

Using frictional force

$F=F_{f}$

$ma=\mu mg$

$a=\mu g$

$\mu=\dfrac{a}{g}$

Put the value into the formula

$\mu=\dfrac{5.29}{9.8}$

$\mu=0.53$

Hence, The coefficient of static friction between the tray and the cup is 0.53.