Question

ghinient MESSAGE My INSTRUCTOR FULL SCREEN « ВАСK NEXT ВCES Chapter 10, Problem 24 08 Your answer is partially correct. Try again. m 49 n 65 m 24 m 40 A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f 1.86 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 4.96 cm. Number 0.4668 Units No units the tolerance is +/-2 96 Click if you would like to Show Work for this question: Open Show Work SHOW HINT m LINK TO TEXT

Please make sure the answer is clear and stated with units. thanks!

Answer 1

Given,

The frequency, f = 1.86 Hz

The amplitude of the motion, A = 4.96 cm = 0.0496 m

We know,

x = A sin $\omega$ t

dx/dt = A $\omega$ cos $\omega$ t

Now,

The acceleration, a = d²x / dt² = - A $\omega$ ² sin $\omega$ t

a = - A $\omega$ ² sin $\omega$ t

$\omega$ = 2$\pi$f = 2$\pi$ * 1.86

$\omega$ = 11.68 rad/s

The maximum acceleration is, a = A $\omega$ ²

a = 0.0496 * 11.68²

= 6.77 m/s²

The maximum frictional force, F_max = $\mu$ mg

ma = $\mu$ mg

The coefficient of static friction, $\mu$ = a /g

$\mu$ = 6.77 / 9.8

= 0.691