A tray is moved horizontally back and forth in simple harmonic motion at frequen

Question

A tray is moved horizontally back and forth in simple harmonic motion at frequency 2.00 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 5.00

A tray is moved horizontally back and forth in simple harmonic motion at frequency 2.00 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 5.00 ´ 10-2 m.

Explanation / Answer

focusing for a second only on the slipping issue, we know slipping will occur when the force acting ont he cup exceeds static friciton, or slipping will occur when

F=ma > u mg or when a > u g

how do we find a? consider the tray as an oscillator in simple harmonic motions; the frequency of motion is given by

f = (1/2pi)sqrt[k/m] where k is the spring constant and m the mass

we are told neither k nor m, but we know their ratio:

f=2 Hz, so sqrt[k/m]=2 pi f and

k/m=(2*2 pi)^2 = 157.9136
now, in simple harmonic motion, we have that

F=ma = kx, so that

a =(k/m) x; we know k/m=157.9136 and we are told that slipping occurs when x=0.05m, therefore, the acceleration needed to make the cup slip is

a=(157.9136)(0.05)=7.8956 m/s/s

since we know that a= ug, this means that
u=a/g=7.8956/9.8=0.80567

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