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

Question

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.02 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 4.48 cm.

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=2.02Hz, so sqrt[k/m]=2 pi f and k/m=(2.02*2 pi)^2 now, in simple harmonic motion, we have that F=ma = kx, so that a =(k/m) x; we know k/m and we are told that slipping occurs when x=0.048m, therefore, the acceleration needed to make the cup slip is a=k/m * x just put all values and calculate a since now we know that a= ug, this means that u=a/g=a/9.8

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.