A box sits on a level platform. a) First, the platform vibrates horizontally in
ID: 2209931 • Letter: A
Question
A box sits on a level platform.
a) First, the platform vibrates horizontally in simple harmonic motion of constant frequency f = 2.18 Hz, We slowly increase the amplitude of the vibration, and find that the box just begins to slide when the amplitude reaches A = 1.12 cm. Find ?s, the coefficient of static friction between the box and the platform.
1
b) Now, the platform is moving up and down in simple harmonic motion of constant amplitude A = 1.12 cm, and the frequency slowly increases. Find the minimum frequency for which the box will just lose contact with the surface at the top of its motion.
2 Hz
Explanation / Answer
A* (omega^2) = meu *g => Coefficient of static friction 'meu' = 0.0112* [(2*3.14*2.18)^2] / 9.8 = 0.2144 .............. (b) A* ((2*pi*freq.)^2) = g => frequency = 4.7 Hz
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