Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of st

Question

Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of static friction between the block and the ramp is 0.340, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

Strategy This is an application of Newton's second law involving an object in equilibrium. Choose tilted coordinates, as in Figure 4.21. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, fs = µsn.

.(a) What is the new coefficient of static friction if the maximum angle turns out to be 31.4°? µs = ?

. (b) Find the maximum static friction force that acts on the block. fs = ?

Please be as detailed as possible, thanks.

Explanation / Answer

along the ramp,
m g sin(theta) - f = 0

m g sin(theta) - us m g cos(theta) = 0

tan(theta) = us

theta = tan^-1(0.340) = 18.8 deg ......Abns

(A) us = tan31.4 = 0.61

(b) fs = m g sin31.4 = us m g cos31.4

fs = 2.80 x 9.8 x sin31.4 = 14.3 N

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