A block with mass 22.8 is placed on an inclined plane with slope angle 27.4 and

Question

A block with mass 22.8 is placed on an inclined plane with slope angle 27.4 and is connected to a second hanging block that has mass 6.6 by a cord passing over a small, frictionless pulley. The coefficient of static friction is 0.48 and the coefficient of kinetic friction is 0.27. If the inclined slope is frictionless, what is the acceleration of the 2 masses? Use up the incline as your + x direction. What is the tension in the cord in this case? Assume friction is down the incline. If friction is down the incline and the 2 blocks are not moving, what is the value of friction? What is the maximum static friction?

Explanation / Answer

case 1: incline is frictionless

The hanging mass(m) will fall and mass on incline(M) will move in +ve x direction.
mg - T = ma ; T - Mg sin 27.4 = Ma
adding and substituting the value, we get a = -1.3 m/s2

T = m(g-a) = 73.26 N

Friction is down the incline, means body has tendency of moving up the incline.
mg - T = ma ; T - Mg sin27.4 - sMg cos27.4 = Ma

Adding and putting the value, a = -4.536 m/s2

T = m (g-a) = 94.62 N

f = T - mg sin27.4 - ma = 27.56 N

For Maximum static friction , acceleration = 0.

f = mg - Mg sin 27.4 = -38.14 N

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