Ramesh V
Last Activity: 15 Years ago
Free body diagrams of two blocks are shown below:
on balancing forces on each block in X and y axis respectively, we have:
for A:
N -150 = 7.5a
f = uN =13a
for B:
0.866R - N =150 + 20a
433 -0.5R -uN = 34.64a here u is coeff. of friction btn 2 blocks
on solving above equations, we have
N = 150 + 7.5a
433 - 47.64a = 0.5R
so, its 750 - 82.5a = 300 + 27.5a
450 = 110a
the acc of the system a = 45 / 11 m/sec2
the coeff. of friction is
u(150 + 20a ) = 13a
so u = 13a / (150+20a)
The coeff. of friction is u = 0.2942 or 9.(3)1/2 / 53.
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