In this near earth situation, the pulley has mass 0.63kg, radius 0.32m, and enco
ID: 2260867 • Letter: I
Question
In this near earth situation, the pulley has mass 0.63kg, radius 0.32m, and encounters a frictional torque of 0.29N*m about the pulley axis. The hanging mass is 3.4kg, the mass on the plane is 1.6 kg, the plane is inclined 57.3 degrees above the horizontal and has kinetic friction coefficient of 0.15. Use dynamics to determine the system acceleration, string tensions, and the speed after the mass on the plane has moved 0.83m.
Explanation / Answer
m1 a = m1 g - T1
==> 3.4 * a = 3.4 * 9.8 - T1
==> T1 = 33.32 - 3.4 * a
m2 a = T2 - m2 g sin(57.3) - u m2 g cos(57.3)
==> 1.6 * a = T2 - 1.6 * 9.8 * 0.84151 - 0.15 * 1.6 * 9.8 * 0.54024
==> 1.6 * a = T2 - 11.924
==> T2 = 11.924 + 1.6 * a
alpha = a/R
==> alpha = a/0.32
I = 0.5 m3 R^2 = 0.5*0.63*0.32*0.32 = 0.032256
R (T1 - T2) = I alpha
==> 0.32 * ((33.32 - 3.4 * a) - (11.924 + 1.6 * a)) = 0.032256 * a/0.32
==> a = 4.02559
==> system acceleration = a = 4.03 m/s2
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string tensions:
T1 = 33.32 - 3.4 * a = 33.32 - 3.4 * 4.02559 = 19.6 N
T2 = 11.924 + 1.6 * a = 11.924 + 1.6 * 4.02559 = 18.4 N
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the speed after the mass on the plane has moved 0.83m:
v = sqrt(2 a d) = sqrt(2*4.02559*0.83) = 2.59 m/s
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