A mass m 1 = 6.7 kg rests on a frictionless table. It is connected by a massless

Question

A mass m1 = 6.7 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m2= 2 kg that hangs freely.

Now the table is tilted at an angle of = 80° with respect to the vertical. Find the magnitude of the new acceleration of block 1.

1- Now the angle is decreased past the “critical” angle so the system accelerates in the opposite direction. If = 34° find the magnitude of the acceleration?

2-

Compare the tension in the string in each of the above cases on the incline:

T at 80° = Tcritical = T at 34°

T at 80° > Tcritical > T at 34°

T at 80° < Tcritical < T at 34°

Explanation / Answer

Theta = 80
let acc of the blocks be a
m1gcos(theta) - T = m1a
T - m2g = m2a
6.7*9.8cos(80) - 2*9.8 = a*[6.7 + 2] ; a = -0.942 m/s/s i.e. a = 0.942 m/s/s in the direction of m2

1. 6.7*9.8cos(34) - 2*9.8 = a*[8.7]; a = 4 m/s/s in the opposite direction

2. Critical case : a = 0, so Tc = m2g = 2*9.8 = 19.6N
Case 80: T = 6.7(9.8cos(80) + 0.942) = 17.71 N
Case 34: T = 6.7(9.8cos(35) - 4) = 26.98 N
Critical angle => cos(theta) = [m1a + T]/[m1g] => theta = 78.32 degress with vertical

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