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PULLEY SAMPLE PROBLEM:

A 3 kg mass and a 5 kg mass are suspended form a frictionless pulleys, as shown in the figure.
a) Determine the system acceleration
b) Determine the tension in the rope.
 

 

SOLUTION:

If we consider these two masses and string as one mass, something like a rug, the solution of the problem will be easier.

Total mass of the system:  m1 + m2 = 3 kg + 5 kg = 8 kg (mass of the string is assumed to be zero.)

Fg = m.g

 Fg1 = (3 kg)(9.8 m/s2) = 29.4 N (weight of the first mass)

 Fg2 = (5 kg)(9.8 m/s2) = 49 N (weight of the second mass)

 Fg1 pulls the system counter-clockwise, while Fg2 pulls the system clockwise.  Since Fg2 is greater than Fg1, the system moves clockwise.

The net force pulling the system clockwise:  Fnet =  Fg2 -  Fg1 =  m2.g - m1.g = g(m2 - m1) = (9.8 m/s2)(5 kg - 3 kg) = 49 N - 29.4 N = 19.6 N

System acceleration:  a = Fnet / (m1 + m2) = g(m2 - m1) / (m1 + m2)

System acceleration:  a = Fnet / (m1 + m2) = g(m2 - m1) / (m1 + m2) = (9.8 m/s2)(5 kg - 3 kg)/ (5 kg + 3 kg) = 19.6 N / 8 kg = 2.45 m/s2

b ) the tension in the rope.

 Fnet =  Fg + T

 T = Fnet - Fg = m.a - m.g = m (a -g)

T = m (a - g)

Since the friction between the rope and the bar is negligible, the tension in the rope will be the same in the every point of the rope.

Let [down] be positive.  We also can assume [up] will be positive.  The result will be the same.

First way to calculate the tension (Consider only m1 and left part of the rope of the bar):

T = m1 (a - g)

a = 2.45 m/s2 [up] (Since m1 is accelerating up) = - 2.45 m/s2

g = 9.8 m/s2 [down] = 9.8 m/s2

T = (3 kg) ( - 2.45 m/s2 - 9.8 m/s2) = - 36.75 N = 36.75 N [up]

Second way to calculate the tension (Consider only m2 and right part of the rope of the bar):

T = m2 (a - g)

a = 2.45 m/s2 [down] (Since m2 is accelerating down) = 2.45 m/s2

g = 9.8 m/s2 [down] = 9.8 m/s2

T = (5 kg) ( 2.45 m/s2 - 9.8 m/s2) = - 36.75 N = 36.75 N [up]

We get the same values for the tension by using by two ways.