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: m_{1} + m_{2} = 3 kg
+ 5 kg = 8 kg (mass of the string is assumed to be zero.)

Fg = m.g

Fg_{1} = (3 kg)(9.8 m/s^{2})
= 29.4 N (weight of the first mass)

Fg_{2} = (5 kg)(9.8 m/s^{2})
= 49 N (weight of the second mass)

Fg_{1 }pulls the system
counter-clockwise, while Fg_{2 }pulls the system clockwise.
Since Fg_{2} is greater than Fg_{1}, the system
moves clockwise.

The net force pulling the system clockwise:
F_{net} = Fg_{2} - Fg_{1} =
m_{2}.g - m_{1}.g = g(m_{2} - m_{1}) = (9.8 m/s^{2})(5 kg - 3 kg) = 49 N
- 29.4 N = 19.6 N

System acceleration:
a = F_{net}
/ (m_{1} + m_{2}) = g(m_{2} - m_{1}) / (m_{1} + m_{2})

System acceleration: a = F_{net}
/ (m_{1} + m_{2}) = g(m_{2} - m_{1}) / (m_{1} + m_{2}) = (9.8 m/s^{2})(5 kg -
3 kg)/ (5 kg + 3 kg) = 19.6 N / 8 kg = 2.45 m/s^{2}

b ) the tension in the rope.

F_{net} = Fg + T

T = F_{net }- 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 m_{1} and left part of
the rope of the bar):

T = m_{1} (a - g)

a = 2.45 m/s^{2} [up] (Since m_{1} is
accelerating up) = - 2.45 m/s^{2}

g = 9.8 m/s^{2} [down] = 9.8 m/s^{2}

T = (3 kg) ( - 2.45 m/s^{2} - 9.8 m/s^{2})
= - 36.75 N = 36.75 N [up]

Second way to calculate the tension (Consider only m_{2} and right part of
the rope of the bar):

T = m2 (a - g)

a = 2.45 m/s^{2} [down] (Since m_{2}
is accelerating down) = 2.45 m/s^{2}

g = 9.8 m/s^{2} [down] = 9.8 m/s^{2}

T = (5 kg) ( 2.45 m/s^{2} - 9.8 m/s^{2})
= - 36.75 N = 36.75 N [up]

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