Question

5. A simple Atwood machine consists of two masses m₁ and m₂ that are connected by a string wound over a pulley, as seen in the figure below. Assume m₂ is larger than m₁. Motion in the upward direction is positive. On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass. Then answer the following questions.

(b) Using the direction rosette indicate the direction for each of the forces acting on m₁ that you identified in part (a). If any of the forces listed below is not applicable, choose 'NA'.

(c) Identify the forces acting on the mass m₂. (Select all that apply.)

(d) Using the direction rosette indicate the direction for each of the forces acting on m₂ that you identified in part (c). If any of the forces listed below is not applicable, choose 'NA'.

Answer 1

The force on m1 are

1. Tension T in the upward

2 Weight m1g in the downward

Now the net force on m1 is T -m1g

Then T -m1g =m1a

Then the acceleration is given by a =(T/m1) -g

2.

Forces acting on m2 are
1. Weight m2g downward
2. Tension T upward
Net downward force on m2 = m2g - T
Therefore m2g - T = m2a
Or a = (m2g - T)/m2
Or a = g - T/m2----------------(2)
From (1) and (2)
T/m1 - g = g - T/m2
Or T/m1 + T/m2 = g + g
Or T(1/m1 + 1/m2) = 2g
Or T(m1+m2)/(m1m2) = 2g
Or T = 2m1m2g/(m1+m2)