Let us consider both loads as points of matter lying at the loads' respective centres of gravity. Load M₂ begins to descend with a linear acceleration a, which is not equal to g, since the presence of the load M₁ causes the rod to exert an upward force F on M₂. Therefore, from the second law of Newton, we have
M₂a = M₂g - F (1)
Load M₁ begins to rise with the same acceleration a, since the presence of load M₂ causes the rod to exert an upward force of F' on M₁. Therefore,
M₁a = -M₁g + F' (2)
Since we can neglect the mass of the rod, the moments of the forces F and F', acting on the rod, relative to the axis of rotation, must be equal and opposite and their sum must accordingly be zero. (Otherwise a rod of infinitely small mass would be given infinitely great angular acceleration.) But the distance of these forces from the fulcrum is the same for both, therefore F = F'.
Then from equations (1) and (2) we get:
a = (M₂ - M₁)g/(M₁ + M₂) ; F = 2M₁M₂g/(M₁ + M₂)
The resultant of forces F and F' is equal and opposite to the force of the rod's pressure on its axis at the initial moment. Consequently this last force equals
R = 4M₁M₂g/(M₁ + M₂) = 3·5 kg.
THE DYNAMICS OF MOTION IN A STRAIGHT LINE 109
The problem we have just considered is in principle no different from problem 42. In this one the role of the pulley was played by the rod, fixed at its mid-point.