Questions: Two blocks of masses m and M are suspended as shown above by strings of negligible mass. If a person holding the upper string lowers the blocks so that they have a constant downward acceleration a, the tension in the string at point P is

Solution Steps

The forces acting on mass \(m\) are the tension \(T\) upwards and the weight \(mg\) downwards. Since the block is accelerating downwards with acceleration \(a\), the net force is downwards. Therefore, \(mg - T = ma\) \(T = mg - ma = m(g - a)\)

The forces acting on point P are the tension in the upper string, \(T'\), acting upwards, and the tension in the lower string, \(T\), acting downwards. The mass of the strings is negligible. Since the system of masses \(m\) and \(M\) is accelerating downwards with acceleration \(a\), point P is also accelerating downwards with acceleration \(a\). \(T' - T = (m+M)a\) \(T' - m(g - a) = (m+M)a\) \(T' = ma + Ma + mg - ma\) \(T' = M a + m g\)

The tension in the string at point \(P\) is the tension in the upper string, \(T'\). \(T' = M a + m g\)

Final Answer