A rope of mass m is sliding down a pulley. What will be the force exerted by the rope on the pulley?
Assumption: Pulley is massless; All surfaces are frictionless.
Let the mass per unit length of the rope be !!dm!!
Let the length and tension on shorter side be !!l_1!! and !!T_1!! respectively,
Let the length and tension on longer side be !!l_2!! and !!T_2!! respectively.
Let the downward acceleration of the rope be !!a!!
The equations of motion of the rope on shorter and longer side respectively are:
!!T_1- (dml_1)g = (dml_1)a!!
!! (dml_2)g - T_2 = (dml_2)a!!
!!T_1 + T_2 = dml_1(a+g) + dml_2(g-a)!!
!!= dm(l_1+l_2)g + dm(l_1-l_2)a!!
!!= mg - dm(l_2-l_1)a!!
Since all quantities in the second term are positive, !!T_1 + T_2!! (net tension) will be less than mg. Therefore the force exerted by the rope on the pulley will be less than mg.
Since the length of the rope is increasing on one side, we can conclude that the centre of mass of the rope is accelerating downwards. Its acceleration cannot be greater than g (as no external force is acting on the system besides gravity) or equal to g (as it is not a freely falling body). Therefore, since the net acceleration is less than g, the difference between its weight and the net downward force is reflected on the force acting between the pulley and the rope, which will be less than mg.
Equation: mg - F = ma
(Since g and a are positive, F will be less than mg)