A rope of negligible mass can support a load of M kg. Prove that the mass of the greatest load which can be raised is equal to ` (M )/(1+ (2h)/(g t^2))`kg, where g is the acceleration due to gravity and h is the height through which the said load rises from rest with uniform acceleration in time t.

Explanation

Since the rope can support a load of M kg, the maximum tension the rope can withstand is given by T = Mg (figure a). Now, if a mass m is raised by the rope with a uniform acceleration a as shown in figure (b), then the net

Upward force on the mass m = (T" - mg) where T. is the tension in the string. Then, from Newton.s second law of motion,
` T. = mg =ma or T. =m(g+a)`
for the greatest value of m
` T.=T therefore m (g +a) = Mg `
` therefore m ( Mg)/(g+a) = (M )/(1 +a//g)`
Now, from the equation of motion, `s=ut + 1/2 at^2`
We get ,`h= 0 +1/2 a t^2 or a = (2h)/(t^2)`
from equation (i ) `m=(M )/(1 +(2h)/(g t^2) ) kg .`