Question

two masses mand M are connected at two ends of an inextensible string .The string passes over a smooth frictionlesd pulley .Other accelration of the masses and the tension in string

Answer 1

We regard first M and m forming a single system. Then,applying Newton’s second law,we have

(M+m) a=F ……………(1) Let F be applied on M. From free body diagram of M we have following equation.

Ma=F-T………………….(2) Therefore,T=F-Ma .Substituting for a from Eq.(1),we get,

T=F-M(F/M+m)=Fm/M+m…………(3).

This tension is the force on m. Then, writing Eq. Of motion for m,

ma=T=Fm/(M+m). But if we assume the force F applied on m then the tension will be T=FM/(M+m).

Answer 2

Answer- The above question is from the chapter 'Laws of Motion'.

Given question: Two masses m and M are connected at two ends of an inextensible string. The string passes over a smooth friction-less pulley. Find other acceleration of the masses and the tension in string.

Answer: If two blocks of masses m and M are connected by light inextensible string passing over a smooth fixed pulley of negligible mass such that M > m,

Forces acting on M block:-

1) Weight, Mg vertically downwards

2) Tension, T vertically upwards

Forces acting on m block:-

1) Weight, mg vertically downwards

2) Tension, T vertically upwards

Let their acceleration be 'a'.

The equations are:-

Mg - T = Ma --- (1)

T - mg = ma --- (2)

Adding both equations, we get,

Mg - mg = Ma + ma

(M - m)g = (M + m)a

a = (M - m)g/(M + m)

Put value of a in equation 1, we get,

Mg - T = M × (M - m)g/(M + m)

T = Mg - M × (M - m)g/(M + m)

T = Mg (M + m - M + m)(M + m)

T = 2Mmg/(M + m)

So,

$\boxed{a = \dfrac{(M - m)g}{(M + m)}}\\\boxed{T = \dfrac{2Mmg}{M +m}}$