Question

Two blocks connected by an inextensible string passing over a light frictionless pulley are
resting on two smooth inclined planes as shown in figure. Determine the acceleration of the
blocks and the tension in the string. Assume the string to be massless.

Answer 1

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

Given question: Two blocks connected by an inextensible string passing over a light friction-less pulley are resting on two smooth inclined planes as shown in figure. Determine the acceleration of the blocks and the tension in the string. Assume the string to be mass-less.

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,

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

Answer 2

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

Given question: Two blocks connected by an inextensible string passing over a light friction-less pulley are resting on two smooth inclined planes as shown in figure. Determine the acceleration of the blocks and the tension in the string. Assume the string to be mass-less.

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,

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

a=

(M+m)

(M−m)g

T=

M+m

2Mmg