Question

Two blocks of mass $M$ and $3 M$ (in $kg$ ) are connected by an inextensible light thread which passes over a light frictionless pulley. The whole system is placed over a fixed horizontal table as shown below. The coefficient of friction between the blocks and table surface is $μ$ . The pulley is pulled by a string attached to its center and accelerated to the left with acceleration $a (m s^{- 2})$ . Assume that gravity acts with constant acceleration $g (m s^{- 2})$ downwards through the plane of the table.

(a) Find the horizontal acceleration of both the blocks. (Assume that both the blocks are moving.)

(b) What is the maximum acceleration $a$ , for which the block of mass $3 M$ will remain stationary?

Accepted Answer

(a) Assuming that, the acceleration of blocks of mass $M$ and $3 M$ are $a_{1}$ and $a_{2}$ respectively and applying the pulley constraint relation, we have,

$a_{pulley} = \frac{a_{1} + a_{2}}{2} = a$ ............(i)

By observing the free-body diagram of block $M$ and balancing the horizontal forces, we get,

$T - μ M g = M a_{1}$ .................(ii)

From the free-body diagram of block of mass $3 M$ and balancing the horizontal forces again, we get,

$T - 3 μMg = 3 {Ma}_{2}$ .................(iii)

From (ii) and (iii),

$- μ M g + 3 μ M g = M a_{1} - 3 M a_{2}$

$\Rightarrow 2 μ g = a_{1} - 3 a_{2}$ ..............(iv)

From (i) and (iv),

$2 μ g = (2 a - a_{2}) - 3 a_{2}$

$\Rightarrow a_{2} = \frac{a - μ g}{2}$ ...................(v)

Again from (i) and (v),

$a_{1} = 2 a - a_{2} = 2 a - (\frac{a - μ g}{2}) = \frac{3 a}{2} + \frac{μ g}{2}$

$\Rightarrow a_{1} = \frac{3 a + μ g}{2}$

(b) Let the acceleration of block of mass $M$ be $a_{1}$ and that of block of mass $3 M$ be $a_{2}$ . Since, $3 M$ to remain at rest $a_{2} = 0 .$

With the help of string constraint, we have,

$a = \frac{a_{1} + a_{2}}{2}$

Acceleration $a$ will be maximum when $a_{1}$ is maximum ( $a_{2} = 0$ ). Acceleration $a_{1}$ is maximum when the tension in the string is maximum.

For the block of mass $3 M$ to be stationary,

$T \leq 3 μ M g$

$T_{\max} = 3 μ M g$

For the block of mass $M$ ,

$T_{\max} - μ M g = M a_{1}$

$\Rightarrow 3 μ M g - μ M g = M a_{1}$

$\Rightarrow 2 μ g = a_{1}$

So, the maximum acceleration $a$ ,

$a = \frac{a_{1} + a_{2}}{2} = \frac{2 μ g + 0}{2} = μ g$

Important Questions on Newton's Laws of Motion II

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