Question

A pulley-rope-mass arrangement is shown in the figure. Find the acceleration of block $m_{1}$ when the masses are set free to move. Assume that the pulley and the ropes are ideal.

Accepted Answer

Method 1: Constraint relations.

For the upper string, the length of string $l_{1}$ not to change and for this string not to slacken, acceleration of $m_{1}$ with respect to the fixed pulley
$=$ acceleration of the movable pulley with respect to the fixed pulley

or $|a_{1}| = |a_{P}| \dots (1) .$

Constraint relations for string $2$

Let us assume that block $m$ is moving up with acceleration $a_{1}$ and blocks $m_{2}$ and $m_{3}$ are moving down with acceleration $a_{2}$ and $a_{3},$ respectively, with respect to ground.

Total length of string $2$ is constant.

Therefore,

$l_{2} = (x_{2} - x_{P_{2}}) + (x_{3} - x_{P_{2}}) + l_{0}^{'} \dots (2)$

where $l_{0}$ is the length of string on pulley $2$ passing through the pulley $P_{2}$

$l_{2} = x_{2} + x_{3} - 2 x_{P_{2}} + l_{0}^{'}$

Differentiating equation $(2)$ with respect to time,

$\frac{d l_{2}}{d t} = 0$ (as total length of string is constant)

$\frac{d l_{0}^{'}}{d t} = 0$ (length of the string over pulley is constant)

Differentiating equation $(2)$ twice with respect to time

$2 a_{1} = a_{3} + a_{2} \dots (3)$

Method 2: The acceleration of block $m_{1}$ and pulley $2$ will be same in magnitude. Now considering pulley $2$ and block $2$ and $3 .$

Change in the length of segment $(1)$ ,

$Δ l_{1} = (+ y_{2}) + (- y_{P_{2}}) = y_{2} - y_{P_{2}}$

Change in the length of segment $(2)$ ,

$Δ l_{2} = (y_{3}) + (- y_{P_{2}}) = y_{3} - y_{P_{2}}$

Total sum change in the segment length of string should be zero.

$Δ l = 0 = [y_{2} - y_{P_{2}}] + [y_{2} - y_{P_{2}}]$

$\Rightarrow Δ l = 0 = y_{2} + y_{3} - 2 y_{P_{2}}$

$\Rightarrow 2 y_{P_{2}} = y_{3} + y_{2} \Rightarrow 2 a_{P_{2}} = a_{3} + a_{2}$

or $2 a_{1} = a_{3} + a_{2}$

This equation is same as equation $(3)$ calculated in method 2.

B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 52: ILLUSTRATION 6.47

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 52: ILLUSTRATION 6.47

Questions from B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 52: ILLUSTRATION 6.47 with Hints & Solutions

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