In the arrangement shown in figure find the tensions in the rope and accelerations of the masses $m_{1}$ and $m_{2}$ and pulleys $P_{1}$ and $P_{2}$ when the system is set free to move. Assume the pulleys to be massless and strings are light and inextensible.

p) T (i) m₁g

q) a₁ (ii) m₂g

r) a₂ (iii) (m₁ - m₂)g

s) ap₁(iv) g downward

t) ap₂ (v) g upward

(vi) 3g downward

p -(ii); q-(vi); r- (iv); s-(v); t-(iv)

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p -(i); q-(vi); r- (v); s-(vi); t-(iv)

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p -(i); q-(iv); r- (iv); s-(vi); t-(iv)

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p -(ii); q-(v); r- (v); s-(iv); t-(vi)

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Solution

The correct option is C
p -(i); q-(iv); r- (iv); s-(vi); t-(iv)

Free body diagrams

Applying the equations of motion for pulleys and blocks

For $m_{1}$ : T - $m_{1} g = m_{1} a_{1}$ ........(i)

For $m_{2}$ : $m_{2} g - T_{2} = m_{2} a_{2}$ .........(ii)

For pulley $P_{1}$ : 2T - T = $0 \times a_{p 1}$ ...(iii)

For pulley $P_{2} : T_{2} - 2 T = 0 \times a_{p 2}$ ....(iv)

From Eq (iii), T = 0; From Eq. (iv), $T_{2} = 0$ , and from Eqs. (i) and (ii), $a_{1} = - g$

Also $a_{2} = g$ . Hence block $m_{1} a n d m_{2}$ both will move down with acceleration g in downward direction.

Calculating $a_{p 1}$ and $a_{p 2}$ :

$a_{p 2}$ will be same as the acceleration of block $m_{2}$ so $a p_{2} = a_{2} = g$

Constraint relations:

Consider the reference line and the position vectors of the pulleys and masses as shown in figure write the length of the rope in terms of position vectors and differentiate it to obtain the relations between accelerations of the masses and pulleys.

·For the length of the string connecting $P_{2}$ and $m_{2}$ not to change and for this rope not to slacken. $a_{p 2} = a_{2} = g$ .

Length of the string connecting $P_{1}$ to $m_{1}$ not to change and for this rope not to slacken:

$l = (x_{1} - x_{p 1}) + x_{p 1} + (x_{p 2} - x_{p 1}) + x_{p 2}$

Differentiating this equation with respect to, t twice,

we get $\frac{d^{2} l}{d t^{2}} = \frac{d^{2} x_{1}}{d t^{2}} - \frac{d^{2} x_{p 1}}{d t^{2}} + 2 \frac{d^{2} x_{p z}}{d t^{2}} \Rightarrow 0 = - a_{1} + a_{p 1} + 2 a_{p z}$

$\frac{d^{2} x_{1}}{d t} = - V$ as $x_{1}$ is decreasing by our assumed acceleration

$\frac{d x_{p 1}}{d t} = - V_{p 1}$ as $P_{1}$ is decreasing by our assumed acceleration

$\frac{d x_{p 2}}{d t} = V_{p 2}$

Now $a_{1} = - g$

$a_{p 2} = a_{2} = g$

$\Rightarrow a_{p 1} = - 3 g$

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Similar questions

In the arrangement shown in figure find the tensions in the rope and accelerations of the masses $m_{1}$ and $m_{2}$ and pulleys $P_{1}$ and $P_{2}$ when the system is set free to move. Assume the pulleys to be massless and strings are light and inextensible.