A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure

Question

A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a fixed horizontal ice-slab and being pulled by 2 Mg force. In this situation. Tension in the string is $\frac{x}{5}$ Mg for $x$ = __________. Neglect mass of the string and friction of the block (bigger mass) with ice slab. (Given g = acceleration due to gravity)

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(6)

$For 4 m$

$2 M g - T = 4 M a$ $. . . (i)$

$For M$

$T - m g = M a$ $. . . (i i)$

$Adding (i) & (ii), we get$

$M g = 5 M a \Rightarrow a = \frac{g}{5}$

$So, T = M a + M g = \frac{M g}{5} + M g = \frac{6}{5} M g$

Solution

Ans.
(6)

$For 4 m$

$2 M g - T = 4 M a$ $. . . (i)$

$For M$

$T - m g = M a$ $. . . (i i)$

$Adding (i) & (ii), we get$

$M g = 5 M a \Rightarrow a = \frac{g}{5}$

$So, T = M a + M g = \frac{M g}{5} + M g = \frac{6}{5} M g$

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Solution

Ans. (6) For 4m 2Mg-T=4Ma ...(i) For M T-mg=Ma ...(ii) Adding (i) & (ii), we get Mg=5Ma⇒a=g5 So, T=Ma+Mg=Mg5+Mg=65Mg

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Ans.
(6)

$For 4 m$

$2 M g - T = 4 M a$ $. . . (i)$

$For M$

$T - m g = M a$ $. . . (i i)$

$Adding (i) & (ii), we get$

$M g = 5 M a \Rightarrow a = \frac{g}{5}$

$So, T = M a + M g = \frac{M g}{5} + M g = \frac{6}{5} M g$