In the given system,if M_2 = 2M_1,the tension | vedclass

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(A) For the initial configuration where $M_1$ is on the incline and $M_2$ is hanging vertically:The tension $T$ is given by $T = \frac{M_1 M_2 (1 + \s

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$T$

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(A) For the initial configuration where $M_1$ is on the incline and $M_2$ is hanging vertically:The tension $T$ is given by $T = \frac{M_1 M_2 (1 + \sin 	heta)}{M_1 + M_2} g$.Given $M_2 = 2M_1$,we substitute $M_2 = 2M_1$ into the equation:$T = \frac{M_1 (2M_1) (1 + \sin 	heta)}{M_1 + 2M_1} g = \frac{2 M_1^2 (1 + \sin 	heta)}{3 M_1} g = \frac{2}{3} M_1 g (1 + \sin 	heta)$.Now,if the positions are interchanged,$M_2$ is on the incline and $M_1$ is hanging vertically:The new tension $T'$ is given by $T' = \frac{M_2 M_1 (1 + \sin 	heta)}{M_2 + M_1} g$.Substituting $M_2 = 2M_1$ into this equation:$T' = \frac{(2M_1) M_1 (1 + \sin 	heta)}{2M_1 + M_1} g = \frac{2 M_1^2 (1 + \sin 	heta)}{3 M_1} g = \frac{2}{3} M_1 g (1 + \sin 	heta)$.Comparing the two expressions,we see that $T' = T$.

In the given system,if $M_2 = 2M_1$,the tension in the string is $T$. If the positions of the blocks are interchanged,what will be the new tension in the string?

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