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(A) Let the tension in the string connected to block $A$ be $T$. The tension in the string connected to block $B$ and the spring balance is $T'$.Since

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$\frac{10}{13} \, kg$

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(A) Let the tension in the string connected to block $A$ be $T$. The tension in the string connected to block $B$ and the spring balance is $T'$.Since the pulley connected to the spring balance is movable,if block $B$ moves down with acceleration $a$,the pulley moves up with acceleration $a/2$.For block $B$ $(2 \, kg)$: $2g - T' = 2a \quad ...(1)$For the movable pulley: $2T' - T = 0 	imes (a/2) \Rightarrow T = 2T' \quad ...(2)$For block $A$ $(5 \, kg)$: $T = 5a_A$. Since the string length is constant,the acceleration of block $A$ is $a_A = a/2$. So,$T = 5(a/2) \Rightarrow a = 2T/5 \quad ...(3)$Substituting $T = 2T'$ into $(3)$: $a = 2(2T')/5 = 4T'/5$.Substituting $a$ into $(1)$: $2g - T' = 2(4T'/5) = 8T'/5$.$2g = T' + 8T'/5 = 13T'/5 \Rightarrow T' = 10g/13$.The reading of the spring balance is $T' = 10/13 \, kg$ (in terms of force equivalent).

If all the pulleys are massless and the string is ideal,find the reading of the spring balance.

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