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Reading of the spring balance = restoring force of the spring
rnSince M1 >M2, there will be an acceleration in the system (which consists of two masses M1, M2, Massless spring balance, massless strings, fixed pulleys).
rnM1g - M2g = (M1 + M2) a
rna= (M1 – M2) g/(M1 + M2)
rnRestoring force, R= T1 =T2
rnThus, R= T1= M1 (g-a) = M1 (g- (M1 – M2) g/(M1 + M2)) =M1g (M1 + M2 – (M1 – M2))/ (M1 + M2)
rn = M1g (2M2)/ (M1 + M2) = 2M1M2g/ (M1 + M2)
rnThe reading of Reading of the spring balance (in unit of mass) is thus = 2M1M2/ (M1 + M2)
rn