Q. 39

# Figure (5-E21) sh Given: i) Given, mass of the man = 60 kg

Let W‘= apparent weight of the man in this case

From the free-body diagram of the man, W‘+ T − 60g = 0

T = 60g W (i)

From the free-body diagram of the box,

T − W‘= 30g = 0 …(ii)

From equation (i), we get: 60g − W‘− W‘− 30g = 0

W= 15g

Hence, the weight recorded on the machine is 15 kg.

(ii) To find his actual weight, suppose the force applied by the men on the rope is T because of which the box accelerates upward with an acceleration a‘ .Here we need to find

Correct weight = W = 60g

From the free-body diagram of the man, T’ + W − 60g − 60a = 0

T 60a = 0

T = 60a (i)

From the free-body diagram of the box, T’ − W − 30g − 30a = 0

T 60g 30g 30a = 0

T = 30a 900 (ii)

From equations (i) and (ii), we get: T’ = 2T’ – 1800 = T’ = 1800 N

So, the man should exert a force of 1800 N on the rope to record his correct weight on the machine

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