Two boys of masses 10 kg and 8 kg are moving along a vertical rope, the former climbing up with acceleration of 2 $ms^{- 1}$. 2 while later coming down with uniform velocity of 2 $ms^{- 1}$. Then tension in rope at fixed support will be(Take $g = 10ms^{- 2}$)

Question

Dr. Abubacker · Accepted Answer

(b) Since, $m_{2}$ moves with constant velocity
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$	herefore\ f_{2} = m_{2}g$
$f_{2} = 8 	imes 10 = 80$N
Since, boy of mass $m_{1}$ moves with acceleration $a = 2ms^{- 2}$ in upward direction
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$	herefore f_{1} = m_{1}g = m_{1}a$
$	herefore f_{1} = m_{1}g = m_{1}a$
$= 10 	imes 10 + 10 	imes \  = 120$ N

Q. Two boys of masses 10 kg and 8 kg are moving along a vertical rope, the former climbing up with acceleration of 2 $ms^{- 1}$. 2 while later coming down with uniform velocity of 2 $ms^{- 1}$. Then tension in rope at fixed support will be(Take $g = 10ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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