Two blocks of masses 2 ,kg and 4 ,kg are hang | vedclass

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(A) In the frame of the elevator,both blocks experience a pseudo force $F_p = ma$ acting downwards,where $a = \frac{g}{2}$.For the $4 \,kg$ block: The

Vedclass · Accepted Answer

$40$

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(A) In the frame of the elevator,both blocks experience a pseudo force $F_p = ma$ acting downwards,where $a = \frac{g}{2}$.For the $4 \,kg$ block: The effective downward force is $W_{eff} = m_1(g + a) = 4(g + \frac{g}{2}) = 4(\frac{3g}{2}) = 6g$.The equation of motion for the $4 \,kg$ block is $6g - T = 4a_{rel}$,where $a_{rel}$ is the acceleration relative to the elevator.For the $2 \,kg$ block: The effective downward force is $W_{eff} = m_2(g + a) = 2(g + \frac{g}{2}) = 2(\frac{3g}{2}) = 3g$.The equation of motion for the $2 \,kg$ block is $T - 3g = 2a_{rel}$.Adding the two equations: $(6g - T) + (T - 3g) = 4a_{rel} + 2a_{rel} \Rightarrow 3g = 6a_{rel} \Rightarrow a_{rel} = \frac{g}{2}$.Substituting $a_{rel}$ into the second equation: $T = 3g + 2(\frac{g}{2}) = 3g + g = 4g$.Given $g = 10 \,m/s^2$,$T = 4 	imes 10 = 40 \,N$.

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