Find the minimum value of mass m that should | vedclass

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(B) For the system to remain at rest,the tension $T$ in the string must be balanced by the limiting friction $f_L$ acting on the $20 \, kg$ block.The

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$20$

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(B) For the system to remain at rest,the tension $T$ in the string must be balanced by the limiting friction $f_L$ acting on the $20 \, kg$ block.The tension $T$ is equal to the weight of the hanging block: $T = 10 \, g$.The normal force $N$ acting on the $20 \, kg$ block is $N = (m + 20) g$.The limiting friction is $f_L = \mu N = 0.25(m + 20) g$.Equating the forces for equilibrium: $T = f_L$.$10 g = 0.25(m + 20) g$.Dividing both sides by $g$: $10 = 0.25(m + 20)$.$10 = 0.25m + 5$.$0.25m = 5$.$m = \frac{5}{0.25} = 20 \, kg$.

Find the minimum value of mass $m$ that should be placed on the $20 \, kg$ block so that the system remains at rest. (Given coefficient of friction $\mu = 0.25$) (in $, kg$)

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