The blocks A and B are arranged as shown in t | vedclass

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(A) For the motion to start,the tension $T$ in the string must be equal to the limiting friction force acting on block $A$.Let $m_A = 10 \, kg$ be the

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$2$

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(A) For the motion to start,the tension $T$ in the string must be equal to the limiting friction force acting on block $A$.Let $m_A = 10 \, kg$ be the mass of block $A$ and $m_B$ be the mass of block $B$.The limiting friction force on block $A$ is $f_L = \mu N = \mu m_A g$,where $\mu = 0.20$ is the coefficient of friction.The tension $T$ in the string is provided by the weight of block $B$,so $T = m_B g$.For the motion to start,$T = f_L$.Substituting the values: $m_B g = \mu m_A g$.$m_B = \mu m_A = 0.20 	imes 10 \, kg = 2 \, kg$.Thus,the minimum mass of $B$ required to start the motion is $2 \, kg$.

The blocks $A$ and $B$ are arranged as shown in the figure. The pulley is frictionless. The mass of $A$ is $10 \, kg$. The coefficient of friction of $A$ with the horizontal surface is $0.20$. The minimum mass of $B$ to start the motion will be...... $kg$.

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