Two blocks A and B of masses 5 ,kg and 3 ,kg | vedclass

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(C) Let the force $F$ be applied to block $A$ $(5 \,kg)$.Since there is no relative slipping,both blocks move with the same acceleration $a$.The total

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$4$

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(C) Let the force $F$ be applied to block $A$ $(5 \,kg)$.Since there is no relative slipping,both blocks move with the same acceleration $a$.The total mass of the system is $M = 5 \,kg + 3 \,kg = 8 \,kg$.The acceleration of the system is $a = \frac{F}{M} = \frac{F}{8}$.For the upper block $B$ $(3 \,kg)$,the only horizontal force causing acceleration is the static friction $f$ between $A$ and $B$.Thus,$f = m_B \cdot a = 3 \cdot \left(\frac{F}{8}\right) = \frac{3F}{8}$.The maximum value of static friction is $f_{max} = \mu \cdot N = \mu \cdot m_B \cdot g = 0.5 \cdot 3 \cdot 10 = 15 \,N$ (taking $g = 10 \,m/s^2$).For no relative slipping,$f \leq f_{max}$,so $\frac{3F}{8} \leq 15$.$3F \leq 120 \implies F \leq 40 \,N$.Since $1 \,kg \,wt. = 10 \,N$,the maximum force in $kg \,wt.$ is $\frac{40}{10} = 4 \,kg \,wt.$

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