Two blocks A and B of masses 4 kg and 6 kg | vedclass

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(A) The force required to make block $A$ slip over block $B$ is the limiting friction force $f_{max} = 12 \ N$.For block $A$ to move with block $B$ wi

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The force required to make block $A$ slip over block $B$ is the limiting friction force $f_{max} = 12 \ N$.For block $A$ to move with block $B$ without slipping,the maximum acceleration $a_{max}$ of the system is determined by the maximum friction force acting on $A$:$f_{max} = m_A \cdot a_{max}$$12 \ N = 4 \ kg \cdot a_{max}$$a_{max} = 3 \ m \ s^{-2}$Now,consider the system of both blocks $A$ and $B$ moving together with acceleration $a_{max}$. The total mass of the system is $M = m_A + m_B = 4 \ kg + 6 \ kg = 10 \ kg$.The maximum horizontal force $F_B$ that can be applied on $B$ is:$F_B = M \cdot a_{max}$$F_B = 10 \ kg \cdot 3 \ m \ s^{-2} = 30 \ N$.

Two blocks $A$ and $B$ of masses $4 \ kg$ and $6 \ kg$ are as shown in the figure. $A$ horizontal force of $12 \ N$ is required to make $A$ slip over $B$. Find the maximum horizontal force $F_B$ that can be applied on $B$ so that both $A$ and $B$ move together. (Take $g=10 \ m \ s^{-2}$) (in $N$)

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