A block of mass M is placed on a horizontal s | vedclass

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Question

(A) Let the acceleration of the system be $a$. Since all three blocks move together,they share the same acceleration $a$. The total mass of the system

Vedclass · Accepted Answer

$\frac{m}{m + m_0 + M}$

Vedclass · Answer

(A) Let the acceleration of the system be $a$. Since all three blocks move together,they share the same acceleration $a$. The total mass of the system is $(M + m_0 + m)$. The driving force is the weight of block $m$,which is $mg$. Applying Newton's second law to the whole system: $mg = (M + m_0 + m)a$,so $a = \frac{mg}{M + m_0 + m}$.Now,consider the block $m_0$. The only horizontal force acting on $m_0$ is the static friction $f$ exerted by block $M$. For $m_0$ to move with acceleration $a$,we must have $f = m_0 a$.The maximum value of static friction is $f_{max} = \mu N = \mu m_0 g$,where $N = m_0 g$ is the normal force between $m_0$ and $M$.For the blocks to move together,the required friction must not exceed the maximum static friction: $f \le f_{max}$.Substituting the expressions: $m_0 a \le \mu m_0 g$.$a \le \mu g$.$\frac{mg}{M + m_0 + m} \le \mu g$.$\mu \ge \frac{m}{M + m_0 + m}$.Thus,the minimum value of $\mu$ is $\frac{m}{M + m_0 + m}$.

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