A block of mass m_{2} is placed on a horizont | vedclass

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(B) Let $N_{1}$ be the normal force between the two blocks and $N_{2}$ be the normal force between the lower block and the table.For the upper block o

Vedclass · Accepted Answer

$1+\frac{m_{2}}{m_{1}}$

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(B) Let $N_{1}$ be the normal force between the two blocks and $N_{2}$ be the normal force between the lower block and the table.For the upper block of mass $m_{1}$, the normal force is $N_{1} = m_{1}g$.For the lower block of mass $m_{2}$, the total downward force is $N_{2} = m_{2}g + N_{1} = (m_{1} + m_{2})g$.The horizontal force $F$ on the upper block creates a frictional force $f_{1}$ on the lower block, where $f_{1} \leq \mu_{1}N_{1} = \mu_{1}m_{1}g$.For the lower block to never move, the maximum static frictional force from the table must be greater than or equal to the maximum possible frictional force exerted by the upper block on the lower block.Thus, $\mu_{2}N_{2} \geq \mu_{1}N_{1}$.Substituting the values, $\mu_{2}(m_{1} + m_{2})g \geq \mu_{1}m_{1}g$.Dividing both sides by $\mu_{2}m_{1}g$, we get $\frac{m_{1} + m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$.Therefore, $\frac{\mu_{1}}{\mu_{2}} \leq 1 + \frac{m_{2}}{m_{1}}$.The maximum possible value is $1 + \frac{m_{2}}{m_{1}}$.

Similar Questions

When $F = 2\,N$,the frictional force between the $10\,kg$ block and the $5\,kg$ block is $..........\,N$. (Given: $\mu_s = 0.1$ between blocks,$\mu_s = 0.3$ between $5\,kg$ block and ground).

The value of $\theta$ is increased gradually from $\theta = 0$. At $\theta = \tan^{-1}(1/2)$,both blocks just start slipping. Then the value of $\mu_2$ is: $(g = 10 \ m/s^2)$

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