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A block of mass $m_{2}$ is placed on a horizontal table and another block of mass $m_{1}$ is placed on top of it. An increasing horizontal force \(F=\alpha t\) is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is $\mu_{i}$ and that between the lower block and the table is $\mu_{2}$, then what is the maximum possible value of $\mu_{1} / \mu_{2} ?$
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Verified Answer
The correct answer is:
$1+\frac{m_{2}}{m_{1}}$
According to the question.
$\therefore \quad N_{2}=m_{2} g+N_{1}=\left(m_{1}+m_{2}\right) g$
: Lower block never moves.
$\therefore \quad f_{r_{2}} \geq f_{n}$
$\quad \mu_{2} N_{2} \geq \mu_{1} N_{1} \Rightarrow \mu_{2}\left(m_{1}+m_{2}\right) g \geq \mu_{1} m_{1} g$
$\Rightarrow \quad \frac{m_{1}+m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$
$\therefore \quad \frac{\mu_{1}}{\mu_{2}} \leq 1+\frac{m_{2}}{m_{1}}$
$\left.\therefore \quad \frac{\mu_{1}}{\mu_{2}}\right|_{\operatorname{man}}=1+\frac{m_{2}}{m_{1}}$
$\therefore \quad N_{2}=m_{2} g+N_{1}=\left(m_{1}+m_{2}\right) g$
: Lower block never moves.
$\therefore \quad f_{r_{2}} \geq f_{n}$
$\quad \mu_{2} N_{2} \geq \mu_{1} N_{1} \Rightarrow \mu_{2}\left(m_{1}+m_{2}\right) g \geq \mu_{1} m_{1} g$
$\Rightarrow \quad \frac{m_{1}+m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$
$\therefore \quad \frac{\mu_{1}}{\mu_{2}} \leq 1+\frac{m_{2}}{m_{1}}$
$\left.\therefore \quad \frac{\mu_{1}}{\mu_{2}}\right|_{\operatorname{man}}=1+\frac{m_{2}}{m_{1}}$
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