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Calculating Friction Using Coeffcients

Two bars of m...

Question

Two bars of masses $m_{1}$ and $m_{2}$ connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to $k$ . The minimum constant force that has to be applied in the horizontal direction to the bar of mass $m_{1}$ in order to shift the other bar is $F_{m i n} = k g (m_{1} + \frac{m_{2}}{x})$ . Find $x$ .

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Solution

Let $x$ be the compression in the spring when the bar $m_{2}$ is about to shift. Therefore at this moment spring force on $m_{2}$ is equal to the limiting friction between the bar $m_{2}$ and horizontal floor. Hence
$k_{s p r i n g} x = k m_{2} g$ ......(1) where k is the coefficient of friction.
The spring has to be extended by x to make the block $m_{2}$ move.
Notice that the bar2 is just about to move, it hasn't started moving. The work done by the constant force
F is a sum of work done on the spring to stretch it by x and moving the bar $m_{1}$ by x.
$∴ F x = \frac{1}{2} k_{s p r i n g} x^{2} + k m_{1} g x$
$\Rightarrow F = \frac{1}{2} (k m_{2} g) + k m_{1} g = k g (m_{1} + \frac{m_{2}}{2})$ where we have substituted
$k_{s p r i n g} x = k m_{2} g$ from (1)
Comparing with $F_{m i n} = k g (m_{1} + \frac{m_{2}}{x})$ , we get $x = 2$

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