Question

Two bars of masses ${m_1}$ and ${m_2}$ connected by a non-deformed light spring rests on a horizontal plane. The coefficient of friction between the bar 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 ${m_1}$ to shift the other bar is ${F_{\min }} = kg\left( {{m_1} + \dfrac{{{m_2}}}{x}} \right)$. Find $x$ .

Answer 1

Hint: The motion is possible due to the friction. When the object moves on a rough surface friction acts in the direction opposite to the motion. Static friction is a type of friction. When an object is at rest, we apply force on it but it does not move at that time static friction works on the object.

Complete step by step solution:
Let $x$ be the length to which spring is compressed when the bar ${m_2}$ is about to shift. At this moment when the motion is about to happen to limit friction works on the system. Hence, we can write the following.
$k{m_2}g = {k_s}x$ …………….(1)
Here, $k$ is the coefficient of friction given in the question and, ${k_s}$ in the spring constant.
In this situation when the bar is about to move, we say that the applied force is equal to the work done on compressing the spring and shifting bar ${m_1}$.
${F_{applied}}x = k{m_1}gx + \dfrac{1}{2}{k_s}{x^2}$ …………..(2)
Let us combine equations (1) and (2), we get the following.
${F_{applied}} = k{m_1}gx + \dfrac{1}{2}{k_s}{x^2}$
${F_{applied}} = k{m_1}g + \dfrac{1}{2}k{m_2}g$
Let us further simplify the above expression.
${F_{applied}} = kg({m_1} + \dfrac{1}{2}{m_2})$ …………………..(3)
Now the minimum force given in the question is below.
${F_{\min }} = kg\left( {{m_1} + \dfrac{{{m_2}}}{x}} \right)$
Let us compare this equation with equation (3) and we can write the values of $x$.
$x = 2$
Hence, the value $x$ is $2$ .

Note:
Static friction is also called self-adjusting force.
Any given surface has a coefficient of friction which fixes the amount of maximum friction that the surface can apply to the given object.
When we apply a force on an object which is at rest we have observed that we continuously increase the applied force and that is counterbalanced by this self-adjusting force.