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 $\mu$. What minimum constant force has to be applied in the horizontal direction to the bar of mass $m_1$ in order to shift the other bar ?

Answer 1

$\left[\left(m_1+\frac{m_2}{2}\right)\mu g\right]$

Answer 2

The problem states that two bars, $m_1$ and $m_2$, are connected by a light, non-deformed spring and rest on a horizontal plane with a coefficient of friction $\mu$. We need to find the minimum constant horizontal force $F$ applied to $m_1$ that will cause $m_2$ to shift.

For $m_2$ to shift, the spring force ($F_s$) acting on it must overcome the maximum static friction force ($f_{s2,max}$) acting on it. The maximum static friction on $m_2$ is $f_{s2,max} = \mu N_2 = \mu m_2 g$. Thus, for $m_2$ to just begin moving, the spring force must be at least $F_s = \mu m_2 g$.

Now consider the forces acting on $m_1$. The applied force $F$ is to the right. The spring force $F_s$ acts on $m_1$ to the left (opposing the extension). The static friction force ($f_{s1}$) on $m_1$ also acts to the left, opposing the applied force $F$. The maximum static friction on $m_1$ is $f_{s1,max} = \mu N_1 = \mu m_1 g$.

The condition for $m_2$ to start moving is $F_s \ge \mu m_2 g$. The force balance on $m_1$ at the point where $m_2$ is about to move is $F = F_s + f_{s1}$.

To find the minimum force $F$ that causes $m_2$ to shift, we consider the scenario where $m_1$ is also at its limit of static friction, meaning $f_{s1} = f_{s1,max} = \mu m_1 g$. This ensures that $m_1$ is capable of producing the necessary spring extension without accelerating significantly itself, or is on the verge of moving.

So, $F_{min} = F_s + f_{s1,max}$. The problem's provided answer is $\left[\left(m_1+\frac{m_2}{2}\right)\mu g\right]$. If we assume this answer is correct, then: $F_{min} = \mu m_1 g + \frac{1}{2}\mu m_2 g$. This implies that the required spring force to initiate the movement of $m_2$ is $F_s = \frac{1}{2}\mu m_2 g$. However, the condition for $m_2$ to shift is $F_s \ge \mu m_2 g$. The derived spring force $F_s = \frac{1}{2}\mu m_2 g$ is insufficient to move $m_2$ according to standard physics principles.

There appears to be an inconsistency between the problem statement and the provided answer, as standard physics derivations lead to $F_{min} = \mu m_1 g + \mu m_2 g = \mu g (m_1 + m_2)$.

Given the instruction to use the provided answer, we present it as is, acknowledging the discrepancy with standard physics. The answer suggests that the force required to stretch the spring to the point where $m_2$ just starts to move is related to $\frac{1}{2}\mu m_2 g$, which is not derivable from the problem as stated using conventional mechanics.