# Two-block system connected to a spring

Say you have two blocks with masses $m_1$ and $m_2$, where $m_1>m_2$. The smaller block sits atop the larger block. The larger block is connected to a spring, which is then connected to a wall a distance $x$ away. What must the coefficient of friction be in order for the smaller block to stay on top of the larger block as the spring pushes the masses back and forth.

So I get

$$f\geq m_1\ddot{x},$$

which says that the friction experiences by the smaller block, must be greater than or equal to the force experienced by the larger block. We know that

$$f=\mu_sN,$$

so it becomes $$\mu_sN \geq m_1\ddot{x}\Rightarrow \mu_s \geq \frac{m_1\ddot{x}}{N}.$$ Is this wrong?

Well, according to the condition given in the question, the smaller mass must move with the larger one. So that means their accelerations must be the same. That makes the first equation $$f = m_1\ddot{x}$$ And as for your second equation, you must understand that friction opposes relative motion. This means that, so far as it can, the frictional force will 'try', so to speak, to make the accelerations equal. So we model it as an adjustable force upto a maximal limit. This makes the second equation $$f \leq \mu_sN$$ Now, so long as $\mu_sN$ is greater than or equal to $m_1\ddot{x}$, static friction will keep the blocks at relative rest. So we get $$\mu_sN \geq m_1\ddot{x}$$ $$\implies \mu_s \geq \frac{m_1\ddot{x}}{N}$$ Looks like your answer was right after all. Note that the limit can be made precise by putting in the maximal value of the larger block's acceleration over the course of its motion.