# Why does this one move faster?

Consider a 2 body system as shown:

Consider the floor to be absolutely smooth and the coefficient of friction for the contact between $$m_1$$ and $$m_2$$ to be $$\mu$$. Now suppose I apply a force $$F$$ that causes the system to move, and that force $$F$$ is applied on the upper block ($$m_1$$). Then, why does it ($$m_1$$) move faster than $$m_2$$? Why does it have a greater acceleration?

• isn't there a need of relation between the two masses ? Since the acceleration of both the masses will depend on the friction between them . – A Student 4ever Sep 9 '20 at 8:06

Considering $$m_1$$ and $$m_2$$ as a system, there is a net horizontal force $$F$$ acting on them so their centre of mass must accelerate with acceleration

$$\displaystyle a_0 = \frac F {m_1+m_2}$$

This is true regardless of whether or not $$m_1$$ moves relative to $$m_2$$.

The frictional force which opposes $$m_1$$ moving relative to $$m_2$$ has a maximum value of $$\mu m_1 g$$. If $$F - \mu m_1 g \le m_1a_0$$ then $$m_1$$ will not move relative to $$m_2$$ and both blocks will accelerate with acceleration $$a_0$$. Substituting the expression for $$a_0$$ above, we see that this condition becomes

$$\displaystyle F - \mu m_1 g \le F \frac {m_1} {m_1+m_2} \\ \displaystyle \Rightarrow F \frac {m_2} {m_1+m_2} \le \mu m_1 g \\ \displaystyle \Rightarrow F \le \mu g (m_1+m_2) \frac {m_1}{m_2}$$

In this case, the frictional force is

$$\displaystyle F \frac {m_2} {m_1+m_2}$$

and we can see that (by Newton's Third Law) this acts on $$m_2$$ to accelerate it with acceleration $$a_0$$ too. In effect, the force $$F$$ is divided between $$m_1$$ and $$m_2$$ in proportion to their masses, so that they both have the same acceleration.

On the other hand, if

$$\displaystyle F > \mu g (m_1+m_2) \frac {m_1}{m_2}$$

then the net force on $$m_1$$ is $$F - \mu m_1 g$$ so $$m_1$$ accelerate with acceleration

$$\displaystyle a_1 = \frac F {m_1} - \mu g$$

and the force on $$m_2$$ is $$\mu m_1 g$$ so $$m_2$$ accelerates with acceleratiion

$$\displaystyle a_2 = \mu g \frac {m_1}{m_2}$$

and we have

$$\displaystyle a_1 > \mu g \frac {m_1+m_2}{m_2} - \mu g \\ \displaystyle \Rightarrow a_1 > \mu g \frac {m_1}{m_2} \\ \displaystyle \Rightarrow a_1 > a_2$$

Also note that the acceleration of the centre of mass of the system is the weighted sum of $$a_1$$ and $$a_2$$ which is

$$\displaystyle \frac {m_1a_1+m_2a_2} {m_1+m_2} = \frac {(F - \mu m_1 g) + \mu m_1 g} {m_1+m_2} = \frac {F} {m_1+m_2} = a_0$$

as we expect.