Two masses with the same acceleration

Problem

Two blocks are pushed together by a horizontal force on a smooth floor: mass $m_1 = 3\ \text{kg}$ is in front, mass $m_2 = 5\ \text{kg}$ is behind (where force $\vv{F}$ is applied). It is known that the contact between the two blocks exerts a force of $9\ \text{N}$. Find the applied force $F$ and the acceleration of the system.

Solution

The two blocks are in contact and move together, so they have the same acceleration $a$. The force $\vv{F}$ is applied to the rear block $m_2$, which in turn pushes the front block $m_1$ through the contact surface.

Analysis of the front block. The only horizontal force acting on $m_1$ is the contact force $F_c = 9\ \text{N}$ exerted by $m_2$. We apply Newton’s second law to block $m_1$ alone: $F_c = m_1\, a$

We obtain the acceleration: $a = \frac{F_c}{m_1} = \frac{9}{3}$

$\ev{a = 3\ \text{m/s}^2}$

Analysis of the whole system. Treating the two blocks as a single body of mass $m_1 + m_2$, the only external horizontal force is $\vv{F}$: $F = (m_1 + m_2)\, a = (3 + 5) \cdot 3$

$\ev{F = 24\ \text{N}}$

A remark on the third law: block $m_2$ pushes $m_1$ forward with $9\ \text{N}$, and by the third law $m_1$ reacts by pushing $m_2$ backward with $9\ \text{N}$. As a check, for $m_2$ we have $F - F_c = m_2 a$, i.e. $24 - 9 = 15 = 5 \cdot 3$: everything is consistent.

Links

Topics: Dynamics Concepts: Newton’s second law · Newton’s third law Objects: Block