Two bodies $A$ and $B$ of mass $5kg$ and $10kg$, in contact with each other rest on a table against a rigid wall (see figure). The coefficient of friction between the bodies and the table is $0.15$. A force of $200N$ is applied horizontally to $A$. What are (a) the reaction of the vertical wall. (b) the action-reaction forces between $A$ and $B$? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between ${\mu }_s$ and ${\mu }_k$.

(a) 

Mass of body $A$, $m_A=5kg$ 

Mass of body $B$, $m_B=10kg$

Applied force, $F=200N$

Coefficient of friction, ${\mu }_s=0.15$

The force of friction is given by the relation:

$f_s=\mu (m_A+m_B)g$

$=0.15(5+10)\times 10$

$=1.5\times 15=22.5N$ leftward

Net force acting on the partition $=200-22.5=177.5N$ rightward

As per Newtons third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be $177.5N$, in the leftward direction.

(b) 

Force of friction on mass $A$:

$F_A=\mu m_{A}g$ 

$=0.15\times 5\times 10=7.5N$ leftward

Net force exerted by mass A on mass $B=200-7.5=192.5N$ rightward

As per Newtons third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., $192.5N$ acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = $177.5N$

The equation of motion for the system of acceleration a,can be written as:

Net force= ($m_A$ + $m_B$) a

a = Net force / ($m_A$ + $m_B$)

= $177.5/(5+10)=177.5/15=11.83$ m/s${}^2$

Net force causing mass A to move:

F${}_A$ = m${}_A$ $a=5\times 11.83=59.15N$

Net force exerted by mass $A$ on mass $B=192.5-59.15=133.35N$ 

This force will act in the direction of motion. As per Newtons third law of motion, an equal amount of force will be exerted by mass $B$ on mass A, i.e., $133.3N$, acting opposite to the direction of motion.