15.
Wtfah
Two bodies A and B of mass 5 kg and 10 kg
in contact with each other rest on a table again
a rigid wall. The coefficient of friction between
the bodies and the table is 0.2, when the force
of 200 N is applied horizontally on A. Which of
the following is correct?
А A
B
200 N
(1) The reaction of wall is 200 N
(2) The action reaction force between A and B
is more than reaction of wall
(3) The action reaction force between A and B
is less than reaction of wall
(4) Both (1) and (3) are correct
Answers
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 μ
a)
Mass of body A, m
A
=5kg
Mass of body B, m
B
=10kg
Applied force, F=200N
Coefficient of friction, μ
s
=0.15
The force of friction is given by the relation:
f
s
=μ(m
A
+m
B
)g
=0.15(5+10)×10
=1.5×15=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
=μm
A
g
=0.15×5×10=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=5×11.83=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.