Body A of mass 5 kg, exerts 5 N force on a other body B of mass 10 kg. the reaction force exerts by B on A will be *
Answers
ANSWER
(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
you can solve it like this.