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Question

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 μs and μk.
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Solution

(a)

Mass of body A, mA=5kg

Mass of body B, mB=10kg

Applied force, F=200N

Coefficient of friction, μs=0.15

The force of friction is given by the relation:

fs=μ(mA+mB)g

=0.15(5+10)×10

=1.5×15=22.5N leftward

Net force acting on the partition =20022.5=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:

FA=μmAg

=0.15×5×10=7.5N leftward

Net force exerted by mass A on mass B=2007.5=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= (mA + mB) a

a = Net force / (mA + mB)

= 177.5/(5+10)=177.5/15=11.83 m/s2

Net force causing mass A to move:

FA = mA a=5×11.83=59.15N

Net force exerted by mass A on mass B=192.559.15=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.


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