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Laws of Motion

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Physics Part I

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Physics

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Class 10 Class 12
Two bodies and of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally toA. What are
(a) the reaction of the partition
(b) the action-reaction forces between 
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
s and μk.


(a)

Mass of body A, mA = 5 kg

Mass of body B, mB = 10 kg

Applied force, = 200 N

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

                 = 22.5 N, leftward 

Net force acting on the partition = 200 – 22.5

                                                 = 177.5 N, rightward 

According to Newton’s 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 177.5 N,directed leftwards.

(b) 

Force of friction on mass A, fA = μmA

                                        = 0.15 × 5 × 10

                                        = 7.5 N leftward 

Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N, rightward 

Now, according to Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A.

i.e., Net force acting on mass A by mass B= 192.5 N 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 = 177.5 N

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

Net force = (mA + mB) a

∴ Acceleration, a = fraction numerator Net space Force over denominator straight m subscript straight A space plus space straight m subscript straight B end fraction 

                       = fraction numerator 177.5 over denominator left parenthesis 5 plus 10 right parenthesis end fraction space equals space fraction numerator 177.5 over denominator 15 end fraction space equals space 11.83 space straight m divided by straight s squared

Net force causing mass A to move,

                 FA = mA

                     = 5 × 11.83

                     = 59.15 N 

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

This force will act in the direction of motion.

According to Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion. 
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