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Newton's Laws of Motion


(6)Newton's third law of motion


  • Statement of newton's third law of motion is " To every action there is always an equal and opposite reaction".
  • Thus,whenever a body exerts force on another then another object exert an equal force on previous body but in opposite direction
  • Force example motion of rocket depends on the third law of motion i.e, action and reaction .Rocket exerts action force on gas jet in backward direction
  • Force of action and reaction acts on different objects i.e,
    Force object 1 exerts on object 2= Force object 2 exerts on object 1
    i.e,
    F12=-F21
    Action=-(Reaction)
  • According to Newtonian mechanics force is always a mutual interaction between the bodies and force always occurs in pairs
  • Equal and opposite mutual forces between two bodies is the basic idea between Newton's third law of motion
  • While considering a system of particles ,internal force always cancel away in pairs i.e consider two particles in a body if F12 and
    F21 are internal forces between particle system 1 and 2 then they add up to give a null internal force.Same way internal forces for the particles

(7) Applying Newton's law of motion


  • Newton's law of motion ,we studied in earliar topics are the foundation of mechanics and now we look forward to solve problems in mechanics
  • In general, we deal with mechanical systems consisting of different objects exerting force on each other
  • While solving a problem choose any part of the assembly and apply the laws of motion to that part including all the forces on the chosen part of the assembly due to remaining parts of the assembly
  • Following steps can be followed while solving the problems in mechanics
    1)Read the problem carefully
    2) Draw a schematic diagram showing parts of the assembly for example it may be a single particle or two blocks connected to string going over pulley etc
    3) Identify the object of prime interest and make a list of all the forces acting on the concerned object due to all other objects of the assembly and exclude the force applied by the object of prime interest on the other parts of the assembly
    4) Indicate the forces acting on the concerned object with arrow and Label each force for example tension on the object under consideration can be labeled by letter T
    5) Draw a free body diagram of the object of interest based on the labeled picture.Free body diagram for the object under consideration shows all the forces exerted on this object by the other bodies.Do not forget to consider weight W=mg of the body while labeling the forces acting on the body
    6) If additional objects are involved draw separate free body diagram for them also
    7)Resolve the rectangular components of all the forces acting on the body
    8) Write Newton second law of equation for the body and solve them to find out the unknown quantities
    9) Do not forget to employ Newton's third law of motion for action reaction pair which results in null resultant force
  • Following solved example would clearly illustrate how to apply Newton's laws of motion following the above given procedure
Solved Example :
Question:
A horizontal forces of magnitude 500N pulls two blocks of masses m1=10 kg and m2=20 kg which are connected by the light inextensible string and lying on the horizontal frictionless surface.Find the tension in the strings and acceleration of each mass when forces is applied on mass
m2? Solution:
Given that force is applied on the block m2 as shown in the figure below
two blocks of masses which are connected by the light inextensible string

Let T be the tension in the string and a be the acceleration of each mass .Now we will draw free body diagrams for each masses


Free body diagram of two blocks of masses which are connected by the light inextensible string

Weights of the blocks m1g and m2g are balanced by their normal reaction R1 and R2 respectivley.The equations of motion of the two massed are found using Newton's second law of motion
m1a=T ...............................(1)
m2a=F-T ............................(2)

Dividing 1 by 2
we get
T=m1F/(m1+m2)
Substituting the given values
T=166.7 N
Using value of T in equation 1 ,we find
a=16.67 m/s2

Above sample problem shows how to solve a typical mechanics problem.Similarly by adopting given procedure we can solve other such problems

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