UNIT-V Dr.G.Elangovan Dean(i/c) University College Of Engineering Presented by Dr.G.Elangovan Dean(i/c) University College Of Engineering Dindugul

Topics To Be Covered Newton’s Law of Motion Work & Energy Method Impulse & Momentum Relation

Newton’s Law Of Motion

Newton’s 1st Law:- Interia:- FORCE :- Every body continues to be in its state of rest or of uniform motion in a straight line unless and until it is acted upon some external force to change the state. It is push or pull, which either changes or tends to change the state of rest or of uniform motion of a body. Interia:- In the absence of an external force, no body can change on it’s own , it’s state of rest or the state of uniform motion along a state line. FORCE :-

Equal forces in opposite directions produce no motion Balanced Force Equal forces in opposite directions produce no motion

Unbalanced Forces Unequal opposing forces produce an unbalanced force causing motion

Force = Mass x Acceleration Force is measured in Newtons Newton’s Second Law The rate of change of momentum of a moving body is directly proportional to the impressed force & take place in the direction of the force applied. Force = Mass x Acceleration Force is measured in Newtons

To every action, there is always an equal & opposite reaction. Newton’s Third Law To every action, there is always an equal & opposite reaction.

FRICTIONAL FORCE When two bodies are in contact with one another, the property of two bodies by virtue of which a force is exerted b/w them at their point of contact to prevent one body from sliding on the other, is called ‘Frictional force’. Its denoted by F. The maximum frictional force developed at the contact area is called as ‘Limiting friction’. Its denoted by the Fm Contd….

The ratio of limiting friction to the normal reaction is known as ‘Co-efficient of friction’, denoted by the symbol ‘µ’. Co-efficient of friction = Limiting Friction Normal reaction µ = Fm NR Fm = µ x NR

Case I: Body moving on rough horizontal surface Consider a body moving on a rough horizontal surface towards right as shown in fig. IF Co-efficient of friction = µ Fm = µ x NR But NR = Weight ‘W’ Fm = µ x W

Case II: Body pulled up on an inclined surface When a body is pulled up on an inclined force as shown In fig. then the frictional Force Fm acts in the opposite (downward) direction. NR –W COS θ = 0 NR = W COS θ Fm = µ x W COS θ

Case III: Body sliding downwards When a body is sliding downwards as shown in fig. then the frictional force Fm acts in the upward direction. Resolving the forces normal to the plane & equated to zero. NR –W COS θ = 0 NR = W COS θ Fm = µ x W COS θ

Relationship b/w three Cases Fm = µ x W Fm = µ x W COS θ

D’Alembert’s Principle D’Alembert’s Principle is an application of Newton’s 2nd law.For the static equilibrium of forces in plane,we have seen the equations to be satisfied are ∑H=0; ∑V=0 and ∑M=0. P= m a P = External force m = Mass of the moving body a = Acceleration of the body

CaseI:- CaseII:- P P=ma ∑H=0 P-ma=0

Case III:- Case IV ∑F=ma ∑F-ma=0

Case V When a body is subjected to two forces P1 and P2 as shown in fig. then the resultant force is P1 - P2 ( assuming P1 > P2 ). Here,P – ma= 0 equation may be written as ∑F=0 ie, P1 - P2 – ma = 0 a = P1 - P2 m

Hence, D’Alembert’s principle States that ” The system of forces acting on a body in motion is in dynamic equilibrium, with the inertia force of the body”.

WORK-ENERGY METHOD

WORK Work is defined as the product of force & displacement of the body. Case I Work done by the force = Force x Distance Moved = P X S S P

Case II UNIT OF WORK In SI system of units, force in Newton & the distance in meters. Unit of work = 1 Nm= 1 joule

ENERGY The capacity of doing work is known as ‘Energy’. Unit of energy is same as that of work. Mechanical energy is classified into 2types: Potential energy Kinetic energy

Potential energy It is the capacity to do work by virtue of position of the body. It is denoted by P.E. The force of attraction or the weight of the body=W=mg Work done by the body = Force X Distance = mg x h P.E = mgh

kinetic energy It is the capacity to do work by virtue of motion of the body. It is denoted by k.E. Kinetic Energy = ½ mv2

Work-Energy Equation ∑Fx . S = W/2g ( v2 - u2 ) Work done = Final Kinetic Energy – Initial Kinetic Energy P.S =Work done by the moving body = Force X Distance (W/2g)v2 = Final Kinetic Energy (W/2g)u2 = Initial Kinetic Energy ∑Fx . S = W/2g ( v2 - u2 )

IMPULSE OF A FORCE When a large force acts for a short period of time, that force is called an impulsive force. It is denoted by the symbol I. It is a vector quantity. t2 I = ʃ F dt t1 Linear Impulse = Force x Time

MOMENTUM :- It is defined as the total quantity of motion contained in a body and is measured as the product of the mass of the body and it’s velocity. (momentum) M= m x v m =mass of the body (Kg) v =velocity (m/s) M = Linear momentum (Kg.m/s)

Impulse- Momentum Equation The Impulse- Momentum equation is also derived from the Newtons second law, t2 ʃ F dt = m (v-u) t1 ʃ F dt = Impulse m (v-u) = Change of momentum Ie, Final momentum – Initial Momentum

Contd….. ie., Impulse = Final momentum – Initial momentum Therefore Impulse = m(v-u) or W/g(v—u) The impulse of the force acting on a particle is equal to the change in the linear momentum of the particle. ie., Impulse = Final momentum – Initial momentum Final momentum = Initial momentum + Impulse

Problems A block of weight 2000 N is in contact with a plane inclined at 30 ˚ to the horizontal. A force ‘P’ parallel to the plane & acting up the plane is applied to the body as shown in fig. The coefficient of friction between the contact surfaces is 0.20.Find (i) The value of P just cause the motion to impend up the plane (ii) The value of P just prevent the motion down the plane. Figure shows two blocks of weight 60N & 140N, placed on two inclined surfaces and connected by an inextensible string. Calculate the acceleration of the system and the tension in the string Take µ = 0.2

3). A 500N block is in contact with a level plane, the co-efficient of friction between two contact surfaces being 0.25. if the block is acted upon by a horizontal force of 1300N, what time it will elapse before the block reaches a velocity of 24 m/s. 4)An aeroplane of mass 8t is flying at a rate of 250kmph, at a height of 2km above the ground level. Calculate the total energy possessed by the aeroplane.

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