Presentation

Made by :• Danyal Haider (01-133132-014)

• Kamran Shah (01-133132-027)

• PRESENTATION TOPICS • Relation between force , mass and

acceleration .• Applications of Newton second law of motion.• Equations of motions .• Introduction to kinetics of particles .

Relation between force , mass and acceleration : Relation between force , mass and acceleration is refered to Newton second law of motion . FORCE :-• Force is the action of one body on another body .

A force tends to move a body in the direction of its action . The action of the force is charaterized by its magnitude , by the direction of its action , and its point of application .

OR

Force is a vector quantity having both magnitude and direction . The magnitude of the force can be measured using a spring

scale. Unit of force is Newton .

MASS :Mass is a measure of Inertia of a body .

Mass is a scalar quantity .Unit of mass is Kilogram (kg) .

ACCELERATION :• The rate of change of velocity is called acceleration .• Acceleration is a vector quantity so it can be changed in two ways:

1. changing its magnitude 2. changing its direction• Acceleration can be positive or negative. If an object’s speed

increases with time, it will have positive acceleration. If an object’s speed decreases with time, it will have negative acceleration that’s also known as deceleration. Si unit of acceleration is meters /second squared .

“NEWTON”

Sir Isaac Newton (4 January 1643 – 31 March 1727) was an English physicist and mathematician. He is famous for his work on the laws of

motion, optics, gravity, and calculus. In 1687, Newton published a book called the Principia Mathematica in which he presents his theory of universal gravitation and three laws of motion.

Newton’s second law:

• A particle will have an acceleration proportional to the magnitude of the resultant force acting on it and in the direction of the resultant force .

• The resultant of the forces acting on a particle is equal to the rate of change of linear momentum of the particle .

• The sum of the moments about “O” of the forces acting on a particle is equal to the rate of change of angular momentum of the particle .

• When a particle of mass m is acted upon by a force the acceleration of the particle must satisfy, F=ma

• If force acting on particle is zero, particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity.

Linear Momentum Conservation Principle:

• If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction .

• Replacing the acceleration by the derivative of the velocity yields ΣF = m(dv/dt) = d/dt (mv) =dL/dtWhere ,L = linear momentumof the particle

SOLUTION:• Resolve the equation of motion for

the block into two rectangular component equations.

• Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. Q # 1

• A 200-lb block rests on a horizontal

• plane. Find the magnitude of the force

• P required to give the block an acceleration

• or 10 ft/s2 to the right. The coefficient

• of kinetic friction between the

• block and plane is u^k = 0.25

SOLUTION:• Resolve the equation of motion for the

block into two rectangular component equations.

ΣFx = ma : Pcos30 - 0.25N = (6.21lb.s2/ft)(10ft/s2)= 6.21lbΣFy = 0 :N − Psin 30° − 200lb = 0• Unknowns consist of the applied force P

and the normal reaction N from the plane. The two equations may be solved for these unknowns.

N = P sin 30° + 200 lb P cos 30° - 0.25(P Sin 30° + 200 lb) = 62.1 lb P = 151 lb

• PRESENTATION TOPICS • Relation between force , mass and

acceleration .• Applications of Newton second law of motion.• Equations of motions .• Introduction to kinetics of particles .

Applications of Newton second law of motion :

• When a force is applied on a body of mass(m), there is a change in velocity w.r.t time , and change in velocity means acceleration is produced in a body .

We know that , Acceleration(a) is directly proportional to Force(F)• In the above figure, if the

person applies lesser force on the vehicle . The less acceleration is produced in a body .

• On the other hand , if the person applies greater force on the vehicle . The more acceleration is produced in a body .

• PRESENTATION TOPICS • Relation between force , mass and

acceleration .• Applications of Newton second law of motion.• Equations of motions .• Introduction to kinetics of particles .

Equations of motionPROOF OF EQUATION OF MOTION.

1st Equation of motion.Vf = vi + at

Proof. Consider a body moving with initial velocity vi .if a force F act on the body and produces an acceleration a in it, then after time t seconds. Its final velocity becomes vf. Then change in velocity of the body is vf - vi, and the rate of change of velocity of the body is equal to the acceleration acting on the body,i.e,a = vf – vi/tvf – vi = atvf = vi + at

2nd Equation of Motion. S = vit + at2/2

Proof. According to the first equation of motion, we have Vf = vi + at eq no (1)If the body has constant acceleration, then its average velocity is given by Vav = vi + vf/2 eq no (2)Distance covered by the body in time t is giver byS = vav * t eq no (3) Putting the mauve of Vav from eq (2) in eq (3) we get S = t * (vi + vf)/2Putting the value of vf from eq,(1) in eq,(4), we get S = t (vi + vi + a t)/2 = 2vit + at2/2 S = vi t + at2/2

Third equation of motion.2as = vf

2 – vi2

Proof.

We know thatS = vav . t eq(1)Vav = vi + vf/2 eq(2)From 1st equation of motionVf = vi + at t= vf – vi/a this is eq (3)Putting the value of vav and t from eq,(2) and (3), in eq,(1),S = (vi + vf)/2* (vf – vi)/a2as = (vf +vf). (vf – vi)2as = vf

2 – vi2

Some Rules to be used in the above three equation.

• If the body starts its motion from rest then use vi=0

• If a moving body stops its motion then vf=0• If the velocity of the body is incressing then its

acceleration is positive and its directed is the same as that of velocity.

• If the velocity of the body is decreasing then its acceleration is negative and its directed is opposite to that of the velocity .

Some Numerical Problems.

Q # 2: What force is required to move a mass of 5kg through 150 meters in 10 sec. if the mass is originally at rest.Given data m=5kg s=150m t=10sec For this we have to find the acceleration acting on the body and according to the gives data we apply second equation of motion.

F=?S=vit+1/2at2

150=0*10*a*100/2

a=3m/sec2

F= ma F=5*3=15N

Equation of motion for a freely falling body.

If a body is moving freely upward of falling downward then its acceleration a will be the acceleration due to gravity.i.e g=9.8m/sec2 and its distance S will show the height h of the body. Then the equation of motion will take the form. Downward motion Vf=vi+at vf=vi+gtS=vit+1/2at2 h=vit+1/2gt2

2as=vf2-vi2 2gh=vf

2-vi2

Upward motionVf=vi+at vf=vi-gtS=vit+at2/2 s=vit-gt2/22as=vf

2-vi2 -2gt=vf

2-vi2

Q # 3:. A 50kg shell of conical shape is dropped from a height of 1000m. It falls freely under gravity and sinks in into the ground .Find (a) Its velocity of the moment it touches the ground.(b). the time it takes to reach the groun.(c). the deceleration of the shell as it penetrates into the ground.

Given data ism=50kgvi=0s=1m using 2nd equation of motion find the vf.2gh=vf

2-vi2

2*(9.8)*1000 = vf2-0

Vf = 140ms-1

(c) for penetrating into ground.For ground the velocity of 140ms-1 is intial velocity. In ground it comes to rest after covering a distance of 1m.Given data isVi’ = 140ms-1Vf’ = 0S = 1ma= ?Using equation motionFrom 3rd equation motion2as = vf

2-vi2

2*a*1 = 0(140)2

a= -19600/2a = -9800ms-1

Now find the t?Using equation of motion.From 1st equation on motion.Vf = vi+gt140 = 0+9.8*tt = 140/9.8= 14.3sect = 14.3sec (c) for penetrating into ground.For ground the velocity of 140ms-1 is intial velocity. In ground it comes to rest after covering a distance of 1m.Given data isVi’ = 140ms-1Vf’ = 0S = 1ma= ?Using equation motionFrom 3rd equation motion2as = vf

2-vi2

2*a*1 = 0(140)2

a= -19600/2a = -9800ms-1

• PRESENTATION TOPICS • Relation between force , mass and acceleration .• Applications of Newton second law of motion.• Equations of motions .• Introduction to kinetics of particles .

Introduction to kinetics of particles Difference between kinematics and kinetics :Kinematics:- kinematics is the study of motion (not focusing upon forces , just the motion) .Kinetics :- kinetics is the study of motion in relation to forces . Introduction to kinetics of particles According to Newton second law, a particle will accelerate when it is subjected to unbalanced forces . kinetics is the study of the relations between unbalanced forces and resulting changes in motion .

Methods of solution of problems of particle kinetics:-There are three basic methods of solution of problems of particle kinetics . These three methods are summarized as follows :

(1)DIRECT APPLICATION OF NEWTON SECOND LAW:-First we applied the Newton second law F=ma to determine the instantaneous relation between force and acceleration they produce . Then solve a problem using x-y coordinates for plane motion problems and x-y-z coordinates for space motion problems . (2)WORK-ENERGY EQUATIONS:-Next, we apply F=ma with respect to displacement and derived the equations for work and energy . These equations enable us to relate the initial and final velocities to the work done during an interval of forces . (3)IMPULSE-MOMENTUM EQUATIONS:-Then we write Newton second law in terms of force equals time rate of change of linear momentum and moment equals time rate of change of angular momentum . Then solve these relations with respect to time and derived the impulse and momentum equations .

SOLUTION: The car travels in a horizontal

circular path with a normal component of acceleration directed toward the center of the path . The forces acting on the car are its weight and a normal reaction from the road surface.

Resolve the equation of motion for the car into vertical and normal components. Solve for the vehicle speed.

Q # 4 :Determine the rated speed of ahighway curve of radius ρ = 400 ftbanked through an angle θ = 18o. Therated speed of a banked highway curveis the speed at which a car shouldtravel if no lateral friction force is to

be exerted at its wheels.

• Resolve the equation of motion forthe car into vertical and normalcomponents.ΣFy = 0 : R Cos θ – W = 0 R = W/Cos θ

ΣFn = man : R sin θ = (W/g) *an

(W/Cos θ)*Sin θ =(W/g)* (v ^2/ ρ )

• Solve for the vehicle speed.

v^2 = g ρ Tan θ =(32.2ft/s^2)(400 ft)Tan 18° v = 64.7 ft/s = 44.1mi/h

SOLUTION:• The car travels in a horizontal circularpath with a normal component ofacceleration directed toward the centerof the path.The forces acting on thecar are its weight and a normalreaction from the road surface.

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