NEWTON'S LAW OF MOTION

Newton’s Law

• Isaac Newton (1642 – 1727)
• Newtonian mechanics – classical mechanics
• Cannot apply for speeds close to speed of light – Quantum mechanics / Einstein theory of relativity.

Sir Isaac Newton.

Newton worked in many areas of mathematics and physics. He developed the theories of gravitation in 1666, when he was only 23 years old. Some twenty years later, in 1686 his three laws of motion in the "Principia Mathematica Philosophiae Naturalis.”

Newton’s Law of Motion

• Newton’s First Law of Motion (Law of Inertia)

-An object at rest will remain at rest, or continues to move with uniform velocity in a straight line unless it is acted by a external force

Fnet = Σ F = 0

• Newton’s Second Law of Motion

-The rate of a change of momentum of an object with time is directly proportional to the net force acting on it.

Fnet = Σ F = ma

• Newton’s Third Law of Motion

-For every action, there is an equal and opposite reaction; action and reaction forces act on different objects

F1 = -F2

Newton’s First Law of Motion

(Law of Inertia)

• In the absence of external forces, an object at rest remains at rest and,
• An object in motion continues in motion with constant speed.
• As long as no net force acts on it.
• When no force acts on an object, F = 0, a = 0.

-“No net force” means:

-No force acts on the object.

-Forces acts on the object, but their sum is zero.

• Inertia is the tendency for object to resist change in its state.
• If the net force on an object is zero, the object moves with constant velocity

Examples:

Newton’s Second Law of Motion

• Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass.

• Force is a vector, so ΣF = ma is true along each coordinate axis.
• Acceleration of object is directly proportional to its force.

• That is, ΣF = ma also

• The object moves in the direction of the net force acting on it.o

F=ma​​

Newton’s Third Law of Motion

action

• Any time a force is exerted on an object, that force is caused by another object.
• This will apply Newton’s Third Law:

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.

reaction

For every action there is an equal and opposite reaction.

Newton’s Third Law of Motion

• Force always occur in pairs.
• For every action force there is an equal and opposite reaction force.
• If two objects interact, the force F1 exerted on object 1 by object 1 is equal in magnitude and opposite direction to the force F2 exerted on object 2 by object 1

F1 = -F2

Normal Force, N

• The force exerted perpendicular to a surface is called the Normal Force.
• Cannot exist on its own.
• Always perpendicular to the surface.
• It is exactly as large as needed to balance the force from the object

N

W

Mass vs. Weight

• Mass is the measure of inertia of an object. In the SI system, mass is measured in kg.
• Mass is not weight.
• Mass is a property of an object.
• Weight is the force exerted on that object by gravity.
• If you go to the moon, whose gravitational acceleration is about 1/6 g, you will weight much less. Your mass, however, will be the same.

Gravitational Force, W

N

Fg=mg

• Attractive force exerted by earth on objects.
• Directed towards the center.
• Magnitude is called the weight

where,

the relationship between N and W

N – W = 0

N = W

N = mg

Fg = W = mg

W

Frictional Force

• When a solid object moves over the surface of another solid object, its motion is always opposed by a retarding force.
• This is called friction. Therefore, friction is the force directed opposite to the direction of motion.
• It is always parallel to the surfaces in contact.

Object at rest:

• F is applied to object but it remains static or does not move.
• F = fs
• This force of friction is called static friction
• F increase, fs also increase as long as object does not move

N

.

F

f

.

F

fs

W

Frictional Force

• If object just about to start moving when F applied.
• The force of static friction reaches maximum.
• If F > fs max, object moves.
• Friction force now called force of kinetic friction, fk.

On a microscopic scale, most surfaces are rough. The exact details are not yet known, but the force can be modeled in a simple way.

We write frictional force as:

µ is the coefficient of friction, and is

different for every pair of surfaces.

f =µN

N

Motion →

.

F

fk

W

Object about to move

Frictional Force

For static friction:

where µ = coefficient of friction, and

FN = normal force

• fs ≤ µsFN
• fs < µsFN if object does not move
• fs = µsFN when object on the verge of slipping

For moving:

fk = µkFN

Free-body Diagram Involving Incline Surface

An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force.

-The normal force is always perpendicular to the surface.

-The friction force is parallel to it.

-The gravitational force points down.

If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero.

Example

• Two blocks A of mass 10 kg and B of mass 30 kg are side by side and in contact with each another. They are pushed along a smooth floor under the action of a constant force F of magnitude 200 N applied to A as shown in figure. Determine

a) The acceleration of the blocks.

Solution

ΣFx=(mA + mB )a

= (10 + 30)a

a = 5.0 m/s2

Example : (horizontal surface / inclined force)

A force of 15.0 N is applied at an angle of 30° to the horizontal on a 0.75 kg block at rest on a frictionless surface. What is the magnitude of the resulting acceleration of the block and the normal force acting on the block?

This is the total acceleration of the block, since it does not move in the y-direction

The sum of forces in the y-direction must be zero. From the free-body diagram

Example: ( inclined surface)

• Draw the vector free body diagram and find the acceleration and normal force of the object.

*no friction involve

Example: Tension (Vertical : with pulley)

• The Atwood’s machine with masses M1 = 0.55 kg and M2 =0.80 kg,

a) What is the acceleration of the system?

b) What is the tension T in the

string?

Since M2 > M1, M2 will move downward

By substituting (2) into (1)

M2(g – a) = M1 (a + g) 1.35 a = 2.4525

a = 1.82 m/s^2

To find tension T on string,

T = M2(g – a) = 0.8 (9.81 – 1.82) = 6.4 N

Tension: Pulley (horizontal and vertical motion)

Consider two blocks connected by a cord that passes over the frictionless pulley as shown in figure,

Tension of the cord is same, therefore

Equation for block m1:

T = m1a

Equation for block m2:

W2 + T1- T = m2a

Equation for block m3:

W3 - T1 = m3a

Exercise 5.1

1. A boy pulls a box of mass 30 kg with a force of 25 N in the direction shown in Fig. 1.

(a) Ignoring friction, what is the acceleration of the box?

(b) What is the normal force exerted on the box by the ground?

fig.1

(Ans: 0.72 ms-2, 2.8 x 102 N)

2. A girl pushes a 25 kg lawn mower as shown in Fig. 2. If F= 30 N and θ = 37,

(a) what is the acceleration of the mower, and

(b) what is the normal force exerted on the mower by the lawn? Ignore friction.

fig.2

(Ans: 0.96 ms-2, 2.6 x 102 N)

3. Draw the vector free body diagram and find the force of the object. Given the mass of the box is 15 kg and moving at constant acceleration 1.2 ms-2.

(Ans: a = -55.5 N (downwards)

4. A block weighing 50 N rests on an inclined plane. Its weight is a force directed vertically downward, as illustrated in Fig. 3.

• Draw the vector free body diagram and,
• Find the components of the force parallel to the surface of the plane and perpendicular to it.

(Ans: F = 30 N)

5.Assume that the three blocks in Figure below move on a frictionless surface and that a 42 N force acts as shown on the 3.0 kg block. Determine:

• the acceleration given this system.
• the tension in the cord connecting the 3.0 kg and the 1.0 kg blocks.

(Ans: a = 7.0 ms-2, T = 21 N) (Servey/c4/q27

6.Determine the tension in the string and force.

a = 2 ms^-2

(Ans: T1 = 94.4 N, T2 = 70.8 N), F = 104.4 N

Example : (inclined surface / horizontal force)

Suppose a 2-kg object is pulled up an inclined plane with a force of 50.0 N. The surface has a coefficient of friction of 0.2. Find the acceleration of the object.

object is along the incline, hence a = ax and ay = 0

Example: (inclined surface / inclined force)

A block of mass 200 kg is pulled along an inclined plane of 30⁰ by a force, F = 2 kN as shown in figure above. The coefficient of kinetic friction of the plane is 0.4. Find the acceleration of the object.

object is along the incline, hence a = ax and ay = 0

Movement of lift

Case I: Lift is not moving

m = 50 kg

ΣFy = ma

N - w = ma

N – 490 = 50(0)

N = 490 N

Case II: lift up

a = 1 ms^-2

ΣFy = ma

N - w = ma

N – 490 = 50(1)

N = 540

Case III: lift down

a = 1 ms^-2

ΣFy = ma

w - N = ma

490 – N = 50(1)

N = 490 – 50

= 440 N

Exercise 5.2

1.Given F1 is 10 N, F2 is 20 N, mass of the box is 2 kg and the friction is 0.2. Find the acceleration of the box.

(Ans: a = 11.52 ms^-2)

2.A box weighing 400 N is pushed horizontally across the floor of a room by a force 900 N as shown in figure below. If the coefficient of kinetic friction between the box and the floor is 0.56, find the acceleration of the box.

(Ans: 16.56 ms^-2)

3. A block of mass 3 kg is pulled by force F along a 40o inclined plane, if F = 100 N and the coefficient of friction μ = 0.3.

• Draw the forces acting on the block.
• What is the acceleration of the block.

(Ans: a = 24.78 ms-2)