Chapter 5 and 6 Newton Laws

-Newton’s Laws of MotionForce is the action (push or pull) that tends to change the state of motion of an object.

An object at rest needs a force to get it moving; a moving object needs a force to change its velocity

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• When several forces act on an object the object accelerates only if the net force (resultant of all forces) acting on it is not equal to zero. That is:

If the net force acting on an object is zero the acceleration is zero and the velocity of the object remains constant.

• This occurs when the body is at rest or moving in a straight line with uniform velocity

Force Classes

(I) Contact force

When a force acts on an object it may be one of two classes:Contact force which is due to physical contact with another object.

Examples of these forces are: 1. The force one exerts when he

pulls on a coiled spring2. The force exerted by a football

player to kick the ball,

(II) Field force

Field force which do not involve physical contact between two objects but acts through empty space.

Examples of field forces are:(1) The gravitational force

between masses(2) The electric force between

electric charges.(3) The magnetic forces exerted

by a magnet on a an iron bar

Resultant of a force system

The effect of any number of forces acting at the same point on a body is the same as the effect of a single force that equal to the vector sum of the forces. This single force is called the resultant or the net force of all these forces.

When two forces F1 and F2acts at the same time on the same point of a body as in figure, the effect of these two forces equal the vector sum or the resultant Fwhich is given by

F = F1 + F2,.

Resultant of a group of forcesIf F1, F2, F3 and so on are the forces,

then the net force or resultant is given by the vector sum of these forces Fnet =F1 + F2 + F3 + ......= Σ Fi

How To calculate Fnet1. resolve the forces F1, F2, F3 and so

on in its x and y- components2. calculate the x and y-components of

the net force as (Fy)net =F1y + F2y + F3y + ...= Σ Fiy

(Fx)net =F1x + F2x + F3x + ...... = Σ Fix

3. Calculate Fnet usingF2

net =(ΣΣΣΣFix )2 + (ΣΣΣΣ Fiy )2

�2�3

�1

F1x =F1 cos �, F2x =F2 cos �2, and F3x =F3 cos �3

F1y =F1 sin �, F2y =F2y sin �2, and F3y =F3 sin �3

Force and inertiaWhen you push on a bowling ball and an automobile the change in the state of motion of the automobile is much less than that of the bowling ball. This is because the mass of the automobile is muchgreater than that of the bowling ball. From this we see that

When the same force acts for the same time on two objects havingdifferent masses the change in motion of the smaller mass will be greater than the change in motion of the larger mass.

InertiaThe tendency of an object to resist the change in its state of motion is called inertia. The body mass is a measure of its inertia.The relation between force, mass and the change in the state of motion is given by Newton's laws of motion.

Newton’s first Law

A body remains in its state of rest or uniform motion in a straight line unless acted by unbalanced external force.

That is if the net force (Fnet) acting on a body is zero the acceleration a of the body is zero and the body is in equilibrium.

If Fnet = 0 then a = 0.

This simple equation means that if a number

Equation for static and dynamic equilibrium

Since the equation

Fnet ==== Σ Σ Σ Σ F = 0is a vector equation in three dimensional

space, then it has three components. That is

ΣΣΣΣFx = 0 ΣΣΣΣ Fy = 0 ΣΣΣΣ Fz = 0

If the net force Fnet on an object is zero, then the sum of the components of the forces in

Example 1A ball with a mass m = 5 kg is suspended

from a ceiling by means of a cord. when a horizontal force F is applied to the ball as in fig the ball hangs in equilibrium with the cord making an angle of 35o with respect to the vertical. What is the magnitude of the force F and what is the tension T.

T

mg

F

350

Solution

� Fx = F - T sin 35 = 0� T sin 35 = F

� Fx = T cos 35 – mg = 0 � T cos 35 = mg

Tan 35 = F/mg F= mg Tan 35 = 5x9.8tan 35 = 34.3 N

Example IIA block with a mass m =10 kg rests on an inclined

plane as shown in the following diagram. What mass M must be attached to the cord in order to prevent the block from sliding down y.

SolutionFor the mass m resolve the weight

xmg into two components one in x- axis mg sin 30 and one in y-axis. mg cos 30Then applying Σ�� = 0, ��T -mg sin 30 = 0T = mg sin 30 (1)and applyingS Fy =0 for the y- components we getN = mg cos 30.For the second mass M we haveT - Mg = 0 T =Mg (2)From 1 and 2 we getMg = mg sin 30M = 10 sin 30 = 5 kg

M

Newton’s second lawWhenever a net force acts on a body it produces an acceleration in the

direction of the force which is proportional to the magnitude of the force and inversely proportional to the mass of the body. this means that

F = m aF is the net force acting on the object.

To Study the motion of the object we need to know the forces acting on the object not the forces the body exerts on other objects.

.

F

F1

F2

mmFnetaa

•The force F produces an acceleration

a = F/mto the mass m in the same direction as F

•The sum of the two applied forces is zero and the acceleration is zero

m- F F

a Fm

The v vector sum of F1 and F2produces a resultant force Fnet��

which is different from either F1 and F2.

a

F net

F2

m

F1

Newton’s third law

Newton’s third law of dynamics tells us that single forces cannot occur; Forces always occur in pairs . We usually call one of them the force and the other the reaction. The third law states that:

For every action force there is an equal but opposite reaction

This means that if object 1 exerts force on object 2, then object 2 exerts an equal but opposite in direction on object 1. That is a force always paired with an equal reaction force. In equation form,

F12 = - F21

• One example of action reaction pair is a block on a horizontal plane. The block exerts a force equal to its weight mg on the plane , the plane exerts an equal but opposite force on the block. Thus, twoforces are acting on the block�mg�downward and N upward. These two forces are equal in magnitude and opposite in direction. therefore,

the net force acting on the block is zero and it has no acceleration

It is always important to distinguish between those forces acting on the body and those forces that the body exerts on other object.

A body moves under the effect of the force acting on it and not the force that the body exerts.

Fig. 5.9, p.124

Solving Problems with Newton’s Laws –Free-Body Diagrams

1. Draw a sketch.

2. For one object, draw a free-body diagram, showing all the forces acting on the object. Make the magnitudesand directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one.

3. Resolve vectors into components.

4. Apply Newton’s second law to each component.

5. Solve.

Solving Problems with Newton’s Laws –Free-Body Diagrams

When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force.

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

For kinetic – sliding –friction, we write:

is the coefficient of kinetic friction, and is different for every pair of surfaces.

Applications Involving Friction, Inclines


Static friction is the frictional force between two surfaces that are not moving along each other. Static friction keeps objects on inclines from sliding, and keeps objects from moving when a force is first applied.

The static frictional force increases as the applied force increases, until it reaches its maximum.

Then the object starts to move, and the kineticfrictional force takes over.


Applications Involving Friction, InclinesAn 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.

The normal force N= mg cos θθθθ , The forces parallel to the plane is mg sin θ.θ.θ.θ.The frictional force f is parallel to the plane.If the block is in equilibrium, then

f = mg sin θθθθ. .Using Fmax = µµµµs N we get

µµµµs N = mg sin θθθθµµµµs mg cos θθθθ = mg sin θθθθ

µµµµs = tan θθθθ

Problem Solving – A General Approach

1. Read the problem carefully; then read it again.

2. Draw a sketch, and then a free-body diagram.

3. Choose a convenient coordinate system.

4. List the known and unknown quantities; find relationships between the knowns and the unknowns.

5. Estimate the answer.

6. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in.

7. Keep track of dimensions.

8. Make sure your answer is reasonable.

Kinematics of Uniform Circular Motion

Uniform circular motion: motion in a circle of constant radius at constant speed

Instantaneous velocity is always tangent to circle.

Kinematics of Uniform Circular MotionLooking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that

Kinematics of Uniform Circular MotionThis acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

Dynamics of Uniform Circular Motion

For an object to be in uniform circular motion, there must be a net force acting on it.

We already know the acceleration, so can immediately write the force:

(5-1)


We can see that the force must be inward by thinking about a ball on a string:


There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.

If the centripetal force vanishes, the object flies off tangent to the circle.

Highway Curves, Banked and Unbanked

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.


If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.


As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than the static.

2. The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.


Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the

horizontal component of the normal force, and no friction is required. This occurs when:

Nonuniform Circular Motion

If an object is moving in a circular path but at varying speeds, it must have a tangentialcomponent to its acceleration as well as the radial one.

Nonuniform Circular Motion

This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

Centrifugation

A centrifuge works by spinning very fast. This means there must be a very large centripetalforce. The object at A would go in a straight line but for this force; as it is, it winds up at B.

Newton’s Law of Universal GravitationIf the force of gravity is being exerted on objects on Earth, what is the origin of that force?

Newton’s realization was that the force must come from the Earth.

He further realized that this force must be what keeps the Moon in its orbit.

Newton’s Law of Universal GravitationThe gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth.

When there is such a disparity in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant.

Newton’s Law of Universal Gravitation

Therefore, the gravitational force must be proportional to both masses.

By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.

In its final form, the Law of Universal Gravitation reads:

where

(5-4)

Newton’s Law of Universal GravitationThe magnitude of the

gravitational constant Gcan be measured in the

laboratory.

This is the Cavendishexperiment.

Gravity Near the Earth’s Surface; Geophysical Applications

Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth:

Solving for g gives:

Now, knowing g and the radius of the Earth, the mass of the Earth can be calculated:

(5-5)

Gravity Near the Earth’s Surface; Geophysical Applications

The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

Satellites and “Weightlessness”

Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.


The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

Satellites and “Weightlessness”Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though!

The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.


More properly, this effect is called apparentweightlessness, because the gravitational force still exists. It can be experienced on Earth as well, but only briefly:

Chapter 5 and 6 Newton Laws

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