chapter dynamics: newton’s laws motion - universiti …zuhairusnizam.uitm.edu.my/lecture...

Chapter 4

Dynamics: Newton’s Laws of Motion

Units of Chapter 4• Force

• Newton’s First Law of Motion

• Mass

• Newton’s Second Law of Motion

• Newton’s Third Law of Motion

• Weight—the Force of Gravity; and the Normal Force• Weight—the Force of Gravity; and the Normal Force

• Solving Problems with Newton’s Laws: Free‐Body Diagrams

• Problem Solving—A General Approach

4‐1 Force

A force is a push or pull An object at rest needs a force to get itA force is a push or pull. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity.

4‐1 Force

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

4‐2 Newton’s First Law of Motion

It may seem as though it takes a force to keep an object moving. Push your book across a table—when you stop pushing, it stops moving.

But now, throw a ball across the room. The ball keeps moving after you let it go, even though you are not pushing it any more. Why?

It doesn’t take a force to keep an object moving in a straight line—it takes a force to change its motion. Your book stops because the force of friction stops it.

4‐2 Newton’s First Law of Motion

Thi i N t ’ fi t l hi h i ft ll d th l f i tiThis is Newton’s first law, which is often called the law of inertia:

Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.

4 2 N t ’ Fi t L f M ti4‐2 Newton’s First Law of Motion

C t l E l 4 1 N t ’ fi t lConceptual Example 4‐1: Newton’s first law.

A school bus comes to a sudden stop, and all of the backpacks on the floor start to slide forward. What force causes them to do that?

4‐3 Mass

Mass is the measure of inertia of an object, sometimes understood as the quantity of matter in the object. In the SI system, mass is measured in kilograms.

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 weigh much less. Your mass, however, will be the same.much less. Your mass, however, will be the same.

4‐4 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.

It takes a force to change either the direction or the speed of an object. More force means more acceleration; the same force exerted on a moreacceleration; the same force exerted on a more massive object will yield less acceleration.


F i t i t l h di t iForce is a vector, so is true along each coordinate axis.

The unit of force in the SI system is the newton(N)(N).

Note that the pound is a unit of force, not of mass, and can therefore be equated to newtons but not to kilograms.newtons but not to kilograms.


Example 4‐2: Force to accelerate a fast car.

Estimate the net force needed to accelerate (a) a 1000‐kg car at ½ g; (b) a 200‐g apple at the same rate.

Example 4‐3: Force to stop a car.

What average net force is required to bring a 1500‐kg car to rest from a speed of 100 km/h within a distance of 55 m?100 km/h within a distance of 55 m?

4‐5 Newton’s Third Law of Motion

Any time a force is exerted on an object that force is caused by another objectAny time a force is exerted on an object, that force is caused by another object.

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.


A key to the correct application of the third law is that the forces are exerted onlaw is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the sameobject.

4‐5 Newton’s Third Law of MotionRocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket.

Note that the rocket does not need anything to “push” againstanything to “push” against.


Conceptual Example 4‐4: What exerts the force to move a car?

Response: A common answer is that the engine makes the car move forward. But it i t i l Th i k th h l d B t if th ti li kis not so simple. The engine makes the wheels go around. But if the tires are on slick ice or deep mud, they just spin. Friction is needed. On firm ground, the tires push backward against the ground because of friction. By Newton’s third law, the groundpushes on the tires in the opposite direction, accelerating the car forward.p pp , g


Helpful notation: the first subscript is the object that the force is being exerted on; theHelpful notation: the first subscript is the object that the force is being exerted on; the second is the source.


Conceptual Example 4 5: Third law clarificationConceptual Example 4‐5: Third law clarification.

Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, “When I exert a forward force g y ,on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must bereaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct?

4‐6 Weight—the Force of Gravity; and the Normal Force

Weight is the force exerted on an object by gravity. Close to the surface of the Earth where the gravitational force is nearly constant theof the Earth, where the gravitational force is nearly constant, the weight of an object of mass m is:

where


An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there?

The force exerted perpendicular to a surface is

called the normal force. It is exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!)the required force gets too big, something breaks!)


Example 4‐6: Weight, normal force, and a box.

A friend has given you a special gift a box of mass 10 0A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a t bltable.

(a) Determine the weight of the box and the normal force exerted on it by the tableforce exerted on it by the table.

(b) Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal forceforce of 40.0 N. Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force ( ) y p pof 40.0 N, what now is the normal force exerted on the box by the table?


Example 4 7: Accelerating the boxExample 4‐7: Accelerating the box.

What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight say 100 0 N?than the box s weight, say 100.0 N?


Example 4‐8: Apparent weight loss.

A 65‐kg woman descends in an elevator that briefly accelerates at 0.20g downward. She stands on a scale that reads in kg. g g

(a) During this acceleration, what is her weight and what does the scale read?

(b) What does the scale read when the elevator descends at a(b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s?

4‐7 Solving Problems with Newton’s Laws: Free‐Body Diagramsy g

1. Draw a sketch.

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

3. Resolve vectors into components.

l ’ d l h4. Apply Newton’s second law to each component.

5 Solve5. Solve.


Example 4‐11: Pulling the mystery box.Suppose a friend asks to examine the 10.0‐kg

box you were given previously, hoping to guess what is inside; and you respond, “Sure, pull the box over to you.” She then pulls thepull the box over to you. She then pulls the box by the attached cord along the smooth surface of the table. The magnitude of the f t d b th i F 40 0 Nforce exerted by the person is FP = 40.0 N, and it is exerted at a 30.0° angle as shown. Calculate

(a) the acceleration of the box, and (b) the magnitude of the upward force FN

exerted by the table on the box.exerted by the table on the box.


Example 4‐12: Two boxes connected by a cord.

Two boxes, A and B, are connected by a lightweight cord and are resting on a g g gsmooth table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force of 40 0 N i li d t th 10 0 k b Fi d ( )40.0 N is applied to the 10.0‐kg box. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes.g


Example 4‐13: Elevator and counterweight (Atwood’s machine).

A system of two objects suspended over a pulley by a flexible cable is sometimes referred to as an A d’ hi H l h f hAtwood’s machine. Here, let the mass of the counterweight be 1000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying four passengers is 1150 kg. For the latter case calculate (a) the acceleration of the elevator and (b) the tension in the cableelevator and (b) the tension in the cable.


Conceptual Example 4‐14: The advantage of a pulley.

A mover is trying to lift a piano (slowly) up to a second‐story apartment. He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano’s 2000‐N weight?


Example 4‐15: Accelerometer.

A small mass m hangs from a thin string and can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make (a) when the car accelerates at a constant a = 1.20 m/s2 and (b) when the car moves at constant velocitym/s , and (b) when the car moves at constant velocity, v = 90 km/h?


Example 4‐16: Box slides down an incline.

A box of mass m is placed on a smooth incline thatA box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of θ = 30°.

4‐8 Problem Solving—A General Approach

1 Read the problem carefully; then read it again1. 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 relationshipsb t th k d th kbetween 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.

Summary of Chapter 4

• Newton’s first law: If the net force on an object is zero it will remain either at rest or• Newton s first law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed.

• Newton’s second law:

• Newton’s third law:

• Weight is the gravitational force on an object.

• Free‐body diagrams are essential for problem‐solving. Do one object at a time, makeFree body diagrams are essential for problem solving. Do one object at a time, make sure you have all the forces, pick a coordinate system and find the force components, and apply Newton’s second law along each axis.

5‐1 Applications of Newton’s Laws Involving Friction

Friction is always present when two solid surfaces slide along each other.

The microscopic details are not yet fully understood.


Sliding friction is called kinetic frictionSliding friction is called kinetic friction.

Approximation of the frictional force:

Ffr = μkFN .

Here, FN is the normal force, and μk is the coefficient of kinetic friction, which is different for each pair of surfaces.


Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table).

Th t ti f i ti l f i bi it d t b t t li i tThe static frictional force is as big as it needs to be to prevent slipping, up to a maximum value.

Ffr ≤ μsFN .

Usually it is easier to keep an object sliding than it is to get it started.


Note that, in general, μs > μk.


Example 5‐1: Friction: static and kinetic.

Our 10.0‐kg mystery box rests on a horizontal floor. The coefficient ofOur 10.0 kg mystery box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude:

(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.


Conceptual Example 5‐2: A box against a wall.

You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving

ti ll ?vertically?


Example 5‐3: Pulling against friction.

A 10.0‐kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration.


Conceptual Example 5‐4: To push or to pull a sled?

Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the y p psame angle θ in each case.


Example 5‐5: Two boxes and a pulley.

Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and anyignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at thecord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right.moves to the right.


Example 5‐6: The skier.

This skier is descending a 30° slope, at constant g p ,speed. What can you say about the coefficient of kinetic friction?


Example 5‐7: A ramp, a pulley, and two boxes.

Box A of mass 10 0 kg rests on a surface inclined at 37° to the horizontal ItBox A, of mass 10.0 kg, rests on a surface inclined at 37 to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. (a) If the coefficient of static friction is 0.40, determine what range of values for masscoefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system.

4. circular motion

10‐1 Angular Quantities

In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”). The radius of the circle is R. All points on a straight line p gdrawn through the axis move through the same angle in the same time. The angle θ in radians is defined:

l

where l is the arc length.

,R


Example 10‐1: Birds of prey—in radians.

A particular bird’s eye can just distinguish objects that subtend an angle no smaller than

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objects that subtend an angle no smaller than about 3 x 10‐4 rad.

(a) How many degrees is this?

(b) (b) How small an object can the bird just distinguish when flying at a height of 100 m?

10‐1 Angular QuantitiesAngular displacement:

Th l l it i d fi d th t t lThe average angular velocity is defined as the total angular displacement divided by time:

The instantaneous angular velocity:


The angular acceleration is the rate at which the angular velocity changes with time:

The instantaneous acceleration:

10‐1 Angular QuantitiesEvery point on a rotating body has an angular velocity ω and a linear velocity v.

They are related:


Conceptual Example 10‐2: Is the lion faster than the horse?horse?

On a rotating carousel or merry‐go‐round, one child sits on a horse near the outer edge and another child sitson a horse near the outer edge and another child sits on a lion halfway out from the center.

(a) Which child has the greater linear velocity?( ) g y

(b) Which child has the greater angular velocity?


Objects farther from the axis of rotation will move faster.


If the angular velocity of a rotating object changes, it has a tangential acceleration:

Even if the angular velocity is constant, each point on the object has a centripetal acceleration:


Here is the correspondence between linear and rotational quantities:


Example 10‐3: Angular and linear velocities and accelerations.

A carousel is initially at rest. At t = 0 it is given a constant angular acceleration α = 0.060 rad/s2, which increases its angular velocity for 8.0 s. At t = 8.0 s, determine the

it d f th f ll i titimagnitude of the following quantities:

(a) the angular velocity of the carousel;

(b) the linear velocity of a child located 2.5 m from the ( ) ycenter;

(c) the tangential (linear) acceleration of that child;

(d) the centripetal acceleration of the child; and(d) the centripetal acceleration of the child; and

(e) the total linear acceleration of the child.


The frequency is the number of complete revolutions per second:

Frequencies are measured in hertz:

The period is the time one revolution takes:


Example 10‐4: Hard drive.

The platter of the hard drive of a computer rotates at 7200 rpm (rpm = revolutions per minute = rev/min).

(a) What is the angular velocity (rad/s) of the platter?

(b) If the reading head of the drive is located 3.00 cm from the rotation axis, what is the linear speed of the point on the platterrotation axis, what is the linear speed of the point on the platter just below it?

(c) If a single bit requires 0.50 μm of length along the direction of motion, how many bits per second can the writing head writemotion, how many bits per second can the writing head write when it is 3.00 cm from the axis?

10‐1 Angular Quantities10 1 Angular Quantities

Example 10‐5: Given ω as function of timeExample 10‐5: Given ω as function of time.

A disk of radius R = 3.0 m rotates at an angular velocity

ω = (1.6 + 1.2t) rad/s,

where t is in seconds. At the instant t = 2.0 s, determine

(a) the angular acceleration, and

(b) (b) h d d h f h l(b) (b) the speed v and the components of the acceleration aof a point on the edge of the disk.

5‐2 Uniform Circular Motion—Kinematics

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

Instantaneous velocity is always tangent to the circle.


Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that

.


This acceleration is called the centripetal, or radial, acceleration, and it points toward the center of the circle.


Example 5‐8: Acceleration of a revolving ballExample 5 8: Acceleration of a revolving ball.

A 150‐g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?What is its centripetal acceleration?


Example 5‐9: Moon’s centripetal acceleration.

The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km and a period T of 27.3 days. Determine the acceleration of the Moon toward the Earth.


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

5 2 U if Ci l M ti Ki ti5‐2 Uniform Circular Motion—Kinematics

Example 5 10: UltracentrifugeExample 5‐10: Ultracentrifuge.

The rotor of an ultracentrifuge rotates at 50,000rpm (revolutions per minute). A particle at the topof a test tube is 6.00 cm from the rotation axis.Calculate its centripetal acceleration, in “g’s.”

chapter dynamics: newton’s laws motion - universiti …zuhairusnizam.uitm.edu.my/lecture...

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