2-3_-_Force

Kinematics in Two or Three Dimensions; Vectors

Copyright © 2009 Pearson Education, Inc.

Units of Chapter 3

• Vectors and Scalars

• Addition of Vectors—Graphical Methods

• Subtraction of Vectors, and Multiplication of a Vector by a Scalar

• Adding Vectors by Components


• Unit Vectors

• Vector Kinematics

• Projectile Motion

• Solving Problems Involving Projectile Motion

• Relative Velocity

3-1 Vectors and Scalars

A vector has magnitude as well as direction.

Some vector quantities: displacement, velocity, force, momentum


A scalar has only a magnitude.

Some scalar quantities: mass, time, temperature

3-2 Addition of Vectors—Graphical Methods

For vectors in one dimension, simple addition and subtractionare all that is needed.


You do need to be careful about the signs, as the figure indicates.


If the motion is in two dimensions, the situation is somewhat more complicated.

Here, the actual travel paths are at right angles to one another; we can find the displacement by

using the Pythagorean Theorem.



Adding the vectors in the opposite order gives the same result:



Even if the vectors are not at right angles, they can be added graphically by using the tail-to-tip method.



The parallelogram method may also be used; here again the vectors must be tail-to-tip.


3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar

In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction.


Then we add the negative vector.

3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar

A vector can be multiplied by a scalarc; the result is a vector c that has the same direction but a magnitude cV. If c is

negative, the resultant vector points in the opposite direction.

rV r

V


the opposite direction.

3-4 Adding Vectors by Components

Any vector can be expressed as the sumof two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.




If the components are perpendicular, they can be found using trigonometricfunctions.


The components are effectively one-dimensional, so they can be added arithmetically.



Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choose x and y axes.

3. Resolve each vector into x and y components.

4. Calculate each component using sines and cosines.


4. Calculate each component using sines and cosines.

5. Add the components in each direction.

6. To find the length and direction of the vector, use:

and .


Example 3-2: Mail carrier’s displacement.

A rural mail carrier leaves the post office and drives 22.0 km


in a northerly direction. She then drives in a direction 60.0°south of east for 47.0 km. What is her displacement from the post office?


Example 3-3: Three short trips.

An airplane trip involves three legs, with two stopovers. The first leg is due east for 620 km;


the second leg is southeast (45°) for 440 km; and the third leg is at 53°south of west, for 550 km, as shown. What is the plane’s total displacement?

3-5 Unit Vectors

Unit vectors have magnitude 1.

Using unit vectors, any vector can be written in terms of its components:

rV


3-6 Vector Kinematics

In two or three dimensions, the displacement is a vector:


vector:


As ∆t and ∆r become

smaller and smaller, the average velocity approaches the instantaneous velocity.



rv

rv

The instantaneous acceleration is in the direction of ∆ = 2 – 1, and is given by:

rv



Using unit vectors,



Generalizing the one-dimensional equations for constant acceleration:


3-7 Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's


influence of Earth's gravity; its path is a parabola.

It can be understood by analyzing the horizontal and verticalmotions separately.




The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

This photograph shows two balls that start to fall at the same time.


that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.


If an object is launched at an initial angle of θ0

with the horizontal, the analysis is similar except that the initial velocity has a vertical component.


3-8 Solving Problems Involving Projectile Motion

Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.



1. Read the problem carefully, and choose the object(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.


4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

5. Examine the x and y motions separately.


6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.

7. Plan how you will proceed. Use the


7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.


Example 3-6: Driving off a cliff.

A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high


horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.


Example 3-7: A kicked football.


Example 3-7: A kicked football.

A football is kicked at an angle θ0 = 37.0°with a velocity of 20.0 m/s, as shown. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball.


A child sits upright in a wagon which is moving to the right at constant speed as shown. The child extends her hand and throws

Conceptual Example 3-8: Where does the apple land?


child extends her hand and throws an apple straight upward (from her own point of view), while the wagon continues to travel forward at constant speed. If air resistance is neglected, will the apple land (a) behind the wagon, (b) in the wagon, or (c) in front of the wagon?


Conceptual Example 3-9: The wrong strategy.

A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away. At the instant the water

balloon is released, the second boy lets go and falls


from the tree, hoping to avoid being hit. Show that he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.


Example 3-10: Level horizontal range.

(a) Derive a formula for the horizontal range R of a projectile in terms of its initial speed v

0and angle

θ0. The horizontal range is defined as

the horizontal distance the projectile


the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y

0. (b)

Suppose one of Napoleon’s cannons had a muzzle speed, v

0, of 60.0 m/s.

At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?


Example 3-11: A punt.

Suppose the football in Example 3–7 was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Set


football travel before hitting the ground? Set x

0= 0, y

0= 0.


Example 3-12: Rescue helicopter drops supplies.

A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain


package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?


Projectile motion is parabolic:

Taking the equations for x and y as a function of time, and combining them to eliminate t, we find y as a function of x:


find y as a function of x:

This is the equation for a parabola.


Examples of projectile motion. Notice the effects of air resistance.


Summary of Chapter 3

• A quantity with magnitude and direction is a vector.

• A quantity with magnitude but no direction is a scalar.

• Vector addition can be done either graphically


• Vector addition can be done either graphically or by using components.

• The sum is called the resultant vector.

• Projectile motion is the motion of an object near the Earth’s surface under the influence of gravity.

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

• 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 it moving; a moving object needs a force to change its velocity.

4-1 Force


Force is a vector, having both magnitudeand direction. The magnitude of a force canbe 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


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.


This is Newton’s first law, which is often called thelaw 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.



Conceptual 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 ofmatter in the object. In the SI system, mass ismeasured in kilograms.

Mass is not weight.


Mass is a property of an object. Weight is theforce exerted on that object by gravity.

If you go to the Moon, whose gravitationalacceleration is about 1/6 g, you will weigh 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 accelerationand force. Acceleration is proportional to force andinversely proportional to mass.

It takes a force to change


It takes a force to changeeither the direction or thespeed of an object. More forcemeans more acceleration; thesame force exerted on a moremassive object will yield less

acceleration.


Force is a vector, so is true along eachcoordinate axis.

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

Note that the pound is a unit


Note that the pound is a unitof force, not of mass, and cantherefore be equated tonewtons but not to kilograms.


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

Estimate the net force needed to accelerate (a) a1000-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-


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

4-5 Newton’s Third Law of Motion

Any time a force is exerted on an object, that force iscaused 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 correctapplication of the third lawis that the forces are

exerted on different objects.Make sure you don’t usethem as if they were acting


them as if they were actingon the same object.


Rocket propulsion can also be explained using Newton’sthird law: hot gases from combustion spew out of the tailof the rocket at high speeds. The reaction force is whatpropels the rocket.

Note that the rocketdoes not need


does not needanything to “push”

against.


Conceptual Example 4-4: What exerts theforce to move a car?

Response: A common answer is that theengine makes the car move forward. But it isnot so simple. The engine makes the wheels


not so simple. The engine makes the wheelsgo around. But if the tires are on slick ice ordeep mud, they just spin. Friction is needed.On firm ground, the tires push backwardagainst the ground because of friction. ByNewton’s third law, the ground pushes on thetires in the opposite direction, acceleratingthe car forward.


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




Conceptual 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 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 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 bygravity. Close to the surface of the Earth, wherethe gravitational force is nearly constant, theweight of an object of mass m is:


where


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

The force exerted perpendicular to a surface is

called the normal force. It is


called the normal force. It isexactly as large as needed tobalance the force from theobject. (If the required forcegets too big, somethingbreaks!)


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

A 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 table.

(a) Determine the weight of the box and the


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

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


Example 4-7: Accelerating thebox.

What happens when a personpulls upward on the box in theprevious example with a force


previous example with a forcegreater than the box’s weight,say 100.0 N?


Example 4-8: Apparent weight loss.

A 65-kg woman descends in an elevatorthat briefly accelerates at 0.20g

downward. She stands on a scale thatreads in kg.


reads in kg.

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

(b) What does the scale read when theelevator descends at a constant speedof 2.0 m/s?

4-7 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


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.


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 the box by the


attached cord along the smooth surface of the table. The magnitude of the force exerted by the person is F

P=

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 F

N

exerted by the table on the box.


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

Two boxes, A and B, are connected


Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force of 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.


Example 4-16: Box slides down anincline.

A box of mass m is placed on a

smooth incline that makes an angleθ 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 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.


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 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, 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.

2-3_-_Force

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