|
Chapter Three Lecture
 |
|
Fun with Vectors
|
|
|
 Vectors will be used in this class throughout the whole year, so it is very important that you learn this
material and become comfortable with it. As I said previously, vectors are quantities that have both a magnitude and a direction. We draw vector quantiti
 es with an arrow. Length = magnitude. Direction points in
direction of vector. Every time we encounter a vector, we must use its direction and it magnitude. A vector is indicated in print by a bold letter v for velocity, or in writing by a letter with an
arrow over it
 . Vectors do
not have a particular location. I can move a vector that points up and is 2 units long anywhere I want. The magnitude of a vector is a scalar and is written with a regular font. The magnitude of
v is v.
Multiplication by Constant
 If I multiply a vector by a constant, that changes the magnitude of the vector. (Draw 2v
and -v).
Addition and Subtraction
How do we add vectors? Combining vectors that act in parallel directions only involves simple
addition or subtraction. There is also a way to draw vectors which adds (& subtracts) them. We put the tail of one vector on the tip of the other and draw
the resultant. If the vectors are in the same direction, simply a linear addition is done. If you draw the lines very accurately, then you can actually measure the result.
Note that you must always put the tail of the one on the tip of the
other. However, this is not always accurate enough so there is another way of adding vectors --- using components. The components are the
part of the vector along different axes, usually chosen to be the x and the y-axis. Every vector can be written as the sum of its components:
A = Ax + Ay. This is vector addition:
How long is Ax and Ay? To answer this we use trigonometric relations.
So look at the triangle. We see that cosq =Ax/A where Ax and A are the magnitude of the vectors Ax and A. Then we get the length, Ax=Acosq. In a similar manner sinq =Ay/A Þ Ay=Asinq. So there are two ways to specify a vector. We can give its magnitude A and its
direction as an angle, or compass direction, or something, or we can give its x and y components.
There are many directions available for everyday motion. Instead of
walking 10 meters directly north then walking 3 meters directly south a person might walk 10 meters directly north and 3 meters directly west. What method is used to handle the directions of vectors if the motions
are perpendicular to each other?
Suppose a boat is traveling in the ocean at a speed of 40 mph
directly north but there is also a current in the water that is 30 mph to the west. The resulting motion of the boat will no longer be directly north
nor will it be directly west. The two velocity vectors (40mph N and 30mph W) will both influence the overall motion of the boat. In fact, we might even be able to surmise the boat will move generally in a northwest
direction, but how do we determine the overall speed of the boat – it is not simply 40 mph. To determine the speed of the boat a geometrical method called the Pythagorean Theorem dealing with the sides of a right
triangle is used.
The Pythagorean Theorem states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squared lengths
of the remaining two sides.
|
 Problem: A boat traveling due
north at 40 mph relative to the shore encounters a cross current of 30 mph flowing to the West. What is the overall velocity of the boat
(speed and direction) due to the combination of these two vectors?
Pythagorean Theorem:
(Result)2 = (Side A)2 + (Side B)2
(R)2 = (40 mph)2 + (30 mph)2
(R)2 = 1600 mph2 + 900 mph2
(R)2 = 2500 mph2
Result = 50 mph to the West of North
|
|
|
|
|
Problem: Find the components of the following vectors A and
B. (Draw components)
 Magnitude of A is 5.00 and
Magnitude of B is 7.00.
Ax = Acosq = (5.00)cos(40)=(5.00)(.766)
Ax = 3.83
Ay = Asinq = (5.00)cos(40)=(5.00)(.642)
Ay = 3.21
Now, watch out because the B is tricky.
Bx= -Bsinq = (-7.00)sin(30)= -(7.00)(.500)= -3.50
By = Bcosq = (7.00)cos(30) = (7.00)(.866) = 6.06
|
|
|
|
How do we add the vectors using components? It is very easy once
you have found the components. (Draw the addition of A and B graphically to make C, then show Cx = Ax + Bx and Cy = Ay + By.)
|
 Problem: What is A + B
Cx = 3.83 + (-3.50) = 0.33
Cy = 3.21 + 6.06 = 9.27.
This is one way to write the answer.
We can also write it with magnitude and direction.
Magnitude:
Direction:
 |
|
|
|
|
|
Problem: A hiker walks 25.0 km due southeast, then 40.0 km in a direction
60° north of east. What are the components of her displacement vectors and how far is she from where she started?
Ax = Acos(-45°) = (25.0 km)(.707) =17.7 km
Ay = Asin(-45° )= (25.0 km)(-.707) =-17.7 km
Bx = Bcos(60°) = (40.0 km)(.500) = 20.0 km
By = Bsin(60°) = (40.0 km)(.866) = 34.6 km
Cx = 17.7 + 20.0 = 37.7 km
Cy = -17.7 + 34.6 = 16.9 km.
(or we can express this as)
at an angle of
 north of east. |
|
|
|
|
 Problem: A skier skis down a hill that
is 770 m long. The vertical drop is 230 m. What is the angle of the hill relative to flat ground?
 |
|
|
|
|
|
Projectile Motion
|
|
|
We have previously discussed motion in a constant gravitational field where the acceleration is constant. We know that:
What about the motion in the x direction? Is there any acceleration?
(Throw a ball). Is the particle changing velocity in the x direction? Amazingly, it is not!! Galileo first showed this.
Demonstration: Drop ball and shoot it.
What is going on in the horizontal direction is independent of what is
going on in the vertical direction. This is always true. The motion in each direction is independent of the motion in the other direction. So we
have general equations, and equations when there is no acceleration in the x direction which is the usual case for objects moving in a gravitational field.
General Equations Equations for x acceleration ax = 0
m/s2
Let's use these equations to solve some problems:
|
|
Problem: A plane must drop a package from 100 m above the ground. The plane is
traveling at 40.0 m/s.
 (a) Where does the package strike the ground? We know that
in the horizontal direction x = vxot. We know vxo but we don't know the distance it goes x or t, the amount of time it takes for the
package to hit the ground. There are no other forces acting on the package in the horizontal direction, and there is no acceleration in that
direction. So how can we find t. We look at what is happening in the
vertical direction. In the vertical direction the package is falling as if it had no horizontal velocity.
And, now how far along the ground can be found by using
(b) What are the horizontal and vertical components of the velocity
just before the package strikes the ground?
|
|
|
|
|
Problem: A golfer hits a ball with an initial speed of 40.3 m/s at an angle
of 32.0° from the horizontal. (a) How long does the golf ball take to reach its maximum height? (b) How far does the ball go and how long is it in the
air? (c) What is the maximum height of the golf ball? (d) What is its speed when it hits the ground? (Draw a diagram)
Vxo = vxcos(32.0°) = (40.3 m/s)(.848) = 34.2 m/s
vyo = vysin(32.0°) = (40.3 m/s)(.530) = 21.4 m/s
(a) Let's look at y-dimension first: At the top of the motion vy =
0m/s, thus:
The 2.18 seconds is the time for the ball to go from the ground up to
its maximum height. This time of 2.18 seconds is also useful because this time is also the same amount it takes the ball to fall from the
maximum height back down to the ground. So, the ball is in the air for
(b) 2´2.18 seconds = 4.36 seconds is the time for the entire trip
Thus the distance along the ground is given by the horizontal velocity component and the entire trip time.
(c) The time to reach the maximum height is the 2.18 seconds, thus
(d) Its final speed is given by
We know vx = vxo = 34.2 m/s
The final velocity is the same as the initial except for the direction of
each. This is true when the final position is at the same level as the initial position. (All of this motion is with no horizontal acceleration,
and assumes no air resistance).
|
|
|
|
|
Problem: A baseball player hits a home run and the ball lands in the left field
seats, 7.6 m above the point at which the ball was hit. The ball lands with a velocity of 49 m/s at an angle of 31° to the horizontal. What is the
initial velocity of the ball when the ball leaves the bat?
We have vx from: cos(31°) = vx/v Þ vx = 49 (m/s)cos(-31°) = 42 m/s
We have vy from: sin(31°) = vy/v Þ vy = 49 (m/s)sin(-31°) = -25 m/s
We have vxo from: ax = 0 Þ vxo =vx = 42 m/s
We need vyo and we have vy and ay = g so we can use
Now using the Pathagorean Theorem
with direction
 |
|
|
|
|
 Problem: A plane is flying horizontally at an
altitude of 4.2km with a speed of 225m/s. When the plane is directly overhead, a projectile is fired at an angle q with a speed of 389 m/s. The projectile hits the
plane. What is the angle q?
What will be the criteria for hitting the plane? The time it takes
the projectile to go to the plane must be the same as the time it takes the plane to go that distance. If the time is the same, what do we know about the
velocities?
The x velocities of the projectile and the plane must be the
same. Remember that what is happening in the x direction is independent of what is happening in y. So all we have to do is set;
|
|
|
Demonstration: Monkey Gun (Not the most politically correct example, but alas…)
Monkey drops from the tree limb just as the gun is fired because the monkey wants to avoid the bullet.
So where will the monkey be after any amount of time?
Horizontal distance is x the bullet's velocity is v.
 . So after that amount of time the bullet will be somewhere along the line yo, but where?
which is the same position the monkey is in.
|
|
Relative Velocities
|
|
|
We use vectors to solve problems involving the two velocities that are relative to each other.
|
 Problem: A person looking out of a
window of a stationary train notices that the drops are falling at a rate of 5.0m/s. When the train is moving, the drops are making an angle of 25° with
the vertical. How fast is the train moving?
Vy = Vcos(25°) = VR = 5.0 m/s
V = (5.0 m/s)/cos(25°) = 5.5 m/s
VT = Vsin(25°) = (5.5 m/s)(.423) = 2.3 m/s
|
|
|
|
|
 Problem: Bike Riding always seems to
give me a headwind, and seldom a tailwind. Why? Wind resistance is the major force making bicycling difficult. Suppose I can pedal 18 mph when there
is no wind. I am traveling east on highway 9 and there is a 15 mph cross wind from the south. You might think that a cross wind wouldn't make it harder
to pedal. But my velocity is a vector and the cross wind is a vector so let's look at the problem like that.
First, it is clear that a headwind will certainly make you go slower.
What about a cross wind?
First draw case with no wind. But I still want to go east, and the
wind is pushing me north, so I have to pedal in such a way that my net velocity is east. This is a right triangle, so I don't even have to
use components.
The vector C shows my pedaling and B is the wind blowing. Since this
is a right triangle we know;
A2 + B2 = C2 or
What if I am trying to go east and I have a partial tailwind blowing
20° east of north? How fast am I going now? I am still pedaling 18 mph.
A + B = C
Ax + Bx = Cx, Ay + By = Cy
Bx=Bsin(20°) = (15.0mph)(.342) = 5.13 mph
By=Bcos(20°) = (15.0mph)(.940) = 14.1 mph
First notice that since C has no component along the y direction that
Ay + By = Cy = 0.0 mph, so
Ay = -By = -14.1 mph.
Then since
Ax2 + Ay2 = A2
Þ Ax =
Cx = C = Ax + Bx = 11.2 + 5.13 = 16.3 mph.
|
|
|
|
So I am still going slower than I would if there was no wind even
though the wind is blowing toward the direction I want to go. Depending on the magnitude of the wind, even partial tail winds can make it harder
to pedal a bicycle. Since any kind of head winds, and partial tail winds can make it harder to pedal, it is much more probable that any wind will
make riding more difficult. It must be coming very much from behind you in order to help you. (The exact angle depends on your velocity and the
velocity of the wind. In the case of me going 18.0 and the wind 15.0, It will start to help me if the angle is greater than 24.6°).
|
