Ch3_GirginL


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= Vectors: Motion and Forces in Two Dimensions - Lesson 1: A and B = October 13, 2011

Vector quantities are often represented by scaled __ [|vector diagrams] __. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction.

Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below.

Representing the Magnitude of a Vector The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale.

 Vectors and Direction A vector quantity is a quantity that is described by magnitude and direction. A scalar quantity is described by its magnitude.Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction.

**Conventions for Describing Directions of Vectors** 
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "__[|tail]__" from east, west, north, or south.
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "__[|tail]__" from due East.

The Pythagorean Theorem The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.

Vector Addition A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result

The direction of a //resultant// vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. Use of Scaled Vector Diagrams to Determine a Resultant The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant. A common Physics lab involves a //vector walk//. Either using centimeter-sized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position.

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors:

20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. SCALE: 1 cm = 5 m



=Vectors: Motion and Forces in Two Dimensions - Lesson 1: C and D = October 17, 2011

Resultants The **resultant** is the vector sum of two or more vectors. It is //the result// of adding two or more vectors together. **A + B + C = R**

In all such cases, the resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. It is the same thing as adding A + B + C + ... . "To do A + B + C is the same as to do R."

In summary, the resultant is the vector sum of all the individual vectors. The resultant is the result of combining the individual vectors together. The resultant can be determined by adding the individual forces together using __[|vector addition methods]__.

Vector Components <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">A __[|vector]__ is a quantity that has both magnitude and direction. __[|Displacement]__, __[|velocity]__,__[|acceleration]__, and __[|force]__ are the vector quantities. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of ashaving two parts. Each part of a two-dimensional vector is known as a **component**. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">To Fido, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward. If the single chain were replaced by two chains. with each chain having the magnitude and direction of the components, then Fido would not know the difference. This is not because Fido is //dumb// (a quick glance at his picture reveals that he is certainly not that), but rather because the combined influence of the two components is equivalent to the influence of the single two-dimensional vector. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;"> <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Each wire exerts a tension force upon the picture to support its weight. Since each wire is stretched in two dimensions (both vertically and horizontally), the tension force of each wire has two components - a vertical component and a horizontal component.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Any vector directed in two dimensions can be thought of as having two different components. The component of a single vector describes the influence of that vector in a given direction.

= Vectors: Motion and Forces in Two Dimensions Lesson 1: E = October 17,2011 Vector Resolution Any vector directed in two dimensions can be thought of as having two components

The two basic methods


 * the parallelogram method
 * __ [|the trigonometric method] __

**Parallelogram Method of Vector Resolution**

The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector.

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 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector: beginning at the __ [|tail] __ of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the __ [|head] __ of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 5) Measure the length of the sides of the parallelogram and __ [|use the scale to determine the magnitude] __ of the components in //real// units. Label the magnitude on the diagram.

The step-by-step procedure above is illustrated in the diagram below to show how a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal may be resolved into two components. (NOTE: because different computer monitors have different resolutions, the actual length of the vector on your monitor may not be 5 cm.)




 * Trigonometric Method of Vector Resolution **

Trigonometric functions can be used to determine the length of the sides of a right triangle if an angle measure and the length of one side are known.

The method of employing trigonometric functions to determine the components of a vector are as follows:

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 * 1) Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the __ [|tail] __ of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the __ [|head] __ of the vector. The sketched lines will meet to form a rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
 * 5) To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.



<span style="color: #800080; display: block; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; text-align: left;">Measuring Angles <span style="color: #800080; display: block; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; text-align: left;">October 14, 2011

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=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Graphical and Analytical Method = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">October 15, 2011

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=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Orienteering Activity = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">October 18, 2011

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Lab Partners: Molly Lambert, Brianna Behrens

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=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Lesson 1: Vectors - Fundamentals and Operations; G & H = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">October 18, 2011

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; font-size: 14px;">**Analysis of a Riverboat's Motion**

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s.

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<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. The two vectors are at right angles to each other. Thus, the __[|Pythagorean theorem]__ can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows:

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">(4.0 m/s)2 + (3.0 m/s)2 = R2 <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">16 m2/s2 + 9 m2/s2 = R2 <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">25 m2/s2 = R2 <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">SQRT (25 m2/s2) = R <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**5.0 m/s = R**

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">The __[|direction]__ of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">tan (theta) = (opposite/adjacent) <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">tan (theta) = (3/4) <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">theta = invtan (3/4) <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**theta = 36.9 degrees**

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**Conclusive statement:** the motion of any object is influence when going against a current or wind.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; font-size: 14px;">Independence of Perpendicular Components of Motion <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. A **component** describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle. This is depicted in the diagram below.



<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">All vectors can be thought of as having perpendicular components.

=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Vectors: Motion and Forces in Two Dimensions - Lesson 2: A and B = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">October 19, 2011

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Part A: <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**__Central Theme__: Projectiles are only acted upon by the force of gravity.**

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Questions: <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">1. Can an object that is thrown vertically (like the example of a cannonball) be a projectile? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Yes, because a projectile is any object that once //projected// or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">2. How can gravity affect a projectile moving horizontally? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">It will cause it to fall below the "Gravity free" path it was set out for.

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<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">3. What would the body diagram of an object moving downwards look like?

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<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">4. How would you describe a projectile that is moving upward? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">It would be moving upward and slowing down because the force of gravity causes it to have a downward force.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Part B <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**__Central Theme__: The force of gravity does not affect the horizontal motion of the projectile.**

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Questions: <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">1. What are the components of the projectiles motion? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Vertical and horizontal motion that are independent of each other.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">2. Can a projectile have a horizontal acceleration? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">No, because vertical forces are the only ones that act upon projectiles, and vertical and horizontal motions are independent of each other.

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">3. What's the difference in motion of a horizontal projectile if gravity was turned off? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">There would be no difference. The projectile would fall the same amount of distance as if it were initially at rest.

=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Vectors: Motion and Forces in Two Dimensions Lesson 2: C = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">October 20, 2011

<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">1. What is the motion of horizontal velocity during the course of trajectory? <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">It remains constant while the vertical velocity changes 9.8 meters per second every second.


 * **Time** || **Horizontal****Velocity** || **Vertical****Velocity** ||
 * 0 s || 20 m/s, right || 0 ||
 * 1 s || 20 m/s, right || 9.8 m/s, down ||
 * 2 s || 20 m/s, right || 19.6 m/s, down ||
 * 3 s || 20 m/s, right || 29.4 m/s, down ||
 * 4 s || 20 m/s, right || 39.2 m/s, down ||
 * 5 s || 20 m/s, right || 49.0 m/s, down ||

2. What happens to the horizontal velocity if a projectile is launched at an angle to the horizontal? The direction will depend on the x and y components of the initial velocity.

3. How would you find the vertical displacement of a projectile? You would use the equation x= Vix(t)

4. What is the equation for the vertical displacement for an angled launched projectile? y = viy • t + 0.5 • g • t2

5. If gravity were able to "shut off", how would the motion of a ball launched from a cannon at 20 m/s be described? It would remain constant at 20 m/s until it wears off its initial inertia.

= "Gourd-o-rama" Final Project = November 1, 2011 Partner: Brianna Behrens



Results: Our project traveled a total distance of 13.75 m in 6.59 s.

Calculations:

Conclusion: Our project, at first, was veering off to the side and hitting the wall when we experimented which was due to the pumpkin being unstable. If we could revise the project we would make the pumpkin more secure in its place. In addition, we would make the project a little lighter since it weight about 2 kilograms by making it smaller and more festive by adding more decorations to it.

=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Ball in Cup Activity = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">

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<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">**Video** <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">media type="file" key="Movie on 2011-10-25 at 12.58.mov" width="300" height="300"

=<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Shoot Your Grade Lab = <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">November 6, 2011 <span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">Lab partners: Molly Lambert & Brianna Behrens

__Hypothesis:__ The ball will be launched from the launcher and will go through five of the tape rings hung from the ceiling and land in the cup.

__Purpose and Rational__ For this lab, our goal was to launch a ball from a launcher at 20 degree angle and have it go through the hoops hung from the ceiling. Our main purpose was to find out approximately where to hand the hoops. To do this, we needed to calculate the height of the launcher from the floor (our delta D for the X component) and the distance that the ball traveled. We were able to get that distance by placing a black sheet of paper on the ground that marked the spot where the ball landed on the floor when it was set at "medium range". Once we figured that out, we got the velocity for the x and y components and used our horizontal distance to find where the hoops would be hung.

__ Materials and Methods __ The materials we used for this lab were a launcher set at "medium range" at 20 degrees, a ball, black string, five tape rings, right-angle clamps, tape measure and a plumb bob. Our methods included changing the positions of the hoops in order for the ball to successfully go through as many as time allowed. At first, the hoops were placed based on our calculations, but our experimental results gave us different coordinates. In addition, another method we had was taping pieces of paper to the hoops to see if the ball actually went through the hoops or if our eyes were fooling us.

__Observations and Data from Initial Velocity__:

__Observations and Data from Performance__: media type="file" key="Movie on 2011-11-08 at 13.24.mov" width="300" height="300" The ball only went through two hoops in this trial.

__Physics Calculations__ :

__ Data Table of Theoretical Measurements __:

__Error Analysis:__ Actual positions of hoops

__Percent Errors for X and Y positions:__

__Samples for % Error:__

__Conclusion:__ After doing our calculations in class and performing the lab, we saw that our hypothesis was not correct. We hypothesized that the ball would go through all five tape rings and land in the cup, but instead, our ball went through only three tape rings. When we calculated our percent error for one of the hoops we found it to be 8.56% error for the x position, and 2.37% error for the y position for one of the hoops. These errors are due to several factors that affected the success of the experiment. One factor was from the air vent in the classroom, which constantly blew in the direction of the strings attached to the hoops, changing their positions. This led to the ball being launched at the new positions of the hoops and missing or hitting a side of the tape. Another factor that increased the errors in this experiment was the launchers inconsistency when launching the ball. Sometimes it would shoot the ball at a higher height than it had previously and sometimes it would shoot it lower. Lastly, another factor that could have caused the errors was the spring inside the launcher not being warm when we launched the ball. At first, we continually changed the positions of the hoops so that the ball could be launched through them, but then the launcher would shoot the ball through the original placements of the hoops. To adress the errors that were brought up in this experiment several things would have been changed. In order to get the best results we would do the experiment in a room without an air vent, or move our equipment somewhere else in the room where the air vent didn't blow on the strings attached to the hoops. By doing this, the hoops will stay in the exact positions that brought the best results. In addition, we would be more careful about the launchers position and angle because it tends to move a little once the ball has been shot. This would also help and improve our results. Finally, we would let the spring warm up and shoot the ball a few times, without changing the position of the hoops. This would eliminate the constant trial and error that went on in this lab. This lab's relevancy to real life can be seen in engineering a catapult because the catapult needs to be set an angle to shoot whatever is on it to the desired position. Also, the engineer would need to figure out the x and y components of the velocity of the catapult. >