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Published byJayde Hare Modified about 1 year ago

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Richard J. Terwilliger by

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Let’s look at some examples.

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We’ll break this GREEN vector up into different RED vectors.

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Here the GREEN vector is broken down into two RED vectors

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The two RED vectors are called the of the GREEN vector.

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Or, the of the two RED vectors is the GREEN vector.

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The same GREEN vector can be broken up into 2 different RED vectors.

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Or two different vectors!

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There are possibilities!

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A little later on we’ll show that two components at right angles are very helpful!

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Here the same GREEN vector is broken up into 3 different component vectors

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Again, the of the three RED vectors is the GREEN vector!

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And, the GREEN vector can be broken into RED vectors

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Let’s see another example.

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This BLUE vector is broken down into several ORANGE vectors.

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The tail of the BLUE vector starts at the same point as the tail of the first ORANGE vector. STARTING POINT

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And they both end at the same point ENDING POINT STARTING POINT

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The of the ORANGE vectors is the BLUE vector. E

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E Or again, the of the BLUE vector are the ORANGE vectors..

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E So what have we learned so far?

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E A vector can be broken down into any number of components. The different arrangements of components are unlimited. Components are one of two or more vectors having a sum equal to a given vector. A resultant is the sum of the component vectors.

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The process of finding the components of a vector given it’s magnitude and direction is called: E

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E is an easy way to find the resultant of several vectors.

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E To show how, let’s go back to two at right angles.. 90 o

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E We’ll sketch in the x-axis so it goes through the TAIL of our original vector. X-axis TAIL

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E X-axis Y-axis TAIL Then we’ll sketch the y-axis so it goes through the TAIL of our original vector, too.

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Y-axis E In the diagram the red vector is called the horizontal or X- component. X-axis

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Y-axis E The blue vector is called the vertical or Y- component. X-axis

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Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

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Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

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Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

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E Knowing the original vector’s magnitude and direction we can solve for Vx and Vy using trigonometry. 90 o 0o0o 180 o 270 o

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E Let’s label the original vector with it’s magnitude and direction. 0o0o 180 o 90 o 270 o

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90 o 270 o E Let’s label the original vector with it’s magnitude and direction. 0o0o 180 o

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90 o 270 o E To determine the value of the X-component (Vx) we need to use COSINE. 0o0o 180 o

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90 o 270 o E Remember COSINE? 0o0o 180 o

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E 0o0o Cosine = Adjacent Hypotenuse 90 o 270 o

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90 o 270 o We need to rearrange the equation: solving for the adjacent side.. Cosine = Adjacent Hypotenuse E 0o0o 180 o

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90 o 270 o Cos = Adj Hyp E 0o0o 180 o Adj = (Hyp) (Cos ) therefore Vx = V cos

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90 o 270 o E 0o0o 180 o Adj = (Hyp) (Cos ) therefore Cos = Adj Hyp Vx = V cos

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90 o 270 o E 0o0o 180 o Vx = V cos Vx = 36 m/s (cos 42 o ) Vx = 26.7 m/s

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90 o 270 o E 0o0o 180 o We can now solve for Vy using Sine.

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90 o 270 o E 0o0o 180 o Sine = Opposite Hypotenuse

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90 o 270 o We are solving for the side opposite the 42 degree angle, Vy, therefore we’ll rearrange the equation solving for the opposite side. E 0o0o 180 o

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90 o 270 o therefore E 0o0o 180 o Vy = V sin Opp = Hyp (Sin ) Sine = Opposite Hypotenuse

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90 o 270 o therefore E 0o0o 180 o Vy = V sin Opp = Hyp (Sin ) Sine = Opposite Hypotenuse

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90 o 270 o E 0o0o 180 o Vy = V sin Vy = 36 m/s (sin 42 o ) Vy = 24.1 m/s

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90 o 270 o E 0o0o 180 o We’ve solved for the horizontal vertical and components of our original vector!

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90 o 270 o E 0o0o 180 o Let’s REVIEW what we did.

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V REVIEW A vector can be broken down into two vectors at right angles A MATHEMATICAL METHOD 90 o

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V REVIEW A MATHEMATICAL METHOD The component that lies on the X-axis is called the horizontal or X - component The X-component is found using Vx = Vcos Vx = Vcos

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V REVIEW A MATHEMATICAL METHOD The component that lies on the Y-axis is called the vertical or Y- component The Y-component is found using Vy = Vsin Vy = Vsin Vx = Vcos

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Now that we know how to break vectors down into their X and Y components we can solve for the resultant of many vectors using:

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Let’s look at a sample problem!

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travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”.

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To solve the problem, using the component method, we’ll carry out the following sequence: 1. List the vectors 2. Solve for the X and Y components of each vector 3. Sum the X components and sum the Y components 5. Solve for the direction of the resultant using tangent 4. Solve for the magnitude of the resultant using the Pythagorean theorem

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First, let’s list all of the vectors changing all of the directions to the number of degrees from east in a counterclockwise direction. 1. List the vectors

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travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”. 90 o 0o0o 180 o 270 o

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travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”. 90 o 0o0o 180 o 270 o d 1 = o d 2 = o d 3 = o d 4 = o 10 o east of north 30 o south of west 71 o east of south 37 o north of east. 4.2 m 3.9 m 2.6 m 1.8 m

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Next, to make things easier to see, let’s set up a table that lists our vectors.. 1. List the vectors

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First label all of the columns Vector (V) X- component Solve Y -component Solve

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Remember the X component is cosine Vector (V) X- component Solve Y -component Solve Vector (V) X- component Solve Y -component Solve (Vcos )

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And the Y component is sine Vector (V) X- component Solve Y -component Solve Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin )

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Next fill in the first column with the different vectors o o o o

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Substitute into the equations for X and then Y o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o Remember to include units!

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Now solve for the value of each X and Y component. Again, remember to include UNITS! Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m

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Sum the X components and then the Y components. Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m Sum ( ) x = 3.48 m Y = 4.49 m

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m Sum ( ) x = 3.48 m Y = 4.49 m The four vectors are now broken down into two.

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m Sum ( ) x = 3.48 m Y = 4.49 m One vector 3.48 m long lies on the X-axis.

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m Sum ( ) x = 3.48 m Y = 4.49 m The other vector, 4.49 m long, lies on the Y-axis.

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Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) Vector (V) X- component Solve Y -component Solve (Vcos ) (Vsin ) o o o o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m 2.53 m 3.84 m m m Sum ( ) x = 3.48 m Y = 4.49 m We’ll plot these two vectors and determine their resultant.

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Y x = 3.48 m Y = 4.49 m We’ll plot these two vectors and determine their resultant. X x = 3.48 m Y = 4.49 m

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Y X x = 3.48 m The resultant of these two vectors is the same as the resultant of our original four vectors!. Y = 4.49 m

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Y X x = 3.48 m The Pythagorean Theorem will be used to determine the MAGNITUDE of the resultant. c 2 = a 2 +b 2 R 2 = X 2 + Y 2 R = X 2 + Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m

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Y X Y = 4.49 m x = 3.48 m We have now found the magnitude of our resultant to be 5.68 m long. c 2 = a 2 +b 2 R 2 = X 2 + Y 2 R = X 2 + Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m

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Y X Y = 4.49 m x = 3.48 m The last step is to find the DIRECTION our resultant is pointing.

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Y X Y = 4.49 m Y = 3.48 m To find the angle we’ll use tangent. Tan = adjacent opposite adj opp = Tan -1 XX YY 3.48 m 4.49 m = Tan -1 = = 52.2 o

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Y X Y = 4.49 m Y = 3.48 m The RESULTANT is o. Let’s go back to our original problem with the results.

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travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”.

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Let’s SUMMARIZE the COMPONENT METHOD

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List all the vectors. Break each vector down into an X and Y component using Cosine for X and Sine for Y. Sum all of the X components and then sum all of the Y components Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant Using Tangent solve for the DIRECTION of the resultant.

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Let’s go over that once again.

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Vector (V) X-component Solve Y-component Solve xx YY o o o o List all the vectors.

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Break each vector down into an X and Y component using Cosine for X and Sine for Y. Vector (V) X-component Solve Y-component Solve o o o o (Vcos ) (Vsin ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o xx YY

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Sum all of the X components and then sum all of the Y components Vector (V) X-component Solve Y-component Solve o o o o (Vcos ) (Vsin ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m Sum ( ) x = 3.48 m 2.53 m 3.84 m m m Y = 4.49 m xx YY Y = 3.48 m Y = 4.49 m

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Vector (V) X-component Solve Y-component Solve o o o o (Vcos ) (Vsin ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m Sum ( ) x = 3.48 m 2.53 m 3.84 m m m Y = 4.49 m Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant c 2 = a 2 +b 2 R 2 = X 2 + Y 2 R = X 2 + Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m xx YY Y = 3.48 m Y = 4.49 m

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Vector (V) X-component Solve Y-component Solve o o o o (Vcos ) (Vsin ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m m 1.70 m Sum ( ) x = 3.48 m 2.53 m 3.84 m m m Y = 4.49 m c 2 = a 2 +b 2 R 2 = X 2 + Y 2 R = X 2 + Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m xx YY Y = 3.48 m Y = 4.49 m Using Tangent solve for the DIRECTION of the resultant. Tan = adjacent opposite adj opp = Tan -1 xx YY 3.48 m 4.49 m = Tan -1 = 52.2 o

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