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Glencoe Physics Ch 4 Remember…. When drawing vectors… length = magnitude (with scale) angle = direction of the vector quantity. When drawing and moving.

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Presentation on theme: "Glencoe Physics Ch 4 Remember…. When drawing vectors… length = magnitude (with scale) angle = direction of the vector quantity. When drawing and moving."— Presentation transcript:

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2 Glencoe Physics Ch 4

3 Remember…. When drawing vectors… length = magnitude (with scale) angle = direction of the vector quantity. When drawing and moving vectors, these two characteristics must be maintained

4 Vector Addition = finding the sum or Resultant of two or more vector quantities

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6 Resultant Vector A Resultant vector is the vector that results from the addition of two or more vectors VAVA VBVB VRVR

7 Graphical Vector Addition 2. Parallelogram method 8 N 4 N 3 N 3 forces act on an object at the same time. F net is not 15 N because these forces aren’t working together. But they’re not completely opposing each either. So how do find F net ? The answer is to add the vectors... not their magnitudes, but the vectors themselves. There are two basic ways to add vectors: 1. Tip to tail method

8 Tip to Tail Method in-line examples Place the tail of one vector at the tip of the other. The vector sum (also called the resultant) is shown in red. It starts where the black vector began and goes to the tip of the blue one. In these cases, the vector sum represents the net force. You can only add or subtract magnitudes when the vectors are in-line! 16 N 20 N 4 N 20 N 16 N 12 N 9 N 12 N 21 N

9 Tip to Tail – 2 Vectors 5 m 2 m To add the red and blue displacement vectors first note: Vectors can only be added if they are of the same quantity—in this case, displacement. The magnitude of the resultant must be less than 7 m (5 + 2 = 7) and greater than 3 m (5 - 2 = 3). 5 m 2 m blue + black Interpretation: Walking 5 m in the direction of the blue vector and then 2 m in the direction of the black one is equivalent to walking in the direction of the red vector. The distance walked this way is the red vector’s magnitude.

10 Commutative Property blue + black black + blue As with scalars (ordinary numbers), the order of addition is irrelevant with vectors. Note that the resultant (black vector) is the same magnitude and direction in each case. If you’ve drawn everything to scale, and drawn the angles correctly, then you can simply measure the resultant vector and (using your scale) determine its magnitude.

11 Vector Properties Addition (Rule A+B = B+A) A + B = C Negative vector A -A just opposite direction Subtraction add negative vector

12 Tip to Tail – 3 Vectors We can add 3 or more vectors by placing them tip to tail in any order, so long as they are of the same type (force, velocity, displacement, etc.). blue + green + black

13 Parallelogram Method 1.Create parallelogram using “copies” of the two vectors 2.Draw Resultant vector from tail of first vector to tip of last vector. 3.You cannot add more than 2 vectors at a time with this method, so…how do you add 3 or 4 vectors with this method??? Note: Opposite sides of a parallelogram are congruent.

14 Comparison of Methods blue + black Tip to tail method Parallelogram method The resultant has the same magnitude and direction regardless of the method used.

15 Opposite of a Vector v - v If v is 17 m/s up and to the right, then -v is 17 m/s down and to the left. The directions are opposite; the magnitudes are the same.

16 Algebraic Solution

17 Pythagorean Theorem 34 m/s 30.814 m/s 25  14.369 m/s Since components always form a right triangle, the Pythagorean theorem holds: (14.369) 2 + (30.814) 2 = (34) 2. Note that a component can be as long, but no longer, than the vector itself. This is because the sides of a right triangle can’t be longer than the hypotenuse.

18 Law of Cosines ab c R A B Law of Cosines: R 2 = A 2 + B 2 - 2 AB cos Θ This side is always opposite this angle. These two sides are repeated. It matters not which side is called A, B, and R, so long as the two rules above are followed. This law is like the Pythagorean theorem with a built in correction term of -2 AB cos Θ. This term allows us to work with non-right triangles. Note if Θ = 90 , this term drops out (cos 90  = 0), and we have the normal Pythagorean theorem. Θ

19 Example Vector Problem A motorboat heads due east at 16 m/s across a river that flows due north at 9.0 m/s. What is the resultant velocity of the boat? If the river is 136 m wide, how long does it take the motorboat to reach the other side?

20 Graphical Solution Draw vectors, tip to tail Using your scale, measure length of R 16 m/s 9 m/s R

21 Solution : Algebraic Method What is the resultant velocity of the boat? A 2 + B 2 = C 2 (9 m/s) 2 + (16 m/s) 2 = R 2

22 Solution: Calculation of time If the river is 136 m wide, how long does it take the motorboat to reach the other side? V = Δd/Δt 16 m/s = 136 m/t 136 m /16 m/s =Δt t = 8.5 s

23 Example Vector Problem An airplane is flying 200 mph at 50°. Wind velocity is 50 mph at 270°. What is the velocity of the plane?

24 0o0o 90 o 180 o 270 o

25 0o0o 90 o 180 o 270 o

26 0o0o 90 o 180 o 270 o

27 0o0o 90 o 180 o 270 o

28 Practice Problems P.67 1-4 Your turn!!

29 4.2 Components of Vectors Any vector directed in two dimensions can be thought of as having two parts Each part of a two-dimensional vector is known as a component. The single two-dimensional vector could be replaced by the two components.

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32 Vector Components 150 N Horizontal component Vertical component AyAy AxAx A Θ A x =horizontal component of vector A A y =vertical component of vector A

33 Component Vectors A x = A cos Θ cos Θ = Adjacent side hypotenuse A y = A sin Θ sin Θ = Opposite side hypotenuse tan Θ = Opposite side Adjacent side Remember…SOH CAH TOA

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35 Finding Components with Trig  v v cos  v sin  Multiply the magnitude of the original vector by the sine & cosine of the angle made with the horizontal. The units of the components are the same as the units for the original vector. Here’s the correspondence: cosine  adjacent side sine  opposite side

36 Note that 30.814 + 14.369 > 34. Adding up vector components gives the original vector (green + red = black), but adding up the magnitudes of the components is meaningless. Component Example 34 m/s -30.814 m/s 205  -14.369 m/s A helicopter is flying at 34 m/s at 25  S of W (south of west). The magnitude of the horizontal component is 34 cos 205   -30.814 m/s. This is how fast the copter is traveling to the west. The magnitude of the vertical component is 34 sin 205   -14.369 m/s. This is how fast it’s moving to the south.

37 v WA = vel. of Wonder Woman w/ resp. to the air v AG = vel. of the air w/ resp. to the ground (and Aqua Man) v WG = vel. of Wonder Woman w/ resp. to the ground (and Aqua Man) Wonder Woman Jet Problem Suppose Wonder Woman is flying her invisible jet. Her onboard controls display a velocity of 304 mph 10  E of N. A wind blows at 195 mph in the direction of 32  N of E. What is her velocity with respect to Aqua Man, who is resting poolside down on the ground? We know the first two vectors; we need to find the third. First we’ll find it using the laws of sines & cosines, then we’ll check the result using components. Either way, we need to make a vector diagram. continued on next slide

38 The 80  angle at the lower right is the complement of the 10  angle. The two 80  angles are alternate interior. The 100  angle is the supplement of the 80  angle. Now we know the angle between red and blue is 132 . Wonder Woman Jet Problem (cont.) continued on next slide 10  32  v WA v AG v WG v WA + v AG = v WG 80  195 mph 304 mph v WG 80  32  100 

39 Wonder Woman Jet Problem (cont.) 195 mph 304 mph v 132  The law of cosines says: v 2 = (304) 2 + (195) 2 - 2 (304) (195) cos 132  So, v = 458 mph. Note that the last term above appears negative, but it’s really positive, since cos 132  < 0. The law of sines says:  sin 132  sin  v 195 = So, sin  = 195 sin 132  / 458, and   18.45  80  This means the angle between green and the horizontal is 80  - 18.45   61.6  Therefore, from Aqua Man’s perspective, Wonder Woman is flying at 458 mph at 61.6  N of E.

40 Wonder Woman Problem: Component Method 32  v WA = 304 mph v AG = 195 mph 10  This time we’ll add vectors via components as we’ve done before. Note that because of the angles given here, we use cosine for the vertical comp. of red but sine vertical comp. of blue. All units are mph. 304 195 103.3343 165.3694 52.789 299.3816 continued on next slide

41 Wonder Woman: Component Method (cont.) 304 195 103.3343 165.3694 52.789 299.3816 103.3343 52.789 165.3694 299.3816 402.7159 mph 218.1584 mph 458.0100 mph Combine vertical & horiz. comps. separately and use Pythag. theorem.  = tan -1 (218.1584 / 402.7159) = 28.4452 .  is measured from the vertical, which is why it’s 10  more than  was. 

42 Comparison of Methods We ended up with same result for Wonder Woman doing it in two different ways. Each way requires some work. You can only use the laws of sines & cosines if: you’re dealing with exactly 3 vectors. (If you’re adding three vectors, the resultant makes 4, and this method won’t work the vectors form a triangle. Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick.

43 END of Ch 4


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