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A Demonstrative Approach. Let’s say we need to find the vertical (V Y ) and horizontal (V X ) components of this here vector. The first step is to find.

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Presentation on theme: "A Demonstrative Approach. Let’s say we need to find the vertical (V Y ) and horizontal (V X ) components of this here vector. The first step is to find."— Presentation transcript:

1 A Demonstrative Approach

2 Let’s say we need to find the vertical (V Y ) and horizontal (V X ) components of this here vector. The first step is to find the bearing of the vector with respect to a reference point, in this case North, (vertical) or the positive y-axis. 60 o Next, we must find the vertical (v y ) and horizontal (v x ) components by using the trigonometric functions, sine and cosine. 15.0 m sin θ = (opp/hyp) VyVy VxVx sin (60 o ) = (V x /15) Using cross multiplication we get V x = 15.0 * sin (60 o ) = 13.0 m cos θ = (adj/hyp) cos (60 o ) = (V y /15) Using cross multiplication we get V y = 15.0 * cos (60 o ) = 7.50 m Thus, V x = 13.0 m and V y = 7.50 m! North

3

4 15.0 m 30 o In this case, the trigonometric functions are switched with respect to the vector components. sin θ = (opp/hyp) cos θ = (adj/hyp)  sin (30 o ) = ( V y / 15.0)  cos (30 o ) = ( V x / 15.0) Notice the trigonometric functions find the opposite component this time. Cross multiplying gives us: V x = 15.0 * cos (30 o ) V y = 15.0 * sin (30 o ) = 13.0 m = 7.50 m Thus, you still have V x = 13.0 m and V y = 7.50 m! As with the ancient art of feng-shui, you must look to the East to find your bearing and thus mark your angle. East VxVx VyVy Whereas in our previous example we had… V x = 15.0 * sin (60º) and V y = 15.0 * cos (60º)

5 Vectors can have negative quantities if they lie below or to the left of your axes! This vector can have either a negative V x or V y component. Can you explain why? Just keep in mind that the magnitude of the vector stays positive.

6 Now we will combine vectors by adding them Step One: Take a deep breath, exhale the nervousness, and pray for wisdom. Step Two: This step requires you to determine a reference point to find the bearing for each of your vectors. REMEMBER: Use the smallest angle with reference to one of the four cardinal points: N, S, E, W Step Three: is where you use your angle and trigonometric functions of sin (θ) and cos ( θ) to find your horizontal, v x, and vertical, v y, components to each vector. Step Four: Having found the horizontal, v x, and vertical, v y, components to each vector, you now add all the horizontal components and vertical components separately. 65° 25° 210° 30° 60° V 1 = 18.0m V 2 = 12.0 m v 1x is adjacent to 25° therefore use COS = 18.0 * cos(25°) = 16.3 m v 1y is opposite to 25° therefore use SIN = 18.0 * sin(25°) = 7.61 m Remember the Great Indian Princess… Soh CahToa v 2x is opposite to 30° therefore use SIN… Remember: use -12.0 since it goes LEFT = - 12.0 * sin(30°) = - 6.00 m v 2y is adjacent to 30° therefore use COS… Remember: use -12.0 since it goes DOWN = - 12.0 * cos(30°) = - 10.4 m V Rx = 10.3 mV Ry = - 2.79 mV Ry = - 2.8 m …SF Step Five: Draw out your new triangle with the resultant v Rx and v Ry components to determine the length of the resultant. (HINT: Always draw the vector with greatest absolute value first.) Then use the Pythagorean Theorem to find the magnitude of your resultant vector, v R. V R = 10.3 2 + 2.8 2 = 10.7 m Step Six: Having drawn out the new triangle with the resultant v Rx and v Ry components, now you will determine the angle. The direction is given as “θ°lesser of greater” or in this case, as South (lesser vector) of East (greater vector). (Remember Princess SohCahToa!) Use the TAN function to set up your equation, then take the ArcTAN as the inverse function. θ Opp Adj tan (θ) = Vy Vx 2.8 10.3 θ= Atan ( ) 2.8 10.3 θ= 15°South of East


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