A Demonstrative Approach

1. A Demonstrative Approach

2. How Do We Decompose a Vector? North Next, we must find the vertical (vy) and horizontal (vx) components by using the trigonometric functions, sine and cosine. Vx Let’s say we need to find the vertical (VY) and horizontal (VX) 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. Vy 15.0 m 60o sin θ = (opp/hyp) sin (60o) = (Vx/15) cos (60o) = (Vy/15) cos θ = (adj/hyp) Using cross multiplication we getVx = 15.0 * sin (60o) Using cross multiplication we getVy = 15.0 * cos (60o) = 7.50 m = 13.0 m Thus, Vx = 13.0 m and Vy = 7.50 m!

3. But wait, Venerable Teacher, what if we used the horizontal axis to find the bearing instead of the vertical axis?

4. Well, Dear Student… As with the ancient art of feng-shui, you must look to the East to find your bearing and thus mark your angle. • In this case, the trigonometric functions are switched with respect to the vector components. • sin θ = (opp/hyp)cos θ = (adj/hyp) • sin (30o) = (Vy / 15.0) • cos (30o) = (Vx / 15.0) You see, young Grasshopper, it matters not which axis you choose as long as your calculations are truthful.--Confucius 15.0 m Vy Notice the trigonometric functions find the opposite component this time. 30o East Vx Cross multiplying gives us: Vx = 15.0 * cos (30o) Vy = 15.0 * sin (30o) = 13.0 m = 7.50 m Whereas in our previous example we had… Vx = 15.0 * sin (60º) and Vy= 15.0 * cos (60º) Thus, you still have Vx = 13.0 m and Vy = 7.50 m!

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

6. θ= Atan ( ) tan (θ) = 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, vx, and vertical, vy, components to each vector. Step Six: Having drawn out the new triangle with the resultant vRx and vRy 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. Step Four: Having found the horizontal, vx, and vertical, vy, components to each vector, you now add all the horizontal components and vertical components separately. Step Five: Draw out your new triangle with the resultant vRx and vRy 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, vR. Soh CahToa Remember the Great Indian Princess… V1 = 18.0m = 18.0 * cos(25°) = 16.3 m = 18.0 * sin(25°) = 7.61 m θ is opposite to 25° therefore use SIN v1y 65° VR = 10.3 2 + 2.8 2 = 10.7 m 25° = - 12.0 * sin(30°) = - 6.00 m = - 12.0 * cos(30°) = - 10.4 m v1x is adjacent to 25° therefore use COS Vy Vx Opp Adj 2.8 10.3 60° VRx = 10.3 m VRy = - 2.8 m …SF VRy = - 2.79 m 30° v2y is adjacent to 30° therefore use COS… Remember: use -12.0 since it goes DOWN 2.8 10.3 θ= 15°South of East 210° is opposite to 30° therefore use SIN… Remember: use -12.0 since it goes LEFT V2 = 12.0 m v2x