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Original Vector. Y component Vector V Y. X component Vector V X. Component – Means to be a piece, or apart, of something bigger A Component Vector is a smaller vector that is apart of a larger one. WHAT’S THE SHAPE?.
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Original Vector Y component Vector VY X component Vector VX Component – Means to be a piece, or apart, of something bigger A Component Vector is a smaller vector that is apart of a larger one WHAT’S THE SHAPE? LOOK AT THE VECTOR,AND THE AXES. WHAT SHAPE CAN YOU MAKE?
ANALYTICAL METHOD • Use trigonometry to break vectors into smaller vectors called Component Vectors. • The X component Vector is along the X axis • The Y component Vector is along the Y axis • Trigonometry deals with angles, and triangles. • This makes addition easier. • Please remember Trig. is your friend. • We can add these component vectors to find the resultant.
TRIGONOMETRY Hypotenuse • SINE = opposite/hypotenuse • COS = adjacent/hypotenuse • TAN = opposite/adjacent • SOH - CAH - TOA Opposite Adjacent
BREAKING INTO COMPONENTS Use Sine and Cosine: Vector (V) y component (VY) Sine = Y/V VY = V*Sine x component (VX) Cos = X/V VX = V*Cos 1) Cosine closes the angle 2) Sine skies, Cosine coasts
First use sine, and cosine to break the vectors into their components Now that the original vector are broken into smaller pieces the original two vectors are gone BREAK INTO COMPONENTS
Resultant When we draw the resultant we get a right triangle, we get a right triangleEVERY TIME WE DO THIS. Now, we can add these vectors easily by the tip tail method Adding the components
Means Angle Y Total (YT) Resultant vector (VR) R X Total (XT) FINDING THE RESULTANT Use Pythagorean Theorem to find the magnitude VR2 = XT2 + YT2 VR = XT2 + YT2 To find the direction use the tangent Tan R = YT/XTR = Tan-1(YT/XT)
VY = 10m *Sin 60 VY = 10m *.866 VY = 8.66 m Vx = 10m *Cos 60 Vx = 10m *.5 Vx = 5 m Vx = 20m *Cos 30 Vx = 20m *.866 Vx = 17.32 m VY = 20m *Sin 30 VY = 20m *.5 VY = 10 m Now that the original vector are broken into smaller pieces the original two are gone Solving a problem 20 meters at 30 degrees + 10 meters at 60 degrees 10 m 60O 20 m 30O First use sine, and cosine to break the vectors into their components
VY = 8.66 m Resultant 10 m + 8.66 m = 18.66 m VY = 10 m 17.32m + 5m = 22.34 m Vx = 5 m Once we have the right triangle we can find the lengths of each side by simply adding the vector length’s together Now we have a regular right triangle and by using trigonometry we can find the resultant VY = 8.66 m Vx = 5 m Vx = 17.34 m VY = 10 m Now that we have these component vectors we can add them Tip Tail and form a right Triangle
C = 846.38m2 = 29.09m The final answer is 29.09 meters at 39.9O a = 22.34m b = 18.66 m C= resultant Resultant 18.66 m q 22.32 m q = tan-1 (b/a) C2 = a2 +b2 q = tan-1(18.66m/22.32m) C2 = (22.32m)2 + (18.66m)2 q = tan-1 (.836) C2 = (498.18m2) + (348.2m2) C2 = 846.38m2 q = 39.9O
WOW, That a lot of work!!(Is there an easier way, to do this?) Yes, there is a much faster and easier way, called the “box method” The box method is a technique that organizes all the work into a few simple steps that anyone can do Lets look at the same problem again using the box method, on the side will be the longer method In the end chose the method that works best for you.
The box method Step 1: Make a box like this one the top, and bottom rows are always done this way You will need 1 row for each Vector. This problem has 2 vectors so we need 2 columns
Then use The Cosine And sine Equations To fill in the next two columns 20m*sin(30O) 10 m 20m*cos(30O) 17.32 m 10m*sin(60O) 8.66 m 10m*cos(60O) 5 m 22.32 m Finally add up the two Columns 18.66 m Step 2: Fill in the vector columns(this sets up the problem) 20 meters at 30 degrees + 10 meters at 60 degrees Write the Vector’s Length and Angle In the Vector column
C = 846.38m2 = 29.09m The final answer is 29.09 meters at 39.9O Using the total columns gets us the right triangle again Vx total = a = 22.34m Vy total = b = 18.66 m C= resultant Resultant 18.66 m q 22.32 m q = tan-1 (b/a) C2 = a2 +b2 q = tan-1(18.66m/22.32m) C2 = (22.32m)2 + (18.66m)2 q = tan-1 (.836) C2 = (498.18m2) + (348.2m2) C2 = 846.38m2 q = 39.9O