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SPH 3U Grade 11 U Physics • This lesson will extend your knowledge of kinematics to two dimensions. • You will be able to solve problems involving displacement in two dimensions using two, more accurate methods SPH 3U Grade 11 U Physics What are we going to cover today? By the end of this lesson, you will be able to: Apply the Trig Method Apply the Comp Method Homework Math Review Right angle triangles (90°)… what math “tools” can we use? Pythagorean Theorem Trig ratio’s S-OH C-AH T-OA Math Review Non-right angle triangle's… what options exist? Sine Law Cosine Law 2 C = 2 A + 2 B - 2ABcosc Allows us to solve problems by looking at parts as right angle triangles Also allows us to solve problems that involve non-right angle triangles Let’s solve yesterdays hiker question using the trigonometric method. SPH 3U Grade 11 U Physics Ex 1: A hiker walks 15 km [N], then 24 km [N60oW], then 21 km [S]. What is his final displacement? Sol’n: This is the diagram 24 km 21 km after we apply vector o 60. addition A Next we break the vector diagram into Triangles What can we do from here? 15 km B Solve for length A and then solve for length B SPH 3U Grade 11 U Physics We apply cosine law to solve for the unknown side 60.o 24 km A 21 km C2 = A2 + B2 - 2ABcosc a a 21 24 2 21 24 cos 60. 2 2 a 513.02 a 22.65 2 2 60.o o SPH 3U Grade 11 U Physics We apply sine law to solve for the angles 60.o 24 km A 21 km 22.65 km sin q sin60. 21 22.65 o q 53.4 o 60.o q SPH 3U Grade 11 U Physics We apply cosine law o 60. again to solve for the final answer 21 km A 24 km 22.65 km 60.o 53.4o 66.6o B a 15 22.65 2 15 22.65 cos 66.6 2 2 a 468.16 a 21.6 2 2 15 km o SPH 3U Grade 11 U Physics We apply sine law o 60. again to solve for the final angle 21 km A 24 km 22.65 km 60.o 53.4o 66.6o B sin q sin66.6o 22.65 21.6 q 74.2o 21.6 km 15 km q Dd = 22 km [N74oW] Break the VECTOR into its component pieces (a y value and an x value) Remember: vector quantities can be represented by directed line segments Dd “All y, no x” v av “No y, all x” r “Some y, some x” If I gave you 5 different “vectors” or movements… how would you go about finding the TOTAL DISPLACEMENT of all 5? Ie. 10m N30ºE = 13m E15ºN = 21m E83ºN = 20m W45ºN = 11m N72ºE = V1 V2 V3 V4 V5 Perhaps with a graph? But what the heck does that tell us? If only we knew the total vertical displacement & the total horizontal displacement We could use Pythagorean Theorem to solve for the total displacement What we can do is figure out the INDIVIDUAL vertical and horizontal displacement of EACH VECTOR and then add them up! If only we knew the total vertical displacement & the total horizontal displacement We could use Pythagorean Theorem to solve for the total displacement What we can do is figure out the INDIVIDUAL vertical and horizontal displacement of EACH VECTOR and then add them up! By breaking up each individual vector (ex. 20 km N45ºE) into a y value (ex. Vertical displacement) and x value (horizontal displacement) we eliminate the need to look at its direction – focusing instead on the magnitude only This effectively turns any set of complicated vectors that might be going in various directions, into a streamlined set of values only going vertical or horizontal (never both) This allows us to use our kinematics in 1D rules to determine the overall vertical and horizontal displacements – which will always form 90º triangles – Pythagorean Theory - EASY! We will let r represent our RESULTANT vector (this is the same as our total displacement) Δdtotal = r r ry = Δdy or the Δ in vertical position q rx = Δdx Or the horizontal displacement What is the only segment of the triangle we automatically know from the start? What determines the names of the rest (think location)? hyp r ryopp ? Ө or 15º adj ? rx Based on this information (hypotenuse and the Ө) how can we figure out the adjacent and opposite lengths? Adjacent Cos Ө = adj / hyp Opposite Sin Ө = opp / hyp cos 15º = adj / 13m Adj = 12.6m sin 15º = opp / 13m Opp = 3.4m Use Pythagorean theory to check ans. Now that we know how we can use the trig formula’s to solve for the adj or opp lengths we can substitute the variables to represent the vertical and horizontal displacement Unknown Value opp opp ry ry sin sin qq hyp hyp r r hyp r ropp y Ө or 15º rx adj Vertical Component ry r sinq hyp ryopp r q Unknown Value adj rx cosq hyp r rx adj Horizontal Component rx r cosq hyp r r ry opp q q rx r cosq rx adj ry r sinq “x-component” of r “y-component” of r i.e. horizontal Δd i.e. vertical Δd Or the adjacent length of the hypotenuse Or the opposite length of the hypotenuse x y r r hyp opp r y qq rxadj What happens if we use (or are provided) the complimentary angle? rx r cosq ry r sinq X-Axis Rule 1. Always take your degree from the x axis! I. If you take it from the y-axis you will have effectively flipped the formula upside, causing cos to represent y (instead of x) and vice versa 2. If the problem involves vectors in multiple directions (i.e. east & west) – determine the degree by counting from the SAME x-axis I. This will calculate vectors with opposite directions (i.e. east vs west) as a positive vs. negative – ensuring an accurate displacement measurement After determining the X & Y values of every vector simply add them up to determine the TOTAL-X & the TOTAL-Y displacement E 45º N E 45º S E 315º S But if we want our formula to do the thinking for us we should use E 315º S This will ensure it’s a negative vertical (south) value but still a positive horizontal (east) value Consider the following vectors: A (rh1) = 5.0 km due East B(rh2) = 3.0 km East 450 North Vector x-component rx y-component ry (rh1) Cos q = ___ km(rx1) (rh1) Sin q = ___ km(ry1) 5.0 cos 0 = 5.0 km 5.0 sin 0 = 0 (rh2) Cos q = ___ km(rx2) (rh2) Sin q = ___ km(ry2) 3.0 cos 45 = 2 .m 3.0 sin45 = 2 .1 km A rh1 B rh2 Resultant rh rx1 + rx2 = rxtotal ry1 + ry2 = rytotal 7.1 km [E] 2.1 km [N] SPH 3U Grade 11 U Physics Every vector consists of two components. • Vector components are to each other • Vector components add together to form the vector rh q rx = r cos q SPH 3U Grade 11 U Physics Now let’s solve use the component method. o Ex 1: A hiker walks 15 km [N], then 24 km [N60 W], Sol’n: then 21 km [S]. What is his final displacement? First, we break each vector down into its components Dd2 x 20.78 km Dd2 y 21 km [S] ? 12 km 60o 24 km [N60oW] rx r cosq ry r sinq 15 km [N] Dd1x 0 Dd3x 0 Dd1y Dd3y 15 km ? 21 km ? SPH 3U Grade 11 U Physics Then we add the x and y comp’s separately DdRx Dd1x Dd2 x Dd3x 0 (20?.78) 0 20.78?km DdRy Dd1y Dd2 y Dd3y (21) ? (12) (15) 6 km ? SPH 3U Grade 11 U Physics Next we draw the comp’s head to tail and use the Pythagorean Theorem DdR q -20.78 km DdR 6 2 (?20.78)2 21.63 km? 6 tan q ? 20.78 o q 16 ? Dd = 22 km [W16oN] SPH 3U Grade 11 U Physics Read pgs 68 – 74 Review tutorials before attempting practice! Questions “Practice” 1 – 2 pg 71 1 – 2 pg 74