Motion - 3 - Cosine Sine Law Vector Addition

University:
High School
Course:
High School Physics
Academic year:

2024
Views:

329

Pages:

17
Author:

1 1

1.5 Adding vectors that form non 90 degree triangles Using Trigonometry (Cosine Law, Sine Law) 1 Law of Cosines - This law can be used for any triangle. - Need two sides and the contained angle (SAS). c b C a (Note: If C = 90̊, then cos 90̊ = 0, equation becomes c2 = a2 + b2) Sample Problem b 12 m C c 32° 14 m a Find the unknown side. The unknown side is side “c” The angle is angle “C” The other two sides are “a” and “b” Law of Sines B a c - Can also be used for any triangle. - Must know 3 of the 4 terms in any two “pairs”. C A b Sample Problem C a 10 cm 75° θ B 11 cm Find θ c b A Label the diagram Choose the right “pairs” 1.6 The “5-step” Plan Using Trig to Solve Vector Problems 18 m [23° E of N] + 24 m [41° S of E] 1. Draw vector diagram (approximate) C a 41° b 24 m 18 m 23° A B c 2. Do “detective” work to find the contained angle. C b 41° 49° a 18 m 24 m 23° 72° 23° B A c 3. Use Cosine Law to find magnitude of the resultant C a 18 m 49° 23° 41° b 24 m 72° 23° A B c C = 25 m (2 significant figures) 4. Use the Sine Law to find the angle C49° 41° a 23° 18 m 23° b 24 m 72° A 88° B c But, the angle from North to the resultant is 23° + 65° = 88° 25 m [88° E of N] 5. Express the resultant using proper notation The displacement is 25 m [88° E of N] or 25 m [2° N of E] You only have to give one … they are both correct. More than 2 vectors: Options! ► Do the “5-step” plan with two vectors at a time until you are left with the resultant. ► Do the “perpendicular component” method, where you find all x & y components of each vector first, and then combine components. ► Which seems more efficient? 1.7 Perpendicular Components Method to Complex (3 or more) Vector Problems - We have learned two methods: Sine & Cosine Laws, and “Perpendicular Components” (PC). - PC is most efficient when you have 3 or more vectors to add. Ex #1. Find the resultant of: 21 m [32º N of E] + 55 m [15º S of W] + 33 m [21º W of S] 21 m 32° x y y x 15° 55 m 33 m 21° m m m y x m - m m m - m m m 47.1433 m θ 33.9147 m c → d = 58 m [36° S of W] Here are the individual vectors: 21 m 15° 32° 55 m 33 m 21° Here’s how the vectors add up to give the resultant: 36° 58 m → d = 58 m [W 36 S] Practice Problem ! y r t u o Y 31.0 m [25.0° E of N] + 59.0 m [15.0° S of E] + 82.0 m [48.0° S of W] x 15° 59 m x y 25° 31 m y x m 15.2521 θ 48° m y 48.1127 m Δd 82 m m Δd N m Δd=50.5 m [72.4° S of E] m m m m ► 6.049 m [18° S of E] + 12.1 m [34° W of N] 1. Vector Diagram 3. Cosine Law to get Magnitude: Obtuse Angle! B a c 12.1 m 4. Sine Law to get angle: But, from the diagram we see that angle is obtuse. 34° A 18° 6.0 m b C 2. Detective work to get contained angle: Angle C = 90 – 18 - 34 = 38° Therefore it should be 180 – 64.9232 = 115.077° Dir = 115.007 – 90 - 18 = 7° 5. Express final answer using proper notation: → d = 8.2 m [7° W of N]