Application of Trigonometry in Biomechanics

1. Application of Trigonometry in Biomechanics Find a distance or displacement given a set of coordinates Separate muscle force into a component causing movement and a component affecting joint stability Analyze projectile motion (baseball, discus, javelin, basketball, etc.) Find the net force acting on an object or body segment Analyze the effect of the weight of a body segment or outside force on movement

2. Trigonometry - Basics Q F - general a + ß + g = 180° A ß g B C a

3. Right triangle - Basics Q F - general a + ß = 90° Pythagorean Theorem: A2 + B2 = C2 90° A ß g B C - hypotenuse a

4. 2000 Super Bowl: St. Louis vs. TennesseeKurt Warner of the St. Louis Rams is standing on his own 20 yard line and is 25 yards from the right sideline. He throws the ball to a receiver that on the opponent’s 35 yard line and is 2 yards from the right sideline when he catches the ball.A) What was the horizontal distance covered by the ball before it was caught?B) If the ball was in the air for 3.5 seconds, what was the average forward or horizontal velocity of the ball during flight?

5. Football Field: 20 yds 35 yds 25 yds 2 yds 100 yds 50 yd line quarterback receiver C B A

6. Given:position 1 (pos1) = 20 yds position 2 (pos2) = 100 yds – 35 yds = 65 yds Dt = 3.5 sec Quarterback 25 yards from sideline (DQ->S ) Receiver 2 yards from sideline (DR->S ) • Find: • The horizontal distance traveled by the football (df) • The average horizontal velocity of the football (vf)

7. C B 25 yds A 2 yds Diagram and Derived Information • A = 65 yds – 20 yds = 45 yds • B = DQ->S - DR->S = 25 yds – 2 yds = 23 yds • C =Horiz. Distance (Range) traveled by the football = df

8. A2 + B2 Formulas: • A) C2 = A2 + B2 • B) df = C = • C) vf = df/Dt Solutions: A) Df = Df = 50.53 yds = 151.61 ft B) vf= 151.61 ft / 3.5 sec vf = 43.32 ft / sec (45 yds)2 + (23 yds)2

9. Right triangle - Basics sin q =length of the opposite side length of the hypotenuse sin a = A/C cos q =length of the adjacent side length of the hypotenuse cos a = B/C tan q =length of the opposite side length of the adjacent tan a = A/B A ß g B a C - hypotenuse

10. Trig. example Justine is performing leg extension exercises. The distance from her knee to her foot = 25 cm. If the position of her leg is 50° below horizontal, what are the horizontal and vertical distances from her knee to her foot? Given: dk->f = 25 cm ß = 50° Find: x (horizontal - side A) and y (vertical - side B)

11. Trig. example Given: dk->f = 25 cm = C ß = 50° Find: x (horizontal - side A) and y (vertical - side B) DIAGRAM: A or x ß = 50° B or y C

12. Trig. example Formula: sin q = opp/hyp cos q = adj/hyp Solutions: cos 50° = x/25 cm x = 25 cm • cos 50° x = 16.1 cm sin 50° = y/25 cm y = 25 cm • sin 50° : y = 19.15 cm A ß = 50° B C

13. Law of cosines can be used for any triangle c2 = a2 + b2 - 2•a•b•cosq Pythagorean theorem Correction for angle q a q b ß a c

14. Law of sines can be used for any triangle The ratio between a side and the sine of the opposite angle is constant for any given triangle a/sina = b/sinß = c/sing a q b ß a c

15. Example The vastus lateralis is pulling with a force of 200 N on the quadriceps tendon. The vastus medialis is pulling with a force of 160 N on the quadriceps tendon. If the forces of the two muscles pull at an angle of 28° with each other, find the net force. Given: FVL = 200 N FVM = 160 N q = 28° Find: Fnet

16. Example Diagram: Formula: c2 = a2 + b2 - 2•a•b•cosq To find net, add, the arrows representing the force “tip to tail” NET is the “shortcut” VL Vm 28° 160N 28° Fnet 200 N

17. Example Formula: c2 = a2 + b2 - 2•a•b•cosq To find net, add, the arrows representing the force “tip to tail” NET is the “shortcut” 28° Fnet Q = 180° - 28° = 152°

18. Example Solution: c2 = a2 + b2 - 2•a•b•cosq c = (2002 + 1602 - 2•200•160•cos 152°) c = (65,600 - 64,000 • (-0.883)) c = (122,112N2) C = 349.44 N 160N Fnet 28° 152° 200 N