1. Math Review Scalar Quantities: (Magnitude only) Mass Mass Volume Volume Density Density Speed Speed Vector Quantities (Magnitude and direction) Force Force Weight Weight Pressure Pressure Torque Torque Velocity Velocity

2. Vector Resolution Vectors may be resolved into perpendicular components. The vector composition of each pair of components yields the original vector. + +_ _ + + + + _ _

3. Vector Composition The composition of vectors with the same direction requires adding their magnitudes.

4. Vector Composition The composition of vectors with the opposite directions requires subtracting their magnitudes.

5. Vector Algebra The tip-to-tail method of vector composition.

6. Mathematical approach Hypotenuse Adjacent Opposite θ Sin θ = opposite/hypotenuse opp = hyp x sin θ Cos θ = adjacent/hypotenuse Adj = Hyp x cos θ Tan θ = Opp/Adj Opp 2 + Adj 2 = Hyp 2 Pythagorean Theorem

7. A long jumper takes off with a velocity of 9 m/s at an angle of 25 o to the horizontal. How fast is the jumper moving in the vertical and horizontal directions? Known: V R = 9 m/s (Hyp); Θ = 25 o Unknown: V v (opp); V H (Adj) V v (opp) = 9 m/s (Hyp) x sin Θ V v = 9 m/s x sin 25 o = 3.8 m/s V H (adj) = 9 m/s (Hyp) x cos Θ V H = 9 m/s x cos 25 o = 8.2 m/s 25 o 9 m/s VvVv VHVH VRVR

8. A high jumper takes off with a vertical velocity of 4.3 m/s and a horizontal velocity of 2.5 m/s, what is the resultant velocity of the jumper? Known: V v (opp) = 4.3 m/s V H (adj) = 2.5 m/s V H (adj) = 2.5 m/s Unknown: V R (hyp); Θ Opp 2 + Adj 2 = Hyp 2 Opp 2 + Adj 2 = Hyp 2 4.3 2 + 2.5 2 = V R 2 4.3 2 + 2.5 2 = V R 2 V R = √ (4.3 2 + 2.5 2 ) = 4.97 m/s V R = √ (4.3 2 + 2.5 2 ) = 4.97 m/s Tan Θ = opp/adj = 4.3/2.5 = 1.72 Tan Θ = opp/adj = 4.3/2.5 = 1.72 Θ = tan -1 (1.72) = 59.8 o Θ = tan -1 (1.72) = 59.8 o 4.3 m/s 2.5 m/s Θ VRVR

9. Chapter 13 – Equilibrium and Human Movement Torque Torque Levers Levers Equations of Static Equilibrium Equations of Static Equilibrium Center of Gravity Center of Gravity

10. Mechanical Strength: measured by the maximum torque that can be voluntarily generated at a certain joint Strength determined by: Absolute force developed by muscle Distance from joint center to tendon insertion (affects moment arm) Angle of tendon insertion (affects moment arm)

11. The moment arm of a force is the perpendicular distance from the force’s line of action to the axis of rotation. Moment arm Force line of action Moment arm Force line of action axis

12. Calculate the elbow joint torque when the biceps generate a force of 100 N at the following angles of attachment: a. 30 o b. 60 o c.90 o d.120 o e. 150 o a.b.c. a. Opp = hyp sin Θ = 100 sin 30 o = 50 N F ⊥ = 100 sin 30 o = 50 N T = 50 N x 0.03m = 1.50 Nm b. = 100 sin 60 o = 86.6 N b. F ⊥ = 100 sin 60 o = 86.6 N T = 86.6 N x 0.03m = 2.6Nm c. T = 100N x 0.03 cm = 3.0Nm 30 o F m = 100 N F ⊥ 60 o F⊥F⊥ 3 cm F⊥F⊥

13. Calculate the elbow joint torque when the biceps generate a force of 100 N at the following angles of attachment: a. 30 o b. 60 o c.90 o d.120 o e. 150 o d.e. d. Adj = hyp cos Θ F ⊥ = 100 cos 30 o = 86.6 N T = 86.6N x 0.03m = 2.6 Nm e. F ⊥ = 100 cos 60 o = 50 N T = 50 N x 0.03m = 1.5 Nm 30 o F m = 100 N 60 o 3 cm

14. Rotary versus stabilizing components

15. 6-21

16. What is a lever? A simple machine consisting of a relatively rigid barlike body that can be made to rotate about an axis or a fulcrum A simple machine consisting of a relatively rigid barlike body that can be made to rotate about an axis or a fulcrum There are first, second, and third class levers There are first, second, and third class levers Levers

17. First Class Lever Applied or motive force F R Resistive force Axis of Rotation or fulcrum

18. Second Class Lever F R

19. Third Class Lever FR

20. FR First class RF Second class FR Third class Mechanical advantage/effectiveness =ME = (Moment arm of applied force)/(Moment arm of the resistance) ME = 1, 1 ME = always > 1 ME = always < 1

21. Third Class levers

22. F R fa ra A force can move a resistance through a large range of motion when the force arm (fa) is shorter than the resistance arm (ra).

23. Law of Equilibrium ∑ T = 0 3 cm 30 cm F 130 N (F x 3) + (-130 x 30) = 0 F = (130 x 30)/3 = 1300 N + -

24. Law of the lever A small force can have a large torque or moment of rotation if the lever arm is long Similarly, the force must be large to achieve the same moment of rotation if the lever arm is short

26. ∑ T = 0 F ext x 5cm – (34kg x 17 cm) – (6 kg x 20 cm) = 0 F ext = ((34 kg x 17 cm) + (6 kg x 20 cm))/5 cm F ext = (578 kgcm + 120 kgcm)/5 cm = 139.6 kg

27. ∑ T = 0 F ext x 5cm – (34kg x 31 cm) – (6 kg x 52 cm) = 0 F ext = ((34 kg x 31 cm) + (6 kg x 52 cm))/5 cm F ext = (1054 kgcm + 312 kgcm)/5 cm = 273.2 kg

28. The importance of levers in the mechanism of injuries ∑ T = 0 (L x l) – (M x m) = 0 (L x l) = (M x m) L = M x (m/l) Therefore, if force M has a lever arm (m) 10x the lever arm (l) of force L, the force in the ligament (L) will be 10 x greater than force M.