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Math Review

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

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Math Review

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

  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 Sin θ = opposite/hypotenuse opp = hyp x sin θ Cos θ = adjacent/hypotenuse Adj = Hyp x cos θ Tan θ = Opp/Adj Opp2 + Adj2 = Hyp2 Pythagorean Theorem Hypotenuse Opposite θ Adjacent

  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? VR 9 m/s Vv 25o VH • Known: VR = 9 m/s (Hyp); Θ = 25o • Unknown:Vv (opp); VH (Adj) • Vv (opp) = 9 m/s (Hyp) x sin Θ • Vv = 9 m/s x sin 25o = 3.8 m/s • VH (adj) = 9 m/s (Hyp) x cos Θ • VH = 9 m/s x cos 25o =8.2 m/s

  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:Vv(opp) = 4.3 m/s VH (adj) = 2.5 m/s Unknown: VR (hyp); Θ • Opp2 + Adj2 = Hyp2 4.32 + 2.52 = VR2 VR =√ (4.32 + 2.52) = 4.97 m/s • Tan Θ = opp/adj = 4.3/2.5 = 1.72 Θ = tan -1 (1.72) = 59.8 o 4.3 m/s VR Θ 2.5 m/s

  9. Chapter 13 – Equilibrium and Human Movement • Torque • Levers • Equations of Static Equilibrium • 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. axis Force line of action Moment arm axis Force line of action Moment arm The moment arm of a force is the perpendicular distance from the force’s line of action to the axis of rotation.

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

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

  14. Rotary versus stabilizing components

  15. 6-21

  16. Levers • 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 • There are first, second, and third class levers

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

  18. Second Class Lever F R

  19. Third Class Lever F R

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

  21. Third Class levers

  22. R F 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 F + 3 cm 130 N 30 cm (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

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

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

  27. 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.

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