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Horizontal force

Horizontal force.

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Horizontal force

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  1. Horizontal force Maurice Greene has a body weight of 850 N. He is running at 5.5 m/sec at the 10 m mark into the race. At the 30 m mark, he is running at a velocity of 9.5 m/sec. If Maurice Greene covers that distance in 2.5 sec, what is the average horizontal force used during that time period Given: v1 = 5.5 m/sec ∆t = 2.5 sec v2 = 9.5 m/sec FBW = 850 N Find: Horizontal force (FH) Diagram:

  2. Horizontal force - running Formula: F = m•a a = ∆v/∆t = (v2 - v1) ∆t Solution: FBW = m•ag m = FBW/ag m = 850 N/9.81 m/sec2 = 86.6 kg FH = m • a = m(v2-v1)/∆t FH = 86.6 kg (9.5 - 5.5 m/sec)/2.5 sec = 138.56 N over the 2.5 sec interval

  3. Newton’s Laws Developed by Sir Isaac Newton (1642 – 1727) Define the mechanical basis of linear kinetics

  4. Newton’s First Law The Law of Inertia “A body will maintain a state of rest or constant velocity unless acted on by an external force that changes that state.” Mass is lazy

  5. Newton’s First Law of Motion • Also called “Newton’s Law of Inertia” • “A body will maintain a state of rest or constant velocity unless acted on by an external force that changes that state.”

  6. Inertia Definition: A resistance to action or a change in the velocity vector (acceleration). Inertia is the tendency of an object to maintain its current state of motion, whether motionless or moving at a constant velocity. Inertia in the linear system is directly related to the mass of the object

  7. Newton’s First Law The Law of Inertia • EXAMPLES: • An offensive lineman will not move unless force is applied • A 90 mph fast-pitch softball pitch will not slow down unless force is applied

  8. Moment of Inertia (I) • The resistance of an object to angular acceleration • I a m x distribution of mass • I increases as the mass increases and/or is distributed farther from axis of rotation

  9. Moment of Inertia • Symbol: I • Formula: I = Smr2 (for a ROD - different formulas are used for different shapes) I = C • m • r 2 • Where: m is the mass of a particle and r is the distance to the axis of rotation (radius of gyration). r axis mass

  10. Moment of Inertia • If “donut” is added to a bat, moving the weight farther from the grip increases to moment of inertia. axis Bat A has a greater moment of inertia than bat B.

  11. Applications - Moment of Inertia • Changing the moment of inertia during a movement can speed up or slow down rotation. • diver in the “tuck” position will spin faster than the “pike” • Rate of spin (w) increases as skater brings arms closer to body

  12. Decreasing the moment of inertia reduces resistance to rotation. Shot Put: shot pressed against the neck reducing I, allowing ( w), thereby ( v), when it’s released. Applications - Moment of Inertia

  13. Increasing the moment of inertia by increasing size of a golf club head or a tennis racket increases resistance against “twisting” of the club or racket in your hand when struck off-center and increases transfer of force . Moment of Inertia & the Sweet Spot “sweet spot” The moment of inertia (I) is the inner ring surface

  14. Newton’s Second LawThe Law of Acceleration “A force applied to a body causes an acceleration of that body of a magnitude proportional to the force, in the direction of the force, and inversely proportional to the body’s mass.” a = F/m F = m•a

  15. The Law of Acceleration • EXAMPLE: • Lance Berkman swings his bat at a fastball. The impact of the bat on the ball accelerates at 9000 m/sec2. What was the force exerted by the bat at impact if the ball has a mass of 0.25 kg • Given: m = 0.25 kg a = 9000 m/sec2 • Find: Force • Diagram:

  16. The Law of Acceleration - example • FORMULA: • F = m•a • Solution: • F = 0.25 kg • 9000 m/sec2 • F = 2250 kg•m/sec2 • F = 2250 N

  17. Momentum The quantity of motion that an object possesses. Momentum is a vector quantity SYMBOL: Mo FORMULA: Mo = m•v where m is the object’s mass and v is the object’s velocity UNITS: MetricEnglish kg•m/sec slugs•ft/sec Units of mass times units of velocity

  18. Angular Momentum If Mo = m • v Then: Angular momentum = I • w Right hand rule - vector Applications: Diver Football spiral Gyroscope Skater Shot put

  19. Conservation of Momentum • In the absence of external factors (such as friction or air resistance) the total momentum of a given system remains constant. • This is a restatement of Newton’s First Law (The Law of Inertia).

  20. Conservation of Momentum • This principle is expressed as follows: Mo1 = Mo2 or, it can be restated (mv)1 = (mv)2 Mo1 is the momentum of a system at one point in time and Mo2 is the momentum at a later point in time.

  21. Conservation of Angular Momentum • L (angular momentum) = I • w • Where I = m•r2 If I is constant, I can be manipulated to or w

  22. Impulse - most neglected concept in mechanics • Impulse is derived from Newton’s Law of Acceleration • Impulse: the ability to change an object’s momentum by the product of force exerted over a period of time. • FORMULA: Impulse = F•∆t = ∆(m•v) • where: F = the magnitude of an external force • and t = the duration of time over which the force is applied.

  23. Impulse • Impulse • Units: metric English N•sec lb•sec

  24. Impulse - Applications • Impulse = F•∆t = ∆(m•v) • We increase an object’s momentum (can apply a large force or time exerted or both): • (1) hurling a discus or shotput • (2) throwing a baseball or softball • (3) hitting a baseball or tennis ball • (4) takeoff in a high jump (create vertical momentum without losing forward momentum) • (5) jump for a rebound in basketball • * flexing the hips, knees, and dorsiflexing the ankle allows for greater force generation (muscle stretch + more force goes to rotating joints) and more time (greater range of motion)

  25. Impulse - Applications • Impulse = F•∆t = ∆(m•v) • (4) takeoff in a high jump (create vertical momentum without losing forward momentum) Area = F • t A Greater ∆ in momentum in A than B. Will jump HIGHER Force B time

  26. Impulse - Applications • Impulse = F•∆t = ∆(m•v) • kicking a field goal • Creating momentum, vector • http://www.nsf.gov/news/mmg/mmg_disp.cfm?med_id=71011 impulse in action Morton Anderson & NSF scientists

  27. Impulse - Applications • Impulse = F•∆t = ∆(m•v) = m•∆v • We decrease an object’s momentum • to reduce impact force we increase the contact time (∆t) • This is called CUSHIONING or DAMPING: • (1) design of the front end of an auto --> crash • (2) catching a baseball or softball • flexion of elbow + shoulder and deformation of mitt • If ∆t is doubled, then F decreases by 50%! • Injury Prevention

  28. Impulse - Applications • Impulse = F•∆t = ∆(m•v) = m•∆v • We decrease an object’s momentum • to reduce impact force we increase the contact time (∆t) • This is called CUSHIONING or DAMPING: • (3) cushioning of running shoewear, protective gear by deformation • If ∆t is doubled, then F decreases by 50%! • Injury Prevention

  29. Impulse - Example • Sam is a rearfoot striker with a body mass of 70 kg. If he runs at a 6 m/sec pace, his foot approaches the ground at a velocity of 4 m/sec before it strikes. Of course after landing (in stance phase) the foot has a velocity of 0 m/sec/ Sam runs with his old shoes (Shoe A) or his new shoes (Shoe B). Shoe A has a collision time of 0.085 sec, and Shoe B has a collision time of 0.110 sec. Find the impact force for each. • Given: m = 70 kg v1 = 4 m/sec • ∆t (A) = 0.085 sec v2 = 0 m/sec. • ∆t (B) = 0.110 sec • Find: greater Force (impact) • Diagram

  30. Impulse - Example • Formula: F • ∆t = m (∆v) • F • ∆t = m (v2 - v1) • Solution: • Shoe A: FA • 0.085 sec = 70 kg (0-4 m/sec) • FA = (70 kg • -4 m/sec)/0.085 sec • FA = -3294.1 N (downward) • Shoe B FB • 0.110 sec = 70 kg (0-4 m/sec) • FB = (70 kg • -4 m/sec)/0.110 sec • FB = -2545.45 N (downward)

  31. Newton’s Third Law:The Law of Action & Reaction “When one body exerts a force an another, the second body exerts a reaction force on the first that is equal in magnitude and opposite in direction.” OR “For every action there is an equal and opposite reaction.” R A

  32. Newton’s Third Law:The Law of Action & Reaction • EXAMPLES: • While standing or sitting on a swiveling chair, • > rotate your upper body in one direction, • chair + lower body rotate in the opposite direction. • If you push against a rigid surface (like a wall) with a force of 50 N, it pushes back with a reaction force of 50 N.

  33. Ground Reaction Forces (GRF) • The reaction force produced by individuals when walking, running, etc. • As foot contacts the ground it produces a reaction force. A R

  34. Ground Reaction Forces (GRF) • In runners, the magnitude of GRF during running is about 3.5 times the runners body weight (400-700 lbs!) for each foot-strike... • Runners who overstride (stride length too long) generate a GRF with significant backwards horizontal component that retard forward motion • and increase impact forces by decreasing ∆t! • F•t = m•∆v

  35. Fx Fx R Fy R Fy If stride length increases too much, the horizontal component (Fx) of the Ground Reaction Force (R) will tend to increase, retarding forward motion.

  36. Work • Product of the force applied against resistance and displacement of the resistance. • Stair climbing • Military press • Cycle ergometer • SYMBOL: W • FORMULA: W = F • d • UNITS: Metric - Joules (J) 1J = 1 N•m • English – foot-lbs

  37. Work • EXAMPLE: Calculate the work involved in raising a weight of 200 N a height of .5 m W = F • d W = (200N)(.5 m) W = 100 N•m = 100 J

  38. Power • The amount of work performed over a given time period. • SYMBOL: P • FORMULA: P = W/Dt = F•d/Dt = F•v • F • v (strength x speed) • UNITS: Metric – watts (W) 1W = 1J/sec English – horsepower

  39. Power • EXAMPLE: Calculate the power generated in lifting a 200 N weight a height of .5 m over a time interval of 2 sec. W = F•d = (200 N)(.5 m) = 100 J P = W/Dt = 100 J/2 sec = 50 W

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