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Chapter 12 – Linear Kinetics. Force The push or pull acting on the body measured in Newtons (N) The relationship between the forces which affect a body,

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Presentation on theme: "Chapter 12 – Linear Kinetics. Force The push or pull acting on the body measured in Newtons (N) The relationship between the forces which affect a body,"— Presentation transcript:

1 Chapter 12 – Linear Kinetics

2 Force The push or pull acting on the body measured in Newtons (N) The relationship between the forces which affect a body, and the state of motion of that body, can be summarized by Newton’s three Laws of Motion: 1. Law of Inertia A body will continue in its state of rest or motion in a straight line, unless a force (i.e., a net or unbalanced force) acts on it 2. Law of Acceleration If net force acting in a body is not zero, the body will experience acceleration proportional to the force applied Σ F = m · aUnits: 1N = (1 kg) (1 m/s 2 ) 3. Law of Action/Reaction 3. Law of Action/Reaction For every action, there is and equal and opposite reaction.

3 Sprinting example Free Body Diagram

4 Ground Reaction Force ΣF = m. a cg (GRF v - W) = m. a cg where: GRF v = vertical ground reaction force W = weight m = body mass a cg is the vertical acceleration of the center of gravity (CG) GRF v If GRFv = W, then ΣF = 0 (no net force) and a cg = 0 If GRFv > W, then ΣF > 0 (net force upwards) and a cg > 0 (positive) If GRFv < W, then < 0 (net force downwards) and a cg < 0 (negative)

5 If GRF v = W, then ΣF = 0 (no net force) and a cg = 0 1. CG Motionless 2. CG moving with a constant velocity If GRF v > W, then ΣF > 0 (net force upwards) and a cg >0 (positive) 1. the speed of the CG is increasing as it moves upward (+ dir) 2. the speed of the CG is decreasing as it moves downward (- dir) If GRF v < W, then < 0 (net force downwards) and a cg < 0 (negative) 1. the speed of the CG is decreasing as it moves upward (+ dir) 2. the speed of the CG is increasing as it moves downward (- dir) Rapid Squat force trace. Rapid Squat force trace.

6 Ground reaction force examples 1. A person whose mass is 75 kg exerts a vertical force of 1500N against the ground. What is the individual’s acceleration in the vertical direction? (Remember to state whether the acceleration is positive or negative 1500 N 736 N Weight = mg = 75 kg x (– 9.81 ms -2 ) = - 736N ΣF = ma 1500 – 736 = 75 a a = (1500 – 736)/75 = 10.2ms -2 a = (1500 – 736)/75 = 10.2 ms -2

7 2. If the same person then reduces the vertical force to 500 N, what acceleration does the person’s CG now have? 736 N 500 N ΣF = ma 500 – 736 = 75 a a = (500 – 736)/75 = ms -2 a = (500 – 736)/75 = ms -2

8 1. Apply ΣF = ma in the horizontal direction direction 2000 (cos 30 o ) = 90 x a h 2000 (cos 30 o ) = 90 x a h a h = (2000 (cos 30 o ))/90 a h = (2000 (cos 30 o ))/90 a h = /90 = ms -2 a h = /90 = ms Apply ΣF = ma in the vertical direction direction 2000 (sin 30 o ) = 90a v 2000 (sin 30 o ) = 90a v a v = (1000 – 882.9)/90 a v = (1000 – 882.9)/90 a v = 1.3 ms -2 a v = 1.3 ms -2 A 90 kg sprinter pushes against the blocks with a force of 2000 N at an angle of 30 o to the horizontal. Neglecting air resistance, what is the acceleration of the sprinter’s CG in both the horizontal and vertical directions? What is the resultant acceleration? 9N ms ms -2 R Resultant acceleration R 2 = R = ms -2

9 Momentum Quantity of motion that a body possesses Quantity of motion that a body possesses Product of mass and velocity M = m·v Product of mass and velocity M = m·v In the absence of external forces, the total In the absence of external forces, the total momentum of a given system remains constant Conservation of Linear Momentum Total momentum before = Total momentum after Total momentum before = Total momentum after

10 Example V= 5 m/s Total momentum before = Total momentum after M L + M R = M Both m L v L + m R v R = m Both v Both (130 kg) (5 m/s) + (85 kg) (-6 m/s) = (130 kg + 85 kg) (v Both ) 650 kg · m/s – 510 kg · m/s(215 kg) (v Both ) 650 kg · m/s – 510 kg · m/s = (215 kg) (v Both ) vBoth =650 kg · m/s – 510 kg · m/s)/215 kg = 0.65 m/s vBoth = (650 kg · m/s – 510 kg · m/s)/215 kg = 0.65 m/s

11 Impulse – Momentum Relationship M Before + ΣF (∆t) = M After where ΣF (∆t) = impulse Example: 90 kg person lands from a jump. Just before impact, vertical velocity v = - 5 m/s. What would be the mean net ground reaction force if it takes 0.1 s to reach zero velocity? M Before + ΣF (∆t) = M After (90 kg) (- 5 m/s) + ΣF (0.1 s) = 90 kg (0 m/s) ΣF = (90 kg) (5 m/s)/ 0.1 s = 4500 N ≈ 5 times body weight What if it took 0.25 s to reach zero velocity? ΣF = (90 kg) (5 m/s)/ 0.25 s = 1800 N ≈ 2 times body weight

12 Impulse in a javelin throw. Elite athletes are able to apply a force over a longer time frame by leaning back and pulling the javelin from behind the body and releasing it far out in front.

13 Impulse in the high jump. Elite jumpers lean back prior to take- off which allows them to spend more time applying force to the ground.


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