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Chapter 12 – Linear Kinetics

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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.

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Sprinting example Free Body Diagram

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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)

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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.

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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

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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 = - 3.15ms -2 a = (500 – 736)/75 = - 3.15 ms -2

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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 = 1732.1/90 = 19.25 ms -2 a h = 1732.1/90 = 19.25 ms -2 2. Apply ΣF = ma in the vertical direction direction 2000 (sin 30 o ) - 882.9 = 90a v 2000 (sin 30 o ) - 882.9 = 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 3. 1.3 ms -2 19.25 ms -2 R Resultant acceleration R 2 = 1.3 2 + 19.25 2 R = 19.29 ms -2

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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

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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

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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

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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.

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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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