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Chapter 12 – Chapter 12 – Linear Kinetics Linear Kinetics

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Page 1: 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,

Chapter 12 – Linear Chapter 12 – Linear KineticsKinetics

Page 2: 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,

ForceForceThe push or pull acting on the body measured in Newtons (N)The push or pull acting on the body measured in Newtons (N)

The relationship between the forces which affect a body, and The relationship between the forces which affect a body, and the state of motion ofthe state of motion of that that body, can be summarized by body, can be summarized by Newton’s three Laws of MotionNewton’s three Laws of Motion::

1.1. Law of Inertia Law of InertiaA body will continue in its state of rest or motion in a straight 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 itline, unless a force (i.e., a net or unbalanced force) acts on it

2. 2. Law of AccelerationLaw of AccelerationIf net force acting in a body is not zero, the body will If net force acting in a body is not zero, the body will experience acceleration proportional to the force appliedexperience acceleration proportional to the force applied

Σ F = m · aΣ F = m · a Units: 1N = (1 kg) (1 m/sUnits: 1N = (1 kg) (1 m/s22))

3.3. Law of Action/Reaction Law of Action/ReactionFor every action, there is and equal and opposite reaction.For every action, there is and equal and opposite reaction.

Page 3: 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,

Sprinting exampleSprinting example

Free Body Diagram

Page 4: 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,

Vertical Ground Reaction Force

Weight

Ground Reaction Force

ΣF = m . acg

(GRFv - W) = m . acg

where:

GRFv = vertical groundreaction force

W = weightm = body mass

acg is the vertical acceleration of the center of gravity (CG)

GRFv

If GRFv = W, then ΣF = 0 (no net force) and acg = 0

If GRFv > W, then ΣF > 0 (net force upwards) and acg > 0 (positive)

If GRFv < W, then < 0 (net force downwards) and acg < 0 (negative)

Page 5: 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,

If GRFv = W, then ΣF = 0 (no net force) and acg = 0

1. CG Motionless 2. CG moving with a constant velocity

If GRFv> W, then ΣF > 0 (net force upwards) and acg >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 GRFv < W, then < 0 (net force downwards) and acg< 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.

Page 6: 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,

Ground reaction force examplesGround reaction force examples

1. A person whose mass is 75 kg exerts a vertical force of 1500N 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 against the ground. What is the individual’s acceleration in the vertical direction? (Remember to state whether the vertical direction? (Remember to state whether the acceleration is positive or negativeacceleration is positive or negative

1500 N

736 N

Weight = mg = 75 kg x (– 9.81 msWeight = mg = 75 kg x (– 9.81 ms-2-2) = - 736N) = - 736N

ΣΣF = maF = ma

1500 – 736 = 75 a1500 – 736 = 75 a

a = (1500 – 736)/75 = a = (1500 – 736)/75 = 10.210.2 msms-2-2

Page 7: 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,

2. If the same person then reduces the vertical force to 500 N, 2. If the same person then reduces the vertical force to 500 N, what acceleration does the person’s CG now have?what acceleration does the person’s CG now have?

736 N

500 N

ΣΣF = maF = ma

500 – 736 = 75 a500 – 736 = 75 a

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

Page 8: 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,

1. Apply 1. Apply ΣΣF = ma in the horizontalF = ma in the horizontal directiondirection

2000 (cos 302000 (cos 30oo) = 90 x a) = 90 x ahh

aahh = (2000 (cos 30 = (2000 (cos 30oo))/90 ))/90

aahh = 1732.1/90 = = 1732.1/90 = 19.25 ms19.25 ms-2-2

2. Apply 2. Apply ΣΣF = ma in the verticalF = ma in the vertical directiondirection

2000 (sin 302000 (sin 30oo) - 882.9 = 90a) - 882.9 = 90avv

aav v = (1000 – 882.9)/90 = (1000 – 882.9)/90

aavv = = 1.3 ms1.3 ms-2-2

A 90 kg sprinter pushes against the blocks with a force of 2000 N at an angle A 90 kg sprinter pushes against the blocks with a force of 2000 N at an angle of 30of 30oo to the horizontal. Neglecting air resistance, what is the acceleration of 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 the sprinter’s CG in both the horizontal and vertical directions? What is the resultant acceleration?resultant acceleration?

9N

3.

1.3 ms-2

19.25 ms-2

R

Resultant accelerationResultant acceleration

RR22 = 1.3 = 1.322 + 19.25 + 19.2522

R = R = 19.29 ms19.29 ms-2-2

Page 9: 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,

MomentumMomentum Quantity of motion that a body possessesQuantity of motion that a body possesses

Product of mass and velocity M = m·vProduct 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 constantmomentum of a given system remains constant

Conservation of Linear MomentumConservation of Linear Momentum

Total momentum before = Total momentum afterTotal momentum before = Total momentum after

Page 10: 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,

Example V= 5 m/s

Total momentum before = Total momentum afterTotal momentum before = Total momentum after

MMLL + M + MRR = M = MBothBoth

mmLLvvLL + m + mRRvvRR = m = mBothBothvvBothBoth

(130 kg) (5 m/s) + (85 kg) (-6 m/s) = (130 kg + 85 kg) (v(130 kg) (5 m/s) + (85 kg) (-6 m/s) = (130 kg + 85 kg) (vBothBoth))

650 kg · m/s – 510 kg · m/s650 kg · m/s – 510 kg · m/s = (215 kg) (v(215 kg) (vBothBoth))

vBoth =vBoth = (650 kg · m/s – 510 kg · m/s)/215 kg = 650 kg · m/s – 510 kg · m/s)/215 kg = 0.65 m/s0.65 m/s

Page 11: 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,

Impulse – Momentum RelationshipImpulse – Momentum Relationship

MMBeforeBefore + + ΣΣF (∆t) = MF (∆t) = MAfter After

wherewhere ΣΣF (∆t) = impulseF (∆t) = impulse

Example: 90 kg person lands from a jump. JustExample: 90 kg person lands from a jump. Justbefore impact, vertical velocity v = - 5 m/s. Whatbefore impact, vertical velocity v = - 5 m/s. Whatwould be the mean net ground reaction force if itwould be the mean net ground reaction force if ittakes 0.1 s to reach zero velocity?takes 0.1 s to reach zero velocity?

MMBeforeBefore + + ΣΣF (∆t) = MF (∆t) = MAfterAfter

(90 kg) (- 5 m/s) + (90 kg) (- 5 m/s) + ΣΣF (0.1 s) = 90 kg (0 m/s)F (0.1 s) = 90 kg (0 m/s)

ΣΣF = (90 kg) (5 m/s)/ 0.1 s = F = (90 kg) (5 m/s)/ 0.1 s = 4500 N 4500 N ≈ 5 times body weight≈ 5 times body weight

What if it took 0.25 s to reach zero velocity? What if it took 0.25 s to reach zero velocity?

ΣΣF = (90 kg) (5 m/s)/ 0.25 s = F = (90 kg) (5 m/s)/ 0.25 s = 1800 N 1800 N ≈ 2 times body weight≈ 2 times body weight

Page 12: 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,

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.

Page 13: 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,

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.