Transcript KIN340-Chapter12

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, 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
ΣF=m·a
Units: 1N = (1 kg) (1 m/s2)
3. Law of Action/Reaction
For every action, there is and equal and opposite reaction.
Sprinting example
Free Body Diagram
Ground Reaction Force
ΣF = m . acg
(GRFv - W) = m . acg
where:
GRFv = vertical ground
reaction force
Weight
GRFv
Vertical Ground
Reaction Force
W = weight
m = body mass
acg is the vertical acceleration of
the center of gravity (CG)
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)
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.
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
Weight = mg = 75 kg x (– 9.81 ms-2) = - 736N
ΣF = ma
736 N
1500 – 736 = 75 a
a = (1500 – 736)/75 = 10.2 ms-2
1500 N
2. If the same person then reduces the vertical force to 500 N,
what acceleration does the person’s CG now have?
ΣF = ma
500 – 736 = 75 a
a = (500 – 736)/75 = - 3.15 ms-2
736 N
500 N
A 90 kg sprinter pushes against the blocks with a force of 2000 N at an angle
of 30o 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?
1. Apply ΣF = ma in the horizontal
direction
9N
2000 (cos 30o) = 90 x ah
ah = (2000 (cos 30o))/90
ah = 1732.1/90 = 19.25 ms-2
2. Apply ΣF = ma in the vertical
direction
3.
1.3 ms-2
R
19.25 ms-2
Resultant acceleration
R2 = 1.32 + 19.252
R = 19.29 ms-2
2000 (sin 30o) - 882.9 = 90av
av = (1000 – 882.9)/90
av = 1.3 ms-2
Momentum

Quantity of motion that a body possesses

Product of mass and velocity
M = m·v
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
Example
V= 5 m/s
Total momentum before = Total momentum after
ML + MR = MBoth
mLvL + mRvR = mBothvBoth
(130 kg) (5 m/s) + (85 kg) (-6 m/s) = (130 kg + 85 kg) (vBoth)
650 kg · m/s – 510 kg · m/s = (215 kg) (vBoth)
vBoth = (650 kg · m/s – 510 kg · m/s)/215 kg = 0.65 m/s
Impulse – Momentum Relationship
MBefore + ΣF (∆t) = MAfter
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?
MBefore + ΣF (∆t) = MAfter
(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
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.
Impulse in the high jump. Elite jumpers lean back prior to takeoff which allows them to spend more time applying force to the
ground.