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Math Review
Scalar Quantities:
Vector Quantities
(Magnitude only)
(Magnitude and direction)
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Mass
Volume
Density
Speed
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Force
Weight
Pressure
Torque
Velocity
Vector Resolution
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Vectors may be resolved into perpendicular
components. The vector composition of each pair
of components yields the original vector.
Vector Composition
The composition of vectors with the same
direction requires adding their magnitudes.
Vector Composition
The composition of vectors with the opposite
directions requires subtracting their magnitudes.
Vector Algebra
The tip-to-tail method of vector composition.
Mathematical approach
Sin θ = opposite/hypotenuse
opp = hyp x sin θ
Opposite
Cos θ = adjacent/hypotenuse
Adj = Hyp x cos θ
θ
Adjacent
Tan θ = Opp/Adj
Opp2 + Adj2 = Hyp2
Pythagorean Theorem
A long jumper takes off with a velocity of 9 m/s at an angle
of 25 o to the horizontal. How fast is the jumper moving in
the vertical and horizontal directions?
VR
Vv
25o
VH
Known: VR = 9 m/s (Hyp); Θ = 25o
Unknown: Vv (opp); VH (Adj)
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Vv (opp) = 9 m/s (Hyp) x sin Θ
Vv = 9 m/s x sin 25o = 3.8 m/s
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VH (adj) = 9 m/s (Hyp) x cos Θ
VH = 9 m/s x cos 25o = 8.2 m/s
A high jumper takes off with a vertical velocity of
4.3 m/s and a horizontal velocity of 2.5 m/s, what is
the resultant velocity of the jumper?
Known: Vv(opp) = 4.3 m/s
VH (adj) = 2.5 m/s
4.3
m/s
VR
Unknown: VR (hyp); Θ
Θ
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2.5 m/s
Opp2 + Adj2 = Hyp2
4.32 + 2.52 = VR2
VR = √ (4.32 + 2.52) = 4.97 m/s
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Tan Θ = opp/adj = 4.3/2.5 = 1.72
Θ = tan -1 (1.72) = 59.8 o
Chapter 13 – Equilibrium and
Human Movement
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Torque
Levers
Equations of Static Equilibrium
Center of Gravity
Mechanical Strength: measured by the maximum torque
that can be voluntarily generated at a certain joint
Strength determined by:
• Absolute force developed by muscle
• Distance from joint center to tendon insertion (affects moment arm)
• Angle of tendon insertion (affects moment arm)
The moment arm
of a force is the
perpendicular
distance from the
force’s line of
action to the axis
of rotation.
axis
Moment arm
Force
line of
action
axis
Moment
arm
Force
line of
action
Calculate the elbow joint torque when the biceps generate a force of
100 N at the following angles of attachment: a. 30o b. 60o c.90o d.120o
e. 150o
a.
Fm = 100 N
a. Opp = hyp sin Θ
F⊥ = 100 sin 30o = 50 N
T = 50 N x 0.03m = 1.50 Nm
F⊥
30o
3 cm
F⊥
b.
60o
b. F⊥ = 100 sin 60o = 86.6 N
T = 86.6 N x 0.03m = 2.6Nm
3 cm
c.
F⊥
3 cm
c. T = 100N x 0.03 cm = 3.0Nm
Calculate the elbow joint torque when the biceps generate a force of
100 N at the following angles of attachment: a. 30o b. 60o c.90o d.120o
e. 150o
Fm = 100 N
d.
30o
F⊥ = 100 cos 30o = 86.6 N
3 cm
T = 86.6N x 0.03m = 2.6 Nm
e.
60o
3 cm
d. Adj = hyp cos Θ
e. F⊥ = 100 cos 60o = 50 N
T = 50 N x 0.03m = 1.5 Nm
Rotary versus stabilizing
components
6-21
Levers
What is a lever?
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A simple machine consisting of a
relatively rigid barlike body that can
be made to rotate about an axis or a
fulcrum
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There are first, second, and third class
levers
Applied or
motive force
F
First Class Lever
R
Resistive force
Axis of Rotation or fulcrum
Second Class Lever
R
F
Third Class Lever
F
R
Mechanical advantage/effectiveness =ME =
(Moment arm of applied force)/(Moment arm of the resistance)
ME = 1, < 1, >1
F
R
First class
R
F
ME = always > 1
Second class
ME = always < 1
F
R
Third class
Third Class levers
A force can move a resistance through a
large range of motion when the force
arm (fa) is shorter than the resistance
arm (ra).
F
R
fa
ra
Law of Equilibrium
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∑T=0
F
+
3 cm
30 cm
(F x 3) + (-130 x 30) = 0
F = (130 x 30)/3 = 1300 N
130 N
Law of the lever
A small force can have a large torque or moment of
rotation if the lever arm is long
Similarly, the force must be large to achieve the
same moment of rotation if the lever arm is short
∑T=0
Fext x 5cm – (34kg x 17 cm) – (6 kg x 20 cm) = 0
Fext = ((34 kg x 17 cm) + (6 kg x 20 cm))/5 cm
Fext = (578 kgcm + 120 kgcm)/5 cm = 139.6 kg
∑T=0
Fext x 5cm – (34kg x 31 cm) – (6 kg x 52 cm) = 0
Fext = ((34 kg x 31 cm) + (6 kg x 52 cm))/5 cm
Fext = (1054 kgcm + 312 kgcm)/5 cm = 273.2 kg
The importance of levers in the mechanism of injuries
∑T=0
(L x l) – (M x m) = 0
(L x l) = (M x m)
L = M x (m/l)
Therefore, if force M has a
lever arm (m) 10x the lever
arm (l) of force L, the force in
the ligament (L) will be 10 x
greater than force M.