Math ReviewMath Review

Scalar Quantities:Scalar Quantities:(Magnitude only)(Magnitude only)

MassMass VolumeVolume DensityDensity SpeedSpeed

Vector QuantitiesVector Quantities(Magnitude and direction)(Magnitude and direction)

ForceForce WeightWeight PressurePressure TorqueTorque VelocityVelocity

Vector ResolutionVector Resolution

Vectors may be resolved into perpendicular components. The vector composition of each pair of components yields the original vector.

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+_

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+

+

+

+

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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 approachMathematical approach

Hypotenuse

Adjacent

Opposite

θ

Sin θ = opposite/hypotenuse

opp = hyp x sin θ

Cos θ = adjacent/hypotenuse

Adj = Hyp x cos θ

Tan θ = Opp/Adj

Opp2 + Adj2 = Hyp2

Pythagorean Theorem

A long jumper takes off with a velocity of 9 m/s at an angle A long jumper takes off with a velocity of 9 m/s at an angle of 25 of 25 oo to the horizontal. How fast is the jumper moving in to the horizontal. How fast is the jumper moving in the vertical and horizontal directions?the vertical and horizontal directions?

Known: VR = 9 m/s (Hyp); Θ = 25o

Unknown: Vv (opp); VH (Adj)

• Vv (opp) = 9 m/s (Hyp) x sin Θ

Vv = 9 m/s x sin 25o = 3.8 m/s

• VH (adj) = 9 m/s (Hyp) x cos Θ

VH = 9 m/s x cos 25o = 8.2 m/s

25o

9 m/sVv

VH

VR

A high jumper takes off with a vertical velocity of 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 4.3 m/s and a horizontal velocity of 2.5 m/s, what is the resultant velocity of the jumper?resultant velocity of the jumper?

Known:Known: VVvv(opp) = 4.3 m/s(opp) = 4.3 m/s

VVHH (adj) = 2.5 m/s (adj) = 2.5 m/s

Unknown: VUnknown: VRR (hyp); (hyp); ΘΘ

OppOpp22 + Adj + Adj22 = Hyp = Hyp22

4.34.322 + 2.5 + 2.522 = V = VRR22

VVR R == √ (4.3√ (4.322 + 2.5 + 2.522) = ) = 4.97 m/s4.97 m/s

Tan Tan ΘΘ = opp/adj = 4.3/2.5 = 1.72 = opp/adj = 4.3/2.5 = 1.72 ΘΘ = tan = tan -1-1 (1.72) = (1.72) = 59.8 59.8 oo

4.3 m/s

2.5 m/s

Θ

VR

Chapter 13 – Equilibrium and Chapter 13 – Equilibrium and Human MovementHuman Movement

TorqueTorque LeversLevers Equations of Static EquilibriumEquations of Static Equilibrium Center of GravityCenter of Gravity

Mechanical Strength: measured by the maximum torque Mechanical Strength: measured by the maximum torque that can be voluntarily generated at a certain jointthat 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 The moment arm of a force is the of a force is the perpendicular perpendicular distance from the distance from the force’s line of force’s line of action to the axis action to the axis of rotation.of rotation.

Moment armForce line of action

Moment arm

Force line of action

axis

axis

Calculate the elbow joint torque when the biceps generate a force of Calculate the elbow joint torque when the biceps generate a force of 100 N at the following angles of attachment: a. 30100 N at the following angles of attachment: a. 30o o b. 60b. 60o o c.90c.90o o d.120 d.120o o e.e.

150150o o

a.a.

b.b.

c. c.

a. Opp = hyp sin a. Opp = hyp sin ΘΘ F⊥ = 100 sin 30 = 100 sin 30oo = 50 N = 50 N

T = 50 N x 0.03m = T = 50 N x 0.03m = 1.50 Nm1.50 Nm

b. b. F⊥ = 100 sin 60 = 100 sin 60oo = 86.6 N = 86.6 N

T = 86.6 N x 0.03m = T = 86.6 N x 0.03m = 2.6Nm2.6Nm

c. T = 100N x 0.03 cm =c. T = 100N x 0.03 cm = 3.0Nm 3.0Nm

30o

Fm = 100 N F⊥

60o

F⊥

3 cm

3 cm

3 cm

F⊥

Calculate the elbow joint torque when the biceps generate a force of Calculate the elbow joint torque when the biceps generate a force of 100 N at the following angles of attachment: a. 30100 N at the following angles of attachment: a. 30o o b. 60b. 60o o c.90c.90o o d.120 d.120o o e.e.

150150o o

d.d.

e.e.

d. Adj = hyp cos d. Adj = hyp cos ΘΘ

FF⊥⊥ = 100 cos 30 = 100 cos 30oo = 86.6 N = 86.6 N

T = 86.6N x 0.03m = T = 86.6N x 0.03m = 2.6 Nm2.6 Nm

e. Fe. F⊥⊥ = 100 cos 60 = 100 cos 60oo = 50 N = 50 N

T = 50 N x 0.03m = T = 50 N x 0.03m = 1.5 Nm1.5 Nm

30o

Fm = 100 N

60o

3 cm

3 cm

Rotary versus stabilizing Rotary versus stabilizing componentscomponents

6-21

What is a lever?What is a lever?

• A simple machine consisting of a A simple machine consisting of a relatively rigid barlike body that can relatively rigid barlike body that can be made to rotate about an axis or a be made to rotate about an axis or a

fulcrumfulcrum

• There are first, second, and third class There are first, second, and third class leverslevers

LeversLevers

First Class LeverFirst Class LeverApplied or motive force F R Resistive force

Axis of Rotation or fulcrum

Second Class LeverSecond Class LeverFR

Third Class LeverThird Class LeverF R

F R

First class

R F

Second class

F R

Third class

Mechanical advantage/effectiveness =ME =

(Moment arm of applied force)/(Moment arm of the resistance)

ME = 1, < 1, >1

ME = always > 1

ME = always < 1

Third Class leversThird Class levers

F Rfa

ra

A force can move a resistance through a A force can move a resistance through a large range of motion when the force large range of motion when the force arm (fa) is shorter than the resistance arm (fa) is shorter than the resistance arm (ra).arm (ra).

Law of EquilibriumLaw of Equilibrium

∑ ∑ T = 0T = 0

3 cm

30 cm

F

130 N

(F x 3) + (-130 x 30) = 0

F = (130 x 30)/3 = 1300 N

+

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Law of the leverLaw of the lever

A small force can have a large torque or moment of A small force can have a large torque or moment of rotation if the lever arm is longrotation if the lever arm is long

Similarly, the force must be large to achieve the Similarly, the force must be large to achieve the same moment of rotation if the lever arm is shortsame moment of rotation if the lever arm is short

∑ ∑ T = 0 T = 0

FFextext x 5cm – (34kg x 17 cm) – (6 kg x 20 cm) = 0 x 5cm – (34kg x 17 cm) – (6 kg x 20 cm) = 0

FFextext = ((34 kg x 17 cm) + (6 kg x 20 cm))/5 cm = ((34 kg x 17 cm) + (6 kg x 20 cm))/5 cm

FFextext = (578 kgcm + 120 kgcm)/5 cm = = (578 kgcm + 120 kgcm)/5 cm = 139.6 kg139.6 kg

∑ ∑ T = 0 T = 0

FFextext x 5cm – (34kg x 31 cm) – (6 kg x 52 cm) = 0 x 5cm – (34kg x 31 cm) – (6 kg x 52 cm) = 0

FFextext = ((34 kg x 31 cm) + (6 kg x 52 cm))/5 cm = ((34 kg x 31 cm) + (6 kg x 52 cm))/5 cm

FFext ext = (1054 kgcm + 312 kgcm)/5 cm = = (1054 kgcm + 312 kgcm)/5 cm = 273.2 kg273.2 kg

The importance of levers in the mechanism of injuriesThe importance of levers in the mechanism of injuries

∑ ∑ T = 0T = 0

(L x l) – (M x m) = 0(L x l) – (M x m) = 0

(L x l) = (M x m)(L x l) = (M x m)

L = M x (m/l)L = M x (m/l)

Therefore, if force M has a lever Therefore, if force M has a lever arm (m) 10x the lever arm (l) of arm (m) 10x the lever arm (l) of force L, the force in the force L, the force in the ligament (L) will be 10 x ligament (L) will be 10 x greater than force M. greater than force M.