Chương 6: 6.1. The Eye Can Be Treated, To A First Approximation, As A Thin - x0002 - Walled Elastic [PDF]

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CHƯƠNG 6 6.1. The eye can be treated, to a first approximation, as a thin_x0002_walled elastic pressure vessel of diameter Dandwall thickness t. Calculate the distensibility of the eye, β = (1/V)(dV/dp), (compare with Equation [4.17]) as a function of D, t, and the Young’s modulus of the sclera/cornea, E. GIẢI: Assume: t: wall thickness D: diameter 𝜏:hoop stress Force balance: 𝜏t𝜋D = p. 𝜏=

𝜋D2

pD 4t

4

= E𝜀 theo định luật (Hooke 𝜏 = E𝜀) D

𝜀 = p( ) 4Et This is the strain due to a pressure p. Now we consider the additional strain, ∆𝜀, due to an incremental change in diameter, ∆D ∆D D ∆𝜀 = = ∆p ( ) 𝛽=

D 1 dV

V dp

≈

4Et 1 ∆V 3 ∆D

V ∆p

=

D ∆p

=

3D 4Et

Where we have used the fact that ∆V~3D2∆D 6.2. The aqueous humor circulates within the eye, flowing in at a constant rate Q in = 2 μl/min, and draining from the eye at Qout. At steady state, Qout = Qin. The eye also acts like an elastic vessel, in that its volume increases if the intraocular pressure p increases. This is expressed by: V = Cp where V is the volume of the eye (normally about 4 cm3) and C is the compliance of the eye (approximately 3 μl/mmHg). Finally, the outflow rate Qout equals p/R, where R is the (effectively constant) resistance to outflow (about 4mmHg min/μl). a) Balancing mass, show that: 𝑑𝑝 𝑝−𝑝𝑠𝑠 + =0 𝑑𝑡 𝑅𝐶 where pss is the steady-state pressure in the eye, equal to RQin. b) If you poke your eye and increase the pressure by an amount δp = 10 mmHg, how long does it take the pressure to return to within 5% of its steady-state value? GIẢI: dV a. Mass balance: = Qin- Qout (1) dt

Given: V = Cp, Qin = Pss/R, Qout = P/R 1|Page

(1) becomes: C

dP

dt dP

=

Pss

-

P

R R P−Pss

⟹ + =0 dt RC b. The mass balance equation derived in a) still applies in this case. Now solve the differential equation: d(P−Pss ) P−Pss d(P−Pss ) dt =⟹ =dt

RC

⟹ ln(P − Pss ) = -

t

RC −t/RC

P−Pss

RC

+ const

⟹ P − Pss = ce ⟹ P = Pss + ce−t/RC , with c = const Initial condition: P(0) = Pss + 𝛿P ⟹ P = Pss + 𝛿Pe−t/RC It take the pressure to return to within 5% of its steady-state value ⟹ P(t) = 1,05Pss ⟹1,05Pss = Pss + 𝛿Pe−t/RC Solve by t: 𝛿P 10 ) = 3.4.ln( ) = 38,6 (min) with Pss = QminR t = RC.ln( 0,05Pss

0,05.2.4

6.3. The cornea is a water-filled connective tissue that we will treat as being flat and of thickness h (Fig. 6.18). Because of the composition of the cornea, it traps positive ions, so that there are “excess” positive ions in the interior of the cornea compared with the surrounding fluid contacting the cornea. This is equivalent to the surface of the cornea acting like a semipermeable membrane that blocks the passage of positive ions. a) When the cornea is completely dehydrated, its thickness is hdry = 220 μm, and the “excess” concentration of positive ions is 0.8 mM, compared with physiological saline. Assuming that no positive ions leave the tissue when it becomes hydrated, write an expression for the “excess” positive ion concentration as a function of corneal thickness, h. b) As the cornea becomes more hydrated, it thickens and fibers in the cornea become stretched. This creates an effective positive pressure within the cornea, p = k(h − h0), with h0 = 345μm and k = 5.5 Pa/μm. Compute the equilibrium thickness of the cornea when it is exposed to physiological saline at 370C and zero pressure (gauge). Note that the universal gas constant R = 8.314 J/(mol K). Becareful about units here: 1mM is 10−3mol/l.

GIẢI: a. Ions inside the cornea must be conserved 2|Page

⟹ hdry cdry = hc, where c = the concentration of “excess” positivetions 220x0.8x10−3

hdry cdry

0.176

⟹c= = = (𝜇M) h h h b. at equilibrium: (p − 𝜋)saline = (p − 𝜋)cornea ⟹ ∆p - ∆𝜋 = 0, where the osmotic pressure is due to the excess positive ions ⟹ k(h − h0 ) - RTC = 0 RTC 0,176 ⟹ h − h0 =0 with c = k RT

2

h

⟹ h - hh0 - .0,176 = 0 k Given h0 = 345𝜇m, k = 5,5 Pa/𝜇m, R = 8,314J/mol.K, T = 310K RT 8,31.310 0,176 With .0,176 = . −3 = 8,247x104 (𝜇m2 ) k 5,5 10 ℎ = −163 2 4 h – 345.h - 8.247.10 = 0 ⟹ { ℎ = 508 because h is high, so h > 0 ⟹ h = 508 (𝜇m) 6.4. In Section 6.2.1 we presented a model of Schlemm’s canal as a compliant channel, as originally developed by Johnson and Kamm. Show that the non-dimensional height of Schlemm’s canal ℎ̃(x) = h(x)/h0 obeys the following equation:

where the parameter 𝛾 is given by: 𝛾 2 =

12𝜇𝑠 2 𝑤ℎ02 𝑅𝑖𝑤

And 𝑥̃ = x/s is the non-dimensional position in the canal Physically, what does 𝛾 2 represent? When 𝛾 ≪ 1 show that the above equation has a solution of the form h3(dh/dx) = constant. GIẢI: h −h(x) IOP−p(x) From the text we have 0 = 1 - h̃(x) = where h̃ = h/h0 . Note that equation h0

̃ dh

E

1 dp

implies that = dx E dx Also for the solution for flow in a thin channel dp 12𝜇𝜑(x) = 3 where 𝜇, w are constant dx

wh (x)

Rearranging this equation and differentiation with respect to x given: d𝜑 w d dp (h3 ) = dx

= = =

12𝜇 dx w d

dx ̃ dh

(h3 E

12𝜇 dx Ewh30 d

)

dx ̃ dh 3 (h̃ ) 12𝜇 dx dx ̃ 2 Ewh30 dh [3h̃2 ( ) 12𝜇 dx

d + h̃3

2h ̃

dx2

] 3|Page

On the other hand: d𝜑 dx

=

IOP−p(x) Riw

=

wRiw h30

1- h̃ =

12𝜇

̃) E(1−h Riw

̃ 2

dh d [3h̃2 ( ) + h̃3

2h ̃

dx2

dx

]

Finally, nothing that x = x̃s, where s in the half-distance between collector channels, we have: 12𝜇s2

̃ 2

̃ ̃ 2 dh ̃3 d 3 (1 − h) = 3h ( ) + h

dx̃2

dx̃

wRiw h0

Physically, 𝛾 2 =

2h ̃

(*)

12𝜇s2 wRiw h30

pressure drop along Schlemm’s canal from x = 0 to s when the canal is open pressure drop across the inner wall/methwork from x = 0 to s 2 when 𝛾 ≪ 1, then the left hand side of (*) can be ignored and the given equation becomes: ̃ 2

2̃

dh d h 0 = 3h̃2 ( ̃ ) + h̃3 ̃ 2 =

⟹

̃ dh h̃3 ̃ dx

⟹ h3

dh dx

dx

dx

d dx

̃

dh (h̃3 ) dx̃

= constant = constant

4|Page

CHƯƠNG 7 7.1. Consider a small spherical bubble of radius R. A. Show that the energy required to expand this sphere by a small amount ∆𝑅 is 2𝜎∆𝑉/𝑅. Here ∆𝑉 is the increase in volume and σ is the interfacial tension. B. Estimate the time-averaged power required to overcome alveolar surface tension during normal breathing. Take R = 150 μm, σ = 25 dynes/cm, and breathing rate= 12 breaths/min. C. Repeat this calculation for the cat, where R = 50 μm and the tidal volume is 20 ml. Compare this calculated value with a rough estimate of the power obtained from Fig. 7.23. (Take beginning of normal inspiration to occur at 100 ml.) Is surface tension or lung tissue elasticity the dominant restoring force in the cat lung?

GIẢI: A. dE = pdV ⟹ ∆E = p∆V E: energy p: gauge pressure of the spere V: volume ∆V is the increase in volume σ is the interfacial tension then at equilibrium place's law staten that, for a spherical bubble off radius R, the difference 2σ between internal air pressure p and the pressure in the surrounding fluid, 7.1: p = ⟹ ∆E =

2σ∆V

R

R

B. R = 150μm = 0,015 cm σ = 25 dynes/cm = 0,00025 N/cm Breathing rate = 12 breaths/min = 1 breath/ 5sec The volume exchanged per breath is the tidal volume, and it is exchanged approximately 12 times per minute. Tidal volume: V = 500ml = 500 cm3 5|Page

𝑃=

𝐸 2𝜎∆𝑉 2𝜎𝑉 2 . 0,00025 . 500 = .N = = 𝑡 𝑅𝑡 𝑅𝑡 0,015 . 5

= 3,3 N.cm/s = 3,3x 10-2 N.m/s = 0,033 J/s = 0,033 W But expiration is passive, therefore the power is expended only during inspiration Assume ½ time per breath (2,5s) goes for inspiration ⟹ 𝑃=

2 . 0,00025 . 500 = 0,067𝑊 0,015 . 2,5

C. R = 50μm = 0,005 cm Tidal volume: V = 20 cm3 2σV 2 . 0,00025 . 20 P= = = 0,8 N.cm/s = 0,008 W Rt

0,005 . 2,5

In graph p-v, the area of a closed loop is work. So choose any small area, the work: W = 20ml.7cmH₂0 = 2.105.686,4466 = 0,0137 (J) And power P =

𝑊 𝑡

=

0,0137 2,5

= 5,4916.10−3 (W)

And now, comparison of the two power, surface tension is the dominant restoring force in the cat lung. 7.3. A balloon is surrounded by a tank of liquid at negative pressure and is connected to the atmosphere by a tube of length L and crosssectional area A (Fig. 7.25). The pressure inside the balloon p oscillates above and below atmospheric pressure causing small changes in the balloon volume V. The elasticity of the balloon is characterized by its compliance C, defined by ∆𝑝 = ∆𝑉/𝐶

6|Page

A. Derive a second-order differential equation for ∆V(t), assuming that (i) the pressure differential along the tube accelerates the air in the tube and is not used to overcome entrance, exit, or tube losses; and (ii) the air density ρ is constant. From the equation, show that the natural frequency of the system is √𝜌𝐿𝐶 B. For a 70 kg man, A/L is approximately 0.001 m. The equivalent value for a 12 kg dog would be approximately [12/70]1/3 of the value, or 5.6×10−4m. The compliance of dog lungs is approximately 0.029l/cm H2O. Estimate the natural frequency of a dog’s breathing using the formula developed in (a). Measurements indicate that dogs with a body mass of 12 kg pant at about 5.3 Hz. Comment briefly on any differences between your answer and the measured frequency. GIẢI:

a. For balloon 1

Δ𝑝 = Δ𝑉 𝑐

1

(1)

or p= (𝑉 − 𝑉0 ) 𝑐

𝑉 = balloon volume, 𝑉0 = balloon volume for 𝑝 = 0 𝑑𝑢 In airway, mass in = 𝜌𝐴𝐿, but 𝐹 = 𝑚 Force is (Pralloon −𝑃atm )𝐴 = 𝜌𝐴𝐿 ⇒ 𝑝 = 𝜌𝐿

𝑑𝑣

𝑑𝑢

𝑑𝑡

𝑑𝑡

, 𝑝 is balloon gauge pressure.

𝑑𝑡

𝑑𝑉

But 𝑈 = 𝑄/𝐴 and 𝑄 = − ⇒ 𝑑𝑡 Combining (1) and (2) gives 𝑑2𝑉 𝑑𝑡 2

+

𝐴 𝜌𝐿𝐶

𝑉=

𝐴 𝜌𝐿𝐶

𝑑𝑈 𝑑𝑡

=−

1 𝑑2𝑉 𝐴 𝑑𝑡 2

⇒𝑝=−

𝜌𝐿 𝑑2 𝑉 𝐴 𝑑𝑡 2

(2)

𝑉0

⇒ 𝑉 = 𝐶1 cos 𝜔𝑡 + 𝐶2 sin 𝜔𝑡 + 𝑉0 where 𝜔 = √

𝐴 𝜌𝐿𝑐

is the natural frequency, 𝐶1 , 𝐶2 are const.

b. 𝐴 = 5,6. 10−4 m, 𝜌 = 1,2 kg/m3 𝐿

C= 0.0296 L/cmH2 O = 2,96 × 10−7 m5 /N 5,6.10−4

⇒𝜔=√

1,2.2,96.10−7 𝜔

frequency 𝑓 =

2𝜋

= 39,7 𝑠 −1

= 6,3 Hz 7|Page

7.4. Alung is inflated withwater and then with air. The pressure–volume curves for these two inflation procedures are shown in Fig. 7.26, with the right-pointing arrow representing inflation and the leftpointing arrow representing deflation. Assume that the lung has 150× 106 identical alveoli, and that alveoli make up a constant 85% fraction of total lung volume. Based on these curves, graph the relationship between surface tension coefficient (in dynes/cm) and alveolar radius (in microns). Your graph should be quantitatively correct. This is best accomplished by choosing some key points from Fig. 7.26, transforming them to suitable values on your graph, and then interpolating by sketching.

8|Page

GIẢI: Pair be pressure air inflation, Psaline be pressure saline inflation ∆P be the change of pressure R is alveolar radius, 𝜎 is surface tension coefficient ∆𝑷 = 𝑷𝒂𝒊𝒓 - 𝑷𝒔𝒂𝒍𝒊𝒏𝒆 This is the pressure needed to overcome surface tension. Pair and Psaline are read from the graph 3 Radius, R is calculated using: 0.85V = 𝜋𝑅3 𝑁 4 Where V is lung volume; N is number of alveoli 1

⟹R=(

3. 0,85𝑉 3 4𝜋𝑁

) =(

3. 0,85𝑉

1 3

1

) = 1,106 . 10−3 𝑉 3 6

4𝜋.150.10

2𝜎

∆p.𝑅

Surface tension banlance: ∆p = ⟹𝜎= 𝑅 2 Graph the relationship between surface tension coefficient (in dynes/cm) and alveolar radius (in microns) Get value R and σ from above table

9|Page

7.5. Figure 7.12 shows that a solution containing lung extract exhibits hysteresis in its surface tension versus area relationship. In other words, the surface tension is higher during inflation of the lung than during deflation. A. By recalling that mechanical work can be expressed as ∫pdV, show that the work required to inflate all the alveoli in the lung against the effects of surface tension can be written as Work = ∫ 𝝈𝒅𝑨 where σ is the surface tension coefficient, A is the aggregate surface area of all alveoli in the lung, and the integral is carried out from minimum surface area (start of inspiration) to maximum surface area (end of inspiration). To show this result, you may assume that the pressure outside the alveoli is constant and equal to 0 (gauge). B.Assume that the surface tension versus area curve for the entire lung over one breathing cycle can be approximated by the shape in Fig. 7.27. Using this information, determine how much energy is dissipated in surface tension hysteresis effects during one breathing cycle.

GIẢI: a. The work required to inflate all the alveoli in the lung under the pressure p: W = ∫ 𝑝𝑑𝑉 2𝜎 But under stated assumptions, p = . 𝑅 Surface area of sphere: S = 4𝜋R2 ⟹ dS = 8𝜋RdR 4 Volume of sphere: V = 𝜋R3 ⟹ dV = 4𝜋R2 dR 3 Assuming every single alveolus is a sphere: 2𝜎 2𝜎 W = ∫ 𝑑𝑉 = ∫ 4𝜋𝑅2 𝑑𝑅 = ∫ 𝜎8𝜋𝑅𝑑𝑅 = ∫ 𝜎𝑑𝑆 for a single alveolus 𝑅

𝑅

Summing over all alveoli, W = ∫ 𝜎𝑑𝐴 b. Enengy disipated is the area under curve on T-A graph 10 | P a g e

Δσ = 50 – 10 = 40 dyne/cm ΔA = 77 – 75 = 2 m2 ⟹ energy: E = Δσ.ΔA = 40.2 = 8.105 dyne.cm = 0,08 J

=

0.04

N/m

7.9. A membrane oxygenator is being designed as part of a heart– lung bypass machine. It must be able to transfer 200 ml/min of O2 into blood flowing at 5l/min. Assume the blood enters the oxygenator with an effective O2 concentration of 0.1 ml O2/ml blood. a. With what O2 concentration should the blood leave the oxygenator? You can solve this question easily by thinking about an overall mass balance. b. One design is to make the oxygenator as a “stack” containing many “units”, as shown Fig. 7.30. Each unit consists of a channel filled with flowing blood, an O2-filled channel, and flat membranes separating the channels. The membranes are 10 cm × 10 cm by 5 μm thick, and the height of each blood-containing channel is 1 cm. The O2-containing channels are filled with 100% O2, which is equivalent to a blood concentration of 0.204 mlO2/ml blood. How many membrane units are needed to supply the required oxygen? The value for Deff of O2 in the membranes is measured as 10−6 cm2/s.

GIẢI:

a. Base on the mass balance (volume): 𝐶𝑖𝑛 . 𝑄𝑏𝑙𝑜𝑜𝑑 + 200 𝑚𝑙/𝑚𝑖𝑛 = 𝐶𝑜𝑢𝑡 . 𝑄𝑏𝑙𝑜𝑜𝑑  𝐶𝑜𝑢𝑡 = 𝐶𝑖𝑛 +

200 𝑚𝑙/𝑚𝑖𝑛 𝑄𝑏𝑙𝑜𝑜𝑑

= 0,1

𝑚𝑙 𝑂2 𝑚𝑙 𝑏𝑙𝑜𝑜𝑑

+

200 𝑚𝑙 𝑂2 /𝑚𝑖𝑛 5000 𝑚𝑙 𝑏𝑙𝑜𝑜𝑑/𝑚𝑖𝑛

 𝐶𝑜𝑢𝑡 = 0,14 𝑚𝑙 𝑂2 / 𝑚𝑙 𝑏𝑙𝑜𝑜𝑑 b. A control volume in the blood channel:

11 | P a g e

Given: C: O2 concentration in blood 𝑄𝑏𝑙𝑜𝑜𝑑 : blood flow rate in the channel J: O2 flux CO2 : pure O2 concentration ∆y: membrane thickness W: membrane width (into the page) 2J.∆x.W Mass balance: Q blood C(x) + 2J.∆x.W = Q blood C(x+Δx) ⟹ C(x+Δx) - C(x) = Qblood

By Fick’s law: J = Deff

C02 − C

In the limit of Δx → 0:

dC

C

⟹ - ∫C 0ut

d(C − C02 ) C − C02

in

Δy

dx

= 2.

Deff Qblood

L 2DeffW

= ∫0

.

C02 − C Δy

.W

dx ⟹ − ln

Cout − C02

Qblood Δy

Cin − C02

=

2Deff W Qblood Δy

L

(1)

Given: Cin = 0,1 cm3 O2/cm3 blood CO2 = 0,204 cm3 O2/cm3 blood Cout = 0,14 cm3 O2/cm3 blood (from part a) Deff = 10−6 𝑐𝑚2 /s W = 10 cm; L = 10 cm Δy = 5.10−4 cm With N is the number of membrane units: Qblood =

Q N

=

5000 ) 60

(

N

=

83,3 N

cm3 /s

Using the result from (1)

⟺ - ln

0,14 − 0,204 0,1 − 0,204

=

2.10−6 .10 cm 83,3 .5 .10−4 N

. 10

⟹ N = 101,2 So that we need 102 units to supply the required oxygen.

12 | P a g e

CHƯƠNG 8 8.1. Describe the main events occurring between the arrival of an action potential at a motor neuron end plate and contraction of the corresponding muscle.

GIẢI: 1. The action potential travels down the neuron to the presynaptic axon terminal. 2. Voltage-dependent calcium channels open and Ca2+ ions flow from the extracellular fluid into the presynaptic neuron’s cytosol. 3. The influx of Ca2+ causes neurotransmitter (acetylcholine)-containing vesicles to dock and fuse to the presynaptic neuron’s cell membrane. 4. Vesicle membrane fusion with the nerve cell membrane results in the emptying of the neurotransmitter into the synaptic cleft; this process is called exocytosis. 5. Acetylcholine diffuses into the synaptic cleft and binds to the nicotinic acetylcholine receptors in the motor end-plate. 6. The nicotinic acetylcholine receptors are ligand-gated cation channels, and open when bound to acetylcholine. 7. The receptors open, allowing sodium ions to flow into the muscle’s cytosol 8. The electrochemical gradient across the muscle plasma membrane causes a local depolarization of the motor end-plate. 9. The receptors open, allowing sodium ions to flow into and potassium ions to flow out of the muscle’s cytosol. 10. The electrochemical gradient across the muscle plasma membrane (more sodium moves in than potassium out) causes a local depolarization of the motor end-plate. 13 | P a g e

11. This depolarization initiates an action potential on the muscle fiber cell membrane (sarcolemma) that travels across the surface of the muscle fiber. 12. The action potentials travel from the surface of the muscle cell along the membrane of T tubules that penetrate into the cytosol of the cell. 13. Action potentials along the T tubules cause voltage-dependent calcium release channels in the sarcoplasmic reticulum to open, and release Ca2+ ions from their storage place in the cisternae. 14. Ca2+ ions diffuse through the cytoplasm where they bind to troponin, ultimately allowing myosin to interact with actin in the sarcomere; this sequence of events is called excitation-contraction coupling. 15. As long as ATP and some other nutrients are available, the mechanical events of contraction occur. 16. Meanwhile, back at the neuromuscular junction, acetylcholine has moved off of the acetylcholine receptor and is degraded by the enzyme acetylcholinesterase (into choline and acetate groups), causing termination of the signal. 17. The choline is recycled back into the presynaptic terminal, where it is used to synthesize new acetylcholine molecules. 8.2. Shown in Fig. 8.29 is a cross-sectional view through muscle, showing actin and myosin filaments. Knowing that muscle can generate a maximum force of 20 N/𝑐𝑚2 , determine the maximum force exerted by each myosin filament. Make and state appropriate assumptions.

GIẢI: In a circle d = 250nm, there are 31 myosin filaments. Call F the force generated by one filament then: 31xF 2 = 20 250 π(

2

.10−7 )

=> F = 20.π(125. 10−7 )2 .

1 31

= 3,167.10−10 (N)

14 | P a g e

8.3. A highly idealized version of part of the tension–length relationship for cardiac muscle is graphed in the left portion of Fig. 8.30. This relationship effectively determines the pumping behavior of the left ventricle, as follows. Increased blood volume within the left ventricle causes stretching of the ventricular wall muscle fibers, which, in turn, causes the contraction of the ventricle to be more forceful. In this way, the left ventricular blood ejection pressure, and thus also the ejected blood volume, increase in response to increased presystolic ventricular volume. Using this information, plot (to scale) the left ventricular blood ejection pressure as a function of presystolic ventricular volume. (The ejection pressure is the maximum pressure achieved during the isovolumetric phase of contraction.) For purposes of this question, you may assume that the left ventricle is a thin-walled right circular cylinder with constant wall thickness of 0.7 cm (see right portion of Fig. 8.30). Valves are located in the top of the cylinder, and the bottom and top of the cylinder are passive (i.e., do not participate in active contraction). The internal diameter of the ventricle when maximum muscle tension occurs is 6 cm. You may assume that tetanic tension is developed during the isovolumetric phase.

GIẢI: Tension-length graph can be fit by T = c( L − L0 ), where L0 is 65% of Lmax Max tension occurs at d =6cm ⟹ R=3cm ⟹ Lmax = 2𝜋.3 = 6𝜋 cm L0 = 0,65Lmax = 3,9𝜋 (cm) c=

106

(6−3,9)𝜋

= 1,516.105 (dynes/cm3 )

Pressure-stress relation for a thin- walled cylinder: 2pR = 2Th with h is wall thickness and the muscle tension, is generated in the hoop direction Th h h ⟹ p = = c(L − L0 ) = c. 2𝜋(R − R 0 ) R

R

R

⟹ p = 2𝜋hc (1 − Therefore

R0 R

R0 R

V0

=√

V

V

) with V = 𝜋R2 h or R = √

𝜋h

V

⟹ p = 2𝜋hc [1 − √ 0] V

V0 = 𝜋R20 h = 𝜋(0.65.3)2 . 10 = 119,5 (cm3 ) Vmax = 𝜋R2max h = 𝜋. 32 . 10 = 282,7 (cm3 ) 119,5 ⟹ pmax = 2𝜋.0,7.1,516.105 (1 − √ ) = 2,333.105 282,7

(dynes/cm2 )

15 | P a g e

We get the equation of p in terms of V 119,5

p = 6,668.105 [1 − √

V

] with V has units of mL and p has units of dynes/ cm2

8.4. A certain muscle is known to behave according to the three element model presented in Section 8.2, with an effective dashpot damping coefficient of 𝜂0 = 2.5Ns/m. When stimulated with a single twitch in an isometric experiment, it produces 80% maximal tension after 40 ms. While keeping the same muscle length, the muscle is then put in series with a spring having 𝑘0 of 200 N/m. What tension is measured in a newisometric experiment 20 ms after a twitch? GIẢI: Since the muscle follows the 3 elements model: T T ’ ’ = 1 − e−k0t⁄𝜂0 or e−k0t⁄𝜂0 = 1 − T0 T0 T With 𝜂0 = 2,5Ns/m; = 0,8; t = 0,04s ⟹ ⟹

−k’0 t

T0

= ln(1 −

𝜂0 k ’0 =

−

2,5 0,04

T T0

)

ln(1 − 0,8) =100,59 (N/m)

Putting a spring in series with muscle changes the spring constant, k, of the system k = k ’0 + k 0 = 100,59+ 200 = 300,59 (N/m) where k ’0 : spring constant of the musde from above k 0 : spring constant of the spring ⟹

T T0

= 1- e−kt⁄𝜂0 = 1- e

−300,59 . 0,02 2,5

= 0,91 = 91%

Tension is 91% of the maximum 8.5. A muscle is supported from a fixed point and has a mass M attached to it (Fig. 8.31). Assume that the muscle can be modeled using a three element model, noting that the arrangement of elements is different than used in Section 8.2. Call the muscle length x, and denote the value of x before the muscle begins to contract by x0. At time t = 0, the active

16 | P a g e

component of the muscle begins to contract and produces a constant tension T0 for duration C. This causes the mass to rise, i.e., causes x to decrease with time. a. Treating the muscle as massless, show that x(t) is given by

b. If 𝑇0 =15N, 𝑘0 = 500 N/m, M =1 kg, 𝜂0 =100Ns/m, and C =0.1 s, calculate how far the mass M will have risen at the end of the contraction (i.e., at t = C). GIẢI: a. Select the positive direction downwards ∑ Fy = Mg - T = M Also, T = T0 +𝜂0 dx

d2 x

dt2 dx dt

+k 0 (x − x̅), ̅x is length of unstretched spring

At rest ( = 0, T0 = 0) dt Mg = T = k 0 (x0 − x̅) Mg ⟹ x̅ = x0 k0

When muscle contracts: dx T = T0 + 𝜂0 +k 0 (x − x0 ) + Mg dt

⟹ Mg - T = -[T0 + 𝜂0 d2 x

𝜂0 dx

dx dt

+ k 0 (x − x 0 )] = M

k0

⟹ 2+ + (x − x 0 ) = dt M dt M Initial Conditions: x − x0 = 0 at t = 0 (1) d (x − x0 ) = 0 at t = 0 (2) dt

Solve for (*): x − x0 = − where r1 = −

𝜂0 2M

T0 k0

T0

d2 x dt2

(*)

M

(1 + C1 er1t + C2 er2t )

(1 + √1 − 4

k0 M

𝜂0

𝜂0 2

2M

); r2 = −

(1 − √1 − 4

k0 M 𝜂0 2

) with C1 , C2 are constants

Apply initial conditions: (1) ⟹ 1+ C1 + C2 = 0 (2) ⟹ C1 r1 + Cr2 = 0 −r r ⟹ C1 = 2 , C1 = 1 r2 −r1

⟹ x − x0 = −

T0 k0

r2 −r1 r1 er2 t −r2 er1 t

(1 +

r2 −r1

)

b. Given: T0 = 15N, k 0 = 500N/m, M= 1kg, 𝜂0 = 100Ns/m Put the number in we get: 17 | P a g e

r1 = −

100

r2 = −

100

2.1

2.1

(1 + √1 − 4

500.1

(1 − √1 − 4

500.1

1002

1002

) = -94,72 (s −1 ) ) = -5,28 (s −1 )

at t = C = 0,1s x − x0 = −

15 500

(1 +

(−94,72)e(−5,28)0,1 −(−5,28)e(−94,72)0,1 (−5,28)−(−94,72)

) = -1,126 (cm)

The mass moves up 1,126cm at the end of contraction

18 | P a g e

CHƯƠNG 9 9.1. Typical compressive stress–strain curves for cortical bone and for trabecular bone of two different densities are shown in Fig. 9.36. Calculate the approximate strain energy density to failure in each case. Strain energy density, U, is a measure of the ability of a material to absorb energy up to fracture and is given by: where 𝜀𝑢 is the ultimate strain at failure. What does your result imply about the function of trabecular versus cortical bone and the consequences of loss of trabecular bone density,as occurs in osteoporosis?

GIẢI: su −sy Eanelastic = = 𝜀u −𝜀y

60−35 0.235−0.03

= 122 MPa for trabecular bone with 𝜌 = 0,9 g/cm3

There are several methods could be used to approximate 𝜀u the strain energy density from 𝜀 the graph. Generally U = ∫0 u 𝜎 d𝜀 For cortical bone, based on area of triangle for the elastic region plus area of trapezoid for the plastic region: 𝜀 1 1 Uc = ∫0 u 𝜎 d𝜀 ≈ sy 𝜀y + (sy + su )(𝜀y + 𝜀u ) 1

2

1

2

Uc = . 165.0,01 + (165 + 180)(0,025 − 0,01) = 3,4 MPa = 3,4 J/cm3 2

2

19 | P a g e

Similarly, for trabecular bone, the strain energy can also be approximated by the area under the 𝜎 − 𝜀 curve For trabecular bone with 𝜌 = 0,9g/cm3 1 1 Ut ≈ sy 𝜀y + (sy + su )(𝜀y + 𝜀u ) 2 1

2

2 1

2

1

= . 35.0,03 + (60 + 35)(0,235 − 0,03) = 10,3 MPa = 10,3 J/cm3 2 2 For trabecular bone with 𝜌 = 0,3g/cm3 1 1 Ut ≈ sy 𝜀y + (sy + su )(𝜀y + 𝜀u ) 1

= . 5.0,04 + (5 + 5)(0,23 − 0,04) = 1,05MPa = 1,05 J/cm3 2 2 The strain energy density at failure is much greater for dense trabecular bone than for cortical bone. This implies that the trabecular bone can absorb a significantly greater amount of energy before it fails than can cortical bone. Further more, for a given level of energy absorption, trabecular bone will generate a lower peak force. Thus, thabealar bone acts in a manner similar to packing foam in that it absorbs energy from impacts. Loss of trabecular bone density, as occurs in osteoporosis, reduces the energy that can be absorbed prior to failure. The result is a higher risk of failure. 9.2. Recall that several experimental studies have demonstrated that the yield strain of trabecular bone is relatively constant over a wide range of apparent densities. Demonstrate that this is the case using the density dependency relationships in Section 9.3 and Hooke’s law, which is valid for trabecular bone virtually up to the yield strain. GIẢI: 𝜌 is the relative density 𝐸 is the relative modulus Es and ρs are the modulus and density of the bone tissue E∗ and ρ∗ are the apparent properties of a trabecular specimen σy∗ is the apparent compressive yield strength of the trabecular bone σys is the compressive yield strength of the tissue matrix itself Trabecular bone mechanical properties can be approximated From equation (9.4) and (9.5): E∗ Es

𝜌∗ 2

𝜎y∗

𝜌s

𝜎ys

= C1 ( ) ;

𝜌∗ 2

= C2 ( ) 𝜌s

where C1 , C2 are constants 20 | P a g e

Assuming Hooke’s law is valid for trabecular bone up to yeild strain. 𝜎y∗ = E∗ 𝜀y∗ where 𝜀y∗ is the apparent yield strain. ⟹ 𝜀y∗ =

𝜎y∗ E∗

=

𝜎ys c2 Es c1

Since 𝜎ys and Es are tissue matrix properties, they can be treated as constants. Therefore, 𝜀y∗ = const and independent of apparent density. 9.3. In this question, we will work through the derivation of Equation (9.5), which states that the compressive strength of trabecular bone is proportional to the square of the relative density, assuming that the microstructure of trabecular bone can be represented by a lowdensity, rod-like model (Fig. 9.8A). a. Begin by showing that the ratio of the apparent density to the tissue density is approximately proportional to (𝑡/𝑙)2 . b. When low-density trabecular bone is loaded in compression, failure occurs when the vertical struts buckle (Fig. 9.37). The critical load, 𝐹𝑐𝑟𝑖𝑡 , at which a strut of length l, Young’s modulus 𝐸𝑠 , and second areal moment of inertia, I, buckles is given by Euler’s formula: Derive a proportional relationship between the moment of inertia, I, of a single strut and the dimensions of the strut, assuming it has square cross-section with dimension t, as shown in Fig. 9.8A. In this case we are interested in the moment of inertia of the cross-sectional area; that is with respect to the axis that runs through the middle of a strut perpendicular to its long axis. c. You should now be able to use Euler’s formula (Equation [9.44]) and the relationships you derived in (a) and (b) to derive Equation (9.5).Equation (9.5). To do so, you will need to relate 𝐹𝑐𝑟𝑖𝑡 to the compressive strength at collapse, σ*, and the cross-sectional area of the unit cell, 𝑙2 ; that is, 𝐹𝑐𝑟𝑖𝑡 ∼ σ* 𝑙2 .

GIẢI: a. Mass density, also known as specific weight, the mass m per unit volume V The formula for mass density is ρ =

𝑚 𝑉

21 | P a g e

ρ∗ is apparent density ρ s is the tissue density 𝜌∗ 𝜌s

m⁄ Vcube

=m

⁄Vstruts

=

Vstruts Vcube

∝

t2 l l3

t 2

=() l

1

b. For a beam with rectangular cross-section, the moment of inertia I = bh3 , where b 12 and h are the width and height, respectively. For a square cross-section, b = h = t therefore, I ∝ t4 c. Use Euler’s formula and the relationships you derived in (a) and (b): E I Fcrit ∝ 𝜎 ∗ l2 ∝ s2 l Substituing I∝ t 4 and rearranging gives: 𝜎∗ Es

∝

t4 l4

𝜌∗ 2

∝( ) 𝜌s

Since 𝜎𝑦𝑠 and 𝐸𝑠 are tissue matrix properties, they can be treated as constants From equation (9.4) :

𝐸∗ 𝐸𝑠

𝜌∗ 2

= 𝐶1 ( ) 𝜌𝑠

=>

σ∗ 𝜎𝑦𝑠

𝜌∗ 2

= 𝐶2 ( ) 𝜌𝑠

where 𝐶2 is the ratio of the changed constant relative to 𝐶1 9.4. Recall the example in Section 8.4.2 about the person lifting a weight and the resulting stresses in the bones of the forearm. Assuming there is a 0.01 mm defect in the radius bone 8 cm from the elbow, determine how many times the subject can lift the weight before his/her bone fractures. Does your calculation likely overestimate or underestimate the actual number of lifts? Why?

GIẢI:

22 | P a g e

In this case, the axial stress due to the bending moment, M(x), and compressive internal force, Fx (x), is given by: σxy =

M(x) 𝐹𝑥 (𝑥) y+ (8.17) Iz 𝐴

Attaching the descartes coordinate system to the figure representing the applied force, we have the static equilibrium equation: 𝐹𝑥 (8𝑐𝑚) = −𝐽𝑅𝐹𝑦 + 𝑇𝑥1 + 𝑇𝑥2 (1) From the internal force diagram, the bending moment 𝑀𝑥 (8𝑐𝑚) = 2323𝑁. 𝑐𝑚 𝐹𝑏𝑖𝑐𝑒𝑝𝑠 is understanded by 𝑇1 𝐹𝑏𝑟𝑎𝑐ℎ𝑖𝑎𝑙𝑖𝑠 is understanded by 𝑇2

23 | P a g e

From the figure and table 8.1, 𝑇𝑥1 and 𝑇𝑥2 can be calculated: 𝑇𝑥1 = 𝑇1 . cos ϴ1 = 𝑇𝑥2 = 𝑇2 . cos ϴ2 =

𝐹𝑏𝑖𝑐𝑒𝑝𝑠 . cos ϴ1 = 𝐹𝑏𝑖𝑐𝑒𝑝𝑠 . co𝑡 ϴ1 = 238,2𝑁. 𝑐𝑜𝑡76ᴼ = 59,4𝑁 sin ϴ1

𝐹𝑏𝑖𝑐𝑒𝑝𝑠 . cos ϴ2 = 𝐹𝑏𝑟𝑎𝑐ℎ𝑖𝑎𝑙𝑖𝑠 . co𝑡 ϴ2 = 232,6𝑁. 𝑐𝑜𝑡63ᴼ = 118,5𝑁 sin ϴ2

From the text: A similar force balance in the x direction gives JRFx = 304N (1): Fx (8cm) = −JRFy + Tx1 + Tx2 = −304.3N + 59.4N + 118.5N = −126.4N By using the formula J =

π(D0 4 −Di 4 ) 32

with the ratio of inner to outer diameters

obtain D0 = 1,40cm ⇒ R 0 = 0,7cm

Di D0

=

1 2

we

We therefore approximate the equivalent outer diameter as 1.4 cm and the inner diameter as 0.7cm. Using these values we can compute Iz = 0,177cm4 and A = 1,15cm2 Maximum tensile stress (that is, on the top surface of the bone) σmax (x; y) =

2323Ncm . 0,7cm 126,4N − = 9077,1 N⁄cm2 = 9,08. 107 Pa 4 2 0,177cm 1,15cm

With x=8cm ; y=0.7cm ⟹ The magnitude of the maximum tensile stress 8cm away from the elbow is 9,08. 107 Pa We define a new property called the fracture toughness, or critical stress intensity factor, Kc We also define the stress intensity factor K = σ√πa Therefore, the crack propagation condition can be rewritten as: K ≥ K c From the text, when K max = K c , fast fracture will occur, choose the lower bound of K c value to be the K max 24 | P a g e

K = K max = K c 3

From table 9.5, K c = 2,2 MN⁄m2

K max = σmax √πa = 2,2 MN⁄m3/2 With a is the crack length just before fast fracture ⇒ 9,08. 107 Pa√πa = 2,2 MN⁄m3/2 ⇒ a = 0,187. 10−3 m 𝑑𝑎

If we are told that a crack has an initial length 𝑎0 , we can integrate Equation (9.18) = 𝑑𝑁 𝑚 𝐶(ΔK) to determine the number of cycles required for the crack to grow to some final size 𝑎𝑓 . Denoting this number of cycles by Nf (9.20) : 𝑁𝑓 =

𝑎1−𝑚/2 𝑚 𝐶 (Δ𝜎)𝑚 . 𝜋 2 . (1

𝑎𝑓 ( ) ( ) 𝑚 𝑎0 𝑓𝑜𝑟 𝑚 ≠ 2 ∗∗ − ) 2 |

Given: m=2,5 Assuming there is a 0,01 mm defect in the radius bone 8cm from the elbow ⟹ 𝑎0 = 0,01𝑚𝑚 Δ𝜎 = σmax − 0 = 9,08. 107 Pa = 90,8MPa C = 2,5. 10−6 𝑚(𝑀N⁄m3/2 )−2,5

25 | P a g e

(∗∗) 𝑁𝑓 =

(0,187𝑥. )1−2,5/2 − (0,01. 10−3 )1−2,5/2 ≈ 45 2,5 2,5 −6 2,5 2,5. 10 (90,8) . 𝜋 2 . (1 − ) 2

Conclude, the subject can lift the weight about 45 times before his/her bone fractures. That calculation is likely underestimate the actual number of lifts, becaue we ignore the micro structure of the bone and the osteons and collagen tibres. They all limit crack propagation. Besides, calculated figures are selected at the lowest level compared to the data table. 9.5. Derive the constitutive Equation (9.29) for the standard linear viscoelastic model in Fig. 2.35

GIẢI:

The spring constant and damping coefficients in the model in figure 2.35 are replace with the corresponding material constants, Young’s modul E0 , E in place of the spring constant k 0 , k1 and viscosity 𝜇 in place of the damping coefficient ղ0 . Let 𝜎 = stress applied; 𝜎0 = stress in the upper leg; 𝜎1 = stress in the lower leg ε: strain of the entire model 𝜀0 : strain of the elastic element with Young’s modulus E0 𝜀1 : strain of the elastic element with Young’s modulus E1 𝜀𝜇 : strain of the viscostic element ε = 𝜀1 and ε = 𝜀0 +𝜀𝜇 ⟹ ε̇ = 𝜀0̇ + 𝜀𝜇̇ We know 𝜎0 = E0 𝜀0 ⟹ 𝜎0̇ = E0 𝜀0̇ also 𝜎0 = 𝜇𝜀𝜇̇ ε̇ =

𝜎0̇ E0

+

𝜎0 𝜇

But 𝜎0 = 𝜎 - 𝜎1 = 𝜎 − E1 𝜀1 = 𝜎 − E1 𝜀 1 1 𝜀̇ = (𝜎̇ − E1 𝜀̇) + (𝜎 − E1 𝜀) E0

𝜇

After rearranging: 𝜎 +

𝜇 E0

𝜎̇ = E1 𝜀 + 𝜇(1 +

E1 E0

𝜀̇) 26 | P a g e

9.6. Show mathematically that the phase lag or internal friction for the standard linear viscoelastic model is a maximum when 𝜔 = (𝜏𝜀 𝜏𝜎 )−1⁄2 where 𝜏𝜀 and 𝜏𝜎 are defined in Equations (9.31) and (9.32). GIẢI: The internal friction will be an extreme when the first devivative of tan𝛿 with respect to 𝜔 is equal t zero. The expression for tan𝛿 is 𝜔(𝜏𝜎 −𝜏𝜀 ) tan𝛿 = 2 1+𝜔 𝜏𝜎 𝜏𝜀

Recalling the quotient rule for differentiation, we get d d𝜔

(tan𝛿 ) = = = =

Setting

d d𝜔

(𝜏𝜎 −𝜏𝜀 )(1+𝜔2 𝜏𝜎 𝜏𝜀 )−𝜔(𝜏𝜎 −𝜏𝜀 )(2𝜔𝜏𝜎 𝜏𝜀 )

(1+𝜔2 𝜏𝜎 𝜏𝜀 )2 (𝜏𝜎 −𝜏𝜀 )+𝜔2 (𝜏𝜎 −𝜏𝜀 )(𝜏𝜎 𝜏𝜀 )−2𝜔2 (𝜏𝜎 −𝜏𝜀 )(𝜏𝜎 𝜏𝜀 ) (1+𝜔2 𝜏𝜎 𝜏𝜀 )2 2 (𝜏𝜎 −𝜏𝜀 )−𝜔 (𝜏𝜎 −𝜏𝜀 )(𝜏𝜎 𝜏𝜀 ) (1+𝜔2 𝜏𝜎 𝜏𝜀 )2 (𝜏𝜎 −𝜏𝜀 )(1−𝜔2 𝜏𝜎 𝜏𝜀 ) (1+𝜔2 𝜏𝜎 𝜏𝜀 )2

(tan𝛿 ) = 0, we get that the internal friction is a maximum when

𝜔 = (𝜏𝜎 𝜏𝜀 )−1/2 9.7. Consider the load–extension (force–deformation) curves shown in Fig. 9.38. These data are from tensile tests on relaxed papillary muscles from the ventricle of the rabbit heart. Note that the tissue exhibits hysteresis owing to viscous losses, as expected of a viscoelastic material. Also notice, however, that the amount of hysteresis is essentially insensitive to the applied strain rate over two orders of magnitude. Is this experimental observation consistent with the predictions made by the standard linear model? If not, how can the model be improved to account for insensitivity to loading frequency?

GIẢI: 27 | P a g e

The standard linear model exhibits internal friction (and there fore hysteresis) over a rather narrow range of frequencies, which is inconsistent with the experimental evidence. To improve the model to account for insensitivity of internal damping to frequency, we would need to add additional elements to model, or alternatively add exponential terms to the govering quation(i.e, relaxation or creep response) where the time constants for each exponential term differ. In doing so, the improved model would have an internal damping peak that was “spread out” (see figure below) and not as sensitive to frequency

28 | P a g e

CHƯƠNG 10 10.1 A jumper executes a standing jump from a platform that is moving upwards with constant speed Vp. (a) Derive a formula for the maximum elevation of the jumper’s center of gravity in terms of the crouch depth, c, the equivalent force to weight ratio, Fequiv/W, the platform speed, Vp, and other relevant parameters. The elevation of the center of gravity is to be measured with respect to a stationary frame of reference (i.e., one not attached to the platform). (b) If the crouch depth is 18 inches, the ratio Fequiv/W is 2, and the platform speed is 5 ft/s, compute the elevation of the center of gravity. GIẢI: F

equiv 2 a. From the text vpushoff = 2gc [ − 1], measured with respect to the platform. The W total velocity is just Vp + Vpushoff . The elevation of the centre h is

2 2

(Vp + vpushoff ) h= = 2g

[Vp + √2gc(Fequiv /W − 1)] 2g

Alternative Developing the equation in the text, with vT = vpushoff + vp 𝑉𝑇 (𝑡) = gt[(Fequiv /W) − 1] + VP (1) ⟹t=

Solution:

VT (t) − Vp g[Fequiv /W−1] 1

z(t) − z0 = gt 2 [Fequiv /W − 1] + VP t (2) 2 t=τ At end of push-off: {z(τ) − z = c + V τ 0 p Intergrate (1) gives:

1

Sub it in (2): c + Vp τ = gτ2 [Fequiv /W − 1] + Vp τ 2

1

⟹ c = gτ2 [Fequiv /W − 1] 2

1

= ( 2

2 VT (τ)−Vp Fequiv

g[

W

Fequiv

) g[

−1]

W

− 1] (From (1): τ =

VT (τ)−Vp

)

g[Fequiv /W−1]

VT (τ) = √2gc [

⟹

Fequiv W

− 1] + Vp

2

V2T (τ)

[√2gc(Fequiv /W−1)+Vp]

So that:

h=

b. Given

Fequiv /W = 2

2g

=

2g

(*)

29 | P a g e

c = 1,5 ft vp = 5 ft/s g = 32,2 ft/s2 (√2(

Put all value into (*): h =

2 32,2ft )(1,5ft)(2−1) +5ft/s] s2

2(32,2 ft/s2 )

= 3,41ft

10.2 Your 160 lbm friend agrees to have his standing jump analyzed. Standing on a force plate, he crouches to lower his center of gravity, then executes a jump. The force plate measurement gives a reading that can be described by the equation 𝜋𝑡

F(t) = 480sin( )+ 160(1-t/τ) 𝜏

where F(t) is in lbf. Here the push-off duration is 180 ms. How high will your friend’s center of gravity be elevated at the peak of his jump? GIẢI: Given W = 160 Ibm τ = 180 ms = 180. 10−3 s πt t πt t F = 480 sin ( ) + 160 (1 − ) = 3W sin ( ) + W (1 − ) (Ibf) τ τ τ τ g = 32,2 ft/s2 W dv Newton’s Second Law of motion: F(t) − W = ma = g dt

⟹

dv dt

=

g W

F(t) − g

⟹v(t)

=

At t = τ:

g W

∫ Fdt − gt + const

v (τ ) = = =

(const→0

since

τ πt t [3W ( ) (1 )] dt − sin + W − ∫ W 0 τ τ 1 gτ∫0  [3sin(πβ) + (1 − β)]dβ − gτ 3 1 gτ (− cosπβ| 10 + 1 − − 1) π 2 6 1 g

V

=

0

at

t = 0)

gτ (β = t/τ)

= gτ ( − ) π

2

6

1

= (32,2 ft/s2 )(180. 10−3 s) ( − ) π 2 = 8,17ft/s The height of my friend’s center of gravity be elevated at the peak of his jump is: h=

v(τ)2 2g

=

(8,17t/s)2 2(32,2ft/s2 )

= 1,04ft

10.3 A 75 kg stunt man executes a standing jump with the aid of a harness and support wire. In addition to the constant 1100N force his legs exert during the push-off phase, the wire has a lift mechanism that applies a force given by 550e−s/L (in Newtons), where L is 0.4 m and s is the distance traveled. The lift mechanism is engaged at the bottom of the 30 | P a g e

crouch (where s = 0) and a safety catch detaches the wire when the stuntman leaves the ground. a. For a crouch depth of 0.4 m, compute the maximal elevation of his center of gravity. Hint: it may be easiest to work from first principles, recalling that a = v dv/dz =0.5.d(v2)/dz. b. If the safety catch fails to disengage, show that his center of gravity is elevated 0.462m at the top of the jump. Hint: use an approach similar to part (a) for the airborne phase. GIẢI:

Force balance method: ∑ 𝐹 = 𝑚𝑎 s is measured from the beginning of the push off vp is the push off speed and c is the crouch depth F dv dv  F0 + 0 e-s/L – W = m = m𝑣 2

 F0 (1 +   

1 2

e-s/L

−

𝑣 𝑚 ∫0 𝑝 d(𝑣 2 ) 2 1 𝑚𝑣𝑝2 = 𝐹0 (𝑠 2 1

1 2

𝑊 F0

= −

dt 1

)= 𝑚

d𝑠 d(𝑣 2 )

2 d𝑠 𝑐 1 F0 ∫0 (1 + e-s/L 2 𝑐 𝐿.e-s/L 𝑊.𝑠 2

𝑚𝑣𝑝2 = 𝐹0 𝑐 [1 −

𝑊 F0

−

+

F0

)|

𝑊 F0

) 𝑑𝑠

𝑠−𝑐= 0

𝐿.(1−𝑒 𝐿 ) 2𝑐

−

] −0,4𝑚



1

𝑚𝑣𝑝2 2

= 1100𝑁. 0,4𝑚 [1 −

(75.9,81)𝑁 1100𝑁

+

0,4𝑚.(1−𝑒 0,4𝑚 ) 2.0,4𝑚

] = 284,768 (𝑁. 𝑚)

Apply conservation of energy in airborne phase: 1  mgh = 𝑚𝑣𝑝2 =284,768 (𝑁. 𝑚) h=

2 284,768 (𝑁.𝑚) 𝑚𝑔

=

284,768 (𝑁.𝑚) (75.9,81)𝑁

= 0,387 (𝑚)

b.When the catch fails to disengage, we can apply the same analysis to the airborne phase. Now the start velocity is vp , the final velocity is zero (at top of jump), and distances are

31 | P a g e

measured from beginning of push-off phase. Note that F0 Asumming the positive direction is from bottom to top. Newton’s second Law: ∑F = ma F0 −S/L e −W 2 1 −S/L W

⟹ ⟹

c+h F0 ∫c  ( e 2

−

F0

1

d(v2 )

2

dS 1

= m

does not act.

0

= m∫v2  d(v 2 )

) dS

2

p

1

= − mvp2 2

= −h = − 0.6472F0 c L

W

2

F0

(− e−S/L −

⟹

L

⟹

(e−(c+h)/L − e−c/L ) +

2 1

⟹

S)| c+h c

2

(e−c/L )(e−h/L − 1) +

= −0,6472c W

h

= 0,6472c

F0 Wh

= 0,6472

F0 L

c L

(*)

From part (a), we have W/F0 = 0,6689 and c/L = 1: 1

h

h

2

L

L

(*) ⟺ e−1 (e−h/L − 1) + 0,6689 = 0,6472 ⟹ e−h/L + 3,637 = 4,519 h

Numerical solution is = 1,156 ⟹ h = 0,462 m. L

So that his center of gravity is elevated 0,462m at the top of the jump. 10.4. Consider the standing high jump, but this time the jumper is on the moon, where the local gravitational field is one sixth that on earth: gmoon = g/6. (a) Using an analysis similar to that developed in Section 10.1.1, derive a formula for the height that a person’s center of gravity can be elevated in the standing high jump on the moon. (b) If J. C. Evandt were to repeat his record-breaking jump on the moon, what bar height could he clear? To compute this height, you may use the same data and assumptions as were used in the text GIẢI: a. During pushoff: +↑ ∑ Fhigh jump = m

dv W dv W = = F − . g moon (1) dt g dt g

• Fequiv represents the average force exerted by the legs during the push-off phase 32 | P a g e

• c is distance the center of gravity is lowered, measured with respect to the elevation of the center of gravity at the instant the feet leave the ground. • h is maximum elevation of the center of gravity, measured with respect to the same reference datum. • z is vertical location of the center of gravity, measured positive upwards from the floor. • v is vertical velocity of the center of gravity, measured positive upwards. • The moon, where the local gravitational field is one sixth that on earth: g moon = g/6 Replace F by Fequiv : g( Fequiv

g

Fequiv g moon dv (2) − )= W g dt dz

v = gt ( − moon) = W g dt ⟹{ F 1 g equiv z = gt 2 ( − moon) 2

W

g

eliminate : Vpushoff 2 Fequiv g moon = C( − ) 2g W g During the airbone phase: 1 mVpushoff 2 = mg moon h 2 Vpushoff 2 Fequiv g moon g ⟹h= =C ( − ) 2g moon g moon W g Fequiv g ⟹ h = C( . − 1) W g moon b. From the text, If J. C. Evandt were to repeat his record-breaking jump on the moon and a typical crouch distance for a 6 foot tall person is 20 inches. Using these values, we obtain Fequiv = 2,7 and C=20 inches W ⟹ h = 20 inches (2,7. 6 − 1) = 304 inches We will take the jumper’s height to be 6 feet, so that his center of gravity was 36 inches off the ground at the end of push-off. We will also assume that during the jump he oriented his body so his center of gravity just cleared the bar, bar height would be : 304 inches + 36 inches = 340 inches = 28′4′′ above ground.

33 | P a g e

10.5 A 150 lbm person is able to jump 22 inches (elevation of center of gravity) if they first crouch so as to lower their center of gravity by 15 inches. What average tension T is present in their Achilles tendon during the push-off phase? See Fig. 10.28 for nomenclature.

GIẢI: A 150 lbm person is able to jump 22 inches ⟹ W= 150 lbm and h= 22 inches They first crouch so as to lower their center of gravity by 15 inches ⟹ C= 15 inches From energy balence: Epotential = Ework − Ecrouch mgh = FC − WC ⟹ Wh = (F − W)C F h F ⟹ h = C. ( − 1) ⟹ = − 1 W C W h 22 inches ⟹ F = W ( + 1) = 150 lbm. ( + 1) = 370 lbf C 15 inches Take moments about ankle since weight of body and compression force due to acceleration acts through there 1 ∑ Mankle = T . 2 inches − 370 lbf . 6 inches = 0 ⇒ T = 888 lbf 2 10.6 Penelope Polevaulter, having read this book, realizes that her optimal polevaulting strategy is to run as fast as possible and push off strongly during take off. (a) If her weight isW, her approach speed is Va, and she pushes off with an effective force Fo over an effective crouch distance c, estimate the net elevation of her center of gravity, h. Make and state relevant assumptions. (b) Assume her approach speed is 9 m/s and she is able to push off with an effective force F0 of two times her body weight with an effective crouch distance of 25 cm. If her center of gravity starts 30 cm from the ground, estimate the height above the floor which her center of gravity can clear. 34 | P a g e

GIẢI: a. Assume: - total conservation of forward kinetic energy to elevation - neglects mass of pole - no pushing on pole in air - no addition of forward kinetic energy added by pushoff 𝑣𝑎 is approach speed, W is weight, Fo is an effective force 𝑊𝑣 2

Forward kinetic energy is KE = 𝑎 2𝑔 KE associated with pushoff can be calculated by 𝑊 𝐾𝐸𝑣𝑒𝑟𝑡 = 𝑔ℎ𝑣𝑒𝑟𝑡 𝑔

𝐹

From equation: h = c[ − 1] 𝑊 𝐹0

⟹ 𝐾𝐸𝑣𝑒𝑟𝑡 = 𝑊c[ − 1] where c is an effective crouch distance 𝑊

⟹ Total energy KE = W(

2 𝑣𝑎

This gives a total height H of: By conservation of energy: 𝑣2 𝐹 ⟹ H = 𝑎 + c[ 0 − 1] 2𝑔

𝑔

𝐹

+ c[ 0 − 1])

2𝑔 𝑊

𝑊

𝑔H = W(

2 𝑣𝑎

2𝑔

𝐹

+ c[ 0 − 1]) 𝑊

𝑊

b. 𝑣𝑎 is approach speed, W is weight, Fo is an effective force, c is an effective crouch distance Fo Given: 𝑣𝑎 = 9𝑚/𝑠 , = 2 ,c = 0,25 m Then: H =

2 𝑣𝑎

2𝑔

𝐹0

𝑊

+ c[ − 1] = 𝑊

(9 𝑚/𝑠)2 9,81𝑚 ) 𝑠2

2(

+ (0,25m)[2 − 1] = 4,38 (m)

If her center of gravity starts 30 cm from the floor, the total height of her centre of gravity is: 4,38m + 0,3m = 4,68 m 10.7. Derive an expression to estimate the distance L attainable in the long jump, in terms of the approach velocity V. Neglect air drag and assume that planting the foot at the beginning of the jump does not generate a vertical force but rather produces the optimal angle for take off. Find L for V = 10 m/s. (Note: you will have to determine the optimum angle.) GIẢI:

35 | P a g e

𝑣𝑥 = v(t)Cos𝜃 = vCos𝜃 𝑣𝑦 = v(t)Sin𝜃

v is the velocity, L is the distance attainable in the long jump, 𝜃 is the optimum angle In the direction Oy Derivative of velocity with respect to time d𝑣y = −g ⟹ vy = −gt + 𝐶1 dt At t = 0, vy (0) = v(0)Sin𝜃 = vSin𝜃 ⟹ vy (𝑡) = −gt + vSin𝜃 At maximum height : vy (t) = 0 Time to reach maximum altitude 0 = −gt + vSin𝜃 vSin𝜃 ⟹t = 𝑔 ⟹ Total time in air is vSin𝜃 T=2 𝑔 ⟹ Total distance travelled is vSin𝜃 𝑣2 𝑣2 L =𝑣𝑥 .T = vCos𝜃.T= vCos𝜃. 2 = 2 Cos𝜃Sin𝜃 = Sin2𝜃 For maximum L, set 𝑑𝐿

2𝑣 2

𝑑𝐿 𝑑𝜃

𝑔

𝑔

𝑔

= 0. The measure of the optimal angle is:

= C𝑜𝑠2𝜃 = 0 ⇒ C𝑜𝑠2𝜃 = 0 ⇒ 𝜃 = 45° 𝑔 Distance traveled when speed reaches 10m/s is 𝑑𝜃

L=

𝑣2 𝑔

Sin2𝜃 =

10𝑚 2 ) 𝑠 9,81𝑚 𝑠2

(

Sin(2.45°) = 10,2𝑚

10.8. Elvis Stojko (former Canadian and world figure skating champion) is going to execute a triple axel jump (three full rotations in the air). He spends 0.95 s in the air for this jump, his mass is 70 kg, and his average radius of gyration about the vertical axis while airborne is 18 cm (Fig. 10.29). Assume that he remains vertical throughout the jump (even though the picture shows this is not quite true). Before starting his set-up for the jump, he is skating in a straight line. The set-up lasts 0.35 s and at the end of the set-up his skates leave the ice. What average moment does the ice exert on him during the set-up? Notice that there is no moment acting on him while he is in the air, so his angular acceleration while airborne is zero. GIẢI:

36 | P a g e

In air: For 3 full rotations , ∆𝜃= 6𝜋 , ∆𝑡𝑎𝑖𝑟 = 0,95s ∆𝜃 6𝜋 ⟹ 𝜔𝑎𝑖𝑟 = = = 19,842 rad/s (const) , since 𝛼 =0 rad/𝑠 2 in air ∆𝑡𝑎𝑖𝑟

0,95

Set up: staring 𝜔= 0 rad/s ending 𝜔= 19,842 rad/s ⟹ ∆𝜔 = 19,842 ∆𝑡𝑠𝑒𝑡𝑢𝑝 = 0,35s 𝛼=

∆𝜔 ∆𝑡𝑠𝑒𝑡𝑢𝑝

=

19,842 0,35

= 56,691 rad/𝑠 2

Given 𝑥𝐺 = 18 cm = 18.10−2 m m = 70 kg Using the equation: 𝑘𝐺2 = 𝑥𝐺2 Using the equation: 𝐼𝐺 = m𝑘𝐺2 = 70 .(18. 10−2 )2 = 2,268 kg.𝑚2 ⟹ M = 𝐼𝐺 . 𝛼= 2,268 . 56,691 = 128,575 N.m 10.9 Asubject of mass 65 kg has her gait analyzed. Suppose that the x component of the force measured by a force plate takes the shape shown in Fig. 10.30 (compare with Fig. 10.23). (a) If the forward velocity of the walker is 2 m/s at heel strike, what is it 0.4 s after heel strike? (b) Estimate the corresponding change in height for the subject’s center of gravity (i.e., from heel strike to 0.4 s later). For purposes of this question you should base your analysis on the walking model described in Section 10.2.1 and should make and state suitable simplifying assumptions.

GIẢI: a.Given: m =65kg, 𝑣𝑥1 = 2 m/s , g= 9,8 m/𝑠 2 t = 0,4 s 37 | P a g e

∑ 𝐹𝑥 = 𝑚𝑎𝑥 = m

𝑑𝑣𝑥 𝑑𝑡

𝑡2

⟹ m 𝑣𝑥2 - m 𝑣𝑥1 = ∫𝑡1 𝐹𝑥 𝑑𝑡= 40

40

1 2

. 200.0,4 = −40 𝑁. 𝑠

⟹ 𝑣𝑥2 = 𝑣𝑥1 - = 2 - = 1,3846 m/s 𝑚 65 b. We will neglect kinetic energy assoclated with motion of legs relative to the centre of gravity. An interchange between potential and kinetic energy: 1 2

1

𝑣𝑥1 2 −𝑣𝑥2 2

2

2𝑔

𝑚𝑣𝑥1 2 + 𝑚𝑔ℎ1 = 𝑚𝑣𝑥2 2 + 𝑚𝑔ℎ2 ⟹ ℎ2 - ℎ1 =

= 0,1063 m

10.10. Consider the force plate data from a gait experiment shown in Fig. 10.31. By considering the motion of the total leg, compute the reaction forces at the hip at the instant shown by the heavy vertical line. The total leg includes the thigh, shank and foot (see Table 10.2). For purposes of this question, consider the greater trochanter as the effective center of rotation for the hip joint. Note that you do not need to compute a reaction moment, only forces, and that the vertical scales on the two force traces are not the same. Use the following data: subject’s mass 60 kg and height 1.7 m; acceleration of the center of gravity for the total leg is 𝑎𝑥 = −0.25 m/s2 and 𝑎𝑦 = −0.75 m/s2; length of the total leg segment is 0.530 times height (see Fig. 10.21). GIẢI: Subject’s mass 60kg and height 1,7m ⟹ m=60kg, h=1,7m Acceleration of the center of gravity for the total leg is 𝑎𝑥 = −0,25 𝑚⁄𝑠 2 and 𝑎𝑦 = −0,75 𝑚⁄𝑠 2 Length of the total leg segment is 0.530 times height ⟹ 𝐿𝑙𝑒𝑔 = 1,7.0,53 = 0,901 m From table 10.2, 𝑡𝑜𝑡𝑎𝑙 𝑙𝑒𝑔 𝑤𝑒𝑖𝑔ℎ𝑡 = 0,161 𝑡𝑜𝑡𝑎𝑙 𝑏𝑜𝑑𝑦 𝑤𝑒𝑖𝑔ℎ𝑡 mass of total leg is m = 0,161 . 60kg = 9,66 kg From figure 10.31, 𝐹𝑥 and 𝐹𝑦 can be approximated: −2,5 𝐹𝑥 = . 200𝑁 = −38,5𝑁 13 43 𝐹𝑦 = . 60𝑘𝑔. 9,81 𝑚⁄𝑠 2 = 408𝑁 62 Under these assumptions, we can apply Newton’s second law in the horizontal and vertical directions to obtain: Fx + R x = mt ax (reference formula 10.18) Fx + R y − mt g = mt ay (reference formula 10.19)

38 | P a g e

⃗⃗ ∑ Fx = max ⇒ −38,5N + R x = −0,25 m⁄s2 . 9,66kg + ⟹ R x = 36,1N +↑ ∑ Fx = may ⇒ 408N + R y − 9,66kg. 9,81 m⁄s2 = −0,75 m⁄s2 . 9,66kg ⟹ R y = −320,5N

39 | P a g e