PHY2053: Lecture 12 - Department of Physics at UF€¦ · PHY2053, Lecture 12, Elastic Forces, Work and Potential Energy; Power Hooke’s Law • Discovered by Robert Hooke • cca

PHY2053: Lecture 12Elastic Forces - Hooke’s LawWork done by elastic forces

Elastic Potential EnergyPower

PHY2053, Lecture 12, Elastic Forces, Work and Potential Energy; Power

Elastic Forces

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Hooke’s Law• Discovered by Robert Hooke• cca 1660 “ceiiinosssttuv” • cca 1676 “Ut tensio, sic vis”

“As the extension, so the force”• Excellent description of many

elastic systems

• Force is proportional to, and in the opposite direction of, the spring extension (compression)

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∆x

mg

Fspring

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

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x [cm]!0.02 0.03 0.04 0.05 0.06

[N]

GF1

1.52

2.53

3.54

4.55

3.9 N/m±k = 96.5 0.2 cm± = 1.0 0x


Example #1: Real Spring

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mass [g] ∆x [cm] fitted ∆x value [cm]

100 2.10 2.01

200 2.90 3.02

300 4.20 4.03

400 5.00 5.04

500 6.10 6.05

K = 97 N/m

x0 = 1 cm


H-ITT: Springs Connected in Series and Parallel

• Two springs can be connected in series or in parallel• All springs have the same spring constant k• If we hang a weight onto the coupled strings, the total

extension of the system will compare to the extension of a single spring (∆x) as follows:

A) series: ∆xs = 4∆x, parallel: ∆xp = ¼∆xB) series: ∆xs = 2∆x, parallel: ∆xp = ½∆xC) series: ∆xs = ∆x, parallel: ∆xp = ∆xD) series: ∆xs = ½∆x, parallel: ∆xp = 2∆xE) series: ∆xs = ¼∆x, parallel: ∆xp = 4∆x

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H-ITT: Springs Connected in Series and Parallel

• Three springs can be connected in series or in parallel• All springs have the same spring constant k• If we hang a weight onto the coupled strings, the total

extension of the system will compare to the extension of a single spring (∆x) as follows:

A) series: ∆xs = 3∆x, parallel: ∆xp = ⅓∆xB) series: ∆xs = ¾∆x, parallel: ∆xp = ¼∆xC) series: ∆xs = ∆x, parallel: ∆xp = ∆xD) series: ∆xs = ¼∆x, parallel: ∆xp = ¾∆xE) series: ∆xs = ⅓∆x, parallel: ∆xp = 3∆x

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Notes on Springs in Series and Parallel:

7twice

��F = m× �a (1)

F = Gm1 m2

r2(2)

�F2,1 (3)

�F1,2 (4)

W = F ∆r cos(θ) (5)

�

i

�Fi = 0 (6)

�

i

Fi∆r cos(θi) = 0 (7)

∆r�

i

Fi cos(θi) = 0 (8)

�

i

Wi = (9)

Ugrav = −Gm1 m2

r(10)

W = Fx ∆x = max ∆x (11)

2 ax ∆x = v2f,x − v2

i,x (12)

ax ∆x =v2

f,x

2−

v2i,x

2(13)

W = max ∆x = m

�v2

f,x

2−

v2i,x

2

�

= mv2

f,x

2−m

v2i,x

2(14)

K = mv2

2(15)

W = Kf −Ki = ∆K (16)

1

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W0→x =�

i

Fi ∆xi = −1

2k x2 (18)

∆Uelastic = −Welastic =1

2k ∆x2 (19)

2


Work done by an elastic force

• Force opposes deformation → work is always negative

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∆rF Recall the definition of work:

W

∆xi

Fi

xF

Important: Force changes with extension (variable force)

[Convention: x = 0 for relaxed spring]


Elastic potential energy• Elastic forces are conservative forces• It makes sense to define “elastic” potential energy• Same convention as for gravity:

• In context of energy conservation:

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W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W =�

i

Fi ∆xi = −1

2k ∆x2 (18)

∆Uelastic = −Welastic =1

2k ∆x2 (19)

2

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W0→x =�

i

Fi ∆xi = −1

2k x2 (18)

∆Uelastic,0→x = −Welastic,0→x =1

2k ∆x2 (19)

∆Uelastic,x1→x2 =1

2k x2

2 −1

2k x2

1 (20)

2

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W0→x =�

i

Fi ∆xi = −1

2k x2 (18)


2k ∆x2 (19)


2k x2

2 −1

2k x2

1 (20)

Ei = Ki + Ui,grav + Ui,elastic [ + WNC ] = Ef = Kf + Uf,grav + Uf,elastic (21)

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Example #3: Ballista

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• A ballista uses torsion springs to store elastic potential energy, reportedly had range > 500 m

• 1/2 talent caliber [1 talent ~ 32 kg]• What was the spring constant k if ballista had a 500m

range? [∆x ~ 2.0 m]• What was the tension in the retracting

cable at full extension? [made of tendons]

A better measure of material properties is gained by normalizing the overall tendon strength by cross-sectional area. Thus, the results shown in Figure 2 were divided by cross-sectional area and the results presented as MegaPascals (MPa) or N/mm2. The results shown in Figure 3 indicate the similarity between posterior and anterior tibialis tendons as well as the higher ultimate stress value for the peroneus longus. Note that both the posterior tibialis and peroneus longus exhibit strength that is sufficiently comparable to anterior tibialis tendons.

Figure 3. Tendon strength in Newtons per cross-sectional area

0

50

100

150

Anterior Tibialis Posterior Tibialis Peroneus Longus

Mpa

A similar study was performed by Pearsall, et al. 17 in which anterior tibialis, posterior tibialis, and peroneus longus tendons were examined. These investigators tested 16 double stranded allografts in each category. As shown in Figure 4 these grafts were all well above the maximal strengths of the native ACL shown in Table 1.

Figure 4. Tendon ultimate tensile strength

0

500

1000

1500

2000

2500

3000

3500

Anterior Tibialis Posterior Tibialis Peroneus Longus

New

tons

(N)

TENDONS FOR ACL GRAFT RECONSTRUCTION-CLINICAL REQUIREMENTS In assessing the requirements of soft tissue grafts used in ACL reconstruction, it is helpful to know the biomechanical requirements and properties of native ligaments. The strength required for normal activities was estimated by Noyes, et al.9 to be 454 N based on the failure strength of the ACL. More detailed analyses were performed by Morrison10-12 regarding the forces that the ACL and PCL (posterior cruciate ligament) are subjected to during activities of daily living. An overview of these data is shown in Figure 1.

By definition, a functional native anterior cruciate ligament would have the requisite strength to carry out these normal activities. The actual strength of isolated cadaveric tendons has been studied extensively. In representative studies presented here, the ultimate tensile strength of the native ACL, defined as the force tissue can tolerate before failure, is reported to range from 658 to 2195 N (Table 1). Note that the lowest ultimate load to failure was 658 N, representative of an extended age group up to 97, which was well above the 454 N load required for ‘daily living’ as reported in the previous paragraph. Thus, these values should be used as a guide for strength of an ACL substitute.

PERONEUS LONGUS AND POSTERIOR TIBIALIS TENDON-BIOMECHANICAL EXPERIENCE

Chowaniec, et al.16 examined the biomechanical properties of 15 anterior tibialis, 15 posterior tibialis, and 13 peroneus longus human allografts, all single stranded. As shown in Figure 2, these grafts exhibited near identical ultimate load to failure and were similar to the strongest native ACL values reported in Table 1. It is also worthwhile to note that these tendons are commonly used in double-stranded configurations.

PERONEUS LONGUS AND POSTERIOR TIBIALIS BIO-IMPLANTS IN KNEE RECONSTRUCTION

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Table 1. Ultimate Tensile Strength of Native ACL in Various Study Groups

Reference Ultimate Tensile Strength (N)

Noyes and Grood13 734 266 to 1730 660Woo, et al. 14 658 129 to 2160 157Rowden, et al. 15 2195 427

0

500

1000

1500

2000

2500

Anterior Tibialis Posterior Tibialis Peroneus LongusNe

wto

ns (N

)

Figure 2. Tendon ultimate tensile strength

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W0→x =�

i

Fi ∆xi = −1

2k x2 (18)


2k ∆x2 (19)


2k x2

2 −1

2k x2

1 (20)


Pav =∆E

∆t=

W

∆t=

F∆r cos θ

∆t= F

∆r

∆tcos θ = F vav cos θ (22)

2

W = Kf −Ki = ∆K (16)

Fx = −k ∆x (17)

W0→x =�

i

Fi ∆xi = −1

2k x2 (18)


2k ∆x2 (19)


2k x2

2 −1

2k x2

1 (20)


Pav =∆E

∆t=

W

∆t=

F∆r cos θ

∆t= F

∆r

∆tcos θ = F vav cos θ (22)

P = F v cos θ (23)

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Power• Recall the unit “furlong”• Related to farming - distance

that a team of oxen could ploughwithout resting

• 4 oxen would plough a furlong in half the time needed for a pair

• Power: how quickly is work getting done?• Unit: 1 Watt = 1 Joule / second [ 1 kWh = 3.6×106 J ]

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Example #4: TGV Speed RecordThe TGV high speed trainholds a speed record of 550 km/h. The traction system can generate 19.6 MW of power.The limiting factor for the train’s top speed is air drag.Compute the force due to air drag at top speed. Compare that to the weight of a killer whale (6 tons)If the air drag force were maintained, what distance would be needed for the train to stop? The mass of the train is 380 tons. [in reality it took 70 km to stop]

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Next LectureReview before Exam 1

Reminder: Exam 1 is next Wendesday8:20 - 10:10 pm

Documents

PHY2053: Lecture 12 - Department of Physics at UF€¦ · PHY2053, Lecture 12, Elastic Forces, Work and Potential Energy; Power Hooke’s Law • Discovered by Robert Hooke • cca