PHY2053: Lecture 12 Elastic Forces - Hooke s Law Work done by elastic forces Elastic Potential Energy Power
Elastic Forces 2
Hooke s Law Discovered by Robert Hooke cca 1660 ceiiinosssttuv cca 1676 Ut tensio, sic vis As the extension, so the force Excellent W = Kdescription f K i = of K many elastic systems Fspring F x = k x x Force is proportional to, and in the opposite direction of, the spring extension (compression) mg 3
Example #1: Real Spring mass [g] x [cm] fitted x value [cm] k = 96.5 ± 3.9 N/m x 0 = 1.0 ± 0.2 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 F G [N] 5 4.5 4 3.5 3 2.5 2 1.5 1 0.02 0.03 0.04 0.05 0.06!x [cm] x0 = 1 cm 4
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 x B) series: xs = 2 x, parallel: xp = ½ x C) series: xs = x, parallel: xp = x D) series: xs = ½ x, parallel: xp = 2 x E) series: xs = ¼ x, parallel: xp = 4 x 5
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 x B) series: xs = ¾ x, parallel: xp = ¼ x C) series: xs = x, parallel: xp = x D) series: xs = ¼ x, parallel: xp = ¾ x E) series: xs = ⅓ x, parallel: xp = 3 x 6
Notes on Springs in Series and Parallel: twice 7
Work done by an F 2,1 F elastic force r F 1,2 Recall the definition of work: W = F r cos(θ) W = K f K i = K Force opposes deformation work is always negative F xi x W F i =0 Important: F x = Force k i x changes with extension (variable force) F i r cos(θ i )=0 Fi W 0 x = i i F i x i = 1 2 kx2 [Convention: x = 0 for relaxed spring] r F i cos(θ i )=0 U = W = 1 k x 2 8
W = Kf Ki i = K Elastic potential energy F x = k x F x = k x Elastic forces are conservative forces It makes sense to W = define F elastic potential energy i x i = 1 0 x = F i x i = 1 2 k 2 i x2 kx2 Same convention as i for gravity: U U elastic W elastic = 1 elastic,0 x = W elastic,0 x = 1 2 k x2 2 k x2 In context of energy conservation: W = K f K i = K (1 W 0 x = i F x = k x (1 F i x i = 1 2 kx2 (1 U elastic,0 x = W elastic,0 x = 1 2 k x2 (1 U elastic,x1 x2 = 1 2 kx2 2 1 2 kx2 1 U elastic,x1 x2 = 1 2 kx2 2 1 2 kx2 1 (2 E i = K i + U i,grav + U i,elastic [+ W NC ]=E f = K f + U f,grav + U f,elastic (2 9
he tive ties on g e y e 1. Newto 1000 500 0 Example #3: Ballista Anterior Tibialis Posterior Tibialis Peroneus Longus A better ballista measure uses of material torsion properties springs is gained to by store elastic potential energy, normalizing reportedly the overall tendon had strength range by crosssectional area. Thus, the results shown in Figure 2 > 500 m 1/2 were divided talent by caliber cross-sectional [1 area talent and the ~ results 32 kg] presented as MegaPascals (MPa) or N/mm2. The What results shown was in Figure the 3 spring indicate the constant similarity k if ballista had a 500m range? between posterior [ x ~ and 2.0 anterior m] tibialis tendons as well as the higher ultimate stress value for the What peroneus was longus. the Note tension that both the in posterior the retracting cable tibialis and at peroneus full extension? longus exhibit strength [made that of is tendons] sufficiently comparable to anterior tibialis tendons. Figure 3. Tendon strength in Newtons per cross-sectional area ry f 150 100 Mpa CL, 50 0 Anterior Tibialis Posterior Tibialis Peroneus Longus PHY2053, A similar Lecture study 12, was Elastic performed Forces, by Work Pearsall, and Potential et al. 17 in Energy; Power 10
W 0 x = i F x = k x (17 U elastic,0 x = W elastic,0 x = 1 2 k x2 U i,grav + U i,elastic F i x i = 1 2 kx2 (18 U elastic,0 x = W elastic,0 x = 1 2 k x2 (19 U elastic,x1 x2 = 1 2 kx2 2 1 2 kx2 1 (20 i = K i + U i,grav + U i,elastic [+ W NC ]=E f = K f + U f,grav + U f,elastic (21 = E t P av = E t U elastic,x1 x2 = 1 2 kx2 2 1 2 kx2 1 = W t = W t Power Recall the unit furlong Related to farming - distance that a team of oxen could plough without 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 10 6 J ] [+ W NC ]=E f = K f + U f,grav + U f = F r cos θ F r cos θ = t t = F r t cos θ = Fv av co = F r t cos θ = Fv av cos θ (22 P = Fvcos θ 11
Example #4: TGV Speed Record The TGV high speed train holds 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] 12
Next Lecture Review before Exam 1 Reminder: Exam 1 is next Wendesday 8:20-10:10 pm