Applications to Physics and Engineering


 Melvyn Pearson
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1 Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics it hs technicl mening tht depends on the ide of force. Intuitively, you cn think of force s describing push or pull on n object for exmple, horizontl push of book cross tble or the downwrd pull of the Erth s grvity on bll. In generl, if n object moves long stright line with position function s(t), then the force F on the object (in the sme direction) is defined by Newton s Second Lw of Motion s the product of its mss m nd its ccelertion: F = m d s dt () In the SI metric system, the mss is mesured in kilogrms (kg), the displcement in meters (m), the time in seconds (s), nd the force in newtons (N = kg m/s ). Thus force of N cting on mss of kg produces n ccelertion of m/s. In the US Customry system the fundmentl unit is chosen to be the unit of force, which is the pound. In the cse of constnt ccelertion, the force F is lso constnt nd the work done is defined to be the product of the force F nd the distnce d tht the object moves: W = Fd work = force distnce () If F is mesured in newtons nd d in meters, then the unit for W is newtonmeter, which is clled joule (J). If F is mesured in pounds nd d in feet, then the unit for W is footpound (ftlb), which is bout.36 J. EXAMPLE: () How much work is done in lifting.kg book off the floor to put it on desk tht is.7 m high? Use the fct tht the ccelertion due to grvity is g = 9.8 m/s. (b) How much work is done in lifting lb weight 6 ft off the ground?
2 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: () How much work is done in lifting.kg book off the floor to put it on desk tht is.7 m high? Use the fct tht the ccelertion due to grvity is g = 9.8 m/s. (b) How much work is done in lifting lb weight 6 ft off the ground? Solution: () The force exerted is equl nd opposite to tht exerted by grvity, so Eqution gives nd then Eqution gives the work done s F = mg = (.)(9.8) =.76 N W = Fd = (.76)(.7) 8. J (b) Here the force is given s F = lb, so the work done is W = Fd = 6 = ftlb Notice tht in prt (b), unlike prt (), we did not hve to multiply by g becuse we were given the weight (which is force) nd not the mss of the object. Eqution defines work s long s the force is constnt, but wht hppens if the force is vrible? Let s suppose tht the object moves long the xxis in the positive direction, from x = to x = b, nd t ech point x between nd b force f(x) cts on the object, where f is continuous function. We divide the intervl [, b] into n subintervls with end points x, x,...,x n nd equl width x. We choose smple point x i in the ith subintervl [x i, x i ]. Then the force t tht point is f(x i ). If n is lrge, then x is smll, nd since f is continuous, the vlues of f don t chnge very much over the intervl [x i, x i ]. In other words, f is lmost constnt on the intervl nd so the work W i tht is done in moving the prticle from x i to x i is pproximtely given by Eqution : Thus we cn pproximte the totl work by W i f(x i) x W f(x i ) x (3) It seems tht this pproximtion becomes better s we mke n lrger. Therefore, we define the work done in moving the object from to b s the limit of this quntity s n. Since the right side of (3) is Riemnn sum, we recognize its limit s being definite integrl nd so W = lim n f(x i ) x = f(x)dx (4) EXAMPLE: When prticle is locted distnce x feet from the origin, force of x + x pounds cts on it. How much work is done in moving it from x = to x = 3?
3 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: When prticle is locted distnce x feet from the origin, force of x + x pounds cts on it. How much work is done in moving it from x = to x = 3? Solution: We hve The work done is 6 3 ftlb. W = 3 [ x (x 3 + x)dx = 3 + x ] 3 = 5 3 In the next exmple we use lw from physics: Hooke s Lw sttes tht the force required to mintin spring stretched x units beyond its nturl length is proportionl to x: f(x) = kx where k is positive constnt (clled the spring constnt). Hooke s Lw holds provided tht x is not too lrge. EXAMPLE: A force of 4 N is required to hold spring tht hs been stretched from its nturl length of cm to length of 5 cm. How much work is done in stretching the spring from 5 cm to 8 cm? 3
4 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: A force of 4 N is required to hold spring tht hs been stretched from its nturl length of cm to length of 5 cm. How much work is done in stretching the spring from 5 cm to 8 cm? Solution: According to Hooke s Lw, the force required to hold the spring stretched x meters beyond its nturl length is f(x) = kx. When the spring is stretched from cm to 5 cm, the mount stretched is 5 cm =.5 m. This mens tht f(.5) = 4, so.5k = 4 = k = 4.5 = 8 Thus f(x) = 8x nd the work done in stretching the spring from 5 cm to 8 cm is W =.8.5 ].8 8xdx = 8 x = 4 [ (.8) (.5) ] =.56 J.5 EXAMPLE: A lb cble is ft long nd hngs verticlly from the top of tll building. How much work is required to lift the cble to the top of the building? Solution: Here we don t hve formul for the force function, but we cn use n rgument similr to the one tht led to Definition 4. Let s plce the origin t the top of the building nd the xxis pointing downwrd s in the figure bove. We divide the cble into smll prts with length x. If x i is point in the ith such intervl, then ll points in the intervl re lifted by pproximtely the sme mount, nmely x i. The cble weighs pounds per foot, so the weight of the ith prt is x. Thus the work done on the ith prt, in footpounds, is ( x) x i } {{ } }{{} force distnce = x i x We get the totl work done by dding ll these pproximtions nd letting the number of prts become lrge (so x ): W = lim x n i x = xdx = x ] =, ftlb EXAMPLE: A tnk hs the shpe of n inverted circulr cone with height m nd bse rdius 4 m. It is filled with wter to height of 8 m. Find the work required to empty the tnk by pumping ll of the wter to the top of the tnk. (The density of wter is kg/m 3.) 4
5 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: A tnk hs the shpe of n inverted circulr cone with height m nd bse rdius 4 m. It is filled with wter to height of 8 m. Find the work required to empty the tnk by pumping ll of the wter to the top of the tnk. (The density of wter is kg/m 3.) Solution: Let s mesure depths from the top of the tnk by introducing verticl coordinte line. The wter extends from depth of m to depth of m nd so we divide the intervl [, ] into n subintervls with endpoints x, x,...,x n nd choose x i in the ith subintervl. This divides the wter into n lyers. The ith lyer is pproximted by circulr cylinder with rdius r i nd height x. We cn compute r i from similr tringles s follows: r i x i = 4 = r i = 5 ( x i ) Thus n pproximtion to the volume of the ith lyer of wter is nd so its mss is V i πri 4π x = 5 ( x i ) x m i = density volume 4π 5 ( x i) x = 6π ( x i) x The force required to rise this lyer must overcome the force of grvity nd so F i = m i g 9.8 6π ( x i ) x 568π ( x i ) x Ech prticle in the lyer must trvel distnce of pproximtely x i. The work W i done to rise this lyer to the top is pproximtely the product of the force F i nd the distnce x i: W i F i x i 568πx i ( x i ) x To find the totl work done in emptying the entire tnk, we dd the contributions of ech of the n lyers nd then tke the limit s n : W = lim 568πx i ( x n i) x = 568πx ( x) dx = 568π = 568π 48 3 ] ( x x + x 3) dx = 568π [5x x3 + x J 5
6 Section 7.5 Applictions to Physics nd Engineering Hydrosttic Pressure nd Force Deepse divers relize tht wter pressure increses s they dive deeper. This is becuse the weight of the wter bove them increses. In generl, suppose tht thin horizontl plte with re A squre meters is submerged in fluid of density ρ kilogrms per cubic meter t depth d meters below the surfce of the fluid. The fluid directly bove the plte hs volume V = Ad, so its mss is m = ρv = ρad. The force exerted by the fluid on the plte is therefore F = mg = ρgad where g is the ccelertion due to grvity. The pressure P on the plte is defined to be the force per unit re: P = F A = ρgd The SI unit for mesuring pressure is newtons per squre meter, which is clled pscl (bbrevition: N/m = P). Since this is smll unit, the kilopscl (kp) is often used. For instnce, becuse the density of wter is ρ = kg/m 3, the pressure t the bottom of swimming pool m deep is P = ρgd = kg/m m/s m = 9, 6 P = 9.6 kp An importnt principle of fluid pressure is the experimentlly verified fct tht t ny point in liquid the pressure is the sme in ll directions. (A diver feels the sme pressure on nose nd both ers.) Thus, the pressure in ny direction t depth d in fluid with mss density ρ is given by P = ρgd = δd (5) This helps us determine the hydrosttic force ginst verticl plte or wll or dm in fluid. This is not strightforwrd problem becuse the pressure is not constnt but increses s the depth increses. EXAMPLE: A dm hs the shpe of the trpezoid shown in the figure below. The height is m nd the width is 5 m t the top nd 3 m t the bottom. Find the force on the dm due to hydrosttic pressure if the wter level is 4 m from the top of the dm. 6
7 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: A dm hs the shpe of the trpezoid shown in the figure below. The height is m nd the width is 5 m t the top nd 3 m t the bottom. Find the force on the dm due to hydrosttic pressure if the wter level is 4 m from the top of the dm. Solution: We choose verticl xxis with origin t the surfce of the wter: The depth of the wter is 6 m, so we divide the intervl [, 6] into subintervls of equl length with endpoints x i nd we choose x i [x i, x i ]. The ith horizontl strip of the dm is pproximted by rectngle with height x nd width w i, where, from similr tringles, nd so 6 x i If A i is the re of the ith strip, then = = = 6 x i = 8 x i ( w i = (5 + ) = ) x i = 46 x i A i w i x = (46 x i ) x If x is smll, then the pressure P i on the ith strip is lmost constnt nd we cn use Eqution 5 to write P i gx i The hydrosttic force F i cting on the ith strip is the product of the pressure nd the re: F i = P i A i gx i (46 x i ) x Adding these forces nd tking the limit s n, we obtin the totl hydrosttic force on the dm: 6 F = lim gx i (46 x i ) x = gx(46 x)dx n 6 ( = 9.8 ) 46x x dx = N ] 6 [3x x3 3 7
8 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: Find the hydrosttic force on one end of cylindricl drum with rdius 3 ft if the drum is submerged in wter ft deep. Solution: In this exmple it is convenient to choose the xes s in figure bove so tht the origin is plced t the center of the drum. Then the circle hs simple eqution, x + y = 9. We divide the circulr region into horizontl strips of equl width. From the eqution of the circle, we see tht the length of the ith strip is 9 (y i ) nd so its re is The pressure on this strip is pproximtely nd so the force on the strip is pproximtely A i = 9 (y i ) y δd i = 6.5 (7 y i ) δd i A i = 6.5 (7 y i ) 9 (y i ) y The totl force is obtined by dding the forces on ll the strips nd tking the limit: F = lim n 6.5 (7 yi ) 9 (yi ) y 3 = 5 (7 y) 9 y dy = y dy 5 y 9 y dy 3 3 The second integrl is becuse the integrnd is n odd function. The first integrl cn be evluted using the trigonometric substitution y = 3 sin θ, but it s simpler to observe tht it is the re of semicirculr disk with rdius 3. Thus F = y dy = 875 π 3 = 7875π, 37 lb 8
9 Section 7.5 Applictions to Physics nd Engineering Moments nd Centers of Mss Our min objective here is to find the point P on which thin plte of ny given shpe blnces horizontlly. This point is clled the center of mss (or center of grvity) of the plte. We first consider the simpler sitution illustrted in the figure below, where two msses m nd m re ttched to rod of negligible mss on opposite sides of fulcrum nd t distnces d nd d from the fulcrum. The rod will blnce if m d = m d (6) This is n experimentl fct discovered by Archimedes nd clled the Lw of the Lever. Now suppose tht the rod lies long the xxis with m t x nd m t x nd the center of mss t x. If we compre the figure bove nd the figure below, we see tht d = x x nd d = x x nd so Eqution 6 gives m (x x ) = m (x x) = m x + m x = m x + m x = x = m x + m x m + m (7) The numbers m x nd m x re clled the moments of the msses m nd m (with respect to the origin), nd Eqution 7 sys tht the center of mss x is obtined by dding the moments of the msses nd dividing by the totl mss m = m + m. 9
10 Section 7.5 Applictions to Physics nd Engineering In generl, if we hve system of n prticles with msses m, m,...,m n locted t the points x, x,...,x n on the xxis, it cn be shown similrly tht the center of mss of the system is locted t where m = x = m i x i = m i m i x i m i is the totl mss of the system, nd the sum of the individul moments M = m (8) m i x i is clled the moment of the system bout the origin. Then Eqution 8 could be rewritten s mx = M, which sys tht if the totl mss were considered s being concentrted t the center of mss, then its moment would be the sme s the moment of the system. Now we consider system of n prticles with msses m, m,...,m n locted t the points (x, y ), (x, y ),..., (x n, y n ) in the xyplne: By nlogy with the onedimensionl cse, we define the moment of the system bout the yxis to be M y = m i x i (9) nd the moment of the system bout the xxis s M x = m i y i () Then M y mesures the tendency of the system to rotte bout the yxis nd M x mesures the tendency to rotte bout the xxis. As in the onedimensionl cse, the coordintes (x, y) of the center of mss re given in terms of the moments by the formuls where m = x = M y m nd y = M x () m m i is the totl mss. Since mx = M y nd my = M x, the center of mss (x, y) is the point where single prticle of mss m would hve the sme moments s the system. EXAMPLE: Find the moments nd center of mss of the system of objects tht hve msses 3, 4, nd 8 t the points (, ), (, ), nd (3, ).
11 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: Find the moments nd center of mss of the system of objects tht hve msses 3, 4, nd 8 t the points (, ), (, ), nd (3, ). Solution: We use Equtions 9 nd to compute the moments: M y = 3 ( ) = 9 M x = ( ) + 8 = 5 Since m = = 5, we use Equtions to obtin x = M y m = 9 5 nd y = M x m = 5 5 = Thus the center of mss is ( 4 5, ). Next we consider flt plte (clled lmin) with uniform density ρ tht occupies region R of the plne. We wish to locte the center of mss of the plte, which is clled the centroid of R. In doing so we use the following physicl principles: The symmetry principle sys tht if R is symmetric bout line l, then the centroid of R lies on l. (If R is reflected bout l, then R remins the sme so its centroid remins fixed. But the only fixed points lie on l.) Thus the centroid of rectngle is its center. Moments should be defined so tht if the entire mss of region is concentrted t the center of mss, then its moments remin unchnged. Also, the moment of the union of two nonoverlpping regions should be the sum of the moments of the individul regions. Suppose tht the region R is of the type shown in first figure below; tht is, R lies between the lines x = nd x = b, bove the xxis, nd beneth the grph of f, where f is continuous function. We divide the intervl [, b] into n subintervls with endpoints x, x,..., x n, nd equl width x. We choose the smple point x i to be the midpoint x i of the ith subintervl, tht is, x i = (x i + x i )/. This determines the polygonl pproximtion to R shown in the second figure bove. The centroid of the ith pproximting rectngle R i is its center C i ( xi, f(x i) ). Its re is f(x i ) x, so its mss is ρf(x i ) x The moment of R i bout the yxis is the product of its mss nd the distnce from C i, to the yxis, which is x i. Thus M y (R i ) = [ρf(x i ) x]x i = ρx i f(x i ) x
12 Section 7.5 Applictions to Physics nd Engineering Adding these moments, we obtin the moment of the polygonl pproximtion to R, nd then by tking the limit s n we obtin the moment of R itself bout the yxis: M y = lim n ρx i f(x i ) x = ρ xf(x)dx In similr fshion we compute the moment of R i bout the xxis s the product of its mss nd the distnce from C i to the xxis: M x (R i ) = [ρf(x i ) x] f(x i) = ρ [f(x i)] x Agin we dd these moments nd tke the limit to obtin the moment of R bout the xxis: M x = lim n ρ [f(x i)] x = ρ [f(x)] dx Just s for systems of prticles, the center of mss of the plte is defined so tht mx = M y nd my = M x. But the mss of the plte is the product of its density nd its re: nd so x = M y m = ρ m = ρa = ρ ρ xf(x)dx f(x)dx = f(x)dx y = M ρ x m = [f(x)] dx = ρ f(x)dx xf(x)dx f(x)dx [f(x)] dx f(x)dx Notice the cncelltion of the ρ s. The loction of the center of mss is independent of the density. In summry, the center of mss of the plte (or the centroid of R) is locted t the point (x, y), where x = A xf(x)dx y = A [f(x)] dx () EXAMPLE: Find the center of mss of semicirculr plte of rdius r.
13 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: Find the center of mss of semicirculr plte of rdius r. Solution: In order to use () we plce the semicircle s in the figure bove so tht f(x) = r x nd = r, b = r. Here there is no need to use the formul to clculte x becuse, by the symmetry principle, the center of mss must lie on the yxis, so x =. The re of the semicircle is A = πr, so y = A = πr r r r [f(x)] dx = = πr r3 3 = 4r 3π πr. (r x )dx = πr The center of mss is locted t the point (, 4r 3π). r r [r x x3 3 ( r x ) dx EXAMPLE: Find the centroid of the region bounded by the curves y = cosx, y =, x =, nd x = π/. ] r 3
14 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: Find the centroid of the region bounded by the curves y = cosx, y =, x =, nd x = π/. Solution: The re of the region is so Formuls give A = π/ cosxdx = sin x] π/ = nd x = A π/ xf(x)dx = π/ x cosxdx = x sin x] π/ π/ sin xdx = π y = A π/ [f(x)] dx = The centroid is ( π, 8 π). π/ cos xdx = 4 π/ ( + cos x)dx = [x + ] π/ 4 sin x = π 8 If the region R lies between two curves y = f(x) nd y = g(x), where f(x) g(x), then the sme sort of rgument tht led to Formuls cn be used to show tht the centroid of R is (x, y), where x = A x[f(x) g(x)]dx y = A { [f(x)] [g(x)] } dx (3) EXAMPLE: Find the centroid of the region bounded by the line y = x nd the prbol y = x. 4
15 Section 7.5 Applictions to Physics nd Engineering EXAMPLE: Find the centroid of the region bounded by the line y = x nd the prbol y = x. Solution: We tke f(x) = x, g(x) = x, =, nd b = in Formuls 3. First we note tht the re of the region is [ ] x A = (x x )dx = x3 = 3 6 Therefore x = x[f(x) g(x)]dx = [ ] x x(x x )dx = 6 (x x 3 3 )dx = 6 A /6 3 x4 = 4 nd y = A The centroid is (, 5). { [f(x)] [g(x)] } dx = [ ] x /6 (x x 4 3 )dx = 3 3 x5 = 5 5 We end this section by showing surprising connection between centroids nd volumes of revolution. THEOREM OF PAPPUS: Let R be plne region tht lies entirely on one side of line l in the plne. If R is rotted bout l, then the volume of the resulting solid is the product of the re A of R nd the distnce d trveled by the centroid of R. Proof: We give the proof for the specil cse in which the region lies between y = f(x) nd y = g(x) nd the line l is the yxis. Using the method of cylindricl shells we hve V = πx[f(x) g(x)]dx = π x[f(x) g(x)]dx (3) = π(xa) = (πx)a = Ad where d = πx is the distnce trveled by the centroid during one rottion bout the yxis. EXAMPLE: A torus is formed by rotting circle of rdius r bout line in the plne of the circle tht is distnce R (> r) from the center of the circle. Find the volume of the torus. Solution: The circle hs re A = πr. By the symmetry principle, its centroid is its center nd so the distnce trveled by the centroid during rottion is d = πr. Therefore, by the Theorem of Pppus, the volume of the torus is V = Ad = ( πr ) (πr) = π r R 5
m, where m = m 1 + m m n.
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