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A 600-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much work is required to pull up only 25 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. o x7- 100+ @ We divide the cable into small parts with length Ax. If x," is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x,. The cable weighs x, - 6x, Ax (64x) force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax→ 0). lim 8-2, 5x, "Ax W = w -1.100€ 6x dx= tar 1100 W₂ - lim (b) The work required to move the top 25 ft of cable to the top of the building is computed in the same manner as part (a). W₁- - 6x dx= 3x * = 3ײ]25. ft-lb ft-lb Every part of the lower 75 ft of cable moves the same distance, namely 25 ft, so the work done is f( 25 distance force 64x150 dx- J₂5 - (Alternatively, we can observe that the lower 75 ft of cable weighs 75-2450 lb and moves uniformly 25 ft, so the work done is 450 -25- The total work done is W₁ + W₂1875 + 13,125 ft-lb. pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the /th part, in foot-pounds, is

A 600-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much work is required to pull up only 25 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. o x7- 100+ @ We divide the cable into small parts with length Ax. If x," is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x,. The cable weighs x, - 6x, Ax (64x) force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax→ 0). lim 8-2, 5x, "Ax W = w -1.100€ 6x dx= tar 1100 W₂ - lim (b) The work required to move the top 25 ft of cable to the top of the building is computed in the same manner as part (a). W₁- - 6x dx= 3x * = 3ײ]25. ft-lb ft-lb Every part of the lower 75 ft of cable moves the same distance, namely 25 ft, so the work done is f( 25 distance force 64x150 dx- J₂5 - (Alternatively, we can observe that the lower 75 ft of cable weighs 75-2450 lb and moves uniformly 25 ft, so the work done is 450 -25- The total work done is W₁ + W₂1875 + 13,125 ft-lb. pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the /th part, in foot-pounds, is

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A 600-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much work is required to pull up only 25 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. 0- 100+ - We divide the cable into small parts with length Ax. If x; is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x;". The cable weighs 6x;* Ax W = lim 74x2 i=1 (6Ax). xi* force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax → 0). *100 n W₁ = ▬▬▬▬▬▬▬▬▬ 6x;'Ax 6x dx = £Ax W₂ = lim ▬▬▬▬▬▬▬▬ 00 - U (b) The work required to move the top 25 ft of cable to the top of the building is computed in the same manner as part (a). 25 125 1.²56 6x dx = 3x2 i=1 = 25 distance 1100 Every part of the lower 75 ft of cable moves the same distance, namely 25 ft, so the work done is 10 . ft-lb 6Ax force ft-lb *100 150 dx = ft-lb. (Alternatively, we can observe that the lower 75 ft of cable weighs 75 2 = 450 lb and moves uniformly 25 ft, so the work done is 450 25 = The total work done is W₁ + W₂ = 1875 + = 13,125 ft-lb. ft-lb.) pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the ith part, in foot-pounds, is
Transcribed Image Text:A 600-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much work is required to pull up only 25 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. 0- 100+ - We divide the cable into small parts with length Ax. If x; is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x;". The cable weighs 6x;* Ax W = lim 74x2 i=1 (6Ax). xi* force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax → 0). *100 n W₁ = ▬▬▬▬▬▬▬▬▬ 6x;'Ax 6x dx = £Ax W₂ = lim ▬▬▬▬▬▬▬▬ 00 - U (b) The work required to move the top 25 ft of cable to the top of the building is computed in the same manner as part (a). 25 125 1.²56 6x dx = 3x2 i=1 = 25 distance 1100 Every part of the lower 75 ft of cable moves the same distance, namely 25 ft, so the work done is 10 . ft-lb 6Ax force ft-lb *100 150 dx = ft-lb. (Alternatively, we can observe that the lower 75 ft of cable weighs 75 2 = 450 lb and moves uniformly 25 ft, so the work done is 450 25 = The total work done is W₁ + W₂ = 1875 + = 13,125 ft-lb. ft-lb.) pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the ith part, in foot-pounds, is
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