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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 15 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. 100- Ar e

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 15 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. 100- Ar e

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter7: Triangles
Section7.1: The Law Of Sines
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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 15 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. 100+ X 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 ith part, in foot-pounds, is (6Ax). X;* = 6x;* Ax force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax→ 0). = W = lim n→∞ Σ6x₁*AX i=1 100 W₁ = 6x dx = W₂= = lim n→∞ (b) The work required to move the top 15 ft of cable to the top of the building is computed in the same manner as part (a). 15 - 6'³e Jo 6x dx = 3x2 3x2715 0 n i=1 X 15 distance 100 Jo Every part of the lower 85 ft of cable moves the same distance, namely 15 ft, so the work done is = 1926000 X ft-lb 6Ax force = X 100 ft-lb 90 dx = X ft-lb. (Alternatively, we can observe that the lower 85 ft of cable weighs 85 2 = 510 lb and moves uniformly 15 ft, so the work done is 510 15 = 7650 The total work done is W₁₂₁ + W₂ = 675 + 7650 = 8,325 ft-lb. ft-lb.) X pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the
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 15 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. 100+ X 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 ith part, in foot-pounds, is (6Ax). X;* = 6x;* Ax force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax→ 0). = W = lim n→∞ Σ6x₁*AX i=1 100 W₁ = 6x dx = W₂= = lim n→∞ (b) The work required to move the top 15 ft of cable to the top of the building is computed in the same manner as part (a). 15 - 6'³e Jo 6x dx = 3x2 3x2715 0 n i=1 X 15 distance 100 Jo Every part of the lower 85 ft of cable moves the same distance, namely 15 ft, so the work done is = 1926000 X ft-lb 6Ax force = X 100 ft-lb 90 dx = X ft-lb. (Alternatively, we can observe that the lower 85 ft of cable weighs 85 2 = 510 lb and moves uniformly 15 ft, so the work done is 510 15 = 7650 The total work done is W₁₂₁ + W₂ = 675 + 7650 = 8,325 ft-lb. ft-lb.) X pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the
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