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able weighing 25 newtor per meter length is suspended hall pulley at B. Calculate the mass m of the attached cylinder which so determine the horizontal distance from A to C. Because of the sm proximation of a parabolic cable. int: point B gives 21 = bint A gives 9 = WIB 2Fn¹ w(100-xB)² 2.F.

able weighing 25 newtor per meter length is suspended hall pulley at B. Calculate the mass m of the attached cylinder which so determine the horizontal distance from A to C. Because of the sm proximation of a parabolic cable. int: point B gives 21 = bint A gives 9 = WIB 2Fn¹ w(100-xB)² 2.F.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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8. A cable weighing 25 newtons per meter of length is suspended from point A and passes over the small pulley at B. Calculate the mass m of the attached cylinder which will produce the sag of 9 m. Also determine the horizontal distance from A to C. Because of the small sag-to-span ratio, use the approximation of a parabolic cable. wx 2Fn Hint: point B gives 21 = point A gives 9 = w(100-xB)² 2Fn Eliminate Fn to obtain x - 350xB+ 17500 = 0 Solving for B, Use the value x = 60.4m the other is not valid for this system, to solve for Fn using the first two equations. Since tension at B is equal to weight, mg. Use equation T² = (mg)² = F2+(wxB)² to solve for mass, m. Answer: m=270 kg. 2Fn Use the equation y = Fat C, to solve for xc. Take note that AO(fromthelowesttosupportA) = 39.6m, finally AC = AO + xc = 79.12m Answer: AC = 2xc = 79.12m since AO = xc A -100 m 12 m C K 9m OB m
Transcribed Image Text:8. A cable weighing 25 newtons per meter of length is suspended from point A and passes over the small pulley at B. Calculate the mass m of the attached cylinder which will produce the sag of 9 m. Also determine the horizontal distance from A to C. Because of the small sag-to-span ratio, use the approximation of a parabolic cable. wx 2Fn Hint: point B gives 21 = point A gives 9 = w(100-xB)² 2Fn Eliminate Fn to obtain x - 350xB+ 17500 = 0 Solving for B, Use the value x = 60.4m the other is not valid for this system, to solve for Fn using the first two equations. Since tension at B is equal to weight, mg. Use equation T² = (mg)² = F2+(wxB)² to solve for mass, m. Answer: m=270 kg. 2Fn Use the equation y = Fat C, to solve for xc. Take note that AO(fromthelowesttosupportA) = 39.6m, finally AC = AO + xc = 79.12m Answer: AC = 2xc = 79.12m since AO = xc A -100 m 12 m C K 9m OB m
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