Your company\'s current bracket system, shown in the left schematic, is designed

Question

Your company's current bracket system, shown in the left schematic, is designed to support a 150 lbf tension in cable AB and a 30 lbf tension in cable CD. Upper-level management has demanded that as the project manager, you need to lower costs by reducing the system to a single cable, such as the one shown in the right schematic. Determine what tension the cable EF needs to be to keep the structure intact (ie, equivalent to the original design), and where to place the ends of the cable (i.e, the positions E and F).

Explanation / Answer

With 3 dimension problems I alwyas find it easier to deal with the forces as vectors so tension in cable AB will be: F_AB(-3x^+6y^-2z^) where x^, y^, z^ are the directions that corrispoind to x, y , x. The tension in cable CD will be: F_CD(-3x^+6y^-2z^). F_AB = 150 to determine the forces in each direction we will normalize vector F_AB and find its force componets in the x, y, z direction. F_ABx^ = F_AB*x/||F_AB|| = (150)(-3)/(sqrt(3^2+6^2+2^2) = -450 / 7 = -64.28lbf F_ABy^ = F_AB*y/||F_AB|| = (150)(6)/sqrt(3^2+6^2+2^2) = 900 / 7 = 128.57lbf F_ABz^ = F_AB*z/||F_AB|| = (150)(-2)/sqrt(3^2+6^2+2^2) = -300/7 = -42.86lbf F_CDx^ = F_CD*x/||F_CD|| = (30)(-3)/sqrt(3^2+6^2+2^2) = -90/7 = -12.86lbf F_CDx^ = F_CD*y/||F_CD|| = (30)(6)/sqrt(3^2+6^2+2^2) = 180/7 = 25.7lbf F_CDx^ = F_CD*z/||F_CD|| = (30)(-2)/sqrt(3^2+6^2+2^2) = -60/7 = -8.57 lbf Now we jsut add the vector components. Tension in EF = [(-64.28 -12.86)x^ + (128.57+25.7)y^ + (-42.86-8.57)z^] T_EF = (-77.14x^ + 154.27y^ - 51.43z^) T_EF = sqrt(77.14^2 + 154.27^2 + 51.43^2)*[-77.14x^ + 154.27y^2 - 51.43z^]/||T_EF|| T_EF = 180 lfb [-.428x^ + .857y^ - .286z^] Loccation of E and F: z direction... you have 42.86 lbf pulling to the right at a distance of 8 ft from the x axis and then you ahve 8.57 lbf pulling to the right at a distance of 2ft fromt the x axis. Your final z location will be the weighted average of the two... Z = (42.86)(8) + (8.57)(2) / (42.86+8.57) = 7 Y=(25.7)(0)+(154.27)(0) / (25.7 + 154.27) = 0 X=(64.28)(3) + (12.86)(7) / (64.28+12.86) = 11/3 = 3.6666 The location of F is (3.66,0,7) The location of E...we know the unit vector for EF (-.428x^+.857y^-.286z^) and we also know that the x component will be 0. With this information we can find the lenght of cable EF then use the unit vector to determine the y and z positions: L * (x^) = x L = -.428 / -3.6666 = 8.567 X = 0 Y = 0 + 8.567 * .857 = 7.34 Z = 7 + 8.567 * -.286 = 4.55 Loccation of E is (0,7.34,4.55) I hope this is clear and helpful.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.