Class 3 Notes

# Class 3 Notes - PROBLEM 2.46 Ropes AB and AC are thrown to...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 2.46 Ropes AB and AC are thrown to a boater whose canoe had capsized. Knowing that a = 25° and that the magnitude of the force FR exerted by the river on the boater is 70 1b, determine the tension (a) in rope AB, (b) in rope AC. SOLUTION Free-Body Diagram: IA”; 40 l 19 :15" lo" 10° 11¢ 2F; = O: — TABcos25° —- TAC cos40° +(701b)cos10° = 0 (l) 2F), = 0: TABsin25° — TACsin40° + (70 1b)sin10° = 0 (2) Solving Equations (1) and (2) simultaneously: (a) TAB = 38.6 lb 4 (b) TAC = 44.3 lb 4 PROBLEM 2.57 A load of weight 400 N is suspended from a spring and two cords that are attached to blocks of weights 3 W and Was shown. Knowing that the constant of the spring is 800 N/m, determine (a) the value of W, (b) the a unstretched length of the spring. SOLUTION Free-Body Diagram At A: First note from geometry; The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are in the ratio 3:45 The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then: _+- 2p; = 0: ﬂow) +£(W)+13FS = 5 W n 01' Fs = 4.4833W and +12% = 0: gm) + %(W)+§§Fs —400N = 0 Then: %(3W) + ;—:(W) + %(4.4833W) — 400~N ~= 0 01' W = 62.841 N and ﬂ=%HMJ 01' (a) l W = 62.8 N 4 PROBLEM 2.57 CONTINUED (b) Have spring force Fs = k(LAB ‘ L0) Where FAB = kAB(LAB ‘ L0) and LAB = (0.360 m)2 + (1.050 m)2 = 1.110m So: 281.74N = 800N/m(1.110 — lO)m or L0 = 758mm 4 PROBLEM 2.70 h A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 800 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C (a) _+. ZFI = 0: TACB(cos3O° — cos50°) — (800 N)cos50° = 0 Hence T AC3 = 2303.5 N TACB = 2.30 kN 4 (b) +1 ZFy = o; TACB(sin3O° + sin50°) + (800 N)sin50° — Q = 0 (2303.5 N)(sin30° + si1150°) + (800 N)sin50° — Q = 0 or Q = 3529.2 N Q = 3.53 kN 4 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern