HW10 - PROBLEM 3.104 Three hikers are shown crossing a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 3.104 Three hikers are shown crossing a footbridge. Knowing that the weights of the hikers at points C, D. and E are 800 N, 700 N, and 540 N, respectively, determine (a) the horizontal distance from A to the line of action of the resultant of the three weights when a = 1.1 m, (b) the value of a so that the loads on the bridge supports at A and B are equal. SOLUTION Ema"? l“) i l ! Have 2F: — WD — WE : R :éf'c: W I"— W I R z —800 N — 700 N — 540 N R = 2040 N or R = 2040N ,1 Have 2M4: 7(800 N)(1.5 m)— (700 N)(2.6 m)— (540 N )(4.25 m) 5 W “l = —R(d) —5315 Nrm = —(2040 N)a‘ and d 2 2.6054 m or d = 2.61 mto the right ofA C (b) For equal reaction forces atA and B, the resultant, R, must act at the center of the span. (b) L F" :3.“ 741R From EMA : 4%?) | 1, .5 —(800 N)(1.5 m)7(700N)(1.5m+a)7(540 N)(1.5m + 2.5a} ‘____ _ V. , = 7(2040 N)(3 m) 3060 + 2050a : 6120 and a = 1.49268 m or a : l.493m{ J PROBLEM 3.116 Two BOO-mm-diameter pulleys are mounted on line shaft AD. The belts B and C lie in venical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A. SOLUTION Equivalent force-couple at each pulley Pulley 3 R3 2 (290 N)(—60320°j + sin20°k) — 430 Nj = —(702.51N)j+ (99.186 N)k Mn 2 44301514290 N)(0.15m)i = —(21 N‘m)i Pulley C RC : (310 N + 430 N)(7sin10"j 4 coslo°k) : —(137.132 N)j — (778.00 N)k "7593mm MC = [480N73ION)(0.15m)i = (25.5 N-m)i Then R = R8 + RC. 2 —(839.69 N)j7(678.81N)k or R = 7(840 N)j— (679 N)k 4 MA = ME ‘l' My + ram x R5 + ’04 X Ry i j k =—(2lN-m)l+(25.5N-m)l+ 0.45 0 0 Nm 0 —702.51 99.186 i j k +0.90 0 O N~m 0 —l37. l 82 5778.00 : (4.5 Nlmli + (655.57 N-m]j — (439.59 N-m)k or M[, = (4.50 N-m)i + (656N-m)j — (440 N-m)k 1 PROBLEM 3.118 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R = (3.9 1b)i + Ryj 7 (1.1 lb)k and the couple Mf‘; : M75 + (1.5 lb-fl)j —(1.1 lb-fi)k,. (b) Find the corresponding values of R). and MI. [ SOLUTION Have EF: B + C : R 2sz BVY+CY:3.91b 01' B) =3.9167CX (1) 217.: C. = R. (2) 25.; C: :71111) (3) Have EMA: rm x B + rm x C + M5 : Mfi i j k i j 11 $1: 0 4.5 +é 4 0 2.0 +(21b-fi)i = M_Ti+(l.51b-ft)j—(1.llb-fi)k BI 0 0 CI C} 71.1 (2 — 0.16666'1C/r)i+(0.37513Y + 0.166667C‘ + 0.36667)j + {0.33333C} )k = Mri +(1.s)j -(1.1)k From i-eoefficient 2 7 0.166667C) : M) (4) j- coefficient 0.3753x + 0.1666676} + 0.36667 = 1.5 (5) k — coefficient 0.33333q, : 71.1 or C1. = —3.3 lb (6) (a) From Equations (1) and (5): 0.375(39 7 C_() + 0.!66667C\ =1.13333 C( = 032917 = 1.5800011: ' 0.20833 From Equation (1): Bl = 3.9 — 1.58000 : 2.321b B :(2.321b)i4 C : (l.5801b)i— (3.301b)j —(1.llb)k 4 (b) From Equation (2): R" = C7, = —3.30 lb or R“ : —(3r301b)j1 From Equation (4): M‘r = —0.166667[73.30) + 2.0 = 2.55001b-f1 or M). :(2.551b-fi)i4 ...
View Full Document

This note was uploaded on 04/13/2009 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

Page1 / 3

HW10 - PROBLEM 3.104 Three hikers are shown crossing a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online