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

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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 ...
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