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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 BOOmmdiameter 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 forcecouple system at A. SOLUTION
Equivalent forcecouple 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 Nm)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
=—(2lNm)l+(25.5Nm)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 Nm]j — (439.59 Nm)k or M[, = (4.50 Nm)i + (656Nm)j — (440 Nm)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 forcecouple 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 lbﬂ)j —(1.1 lbﬁ)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 +(21bﬁ)i = M_Ti+(l.51bft)j—(1.llbﬁ)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 ieoefﬁcient 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 = 2.55001bf1 or M). :(2.551bﬁ)i4 ...
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 Spring '08
 Jenkins
 Statics, Force, Pulley, $1, 2.6 m

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