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Unformatted text preview: PROBLEM 3.26 The arms AB and BC of a desk lamp lie in a vertical plane that forms an angle of 30° with the xy plane. To reposition the light, a force of
magnitude 8 N is applied at C as shown. Determine the moment of the force about 0 knowing that AB = 450 mm, BC = 325 mm, and line
CD is parallel to the z axis. SOLUTION
N31; 55/1: Have MO : r00 x FC 8 —f where (r00) = (ABE + BCn)c0530° 8c
7 A8,: = (0.450 m)sin45° : 0.31820m 4: RC” = (0.325 m)sin50" = 0.24896 m @ (r(.,0)y = (0A). + AB}, 4 BCy) = 0.150 m + (01450 m)cos45° —(0.325 m)c0550° = 0.25929 m
(rm0): = (AB; + Bsz)sin30°
= (0.31820 m + 0.24896 m)sin30° = 0.28358 m
or rm = (0.49118 m)i + (0.25929 m)j + (0.28358 m)k
(FCL : 7(8N)cos45°sin20° z —1.93476N
(Fay = —(8 N)sin45” = —5.6569 N
(1%): : (8 N)cos45“c0320° = 5.3157 N or FC = —(l.93476 N)i —(5.6569 N)j+(5.3157 N)k i j R
M0 = 0.49118 0.25929 0.28358 Nm
—l.93476 —5.6569 53157
= (2.9825 Nm)i— (3.1596Nm)j — (2.2769 N‘m)k or M0 : (2.98 Nm)i — (3.16N.m)j  (2.28 Nm)k 4 PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from point C to a
line drawn through points A and B. Problem 3.24: A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof at 228 N force directed
along BA. Determine the moment about C of that force. IMC= Fwd d = perpendicular distance from C to line AB.
MC = 1311C x FBA MC 2 (0.96 m)i ~ (0.12 m)j + (0.72 m)k (—(0.1m)i + (1.8 m)j — (0.6)k) F374 = 134F821 = 2 2 2
(0.1) + (1.8} +(0.6) m (228 N) = —(12.0 N)i + (216 NH  (72 N)k 1 j k
MC : 0.90 —0.12 0.72 No:
—12.0 216 —72 = —(l46.88 Nm)i — (60.48 Nm)j + (205.92 Nm)k and (MCI: (146.88)2 + (150.43)2 + (205.92)2 = 260.07 Nm 260.07Nm : (228 N)d
d = 1.14064m
or d =1.141m‘ PROBLEM 3.34 Determine the value of a which minimizes the perpendicular distance
W from point C to a section of pipeline that passes through points A and B. Assuming a force F acts along AB, iMCi = errC x Fi 2 Fid) where
d = perpendicular distance from C to line AB : F(0.57437)i + (0.50257)j — (0.64616)k
WC = (1m)i — (2.8m)j #(a  3m)k
i j k
MC : 1 —2.8 3 — a F
0.57437 0.50257 —0.64616 = [(030154 + 0.50257a)i + (2.3693 — 0.57437a) j + 2.11031413
Since iMCi = 113% x F2] or ‘rm x F2. 2 (dF)2
(0.30154 + 0.5025711)2 + (2.3693 — 0.5743777)Z + (2.1108)2 : d2 Setting 1(d2) = 0 to ﬁnd ato minimize a‘
do 2(0.50257)(0.30154 + 0.5025712) + 2(—0.57437)(2.3693 — 0.574372) = 0 Solving a = 2.0761 m or 0:2.08m4 J ...
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 Spring '08
 Jenkins
 Statics

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