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Unformatted text preview: PROBLEM 3.26 The aims 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 {ah—A8" 5% M0 = 1'00 X FC
B (rook = (AB,z + BCn)cos30°
Asz = (0.450 m)sin45° = 0.31820 111
BC” = (0.325 m)sin50° = 0.24896 m
(rC,0)y = (OAy + ABy — BCy) = 0.150 m + (0.450 m)cos45°
—(0.325 m)cos50° = 0.25929 m
(rC,0)Z = (A3” + BCu)sin30°
= (0.31820 m + 0.24896 m)sin30° = 0.28358 m r00 = (0.49118 m)i + (0.25929 m) j + (0.28358 m)k (FC x = —(8 N)cos45°sin20° = —1.93476 N (Fc)y —(8 N)sin45° = —5.6569 N
(Fe). = (8 N)cos45°cos20° =.5.3157 N
FC = —(1.93476 N)i — (5.6569 N) j + (5.3157 N)k
i j k
.. M0 = 0.49118 0.25929 0.28358 Nm
—1.93476 —5.6569 5.3157
= (2.9825 Nm)i — (3.1596 Nm)j — (2.2769 N»m)k 01' M0 = (2.98 N.m)i — (3.16 Nm)j — (2.28 Nm)k 4 SOLUTION Have where \/ PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from point A to
portion DE of cable DEF. Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 1b. Determine the moment
aboutA of the force exerted by (a) the cable at D, (b) the cable at C. MA = TDEd d = perpendicular distance from A to line DE.
MA = rE/A X TDE
rm = (92 in.) j (24 in.)i + (132 in.) j — (120 in.)k
J(24)2 + (132)2 + (120)2 in. (360 lb) TDE 9"DE T DE = (48 1b)i + (264 lb) j — (24o lb)k i j k
. MA: 0 92 o Nm
43 264 —240 = —(22,0801bin.)i —(4416lb.in.)k and PROBLEM 3.31 CONTINUED MA] = 1[(22,080)2 + (4416)2 = 22,5171bin. 22,517 lb~in. = (3601b)d d = 62.548 in. or d=5.21ﬁ< PROBLEM 3.34 Determine the value of a which minimizes the perpendicular distance
\r from point C to a section of pipeline that passes through points A and Bi Assuming a force F acts along AB,
MC = (rm x F] = F(d) where d = perpendicular distance from C to line AB (8m)i + (7m)j —(9 m)kF
(3)2 + (7)2 + (9)2 m F=kABF= = F(0.57437)i + (0.50257) j — (0.64616)k
rm = (1 m)i — (2.8 m)j —(a —3m)k
i j k
MC = 1 —2.8 3—a F
0.57437 0.50257 —0i64616 = [(0.30154 + 0.50257a)i + (2.3693 — 0.57437a) j + 2.1108k]F
Since MC = ‘rmc x le or er/C >< le = (dF)2 (0.30154 + 0.50257a)2 + (2.3693 — 0.57437a)2 + (2.1108)2 = d2 Setting 1M) = 0 to ﬁnd a to minimize d
da 2(0.50257)(0.30154 + 0.50257a) + 2(—0.57437)(2.3693 — 0.57437a) = 0 Solving a = 2.0761 In or a = 2.08m4 ________—_—____—_ ...
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 Spring '08
 Jenkins

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