HW7 - PROBLEM 3.26 The arms AB and BC ofa desk lamp lie in...

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Unformatted text preview: PROBLEM 3.26 The arms AB and BC ofa desk lamp lie in a vertical plane that forms an angle of 300 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. Have M0 : r00 x Ff where (1-0.0) : (A3“ + BCA_,)cos30° X AB, : (0.450 m)sin 45° : 0.31820 m x. SCH = (0.325 m)sin50° = 0.24896 m (1400)“ : (021V + AB]. 7 BCI) = 0.150 m + (0.450 m)cos45° —(0.325 m)cos50° = 025929 In (rm): = (A3” + Bsz)sin30° = (0.3]820 m + 0.24896 m)sin30° = 0.28358 m or rm) = (0.49118 m)i + (0.25929 m)j 4 (0.28358 m)k (FHA a —(3 N)cos45°sin20° = 71.93476 N (FCL = —(8 N)sin45° = —5.6569 N (Ff): = (s N)cos45°coszo° = 5.3157 N or 1+} 2 {1.93476 N)i 4 (5.6569 N)j + (5.3157 N)k i j k M0 : 0.49118 0.25929 0.28358 Nvm 71.93476 75.0509 53157 = (2.9825 N-m)i — (3.1596 N-m)j 4 (2.2769 N .m)k or M0 = (2.98 Nrm)i — (3.16N‘m)j — (2.23N-m)k 4 PROBLEM 3.30 In Problem 324, 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 2: 228 N force directed along BA. Determine the moment about C of that force. 5 Have (MfleBAd where d : perpendicular distance from C to line AB. \q MC = rAfC x Fm rm = (0196 m)i 7 (0.12 m)j + (0.72 ka FHA : lLrAFLr/r = , 2 7 (0.1)‘ + (1.3) + (0.6)‘ rn (228 N} : _(12.0 N)i + (216 N)i - (72”)1‘ i j k ML. 2 0.95 70.12 0.72 N-m ~12.0 216 772 = 4146.38 N‘m)i 7 (60.48 N-m)j + (205.92 N-m)k 7 and (MN: (146.38)2 + (60.48)2 + (20592) = 260.07 N-m 260.07N-m : (228N)d d =1.14064m or d = 1.141111‘ PROBLEM 3.34 5 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. SOLUTION Assuming a force F acts along AB, ‘_ |I\'1('|=lr.-ri("C Fl = Fldl a . ‘ ‘ ‘C where C / rA/c d : perpendicular distance from C to line AB F : 1431: = (8 m)l7+ (7 In); — (97m)kF (8)" + (7)‘ + (50‘ m : F(0.57437)i + (0.50257)j 7 (0.64616}k r435 = (1m)i —(2.8m)j — (a — 3 m)k i j k 2114f: 1 —2.8 37a F 0.57437 0.50257 —0.64616 2 [(030154 + 0.50257a)i + (2.3693 — 0,57437a)j +2.1108k]F Since lMCl : ‘rlflc x F2 or |rMn x F2! : (de (0.30154 + 0.50257a)2 + (2.3693 — 0.5mm)2 + (2.1108)2 = d2 Setting i(0'3) = 0 to find a to minimize d da 2(0.50257)(0.30154 + 0.50257“) + 2(70.57437)(2.3693 7 0.57437a) : 0 Solving a : 2.0761m or a:2.08m4 ...
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HW7 - PROBLEM 3.26 The arms AB and BC ofa desk lamp lie in...

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