{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW7 - PROBLEM 3.26 The arms AB and BC of a desk lamp lie in...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 (rm-0): = (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 N-m —l.93476 —5.6569 53157 = (2.9825 N-m)i— (3.1596N-m)j — (2.2769 N‘m)k or M0 : (2.98 N-m)i — (3.16N.m)j - (2.28 N-m)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 N-m)i — (60.48 N-m)j + (205.92 N-m)k and (MCI: (146.88)2 + (150.43)2 + (205.92)2 = 260.07 N-m 260.07N-m : (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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online