also supported by two cables. The weight of the pole is 40 lb.
Forces F1 and F2 are applied to the pole at point D. F1 is + 120j
lbs. and F2 is -340k llbs.
(a) Draw the free body diagram of the pole, make a list of all
of the coordinates for each point, and write your
assumptions.
(b) Write the full set of equilibrium equations for the pole.
(c) solve for the tension in the cable CG and DF and for the
reaction forces at A. The two cables are separate.
Augmented Matrix for RREF. When you copy your Augmented
Matrix into MatLab, remember to only use the same number of
equations (rows) as you have unknown values.
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AUGMENTED MATRIX FORM AND MAGNITUDES OF ||TCG|| ||TDF|| ||A||
Augmented Matrix for RREF. When you copy your Augmented Matrix into MatLab, b j j Show the results below as MAGNITUDES. Don't forget the units. Description Calculation Units |||TCG|| = |||TDF|| = |||A||=
A An' + Ayj + Azk = E c (0,8,2) D(-2, 0, 7) € (4,0, 10) A (0,0,4) B (0, 4, 3) ится TCE 3 AY. √p(-2-0)i+ (0-8√j+ (7-2) k DO NOT COPY PASTED THIS SHIT!!! √639²76-82²75²2 ULD = ·8·20 741 -0.8 3j+0.52K - FCD = -0.2071 tcp 1-6.83 tcpj+ 0.52 TCDK. YCE = (4-0)i + (0-8)j + (10-2) K =4i-8j+8R 2 UCE +1- & +8k = +1 -1/³+²/3 k биг +(-8)2+82 тсе 2 33 TCE 1 - 3/3 TCEJ + 2/3² TCE K F₁ = (401) 16 F2 = (-30 K) (6 w = (-25k) 16 {F = 0 => F₁+F₂ +w+TCE + TCP+A=0 40i+(-30k) + (25k) + (Anit Agi + Azk) + (-0-207 тебі - 0'83 Терј+0:52 тор к) + (TCEI - 3/3TCEJ + ² TCEK ) = 0 (40+ An-0.201TeD + 1/3 TCE) i +(AY-0.83 TCD - 3/3 TCE) j إلى ^ AZ
Solution Given that. O Weight of poles = 8016. NOR THIS # In vector form is given below A = 1 = Axi + Ay? + Azk c€0,8₁2) D (-2,0,7) E (4,0, 10) A (0,0,0) B(0,4,8) YcD = E2-0⁹ + (0-8)j + (7-2) 15 NCD -29-8j+515 11-4 # Now calculate yer value in vector form VCD -0.2074 -0.83j+0.521 -21-8j+5k √(-2)² + (-8)² +5² FOO = -0.20++ CD 9 - 0.83+CDj +0.52+ CDK = (4-0) +-8)j + (10-2) K = 4° -8j +8k UCE = +-8j +8k 29-2+1 V.4²+(-83²+8² THE = TCE₂1-2/2 TCE 9 +. + TCE K +₁ = (409) 16 f₂ = (-30k) Ib, w = (-225 k) Ib + EF=0,₂ => F₁+fg+w+ TCE + Tep + A = 0 40P+(-30k) + (25k) + (Axi²+ Ays + Azk) + (-0·20-7¹TD³- 0.83 Tepj + 0.52 TCDK) + ( 3 TC³ - TC + TCE k) = YCE =
4 a) A = Ax' + Ay + Azk CEO, 8, 2) D(-2,0,7.) E (4,0,10) AC 0, 0,4) B (0.4,3) Yep (2-0)i + (0-8)j + (7-2) K 5 = -21-85 +5k UcD = -21-8j+5k √(-2)²+(-8)² +5² UCD = -0.20741-0-83j+0.52k UCD -0.0072 Toni A 93+TCDi+ 0.50 #CDK OR THIS! YCE (9-0)1 + (0-8)J + (10-2) K 4i-8j+8k UCG= +4i-8j+k √4²7 (8)²+82 ki - 12/10 K ce=Tai-Taj + / TCEK -3 F₁ = (401) 16 F₂ = (-301) 16 w = (-25k) 16 <F 2 O=> F₁+F₂+w+TCE + TCP +A=0 40i + (-30k) +(25k) + (Axi + Ayj+Azk) + G0.207 TCDi - 0.83 Tepj +0.52 TODK) + ( Teri - Teai + 1/2/32 Teak) = 0