Question

Problem 2 F 490 mm E A D Z L 240 mm х L = 426 mm h = 186 mm Tension at DF = 103 N Determine the moment of the tension of the rope applied at D about the axis AE. Answer:

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Answer 1

МЕ solution y Here coordinate of cell the points are shown Dhan F (0,490, o) mm in ig. MAE TDF = 103 N. = 10,oo) (426, 186, 0 mm E D A (0, 0, 240) Z D (426, 186, 240 mm (426,0,240) mm → Now For moment about AE MAE MA A AE .: MAE TAE (EDIA XF) umo

Now DF F 10 For E = Foore value * direction E = = 103 (DF) unit vector so o X (- Byg it tary-sevg (of - x)" + (9E-9) 9+ (2, - Zo) i (0 - 426) † + + (490-186) 9 + Fotoli (0-240) û 3049 - 240 + 304 3- - 4261 + 304 J DE IDEI J64265+ (304)²+(-2400 DF mit vector +0.588 ) - 0.416 k so E (103)(-0.7399 (+0.5289 -0.416 ÅR 11 426 it 1 2 mit vector in DE 240K -0.7399 F = -76.8097 1 54.384 U 42.848

160 Now ro-ra 2 DIA =(426, 156, 840) -(0,0,240) rola 426 ; 186 J + ok mm 1 EDIA so 0.426 o + 0. 186 j + A AE unit vector in AĚ TAE IAE AE (426, 186,) - 10,0,240 ) 11 +0. 186 j 0.24 km . AE 0.426 ; TAE 0, 426 + 0.1869 + (-0,247 ² J 10,4265² 10,426 )² + (0,186)² + (0.2472 0.8143 ît ŷ - 0.4558 TAE AE 0.3555 using equation ①, ② and €

MAE 11 TAE CRIA TAEZ 71 DAEY XDIA YDIA EDIA 0.8143 0.3555 -0,45581 10.486 0.186 - 76.8097 54.384 -42.848 Fx Fy F2 MAE (0.8443) (0.1961 - 42,848)) - (0.3555) (0.426)(-42.548) + (-0,4558) [CO. 4R6) ( 54.384) – (0,186) (-76.209) MAE - 30.00 636 + 6.48902 - 17.132 MAE 40.650 Nom. - 40.650 Nom um so the moment of the rope applied at Dis temsion of