Question

Part A - The x Component of the Center of Gravity of All the Components Learning Goal: To determine the center of gravity of a composite body using the principle of superposition. The layout shown is a representation of a machine shop. Components 1 and 2 have masses of mi = 270kg and m2 = 205kg, respectively. Component 3 must be treated as a distributed load, which is w = 170 kg/m”, determined by the area of contact between the component and the shop floor. The dimensions shown have been measured to be a=0.750 m, b= 3.00 m, c=1.50 m, d=0.750 m. e= 3.80 m, and f= 1.10 m. These dimensions represent the x and y components of the locations of the centers of gravity for the respective components. Component 3 has z, y cross-sectional dimensions of I3 = 1.50 m and y3 = 0.200 m. Assun experience uniform weight distribution in all principle directions. The dimensions a through f locate the centroid of the respective component from the y - z plane and 2 - plane.(Figure 1) Determine the component of the center of gravity of all the components. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) Value Units Submit Part B - The y Component of the Center of Gravity of All the Components Determine the y component of the center of gravity of all the components. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) Figure < 1 of 1 > j = y = Value Units Submit Part C - The z Component of the Center of Gravity of All the Components If the heights of the components are given as 21 = 1.95 m, 2 = 0.750 m, and z3 = 0.150 m, determine the 2 component of the center of gravity of all the components. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) CH HÅ* O O ?

Answer 1

ANSWER:
Given that ті, - 270 kg т), - 205 kg m3- 170 х А. А- 1x0-20 m3 - [ЧоYIxo-26 34, - Ца. В kg. а: 0.15от b: 300 mm С. 150 т. dѕ о 15от e-38о 10 Р: I-Iom Ма= 150 3 - 0 2.00т- Parta: Хcy) - m, y, en), 15-- 13 Ew twt'u (2tg+)л?» (9 ғо) Хwtoxu : Ewtowt'uo (sl+sts1.0)xh6+(&tsto) S0%+sbox L, - 10+25+34 - - Әохо -154 205x315+34x525 50°.

= 202.57768754178:5 509. 1149.75 509 | xcm - 225ml Poslb Ycm = mig, tmayz / m3Y₃ mitmz+m3 = m,ft maxldte)4m3d mitmetmy. : 270X1.10+205x(0.7573-80)+34x0.75 210+205734 : 270xl-10 4205x4.55+25.5 509 297+932.15-25-5 509. 125525 509 Yom - 2.46 m Partei q suurt 7 e out in 'n mamy tmg 27011957 205(0,25)+31(045) 270+205+34

- 270X0975+20580375 +34X0075 270+205434 - 26325+76875+255 509 342675 509 Zom: 0.64m