Question

Composite Bodies 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 mı = 300kg and m2 = 225kg, respectively. Component 3 must be treated as a distributed load, which is w = 165 kg/m², determined by the area of contact between the component and the shop floor. The dimensions shown have been measured to be a = 0.650 m. b = 3.00 m, c= 1.50 m d = 0.650 m, e = 4.00 m, and f= 1.50 m. These dimensions represent the r and y components of the locations of the centers of gravity for the respective components. Component 3 has x, y cross-sectional dimensions of 23 = 1.50 m and yz = 0.200 m. Assume the components 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 I - z 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) CHAR O 2 ? * = Value Units Figure < 1 of 1 > 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) CHA ROB? Value Units j =

Answer 1

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given m= Зоока 225 49.5K ma =Sx0x65 T Mi a1 t mn + M3 33 300x O650t 225x CO 650+3)+49 S(0.63+t 30D225+49.5 2123 m YE mitm Y2 t m3da mitm2tmg b) V300x 50 +225 x (0 65+4)+49 sx 0'65 30D 225-t 49 S .66 m m m+ mg) mitmtme 495(2 300x( 25x 3Do 225 49 S O.612m