Question

003 (part 1 of 2) 10.0 points A carpenter's square of uniform density has the shape of an L, as shown in the figure. Assume: A (x, y) coordinate frame with the origin at the lower left corner of the car- penter's square. The x-axis is horizontal and to the right. The y-axis is vertically upward. Given: In the figure, B = 14 cm, C = 3.5 cm, D = 4.1 cm, E= 18 cm. Because the square is uniform in thickness and has a small thickness, we can assume that the weight of each segment of the square is proportional to its area. What is the x-coordinate of the center of gravity? Answer in units of cm. 004 (part 2 of 2) 10.0 points What is the y-coordinate of the center of grav- ity? Answer in units of cm.

Answer 1

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Here, given Carpentex's square & B = 14 cm, c = 3.5cm 2 ; D = 4.1 cm , 6 =18 cm fos finding the y' condinate of the centre of gravity. Alvea of 'I' part 0 il ni= Dxe = 4.18 18. Coins = 73.8 m2 Area of II pard Az = (3-D) XC = (14 – 4.1)*3.5 = 34.65 cm? Nowi OR cenbe gets (X sya) fes part-I * (0,0) = [2 :$ = {2-05, 9] centre of gravity (22.1992) fw Part I (X,Y) = (B- 14.95 2.75 .. x-coordinate of a Centre of gravity; of whole part R XIA14 A2X2 2.05 473.8 + u.gs x 34.65 C AIXAz 73.8+ 34.65 X = XI Art AzXz2 - - 5x = 2.977 cm NOW g-cardinate of the centre of gravity y y, Alt Yz Az 9x 73.8+ 1.75*34.65 - Ait Az =- 73.8 +94.65 18 = 6.684cm