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# oldhw 20 - alexander(jra2623 oldhomework 20 Turner(92510 1...

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alexander (jra2623) – oldhomework 20 – Turner – (92510) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (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 = 12 cm, C = 3 . 7 cm, D = 5 cm, E = 20 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. D E B C What is the x -coordinate of the center of gravity? Correct answer: 3 . 73431 cm. Explanation: Basic Concepts: Center of mass: vectorr c = i m i vectorr i i m i . Solution: Take the origin to be at the outside corner of the L . Divide the carpenter’s square into two rectangles (simple shapes) and find the area and the center of mass of each piece. Each piece will behave as if it were a point particle with mass equal to the mass of the piece and located at the piece’s center of mass. Thus, the center of mass of the whole square will be vectorr c = m top vectorr c top + m bot vectorr c bot m top + m bot , where m top and m bot are the masses of the two pieces and r c top and r c bot are the centers of mass of the two pieces. The mass of each piece is proportional to its area, so if we let σ represent the uniform mass per area of the pieces, m top = σ A top and m bot = σ A bot . For objects where the mass is uniformly distributed, ( i.e. , the density is uniform for thick objects or the mass per area is uni- form for thin objects) the center of mass is at the same position as the object’s geomet- ric center. This means that the ( x cm , y cm ) coordinates of the left/top (5 cm × 20 cm) rectangle are parenleftbigg D 2 , E 2 parenrightbigg , or (2 . 5 cm , 10 cm), and the ( x cm , y cm ) coordinates of the bottom (3 . 7 cm × 7 cm) rectangle are parenleftbigg B + D 2 , C 2 parenrightbigg , or (8 . 5 cm , 1 . 85 cm), since B - D 2 + D = B + D 2 . Thus, A top = D × E = (5 cm) × (20 cm) = 100 cm 2 A bottom = C × W = (3 . 7 cm) × (7 cm) = 25 . 9 cm 2 x cm = σ bracketleftbigg A top D 2 + A bottom B + D 2 bracketrightbigg σ ( A top + A bottom ) = (100 cm 2 ) (2 . 5 cm) (100 cm 2 ) + (25 . 9 cm 2 ) + (25 . 9 cm 2 ) (8 . 5 cm) (100 cm 2 ) + (25 . 9 cm 2 ) = 470 . 15 cm 3 125 . 9 cm 2 = 3 . 73431 cm . 002 (part 2 of 2) 10.0 points What is the y -coordinate of the center of grav- ity? Correct answer: 8 . 32339 cm. Explanation:

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alexander (jra2623) – oldhomework 20 – Turner – (92510) 2 y cm = σ bracketleftbigg A top E 2 + A bottom C 2 bracketrightbigg σ ( A top + A bottom ) = (100 cm 2 ) (10 cm) (125 . 9 cm 2 ) + (25 . 9 cm 2 ) (1 . 85 cm) (125 . 9 cm 2 ) = (1047 . 91 cm 3 ) (125 . 9 cm 2 ) = 8 . 32339 cm .
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oldhw 20 - alexander(jra2623 oldhomework 20 Turner(92510 1...

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