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Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h (- 1). The constants h %3D and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Ax y(x) The column at x has a mass Am = (density volume) y(x) pt Ax. You add all the Am values to get %3! the total mass M. The sum becomes an integral: M = pt y(x) da %3D For the center of mass, you add each column's x Am, and divide by M: pt ry(z) dr M Calculate xc. The only quantity you'll need is e = 2 m.

Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h (- 1). The constants h %3D and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Ax y(x) The column at x has a mass Am = (density volume) y(x) pt Ax. You add all the Am values to get %3! the total mass M. The sum becomes an integral: M = pt y(x) da %3D For the center of mass, you add each column's x Am, and divide by M: pt ry(z) dr M Calculate xc. The only quantity you'll need is e = 2 m.

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rigid body statics

quizzes/ 1097144/take Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h (- 1). . The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Δx y(x) The column at x has a mass Am = (density * volume) = y(x) pt Ax. You add all the Am values to get the total mass M. The sum becomes an integral: M = pt | y(x) da For the center of mass, you add each column's x Am, and divide by M: pt zy(z) de M Calculate xc. The only quantity you'll need is e = 2 m.
Transcribed Image Text:quizzes/ 1097144/take Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h (- 1). . The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Δx y(x) The column at x has a mass Am = (density * volume) = y(x) pt Ax. You add all the Am values to get the total mass M. The sum becomes an integral: M = pt | y(x) da For the center of mass, you add each column's x Am, and divide by M: pt zy(z) de M Calculate xc. The only quantity you'll need is e = 2 m.
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