121616949-math.215 - the denominator has a very familiar...

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9.6 Center of Mass 201 If we set this to zero and solve for ¯ x we get ¯ x = 6. In general, if we divide the beam into n portions, the mass of weight number i will be m i = (1 + x i )( x i +1 - x i ) = (1 + x i x and the torque induced by weight number i will be ( x i - ¯ x ) m i = ( x i - ¯ x )(1 + x i x . The total torque is then ( x 0 - ¯ x )(1 + x 0 x + ( x 1 - ¯ x )(1 + x 1 x + · · · + ( x n - 1 - ¯ x )(1 + x n - 1 x = n - 1 X i =0 x i (1 + x i x - n - 1 X i =0 ¯ x (1 + x i x = n - 1 X i =0 x i (1 + x i x - ¯ x n - 1 X i =0 (1 + x i x. If we set this equal to zero and solve for ¯ x we get an approximation to the balance point of the beam: 0 = n - 1 X i =0 x i (1 + x i x - ¯ x n - 1 X i =0 (1 + x i x ¯ x n - 1 X i =0 (1 + x i x = n - 1 X i =0 x i (1 + x i x ¯ x = n - 1 X i =0 x i (1 + x i x n - 1 X i =0 (1 + x i x . The numerator of this fraction is called the moment of the system around zero, and
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Unformatted text preview: the denominator has a very familiar interpretation. Consider one term of the sum in the denominator: (1 + x i )Δ x . This is the density near x i times a short length, Δ x , which in other words is approximately the mass of the beam between x i and x i +1 . When we add these up we get approximately the mass of the beam. Now each of the sums in the fraction has the right form to turn into an integral, which in turn gives us the exact value of ¯ x : ¯ x = Z 10 x (1 + x ) dx Z 10 (1 + x ) dx ....
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