ch12-p007

# ch12-p007 - 7 We take the force of the left pedestal to be...

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7. We take the force of the left pedestal to be F 1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F 2 and denote its position as x = d . W is the weight of the diver, located at x = L . The following two equations result from setting the sum of forces equal to zero (with upwards positive), and the sum of torques (about x 2 ) equal to zero: 12 1 0 () 0 FFW Fd W L d +−= +− = (a) The second equation gives 1 3.0m (580 N)= 1160 N 1.5m Ld FW d §· =− ¨¸ ©¹ which should be rounded off to 3 1 1.2 10 N F × . Thus, 3 1 || 1 . 2 1 0 N . F (b) Since F 1 is negative, indicating that this force is downward.
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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