newCh11_solutions

newCh11_solutions - L = 0, F R = Mg. Same reasons as above....

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11-1) When you’re done with the individual torques, you can just algebraically add them since they refer to the same axis: 11-2) a) b) You just could guess, but we’ll give some reasoning so we’re not being fair to the question but we can’t help ourselves: Standing on the left end: F L = Mg, F R = 0. Why? Consider the torque about the left end of the board. If F R is non-zero, there will be a net torque, regardless of the value of F L . Therefore, F L must support all the weight of the person. Standing on the right end: F
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Unformatted text preview: L = 0, F R = Mg. Same reasons as above. Standing in the middle: F L = Mg/2, F R = Mg/2. Consider the torque about the middle of the board. The forces from the columns must then be equals, or else there would be a net torque about the middle of the board. Since they are equal, they must each support half the weight of the person. continued e) x = 0: R L F = Mg = 0 L F = Mg(1 - ) = Mg L x = L: R L L F = Mg = Mg L L F = Mg(1 - ) = 0 L x = L/2: R L L/2 F = Mg = Mg/2 L L/2 F = Mg(1 - ) = Mg/2 L...
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newCh11_solutions - L = 0, F R = Mg. Same reasons as above....

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