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A uniform beam resting on two pivots has a length L=6.00 m and mass M=90.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot located a distance l=4.00 m from the left end exerts a normal force n2. A woman of mass m=55.0 kg steps onto the left end of the beam and begins walking to the right as in the figure. The goal is to find the woman’s position when the beam begins to tip.

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I have already solved for the tipping point which is about 5.64 meters using the torque equation. I have also graphed the n1 and n2 as functions of x. enter image description here

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Please correct me if I'm wrong in any of these things, but how do I graph the normal force after the tipping point? I am very confused on what equations I am going to use

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First of all, we'll assume that we're looking at the point in time just as the beam starts tipping, or things get needlessly complicated. It's no longer an equilibrium situation, so the sum of forces and torques do not sum to zero. But it's obvious the n1 force will be zero. For the rest, you need to write up the equations for the angular acceleration and linear acceleration of the beam (as a function of x) and their geometrical relationship to each other. This should allow you to solve for the n2 force.

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