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A beam resting on two pivots has a length ofL = 6.00 m and mass M = 91.0 kg. The pivot under the left end
A beam resting on two pivots has a length ofL = 6.00 m and mass M = 91.0 kg. The pivot under the left end exerts a normal force nl on the beam, and the second pivot placed a distance! = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 61.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to nd the woman's position when the beam begins to tip. I. I]! L_,_4 Q (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin. (Submit a file with a maximum size of 1 MB.) Choose File No le chosen This answer has not been graded yet. (b) Where is the woman when the normal force n1 is the greatest? \"Em (c) What is n1 when the beam is about to tip? N (d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. N (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. x = m (f) Check the answer to part (e) by computing torques around the first pivot point. x = m Except for possible slight differences due to rounding, is the answer the same? 0 Yet
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