A beam resting on two pivots has a length of L = 6.00 mand mass M = 101.1 kg. Th

Question

A beam resting on two pivots has a length of L = 6.00 mand mass M = 101.1 kg. Thepivot under the left end exerts a normal forcen1 on the beam, and the second pivot placed adistance l = 4.00 m from the left end exerts a normalforce n2. A woman of mass m =51.3 kg steps onto the left end of thebeam and begins walking to the right as in the figure below. Thegoal is to find the woman's position when the beam begins to tip.(a) Sketch a free-body diagram, labeling the gravitational andnormal forces acting on the beam and placing the woman xmeters to the right of the first pivot, which is the origin. (Dothis on paper. Your instructor may ask you to turn in thiswork.)
(b) Where is the woman when the normal force n1is the greatest?(c) What is n1 when the beam isabout to tip?(d) Use the force equation of equilibrium to find thevalue of n2 when the beam is about to tip.(e)Using the result of part (c) and the torque equilibrium equation,with torques computed around the second pivot point, find thewoman's position when the beam is about to tip.
(f) Check the answer to part (e) by computing torques around thefirst pivot point.

Explanation / Answer

I'm working on this same homework assignment and I finally figuredthis one out. I'm just going to answer the ones that requirecalculations. c) 0. n1 is equal to 0 at this point because it is no longerin contact with the pivot point. d) Like the problem says, use the force equilibrium equation. Wwoman = (9.8x51.3) Wbeam = (9.8x101.1) F = n1 + n2 - Wwoman - Wbeam = 0 Since n1 is equal to 0, the equation is: n2 = Wwoman + Wbeam. Adding these weights will get you n2. e) Remember that torque is the force times the length of theposition vector. L/2=3 4(n2) = (Wwoman)X + (Wbeam)(3) Plug in the values and solve for X f) This answer is the exact same as the answer for e.

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