A uniform aluminum beam 9.00 m long, weighing 276 N, rests symmetrically on two

Question

A uniform aluminum beam 9.00 m long, weighing 276 N, rests symmetrically on two supports 4.75 m apart (see figure). A boy weighing 660 N starts at point A and walks toward the right.

A uniform aluminum beam 9.00 m long, weighing 276 N, rests symmetrically on two supports 4.75 m apart (see figure). A boy weighing 660 N starts at point A and walks toward the right. In the same diagram construct two graphs showing the upward forces FA and FB exerted on the beam at points A and B as functions of the coordinate x of the boy. Let 1 cm = 100 N vertically, and let 1 cm = 1.00 m horizontally. (Do this on paper. Your instructor may ask you to turn in this work.) From your diagram, how far beyond point B can the boy walk before the beam tips? How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?

Explanation / Answer

Assumption # 1: The 9m beam is center along the 5m-wide supports, i.e. support A is 2m from the left edge and support B is 2m from the right edge.

B) The easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0

So we can use pt. B as the pivot point and determine x:

viewing counter-clockwise torque as negative:

T = 0 = -Wbeam * xbeam + Fboy * xboy

T = 0 = -300(7-4.5) + 600 * xboy

600 xboy = 750

xboy = 1.25 m (to the right of support B) OR
xboy = 8.25 m (from support A)
support B = 7.0 m (from support A)

C) Again, the easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0

So we can use pt. B as the pivot point and determine x:
x = distance of pt B from the right edge.

Since the center-of-mass of the beam is 4.5m (middle of beam), its distance from pt B = (4.5 - x)

viewing counter-clockwise torque as negative:

T = 0 = -Wbeam * xbeam + Fboy * xboy

T = 0 = -300(4.5-x) + 600 * x

900x = 1350

xboy = 1.50 m (from the right edge) OR
xboy = 7.50 m (from support A)

"That boy shouldn't be walking around in a beam. He could fall! :-) "

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