A uniform aluminum beam 9.00 m long, weighing 270 N, rests symmetrically on two

Question

A uniform aluminum beam 9.00 m long, weighing 270 N, rests symmetrically on two supports 5.15 m apart (see figure). A boy weighing 660 N starts at point A and walks toward the right.

(a) 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.)

(b) From your diagram, how far beyond point B can the boy walk before the beam tips?
____m

(c) 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?
_____ m

Explanation / Answer

Assumption # 1: The 9m beam is center along the 5.15m-wide supports, i.e. support A is (9-5.15/2) = 1.925 from the left edge and support B is 1.925m 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 = -270(7.05-4.5) + 660 * xboy

600 xboy = 688.5

xboy = 1.1475 m (to the right of support B)

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 = -270(4.5-x) + 660 * x

xboy = 1.30 m (from the right edge)

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