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 4.65 m apart (see figure). A boy weighing 558 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. Your instructor may ask you to turn in this work.) (b) From your diagram, how far beyond point B can the boy walk before the beam tips? Incorrect: Your answer is incorrect. 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

Given,

Wbeam = 276 N

xbeam = 4.65 m

Wboy = 558 N

B) Since, the forces and torque at support A is zero, i.e. FA= 0, TA = 0

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

viewing counter-clockwise torque as negative:

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

T = 0 = -276 x (7-4.5) + 558 x xboy

xboy = 690/558 =1.236 m

xboy = 1.236 m (to the right of support B)
xboy = 8.236 m (from support A)
support B = 7.0 m (from support A)

C) Again, the forces and torque at support A is zero, i.e. FA = 0, TA = 0

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

Since the center-of-mass of the beam is 4.65m (middle of beam), its distance from point B = (4.65 - x)

viewing counter-clockwise torque as negative:

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

T = 0 = -276 x (4.65-xboy) + 558 x xboy

xboy = 1283.4 / 834

xboy = 1.53 m (from the right edge) OR
xboy = 7.47 m (from support A)

(Please rate my answer if you find it helpful. Good luck...)

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