quick question,(adsbygoogle = window.adsbygoogle || []).push({});

A uniform, aluminum beam 9.00 m long, weighting 300 N, rests symmetrically on two supports 5.00 m apart. A boy weighing 600 N starts at point A and walks toward the right.

How far beyond point B can the boy walk before the beam tips?

ans=1.25 m

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?

this second part is the part im stuck on.

my equations are

all of these are the sum of the forces in that direction

Fx=0

no forces are acting in this direction

Fy=(F(support A)+F(support B))-(F(gravity on bar)+F(gravity on boy))

force of gravity of bar is at its center of mass between the two blocks

Tz=(F(gravity on bar)*(distance)-F(gravity on boy)*(distance))

rotation point is set around point b

i was able to solve for the first part of this problem however for the second part it appears that i have 2 unknowns with 1 effective equation so im not really sure how to go.

i have included a figure but chances are it will take some time for it to come up.

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# Homework Help: Static equilibrium boy walking on beam

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