## [SOLVED] Equilibrium problem

1. The problem statement, all variables and given/known data

A uniform beam of mass 76 kg and length 3 m rests on two pivots, one at the left edge and one 2.4 m from the left edge. How far to the right of the right pivot can a mass of 150 kg be placed without the beam tipping?

2. Relevant equations

3. The attempt at a solution

net forces in y direction, Fy : F1(pivot on egde) - Fg(weight of board) +F1(2nd pivot) - mg

Sum of torque(im torquing around the rigt piovt): F1(2.4m) - Fg(.9m) -mg(x)

the problem im having is that i have two unknown forces from the pivots, and when i solve for one of the forces, the results cancel out and x = 0, i dont what i am supposed to assume?
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 Recognitions: Homework Help Science Advisor F1 and mg would both cause clockwise rotation, so they should have the same sign.
 which F1 are you talking about? and which mg?

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## [SOLVED] Equilibrium problem

Try three equations - three unknowns system (x, normal force one, normal force two)

You can pick first torque center at N1/N2, second at 150g mass .. and this give 2 eqns ..

and third is simply adding all forces ...

F1(pivot on egde) - Fg(weight of board) +F1(2nd pivot) "+" mg .. i think as said by chi meson

Mentor
Blog Entries: 10
 Quote by Oomair which F1 are you talking about? and which mg?
He meant the ones in your torque equation.
 Mentor Blog Entries: 10 What is F1 (on the left edge pivot) when the 150 kg is placed right at the tipping point?

Recognitions:
Homework Help