[SOLVED] Equilibrium problem

Oomair

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?

Chi Meson

F1 and mg would both cause clockwise rotation, so they should have the same sign.

Oomair

which F1 are you talking about? and which mg?

rootX

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

Redbelly98

He meant the ones in your torque equation.

Redbelly98

What is F1 (on the left edge pivot) when the 150 kg is placed right at the tipping point?

Chi Meson

One step further, is F1 needed at all?

Oomair

i got it, thanks for the help