# Equilibrium problem and a beam of mass

Oomair
[SOLVED] Equilibrium problem

## Homework Statement

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?

## 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 I am 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 don't what i am supposed to assume?

Homework Helper
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

Staff Emeritus
Homework Helper
which F1 are you talking about? and which mg?

He meant the ones in your torque equation.

Staff Emeritus