Equilibrium problem and a beam of mass

  • Thread starter Oomair
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  • #1
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[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?


Homework Equations





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?
 

Answers and Replies

  • #2
Chi Meson
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F1 and mg would both cause clockwise rotation, so they should have the same sign.
 
  • #3
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which F1 are you talking about? and which mg?
 
  • #4
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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
 
  • #5
Redbelly98
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which F1 are you talking about? and which mg?
He meant the ones in your torque equation.
 
  • #6
Redbelly98
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What is F1 (on the left edge pivot) when the 150 kg is placed right at the tipping point?
 
  • #7
Chi Meson
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What is F1 (on the left edge pivot) when the 150 kg is placed right at the tipping point?
One step further, is F1 needed at all?
 
  • #8
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i got it, thanks for the help
 

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