Balancing Forces on a Wooden Board: Determining the Point of Equilibrium

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To determine the point of equilibrium on a 4.00 m wooden board weighing 180 N, the upward force of 80 N applied at one end affects the balance. The calculations involve setting the sum of torques equal to zero, with the center of gravity located at 2.0 m from one end. The correct setup for the torque equation considers the distances from the applied forces to the pivot point. The solution reveals that the other person must lift at a distance of 3.6 m from the point where the 80 N force is applied. This analysis highlights the importance of torque and balance in understanding forces acting on the board.
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Two people are carrying a uniform wooden board that is 4.00 m long and weighs 180N. If one person applies an upward force equal to 80 N at one end, at what point does the other person lift?
m (distance from point of application of 80 N force)


I am thinking because of the the fact that the question gives the forces and a distance that I should use the Ʃ torque =0 but I am not sure how to set up the equation, (200)(4)=800 and and either 80x or 80(2)=160 which would create either 800+80x=0 but I don't think that is right at all.


The answer is 3.6 meters but I have yet to figure out how to reverse engineer it for the problem
 
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First, if one person lifts 80 N, what does the other person have to lift.
You know where the center of gravity of the board is, yes, and it weighs 180 N.
So pick a point and figure out the torques about that point.
 
So it should be 180N(4.0m)=720 so 720-80x-100?
 
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