Question

two people are carrying a uniform wooden board, oriented horizontally, that is 3.00 m long and weighs 160 n. if one person applies a 60-n upward force at one end, at what position along the board is the other person supporting it? answer: 0.60 m from the other end.

Answer 1

The solution involves setting the torques equal but with the correct equation: 60N⋅x - 160N⋅(3m-x) = 0,then x = 4.00 m.

The other person is supporting the board 4.00 m from the end where the first person is applying the upward force.

Torque due to weight (τ1) + Torque due to upward force (τ2) = 0

Therefore, the equation should be written as:

W * (L/2) - F1 * x = 0

Substituting the given values:

160 N * (3.00 m / 2) - 60 N * x = 0

Solving for x:

240 N - 60 N * x = 0

60 N * x = 240 N

x = 240 N / 60 N

x = 4.00 m

Therefore, the distance from the end where the other person is supporting the board is 4.00 meters, as you correctly pointed out. I apologize for the previous incorrect solution.