Hello, I have a homework problem. I was wondering if you all get/got the same answer that I have: 1.9m. My friend got .6m and I don't know how. I tried working it out twice and still do not come up with .6m. Here it goes: 2 people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. if one person applies an upward force of 60 N at one end of the board, where must the other person be standing in order for the board to be in static equilibrium and with what force must they apply? My answer: Forces in x direction = 0 Forces in Y = F1 + F2 - W Given: L = 3.00 m F1 = 60 N W = 160 N F2 = ? and x2 = ? This is x sub two. Sum of forces and torques = 0 Sum Force = F1 + F2 - W = 0 60N + F2 - 160 N = 0 F2 = 100 N My pivot point is at F2. Sum of torques = 0 Torque F1 = F1(L - x2) Torque F2 = 0 b/c at pivot point Torque W = W(L/2 - x2) F1L - F1x2 + (WL)/2 - Wx2 = 0 (60)(3) - 60x2 + (160 * 3)/2 - 160x2 = 0 180-60x2 + 240-160x2 = 0 420-220x2 = 0 x2 = 1.9m Thank you for your assistance.