1. The problem statement, all variables and given/known data a uniform beam 5 meters long and weighing 60N carries a load of 80N at one end and 180N on the other end. if it is to beheldhorizontal while resting on a wooden saw horse 1.3 meters from the heavier laod, where would you have to hold it if you are exerting 100N of force? 2. Relevant equations Sum Torque counterclockwise= Sum of Torque clockwise Torque= Fd 3. The attempt at a solution from this problem, I've drew a diagram, which showed 5 different forces. Fsmall-end=80N,Fbig-end=180N, Fme=100N,Fbeam=60N,Fwooden-saw-horse=is unknown but is 1.3m from the big end. The 2 Forces that going up is 110N and Fwoodensaw-horse. The 3 Forces going down is 80N, 60N, and 180N. My teacher mentioned that if we use the Torque equations then we would have to choose a Fulcrum. I chose the wooden saw horse point as a Fulcrum, since the F is unknown there. So I did the whole problem using the Torque equation and the answer came out to be d=1.34m. That means that I, exerting a force of 100N would have to hold the beam at 1.34m from the small end to keep the beam horizontal(balanced). Is my answer correct? or am i missing something and did it wrong? thank you for the help!