1. The problem statement, all variables and given/known data The uniform bar shown below weighs 40N and is subjected to the forces shown. Find the magnitude, direction, and location of the force needed to keep the bar in equilibrium. (the L's in the diagram stand for "length") 2. Relevant equations ΣFx , ΣFy , and ΣT (torque) 3. The attempt at a solution The question says a force is needed to keep the bar in equilibrium so I used both translational and rotational equilibrium formulas for this question (because they are the only equilibrium formulas we have learned so far). I first went through the translational equilibrium calculations (ΣFx, and ΣFy) to find x and y components. Translational Equilibrium: ΣFy=may ΣFy=0 50-60-70-40+(80sin(30))=0 50-60-70-40+(40)= -80 -80 = 0 therefore the unknown vertical component is = (+ or positive) 80N ΣFx=max ΣFx=0 -(80cos(30))=0 -69.2820323=0 -69 = 0 therefore the unknown horizontal component is = (+) 69N Next I used the parallelogram method of vector addition to find the resultant and the answer I got for the resultant was 106N. I then went to go find the angle by using tan and the angle turned out to be 49(degrees). north of east. Now that I have found the magnitude and direction of the unknown force, the only thing left to find is its location and this is where I am having trouble. I tried using the rotational equilibrium equation and this is what I got: ΣT=0 T(counterclockwise)=T(clockwise) (80sin(30))(l1 or lever arm 1) + 50(l3 or lever arm 3)= Mg(weight of beam)*(l2 or lever arm 2) + 60(l3 or lever arm 3) + 70(l3 or lever arm 3) = (80sin(30))(1.00) + 50(0.2) = (40(0.5) + 60(0.2) + 70(0.2) = 50 = 46 I have found this now but I am unsure as to how to use this to find the location of the unknown force needed to keep the bar in equilibrium.