![]() ![]() ![]() Taking the sum of moments around the same point as before, the moment arm of the two diagonal forces are zero, but the 100# force will cause a clockwise rotation. The force system on the right is not in moment equilibrium. Ibrium has been established using this single point, the sum of the moments for that force system will be zero for any point on that plane. For each force, the moment arm is equal to zero. Take the sum of the moments at their point of intersection. The system on the left is in moment equilibrium because it is a concurrent force system. Now solve for the sum of moments equation. Sum F y = 100k - 3/5 (60) - 4/5 (80) = 100 - 36 - 64 = 0īoth systems satisfy the sum of forces equations for equilibrium. Now, using the components, solve for the sum of forces equations. ![]() Therefore, the side marked "ģ" has a value of 3/5 of the value of the diagonal and the side marked "4" is equal to 4/5 the value of the diagonal. From observation, each diagonal is the "5" side of a 3-4-5 triangle. The simplest way to solve these force systems would be to break the diagonal forces into their component pars. In order for a system to be in equilibrium, it must satisfy all three equations of equilibrium,īegin with the sum of the forces equations. Whether or not equilibrium has been satisfied. ![]()
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