'F_{x}T^{1}xT^{2}xT^{3}x0'F_{y}T^{1}yT^{2}yT^{3}y0'F_{y}(T_{1}sin30- )(T_{2} sin180- )(T_{3} sin270- )0(T_{1})(0.5) (T_{2})(0) (125 lbf)(1) 00.5T_{1}125 lbf 00.5T_{1}125 lbfT_{1}250 lbfFORCE EQUILIBRIUMApplication of Newton's LawsCP-04 Page 14Rev. 0Figure 7 Free-Body DiagramThe tension in a cable is the force transmitted by the cable. The tension at any point in the cablecan be measured by cutting a suitable length from it and inserting a spring scale.Solution:Since the object and its supporting cables are motionless (i.e., in equilibrium), we knowthat the net force acting on the intersection of the cables is zero. The fact that the netforce is zero tells us that the sum of the x-components of T , T , and T is zero, and the1 2 3sum of the y-components of T , T , and T is zero.1 2 3The tension T is equal to the weight of the object, 125 lbf. The x and y components of the3tensions can be found using trigonometry (e.g., sine function). Substituting known values into thesecond equation above yields the following.A simpler method to solve this problem involves assigning a sign convention to the free-bodydiagram and examining the direction of the forces.