jF_{y}(T_{1}sin 30- ) 125 lbf 00.5 T_{1}125 lbfT_{1}250 lbfjForces F_{1}F_{2}N W 0Application of Newton's LawsFORCE EQUILIBRIUMRev. 0Page 15CP-04Figure 8 Free-Body DiagramBy choosing (+) as the upward direction and (-) as the downward direction, the student candetermine by examination that 1) the upward component of T is + T sin 30 , 2) the tension T1 1 3ois -125 lbf, and 3) T has no y- component. Therefore, using the same equation as before, we2obtain the following.If the sum of all forces acting upon a body is equal to zero, that body is said to be in forceequilibrium. If the sum of all the forces is not equal to zero, any force or system of forces capableof balancing the system is defined as an equilibrant.Example:A 2000 lbm car is accelerating (on a frictionless surface) at a rate of 2 ft-sec. What forcemust be applied to the car to act as an equilibrant for this system?Solution:a.Draw a free-body diagram.b.A Force, F , MUST be applied in the opposite direction to F such that the2 1sum of all forces acting on the car is zero.