1 Free Body Diagrams

In practice, a problem in engineering mechanics is derived from an actual physical situation. A sketch showing the physical conditions of the problem is known as a space diagram. The methods of analysis apply to a system of forces acting on a particle. A large number of problems involving actual structures, however, can be reduced to problems concerning the equilibrium of a particle.

This is done by choosing a significant particle and drawing a separate diagram showing this particle and all the forces acting on it. Such a diagram is called a free body diagram. 

For Example: - Consider the 75-kg crate shown in the space diagram of Fig (a).

Solution: - This crate was lying between two buildings, and it is now being lifted onto a truck, which will remove it. The crate is supported by a vertical cable, which is joined at A to two ropes which pass over pulleys attached to the buildings at B and C.  It is desired to determine the tension in each of the ropes AB and AC. In order to solve this problem, a free-body diagram showing a particle in equilibrium must be drawn. Since we are interested in the rope tensions, the free-body diagram should include at least one of these tensions or, if possible, both tensions. Point A is seen to be a good free body for this problem. The free-body diagram of point A is shown in F ig (b).  It shows point A and the forces exerted on A by the vertical cable and the two ropes. The force exerted by the cable is directed downward, and its magnitude is equal to the weight W of the crate. We write

W = mg = (75 kg)*(9.81 m/s2) = 736 N

 And indicate this value in the free-body diagram. The forces exerted by the two ropes are not known. Since they are respectively equal in magnitude to the tensions in rope AB and rope AC, we denote them by TAB and TAC and draw them away from A in the directions shown in the space diagram. No other detail is included in the free- body diagram. Since point A is in equilibrium, the three forces acting on it must form a closed triangle when drawn in tip-to-tail fashion. This force triangle has been drawn in F ig (c).  The values TAB and TAC of the tension in the ropes may be found graphically if the triangle is drawn to scale, or they may be found by trigonometry. If the latter method of solution is chosen, we use the law of sines (Lami’s Theorem) and write

TAB/sin 60° = TAC/sin 40° = 736N/ sin 80°

TAB = 647 N and TAC = 480 N

 When a particle is in equilibrium under three forces, the problem can be solved by drawing a force triangle. When a particle is in equilibrium under more than three forces, the problem can be solved graphically by drawing a force polygon.