Force/Free Body Diagrams Supports are translated into forces and moments in a free body diagrams. The following are three common supports and the forces and moments used to replace them. Roller: Pin Connection: Fixed Support: Fy Fx Mo

Reaction Forces Let’s suppose some campers have pitched their tents and are ready to prepare their supper. They build a structure to support their pot over the campfire to cook: Point A Point B Point C P FB FC P Free-Body Diagram of Point A

Reaction Forces The weight of the pot, P, is 50 lbs. The pot hangs from a rope that bisects the symmetric structure that is 5 ft tall and 3 ft wide. We need angle theta in order to solve for the x and y components. Point A Point B Point C P 5 ft 6 ft tan  = opp / adj  = tan-1(opp / adj)  = tan-1 (5 ft / 3 ft)  = 59 5 ft 3 ft  P FBX FBY FCY FCX

Reaction Forces FBX P FBY FCY FCX Solve for the scalar, x and y components of each force: FBX FBY FB 59 FCX FCY FC 59 FCX = FC cos(59) FCY = FC sin(59) FBX = FB cos(59) FBY = FB sin(59)

Reaction Forces FBX P FBY FCY FCX Sum the forces in the positive x and y directions. Assume the positive y-direction is up (+), and the positive x direction is to the right (). ***Remember the forces will sum to zero because we are assuming that the structure is in static equilibrium.

Reaction Forces Solve for the two unknowns using these two equations.   FX = 0; FBX - FCX = 0 FB cos(59) - FC cos(59) = 0 + FY = 0; FBY + FCY - P = 0 FB sin(59) + FC sin(59) - 50 = 0 + FBX P FBY FCY FCX

Reaction Forces Equation 1: FB cos(59) - FC cos(59) = 0 Equation 2: FB sin(59) + FC sin(59) - 50 = 0 Equation 1 shows that FB equals FC : FB cos(59) = FC cos(59) cos(59) cos(59) Replace FC with FB in Equation 2 since they are the same: FB sin(59) + FB sin(59) - 50 = 0 2 FB sin(59) = 50 2 sin (59) 2 sin (59) FB = 29.16 lbs

Reactions at the supports of the structure Reaction Forces If FB = 29.16 lbs, then FC = 29.16 lbs because of Equation 1. Solve for the scalar x and y components: FBX = FB cos(59) = 29.16 lbs * cos (59) = 15 lbs FBY = FB sin(59) = 29.16 lbs * sin (59) = 25 lbs These scalar quantities are the same for the components of FC. 15 lbs 25 lbs 50 lbs Reactions at the supports of the structure 50 lbs 15 lbs 25 lbs