Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium of Rigid Bodies Draw the free body diagram ( replace the supports with reactions)

Eqlb. Rigid BodiesJacob Y. Kazakia © Types of reactions - 1 Roller support : replace with a reaction normal to AB A B Frictionless floor : again, replace with a reaction normal to AB A B Rigid member : replace with a reaction in the direction of the member

Eqlb. Rigid BodiesJacob Y. Kazakia © Types of reactions - 2 A B Frictionless slider : replace with a reaction normal to AB Pin support : replace with two reactions, horizontal & vertical fixed support : replace with two reactions and a couple

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D Where O is a convenient point. Try to choose it so that each equation has only one unknown. OR

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D – example mm 150 mm 30 o 400 N A C D B Determine : a) tension in cable b) reaction at C 150 mm 30 o 400 N A C B T AD C y C x D 200 mm Angle ACD is equal to 90o + 30o or 120o. The triangle ACD is isosceles. Hence angle CAD is 30o Resolve T AD into two components one in the direction of AB and the other normal to AB. Take moments about C

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D – example 1-cont. 150 mm 30 o A C B T AD C y C x D 200 mm Moments about C: (T AD sin30o) 200 – (150 sin30o) 400 = 0 We used the red components of T AD 400 N Sum of forces in the x, y direction: ( use the green components of T AD ) T AD cos60o + C x – 400 = 0 -T AD sin60o + C y = 0 We solve to obtain T AD = 300 N, C x = = 250 N, and C y = 300 sin60o = N

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D – example 1- better method. 150 mm 30 o A C B T AD C y C x D 200 mm Moments about C: (T AD sin30o) 200 – (150 sin30o) 400 = 0 We used the red components of T AD 400 N Take moments about point D (use T AD itself) 400 ( 200 – 75) – C x 200 = 0 --> C x = mm Take moments about point A C x 100 – 400 ( ) + C y 200 cos30o = 0 --> C y = N

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D – example 2 2 kN 1.5 m Determine the reactions at the supports 6 kN 2 kN 6 kN A y BxBx B y 2 m Sum of moments about B: 2 x 6 – 3 x 2 – 1.5 x 2 – 2 x A y = 0, A y = 1.5 kN Sum of forces in x – dir : B x = 4 kN Sum of forces in y dir. B y = 4.5 kN

Eqlb. Rigid BodiesJacob Y. Kazakia © Equilibrium in 2D – example 3 15 kg.25m 350 mm Obtain reactions A x A y B x From sum of forces in x dir.  A x = B x = T From sum of forces in y dir.  A y = N Sum of moments ( about A ) .35 T = 0 T = N 15 x 9.81 = N FBD