Equilibrium of Rigid Bodies Draw the free body
Equilibrium of Rigid Bodies Draw the free body diagram ( replace the supports with reactions) Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 1
Types of reactions - 1 A A B B Roller support : replace with a reaction normal to AB Frictionless floor : again, replace with a reaction normal to AB Rigid member : replace with a reaction in the direction of the member Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 2
Types of reactions - 2 A Frictionless slider : replace with a reaction normal to AB B Pin support : replace with two reactions, horizontal & vertical fixed support : replace with two reactions and a couple Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 3
Equilibrium in 2 D Where O is a convenient point. Try to choose it so that each equation has only one unknown. OR Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 4
Equilibrium in 2 D – example 1 A 200 mm 150 mm Determine : a) tension in cable b) reaction at C 30 o C B 200 mm 400 N D A 200 mm Cy C 150 mm Cx 30 o B T AD D Eqlb. Rigid Bodies Angle ACD is equal to 90 o + 30 o or 120 o. The triangle ACD is isosceles. Hence angle CAD is 30 o 400 N Resolve T AD into two components one in the direction of AB and the other normal to AB. Take moments about C Jacob Y. Kazakia © 2002 5
Equilibrium in 2 D – example 1 -cont. A 200 mm Cy C Moments about C: (T AD sin 30 o) 200 – (150 sin 30 o) 400 = 0 150 mm Cx 30 o B T AD D 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 cos 60 o + C x – 400 = 0 -T AD sin 60 o + C y = 0 We solve to obtain T AD = 300 N, C x = 400 - 150 = 250 N, and C y = 300 sin 60 o = 259. 8 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 6
Equilibrium in 2 D – example 1 - better method. 200 mm Cy A 200 mm C Moments about C: (T AD sin 30 o) 200 – (150 sin 30 o) 400 = 0 150 mm Cx 30 o B T AD D 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 = 250 Take moments about point A C x 100 – 400 ( 100 + 75) + C y 200 cos 30 o = 0 --> C y = 259. 8 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 7
Equilibrium in 2 D – example 2 6 k. N Determine the reactions at the supports 2 k. N 1. 5 m 2 k. N 6 k. N 1. 5 m 2 k. N 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 k. N Sum of forces in x – dir : B x = 4 k. N Eqlb. Rigid Bodies Ay Bx By Sum of forces in y dir. B y = 4. 5 k. N Jacob Y. Kazakia © 2002 8
Equilibrium in 2 D – example 3 15 x 9. 81 = 147. 2 N Obtain reactions. 25 m Bx FBD 15 kg 350 mm Ax Ay From sum of forces in x dir. A x = B x = T From sum of forces in y dir. A y = 147. 2 N Sum of moments ( about A ) . 35 T -. 25 147. 2 = 0 T = 105. 1 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002 9
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