1. Equilibrium of Rigid Bodies
Elementary Engineering Mechanics
Equilibrium of Rigid Bodies
Draw the free body diagram ( replace the supports with reactions)
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002

2. Elementary Engineering Mechanics
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 Bodies
Jacob Y. Kazakia © 2002

3. Elementary Engineering Mechanics
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

4. Elementary Engineering Mechanics
Equilibrium in 2D
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

5. Equilibrium in 2D – example 1
Elementary Engineering Mechanics
Equilibrium in 2D – example 1
200 mm A
150 mm
Determine : a) tension in cable
b) reaction at C
30 o C B
400 N
200 mm D
Angle ACD is equal to 90o + 30o or 120o. The triangle ACD is isosceles. Hence angle CAD is 30o
200 mm A
C y
150 mm
C x
30 o C
Resolve T AD into two components one in the direction of AB and the other normal to AB.
Take moments about C B
400 N
T AD D
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002

6. Equilibrium in 2D – example 1-cont.
Elementary Engineering Mechanics
Equilibrium in 2D – example 1-cont.
Moments about C:
(T AD sin30o) 200 – (150 sin30o) 400 = 0
We used the red components of T AD
200 mm A
C y
150 mm
C x
30 o C B
400 N
T AD
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 D
We solve to obtain T AD = 300 N, C x = = 250 N, and
C y = 300 sin60o = N
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002

7. Equilibrium in 2D – example 1- better method.
Elementary Engineering Mechanics
Equilibrium in 2D – example 1- better method.
Moments about C:
(T AD sin30o) 200 – (150 sin30o) 400 = 0
We used the red components of T AD
200 mm A
C y
150 mm
C x
30 o C
200 mm B
400 N
T AD
Take moments about point D (use T AD itself)
400 ( 200 – 75) – C x 200 = > C x = 250 D
Take moments about point A
C x 100 – 400 ( ) + C y 200 cos30o = > C y = N
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002

8. Equilibrium in 2D – example 2
Elementary Engineering Mechanics
Equilibrium in 2D – example 2
6 kN
Determine the reactions at the supports
2 kN
1.5 m
6 kN
2 kN
1.5 m
2 kN
2 kN
2 m Bx
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
A y
B y
Sum of forces in y dir.
B y = 4.5 kN
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002

9. Equilibrium in 2D – example 3
Elementary Engineering Mechanics
Equilibrium in 2D – example 3
Obtain reactions
15 x 9.81 = N
.25m
.25m
B x
FBD
15 kg
350 mm
A x
A y
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
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002