1. 5.3 Equations of Equilibrium
For equilibrium of a rigid body in 2D,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
∑Fx and ∑Fy represent the algebraic sums of the
x and y components of all the forces acting on
the body
∑MO represents the algebraic sum of the couple
moments and moments of the force components
about an axis perpendicular to x-y plane and
passing through arbitrary point O, which may lie
on or off the body

2. 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
For coplanar equilibrium problems, ∑Fx = 0; ∑Fy
= 0; ∑MO = 0 can be used
Two alternative sets of three independent
equilibrium equations may also be used
∑Fa = 0; ∑MA = 0; ∑MB = 0
When applying these equations, it is required
that a line passing through points A and B is not
perpendicular to the a axis

3. 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
Consider FBD of an arbitrarily shaped body
All the forces on FBD may be
replaced by an equivalent
resultant force
FR = ∑F acting at point A and a
resultant moment MRA = ∑MA
If ∑MA = 0 is satisfied, MRA = 0

4. 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
If FR satisfies ∑Fa = 0, there is no
component along the a axis and
its line of axis is perpendicular
to the a axis
If ∑MB = 0 where B does not lies
on the line of action of FR, FR = 0
Since ∑F = 0 and ∑MA = 0, the
body is in equilibrium

5. 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
A second set of alternative equations is
∑MA = 0; ∑MB = 0; ∑MC = 0
Points A, B and C do not lie on the
same line
Consider FBD, if If ∑MA = 0, MRA = 0
∑MA = 0 is satisfied if line of action of FR passes
through point B
∑MC = 0 where C does not lie on line AB
FR = 0 and the body is in equilibrium

6. 5.3 Equations of Equilibrium
Procedure for Analysis
Free-Body Diagram
Establish the x, y, z coordinates axes in any
suitable orientation
Draw an outlined shape of the body
Show all the forces and couple moments
acting on the body
Label all the loadings and specify their
directions relative to the x, y axes

7. 5.3 Equations of Equilibrium
Procedure for Analysis
Free-Body Diagram
The sense of a force or couple moment
having an unknown magnitude but
known line of action can be assumed
Indicate the dimensions of the body
necessary for computing the moments
of forces

8. 5.3 Equations of Equilibrium
Procedure for Analysis
Equations of Equilibrium
Apply the moment equation of equilibrium
∑MO = 0 about a point O that lies on the
intersection of the lines of action of the two
unknown forces
The moments of these unknowns are zero
about O and a direct solution the third
unknown can be obtained

9. 5.3 Equations of Equilibrium
Procedure for Analysis
Equations of Equilibrium
When applying the force equilibrium ∑Fx = 0
and ∑Fy = 0, orient the x and y axes along
the lines that will provide the simplest
resolution of the forces into their x and y
components
If the solution yields a negative result scalar,
the sense is opposite to that was assumed
on the FBD

10. 5.3 Equations of Equilibrium
Example 5.6
Determine the horizontal and vertical
components of reaction for the beam loaded.
Neglect the weight of the beam in the
calculations.

11. 5.3 Equations of Equilibrium
Solution
FBD
600N force is represented by its x and y components
200N force acts on the beam at B and is
independent of the
force components
Bx and By, which
represent the effect of
the pin on the beam

12. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
+ → ∑ M B = 0;
600 cos 45o N − Bx = 0
Bx = 424 N
A direct solution of Ay can be obtained by applying
∑MB = 0 about point B
Forces 200N, Bx and By all create zero moment about
B

13. 5.3 Equations of Equilibrium
Solution
∑ M B = 0;
100 N (2m) + (600 sin 45o N )(5m) − (600 cos 45o N )(0.2m) − Ay (7 m) = 0
Ay = 319 N
+ ↑ ∑ Fy = 0;
319 N − 600 sin 45o N − 100 N − 200 N + B y = 0
B y = 405 N

14. 5.3 Equations of Equilibrium
Solution
Checking,
∑ M A = 0;
− (600 sin 45o N )(2m) − (600 cos 45o N )(0.2m) − (100 N )(5m)
− (200 N )(7m) + B y (7m) = 0
B y = 405 N

15. 5.3 Equations of Equilibrium
Example 5.7
The cord supports a force of 500N and wraps over
the frictionless pulley. Determine the tension in the
cord at C and the horizontal
and vertical components at
pin A.

16. 5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
Principle of action: equal but opposite reaction
observed in the FBD
Cord exerts an unknown load
distribution p along part of
the pulley’s surface
Pulley exerts an equal but
opposite effect on the cord

17. 5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
Easier to combine the FBD of the pulley and
contracting portion of the cord so that the
distributed load becomes internal to the
system
and is eliminated from the
analysis

18. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
∑ M A = 0;
500 N (0.2m) + T (0.2m) = 0
T = 500 N
Tension remains constant as cord
passes over the pulley (true for
any angle at which the cord is
directed and for any radius of
the pulley

19. 5.3 Equations of Equilibrium
Solution
+ → ∑ Fx = 0;
− Ax + 500 sin 30o N = 0
Ax = 250 N
+ ↑ ∑ Fy = 0;
Ay − 500 N − 500 cos 30o N = 0
Ay = 933N

20. 5.3 Equations of Equilibrium
Example 5.8
The link is pin-connected at a and rest a
smooth support at B. Compute the horizontal
and vertical components of reactions at pin A

21. 5.3 Equations of Equilibrium
Solution
FBD
Reaction NB is perpendicular to the link at
B
Horizontal and vertical
components of reaction
are represented at A

22. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
∑ M A = 0;
− 90 N .m − 60 N (1m) + N B (0.75m) = 0
N B = 200 N
+ → ∑ Fx = 0;
Ax − 200 sin 30o N = 0
Ax = 100 N

23. 5.3 Equations of Equilibrium
Solution
+ ↑ ∑ Fy = 0;
Ay − 60 N − 200 cos 30o N = 0
Ay = 233N

24. 5.3 Equations of Equilibrium
Example 5.9
The box wrench is used to tighten the bolt at
A. If the wrench does not turn when the load
is applied to the handle, determine the
torque or moment applied to the bolt and the
force of the wrench on the bolt.

25. 5.3 Equations of Equilibrium
Solution
FBD
Bolt acts as a “fixed support” it exerts force
components Ax and Ay and a torque MA on
the wrench at A

26. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
∑ M A = 0;
 12 
M A − 52  N (0.3m) − (30 sin 60o N )(0.7 m) = 0
 13 
M A = 32.6 N .m
+ → ∑ Fx = 0;
5
Ax − 52  N + 30 cos 60o N = 0
 13 
Ax = 5.00 N

27. 5.3 Equations of Equilibrium
Solution
+ ↑ ∑ Fy = 0;
 12 
Ay − 52  N − 30 sin 60o N = 0
 13 
Ay = 74.0 N
CCW ↑ ∑ M y = 0;
 12 
M A − 52  N (0.3m ) − (30 sin 60° N )(0.7 m ) = 0
 13 
M A = 32.6 Nm

28. 5.3 Equations of Equilibrium
Solution
Point A was chosen for summing the moments as
the lines of action of the unknown forces Ax and
Ay pass through this point and these forces are
not included in the moment summation
MA must be included
Couple moment MA is a free vector and
represents the twisting resistance of the bolt on
the wrench

29. 5.3 Equations of Equilibrium
Solution
By Newton’s third law, the wrench exerts an
equal but opposite moment or torque on the
bolt
For resultant force on the wrench,
FA = (5.00)2 + (74.0)2 = 74.1N
For directional sense,
74.0 N
−1
θ = tan = 86.1o
5.00 N
FA acts in the opposite direction on the bolt

30. 5.3 Equations of Equilibrium
Solution
Checking,
∑MC = 0;
12N(0.4m) + 32.6N.m − 74.0N(0.7m) = 0
52 
13
19.2N.m + 32.6N.m − 51.8N.m = 0

31. 5.3 Equations of Equilibrium
Example 5.10
Placement of concrete from the
truck is accomplished using the
chute. Determine the force that the
hydraulic cylinder and the truck
frame exert on the chute to hold it in
position. The chute and the wet
concrete contained along its length
have a uniform weight of 560N/m.

32. 5.3 Equations of Equilibrium
Solution
Idealized model of the chute
Assume chute is pin connected to the frame
at A and the hydraulic cylinder BC acts as a
short link

33. 5.3 Equations of Equilibrium
Solution
FBD
Since chute has a length of 4m, total supported
weight is (560N/m)(4m) = 2240N, which is
assumed to act at its midpoint, G
The hydraulic cylinder exerts a horizontal force
FBC on the chute
Equations of Equilibrium
A direct solution of FBC is obtained by the
summation about the pin at A

34. 5.3 Equations of Equilibrium
Solution
∑ M A = 0;
− FBC (0.5m) + 2240 cos 30o N (2m)
+ 2240 sin 30o N (0.0625m) = 0
FBC = 7900 N
+ → ∑ Fx = 0;
− Ax + 7900 N = 0
Ax = 7900 N

35. 5.3 Equations of Equilibrium
Solution
+ ↑ ∑ Fy = 0;
Ay − 2240 N = 0
Ay = 2240 N
Checking,
∑ M B = 0;
− 7900 N (0.5m) + 2240 N (1cos 30o m) +
2240 cos 30o N (1m) + 2240 sin 30o N (0.0625m) = 0

36. 5.3 Equations of Equilibrium
Example 5.11
The uniform smooth rod is subjected to a force and
couple moment. If the rod is supported at A by a
smooth wall and at B and C either at the top or
bottom by rollers, determine
the reactions at these supports.
Neglect the weight of the rod.

37. 5.3 Equations of Equilibrium
Solution
FBD
All the support reactions act normal to the
surface of contact since the contracting surfaces
are smooth
Reactions at B and C are
acting in the positive y’
direction
Assume only the rollers
located on the bottom of
the rod are used for support

38. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
∑ M A = 0;
− B y ' (2m) + 4000 N .m − C y ' (6m) + (300 cos 30o N )(8m) = 0
+ → ∑ Fx = 0;
C y ' sin 30o + B y ' sin 30o − Ax = 0
+ ↑ ∑ Fy = 0;
− 300 N + C y ' cos 30o + B y ' cos 30o = 0

39. 5.3 Equations of Equilibrium
Solution
Note that the line of action of the force
component passes through point A and this
force is not included in the moment equation
B y ' = −1000.0 N = −1kN
C y ' = 1346.4 N = 1.35kN
Since By’ is negative scalar, the sense of By’ is
opposite to shown in the FBD

40. 5.3 Equations of Equilibrium
Solution
Top roller at B serves as the support rather
than the bottom one
1346.4 sin 30o N − 1000.0 sin 30o N − Ax = 0
Ax = 173N

41. 5.3 Equations of Equilibrium
Example 5.12
The uniform truck ramp has a weight of
1600N ( ≈ 160kg ) and is pinned to the body
of the truck at each end and held in position
by two side cables.
Determine the tension
in the cables.

42. 5.3 Equations of Equilibrium
Solution
Idealized model of the ramp
Center of gravity located at the midpoint since the
ramp is approximately uniform
FBD of the Ramp

43. 5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
∑ M A = 0;
− T cos 20o (2 sin 30o m) + T sin 20o (2 cos 30o )
+ 1600 N (1.5 cos 30o ) = 0
T = 5985 N
By the principle of transmissibility, locate T at C
d 2m
=
sin 10 sin 20o
o
d = 1.0154m

44. 5.3 Equations of Equilibrium
Solution
Since there are two cables supporting the
ramp,
T’ = T/2 = 2992.5N