2. Low 39% High 112% Average 82% Mode 67%
Quiz 1 Solutions
Low 39% High 112% Average 82% Mode 67%

3. Chapter 5: EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS
5.1 & 5.2 Objectives:
a) Identify support reactions
b) Draw a free-body diagram.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

4. CONDITIONS FOR RIGID-BODY EQUILIBRIUM
(Section 5.1)
In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces).
Forces on a particle
For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.
 F = 0 (no translation)
and  MO = 0 (no rotation)
Forces on a rigid body
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

5. The truck ramp has a weight of 400 lb.
APPLICATIONS
The truck ramp has a weight of 400 lb.
The ramp is pinned to the body of the truck and held in the position by the cable.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

6. APPLICATIONS (continued)
Two smooth pipes, each having a mass of 300 kg, are supported by the tines of the tractor fork attachment.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

7. THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
1) Create an idealized model
2) Draw a free-body diagram (FBD) showing all the external (active and reactive) forces.
3) Apply the equations of equilibrium to solve for any unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

8. FREE-BODY DIAGRAMS (Section 5.2)
Free-body diagram (FBD)
1. Draw an outlined shape. Here the body is isolated or cut “free” from constraints.
Idealized model
Show all the external forces and couple moments.
a) applied loads, b) support reactions, c) the weight of the body.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

9. Label loads and dimensions on the FBD:
FREE-BODY DIAGRAMS
Idealized model
Free-body diagram
Label loads and dimensions on the FBD:
Known forces and couple moments should be labeled with their magnitudes and directions.
For unknown forces and couple moments, use letters like Ax, Ay, MA, etc..
Indicate any necessary dimensions.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

10. SUPPORT REACTIONS IN 2-D
Support reactions are given in your textbook (Table 5-1).
Rules: If a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction.
If rotation is prevented, a couple moment is exerted on the body in the opposite direction.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

11. Draw: An idealized model and free- body diagram of the foot pedal.
EXAMPLE
Given: The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is
20 lb.
Draw: An idealized model and free- body diagram of the foot pedal.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

12. Think about it…
If the directions of the force and the couple moments are both reversed, what will happen to the beam?
A) The beam will lift from A. B) The beam will lift at B. C) The beam will be restrained.
D) The beam will break.
Answer
2. A A
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

13. Example
Draw a FBD of member ABC, which is supported by a smooth collar at A, roller at B, and link CD.
Look at Table 5-1
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

14. FBD Example (continued)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.1,5.2

15. Think About It… Answers: 1. C 2. B B
A rigid body is subjected to forces as shown. This body can be considered as a ______ member.
A) Single-force B) Two-force
C) Three-force D) Six-force
Answers:
1. C
2. B B
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

16. Think About It… Answers: C 1. C 2. B
The three scalar equations  FX =  FY =  MO = 0, are ____ equations of equilibrium in two dimensions.
A) Incorrect B) The only correct
C) The most commonly used D) Not sufficient
Answers:
1. C
2. B C
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

17. 5.3 EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS
Objectives:
a) Apply equations of equilibrium to solve for unknowns, and,
b) Recognize two-force members.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

18. APPLICATIONS
A 850 lb of engine is supported by three chains, which are attached to the spreader bar of a hoist.
You need to check to see if the breaking strength of any of the chains is going to be exceeded. We can determine the force acting in each of the chains
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

19. EQUATIONS OF EQUILIBRIUM (Section 5.3)
A body is subjected to a system of forces that lie in the x-y plane. When in equilibrium, the net force and net moment acting on the body are zero (as discussed earlier in Section 5.1). Here:
 Fx =  Fy = 0  MO = 0
where point O is any arbitrary point.
Note that these equations are the ones most commonly used for solving 2-D equilibrium problems.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

20. TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4)
The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B).
If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

21. EXAMPLE OF TWO-FORCE MEMBERS
In the cases above, members AB can be considered as two-force members, provided that their weight is neglected.
This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

22. STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS
1. If not given, establish a suitable x - y coordinate system.
2. Draw a free body diagram (FBD) of the object under analysis.
3. Apply the three equations of equilibrium to solve for the unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

23. IMPORTANT NOTES
1. If there are more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics.
2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force, then solving  FX = 0 first allows us to find the horizontal unknown quickly.
3. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

24. Given: The 4kN load at B of the beam is supported by pins at A and C .
EXAMPLE A
Given: The 4kN load at B of the beam is supported by pins at A and C .
Find: The support reactions at A and C.
Plan:
1. Put the x and y axes in the horizontal and vertical directions, respectively.
2. Determine if there are any two-force members.
3. Draw a complete FBD of the boom.
4. Apply the E-of-E to solve for the unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

25. EXAMPLE A (Continued) FBD of the beam:AX AY A
1.5 m C B
4 kN FC
45°
Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using E-of E, we get,
+ MA = FC sin 45  1.5 – 4  3 = 0
Fc = kN
 + FX = AX cos 45 = 0; AX = – kN
 + FY = AY sin 45 – 4 = 0; AY = – kN
Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

26. Find: Support reactions at B and C.
GROUP PROBLEM SOLVING
Given: The jib crane is supported by a pin at C and rod AB. The load has a mass of 2000 kg with its center of mass located at G.
Assume x = 5 m.
Find: Support reactions at B and C.
Plan:
a) Establish the x – y axes.
b) Draw a complete FBD of the jib crane beam.
c) Apply the E-of-E to solve for the unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

27. GROUP PROBLEM SOLVING (Continued)
FBD of the beam
2000(9.81) N
FAB
5 m Cx 5 4 3 Cy
4 m
0.2 m
First write a moment equation about Point C.
Volunteer Needed!
+  MC = (3 / 5) FAB 4 + (4 / 5) FAB 0.2 – 2000(9.81)  5 = 0
FAB = N = 38.3 kN
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

28. GROUP PROBLEM SOLVING (Continued)
FBD of the beam
2000(9.81) N
FAB
5 m Cx 5 4 3 Cy
4 m
0.2 m
FAB = N = 38.3 kN
Now solve the FX and FY equations.
 FX = Cx – (4 / 5) = 0
 FY = – Cy + (3 / 5) – 2000(9.81) = 0
Volunteer Needed!
Solving these two equations, we get Cx = N or kN and Cy = N or kN
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 5.3,5.4

29. 3-D FREE-BODY DIAGRAMS, EQUILIBRIUM EQUATIONS, CONSTRAINTS AND STATICAL DETERMINACY
5.5 Objectives:
a) Identify support reactions in 3-D and draw a free body diagram, and,
b) apply the equations of equilibrium.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

30. APPLICATIONS
Ball-and-socket joints and journal bearings are often used in mechanical systems. To design the joints or bearings, the support reactions at these joints and the loads must be determined.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

31. APPLICATIONS (continued)
The tie rod from point A is used to support the overhang at the entrance of a building. It is pin connected to the wall at A and to the center of the overhang B.
If A is moved to a lower position D, will the force in the rod change or remain the same? By making such a change without understanding if there is a change in forces, failure might occur.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

32. SUPPORT REACTIONS IN 3-D (Table 5-2)
Support reactions are given in your text book (Table 5-2).
As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the opposite direction is developed on the body. Similarly, if rotation is prevented, a couple moment is exerted on the body by the support.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

33. IMPORTANT NOTE
A single bearing or hinge can prevent rotation by providing a resistive couple moment. However, it is usually preferred to use two or more properly aligned bearings or hinges. Thus, in these cases, only force reactions are generated and there are no moment reactions created.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

34. EQUATIONS OF EQUILIBRIUM (Section 5.6)
As stated earlier, when a body is in equilibrium, the net force and the net moment equal zero, i.e.,  F = 0 and  MO = 0 .
These two vector equations can be written as six scalar equations of equilibrium (E of E). These are
FX =  FY =  FZ = 0
MX =  MY =  MZ = 0
The moment equations can be determined about any point. Usually, choosing the point where the maximum number of unknown forces are present simplifies the solution.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

35. CONSTRAINTS AND STATICAL DETERMINACY
(Section 5.7)
Redundant Constraints: When a body has more supports than necessary to hold it in equilibrium, it becomes statically indeterminate.
A problem that is statically indeterminate has more unknowns than equations of equilibrium.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

36. IMPROPER CONSTRAINTS
Here, we have 6 unknowns but there is nothing restricting rotation about the AB axis.
In some cases, there may be as many unknown reactions as there are equations of equilibrium.
However, if the supports are not properly constrained, the body may become unstable for some loading cases.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

37. Find: Reactions at the thrust bearing A and cable BC.
EXAMPLE B
Given: The rod, supported by thrust bearing at A and cable BC, is subjected to an 80 lb force.
Find: Reactions at the thrust bearing A and cable BC.
Plan:
a) Establish the x, y and z axes.
b) Draw a FBD of the rod.
c) Write the forces using scalar equations.
d) Apply scalar equations of equilibrium to solve for the unknown forces.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

38. EXAMPLE B (continued) FBD of the rod:
Applying scalar equations of equilibrium in appropriate order, we get
F X = AX = 0 ; AX = 0
F Z = AZ + FBC – 80 = 0 ;
M Y = – 80 ( 1.5 ) + FBC ( 3.0 ) = 0 ;
Solving these last two equations: FBC = 40 lb, AZ = 40 lb
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

39. EXAMPLE (continued) Now write scalar moment equations about point A
= 40 lb
Now write scalar moment equations about point A
∑M X = ( MA) X (6) – 80 (6) = 0 ; (MA ) X= 240 lb ft
M Z = ( MA) Z = 0 ; (MA ) Z= 0
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

40. Find: The reactions at all the supports for the loading shown.
GROUP PROBLEM SOLVING
Given: A rod is supported by smooth journal bearings at A, B, and C. Assume the rod is properly aligned.
Find: The reactions at all the supports for the loading shown.
Plan:
a) Draw a FBD of the rod.
b) Apply scalar equations of equilibrium to solve for the unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

41. GROUP PROBLEM SOLVING (continued)
A FBD of the rod:
Applying scalar equations of equilibrium
F Y = 450 cos 45 + CY = 0 ; CY = – 318 N
M Y = CZ (0.6) – 300 = 0 ; CZ = 500 N
M Z = – BX ( 0.8 ) – ( – 318 ) ( 0.6 ) = 0 ; BX = 239 N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

42. GROUP PROBLEM SOLVING (continued)
∑ M X = BZ ( 0.8 ) – 450 cos 45 (0.4) – 450 sin 45 ( )
+ 318 ( 0.4 ) ( ) = 0 ; BZ = – 273 N
F X = AX = 0 ; AX = – 239 N
F Z = AZ – ( – 273 ) – 450 sin 45 = 0 ; AZ = N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections