1. 1

2. Chapter Objectives Chapter Outline
Concept of the free-body diagram (FBD) for a particle
Solve particle equilibrium problems using the equations of equilibrium
Chapter Outline
Condition for the Equilibrium of a Particle
The Free-Body Diagram
Coplanar Systems
Three-Dimensional Force Systems 2

3. 3.1 Condition for the Equilibrium of a Particle
Particle at equilibrium if
- At rest
- Moving at constant a constant velocity
Newton’s first law of motion
∑F is the sum of all forces acting on a particle
Newton’s second law of motion ∑F = ma
If a particle follows the Newton's first law of motion, then the particle is either at rest or moving at a constant velocity. The particle is said to be in equilibrium 3

4. 3.2 The Free-Body Diagram
It is a diagram representing all forces (∑F) acting on a particle
The particle is free from all surrouding subjects and draw all external forces on the particle, such as weights, forces from contacts etc.
There are mainly two types of connections and mainly for this course
Spring
Cables and Pulleys 4

5. 3.2 The Free-Body Diagram Spring
Linear elastic spring : the extension is proportional to and exerting force
a spring constant or stiffness k is used to describe how much elasticity the spring contains, e.g. high k means more stiffness 5

6. 3.2 The Free-Body Diagram Cables and Pulley
A cable or cord) is assumed to be weightless and inextensible
The direction of a force is along a cable, and in tension
The tension force is constant when in equilibrium
For any angle θ, the tension T is constant througout the entire calbe 6

7. 3.2 The Free-Body Diagram To construct an FBD 1. Sketch a system
2. Sketch all forces
- Active forces is the force to move a particle
- Reactive forces is a force resisting the movement of a particle, i.e. due to the attachments (e.g. a hook)
3. Lable all forces (if known) with magnitude and direction 7

8. Example 3.1
The sphere has a mass of 6kg and is supported by a system shown. Draw a free-body diagram of the sphere, the cord CE and the knot at C. 8

9. Solution
FBD at Sphere
Two forces acting, weight and the force on cord CE.
Weight of 6kg (9.81m/s2) = 58.9N
Cord CE
Two forces acting: sphere and knot
Newton’s 3rd Law: FCE is equal but opposite
FCE and FEC pull the cord in tension
For equilibrium, FCE = FEC 9

10. Solution FBD at Knot 3 forces acting: cord CBA, cord CE and spring CD
The weight of the sphere is not acting directly onto C but acting through the cable CE 10

11. 3.3 Coplanar Systems
All forces acting on a particle is in the x-y plane
We can split the forces in i and j components and consider for each direction in equilibrium
For a scalar notion, the sum of forces in x and y direction must be zero if a particle is in equilibrium 11

12. 3.3 Coplanar Systems Procedure 1. Free-Body Diagram
- Define where x, y is going to be located
- Label all unknown and known forces with magnitude and/or directions
2. Apple an equilibrium equation
- If the system contains a spring, use F = ks for a force acting by a spring.
- Equations of equilibrium
∑Fx = 0 ∑Fy = 0 12

13. Example 3.4
Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is L’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m. 13

14. Solution
FBD at Point A
Three forces acting, force by cable AC, force in spring AB and weight of the lamp.
If force on cable AB is known, stretch of the spring is found by F = ks. 14

15. Solution 15

16. 3.4 Three-Dimensional Force Systems
For particle equilibrium
∑F = 0
Split all forces into i, j and k components
∑Fxi + ∑Fyj + ∑Fzk = 0
The sum of forces in each direction, x, y, z
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0 16

17. 3.4 Three-Dimensional Force Systems
Procedure for Analysis
Free-body Diagram
Define the location of x, y, z
Label all unknown & known forces with magnitude and direction if known
Equations of Equilibrium
Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 or
in a Cartesian vector notation ∑F = 0
and the forces in each direction i, j, k must be 0
If the answer is negative, then the direction of a force is opposite to what it is defined in the FBD 17

18. Example 3.7
Determine the force developed in each cable used to support the 40kN crate 18

19. Solution FBD at Point A Equations of Equilibrium A (0, 0, 0)
C (-3, 4, 8)
D (+, 0, 0)
FBD at Point A
Equations of Equilibrium 19

20. Solution 20