Chapter 3
Equilibrium of a Particle
Condition for the Equilibrium
of a Particle
• A particle is said to be in equilibrium if:
– remains at rest (originally at rest)
– has a constant velocity (originally in motion)
• Newton’s first law of motion → ΣF=0
• Newton’s second law of motion → ΣF=ma → a=0
Question: Does moving with constant speed implies static
conditions?
This is an example of a 2-D or
coplanar force system.
If the whole assembly is in
equilibrium, then particle A is
also in equilibrium.
To determine the tensions in
the cables for a given weight
of cylinder, you need to
learn how to draw a freebody diagram and apply the
equations of equilibrium.
THE WHAT, WHY, AND HOW OF A
FREE-BODY DIAGRAM (FBD)
Free-body diagrams are one of the most important things for
you to know how to draw and use for statics and other subjects!
What? - It is a drawing that shows all external forces
acting on the particle.
Why? - It is key to being able to write the equations of
equilibrium—which are used to solve for the unknowns
(usually forces or angles).
How?
1. Imagine the particle to be isolated or cut free from its
surroundings.
2. Show all the forces that act on the particle.
Active forces: They want to move the particle.
Reactive forces: They tend to resist the motion.
3. Identify each force and show all known magnitudes and
directions. Show all unknown magnitudes and / or
directions as variables.
y
FBD at A
FD A
Note : Cylinder mass = 40 Kg
A
FB
30˚
x
FC = 392.4 N (What is this?)
EQUATIONS OF 2-D EQUILIBRIUM
FBD at A
FD A
y
A
Since particle A is in equilibrium, the
net force at A is zero.
FB
30˚
x
So FB + FC + FD = 0
or
A
F=0
FC = 392.4 N
FBDatat
FBD
AA
In general, for a particle in equilibrium,
 F = 0 or
 F x i +  Fy j
=
0
= 0i + 0j
(a vector equation)
Or, written in a scalar form,
 Fx = 0 and  Fy = 0
These are two scalar equations of equilibrium (E-of-E).
They can be used to solve for up to two unknowns.
EQUATIONS OF 2-D EQUILIBRIUM (continued)
y
FBD at A
FB
A
FDA
30˚
x
FC = 392.4 N
Note : Cylinder mass = 40 Kg
Write the scalar E-of-E:
+ →  Fx = FB cos 30º – FD =
0
+   Fy = FB sin 30º – 392.4 N = 0
Solving the second equation gives: FB = 785 N
From the first equation, we get: FD = 680 N
←
SIMPLE SPRINGS
(linearly elastic spring)
the length of the spring will
change in direct proportion to
the force F acting on it.
Elongation or
compression
Spring Force = spring constant (or stiffness) x deformation of spring
or
F=k*s
CABLES AND PULLEYS
Unless otherwise stated, all cables (or cords):
1) Negligible weight
2) Can support only tension (or pulling force)
3) Force always acts in the direction of
the cable
T1
T2
With a frictionless pulley and cable
T1 = T2
for any θ
Example 3-1: The sphere in Fig. 3–3 a
has a mass of 6 kg and is supported as
shown. Draw a free-body diagram of
the sphere, the cord CE , and the knot
at C .
Sphere
Cord CE
Knot
EXAMPLE
Given: The box weighs 550 N and
geometry is as shown.
Find:
The forces in the ropes AB
and AC.
Plan:
1. Draw a FBD for point A.
2. Apply the E-of-E to solve for the forces in ropes AB
and AC.
EXAMPLE (continued)
y
FB
FC
5
30˚
FBD at point A
3
4
A
x
FD = 550 N
Applying the scalar E-of-E at A, we get;
+→  F x = FC (4/5) – FB cos 30°= 0
+↑  F y = FB sin 30° + FC (3/5) - 550 N = 0
Solving the above equations, we get;
FB = 478 N
and FC = 518 N
3–19 The unstretched length of spring AB is 3 m. If the block is held in the
equilibrium position shown, determine the mass of the block at D .
Chapter 3 Part 2
THE EQUATIONS OF 3-D EQUILIBRIUM of a Particle
When a particle is in equilibrium, the vector
sum of all the forces acting on it must be
zero ( F = 0 ) .
This equation can be written in terms of its
x, y, and z components. This form is written
as follows.
( Fx) i + ( Fy) j + ( Fz) k = 0
This vector equation will be satisfied only when
 Fx = 0
 Fy = 0
 Fz = 0
These equations are the three scalar equations of equilibrium.
They are valid for any point in equilibrium and allow you to
solve for up to three unknowns.
EXAMPLE
Given: The four forces and
geometry shown.
Find: The force F5 required to
keep particle O in
equilibrium.
Plan:
1) Draw a FBD of particle O if needed.
2) Write the unknown force as
F5 = {Fx i + Fy j + Fz k} N
3) Write F1, F2 , F3 , F4 , and F5 in Cartesian vector form.
4) Apply the three equilibrium equations to solve for the three
unknowns Fx, Fy, and Fz.
EXAMPLE (continued)
Solution:
F1 = {300(4/5) j + 300 (3/5) k} N
F1 = {240 j + 180 k} N
B
F2 = {– 600 i} N
F3 = {– 900 k} N
Point B (3,-4,6)
F4 = F4 (rB/ rB)
=
200 N [(3i – 4 j + 6 k)/(32 + 42 + 62)½]
= {76.8 i
F5 = { Fx i
– 102.4 j + 153.6 k} N
+ Fy j + Fz k} N
EXAMPLE (continued)
Equating the respective i, j, k components to zero, we have
 Fx = 76.8 – 600 + Fx
 Fy =
= 0 ; solving gives Fx = 523.2 N
240 – 102.4 + Fy = 0 ; solving gives Fy = – 137.6 N
 Fz = 180 – 900 + 153.6 + Fz = 0 ; solving gives Fz = 566.4 N
Thus, F5 = {523 i – 138 j + 566 k} N
Using this force vector, you can determine the force’s
magnitude and coordinate direction angles as needed.
3–67 The shear leg derrick is used to haul the
200-kg net of fish onto the dock. Determine the
compressive force along each of the legs AB
and CB and the tension in the winch cable DB.
Assume the force in each leg acts along its axis.
Solution:
A(2, 0, 0)m
C(-2, 0, 0)m
D(0, -5.6, 0)m B(0,4,4)m
Suggested Problems:
All Problems