### engr2320_3_1-4

```EQUILIBRIUM OF A PARTICLE, THE FREE-BODY
DIAGRAM & COPLANAR FORCE SYSTEMS
Today’s Objectives:
Students will be able to :
a) Draw a free body diagram (FBD), and,
b) Apply equations of equilibrium to solve
a 2-D problem.
In-Class Activities:
• Applications
• What, Why and How of
a FBD
• Equations of Equilibrium
• Analysis of Spring and
Pulleys
• Concept Quiz
• Group Problem Solving
• Attention Quiz
1) When a particle is in equilibrium, the sum of forces acting
on it equals ___ . (Choose the most appropriate answer)
A) A constant
D) A negative number
B) A positive number
E) An integer
C) Zero
2) For a frictionless pulley and cable, tensions in the cable
(T1 and T2) are related as _____ .
A) T1 > T2
B) T1 = T2
C) T1 < T2
D) T1 = T2 sin 
T1
T2
APPLICATIONS
The crane is lifting a load. To
decide if the straps holding the
load to the crane hook will fail,
you need to know the force in the
straps. How could you find the
forces?
Straps
APPLICATIONS
(continued)
For a spool of given
weight, how would you
find the forces in cables
AB and AC ? If designing
one, you need to know the
forces to make sure the
rigging doesn’t fail.
APPLICATIONS
(continued)
For a given force exerted on the boat’s towing pendant, what are
the forces in the bridle cables? What size of cable must you use?
COPLANAR FORCE SYSTEMS
(Section 3.3)
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 the cylinder, you need to
learn how to draw a free body
diagram and apply 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.
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
FB
30˚
FD A
A
x
FC = 392.4 N (What is this?)
Note : Cylinder mass = 40 Kg
EQUATIONS OF 2-D EQUILIBRIUM
FBD at A
y
FB
Since particle A is in equilibrium, the
net force at A is zero.
30˚
FD A
A
x
A
FC = 392.4 N
So FB + FC + FD = 0
or  F = 0
FBD at A
In general, for a particle in equilibrium,
 F = 0 or
 Fx i +  Fy j = 0 = 0 i + 0 j (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.
EXAMPLE
y
FBD at A
FB
FD A
A
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 ←
SPRINGS, CABLES, AND PULLEYS
T1
T2
Spring Force = spring constant *
deformation, or
F=k*s
With a frictionless pulley,
T1 = T2.
EXAMPLE
Given: Cylinder E weighs
30 lb and the
geometry is as
shown.
Find: Forces in the cables
and weight of
cylinder F.
Plan:
1. Draw a FBD for Point C.
2. Apply E-of-E at Point C to solve for the unknowns (FCB &
FCD).
3. Knowing FCB , repeat this process at point B.
EXAMPLE
(continued)
y
FCD
x
30
15
FBC
A FBD at C should look like the one at
the left. Note the assumed directions for
the two cable tensions.
30 lb
The scalar E-of-E are:
+   Fx = FBC cos 15º – FCD cos 30º = 0
+   Fy = FCD sin 30º – FBC sin 15º – 30 = 0
Solving these two simultaneous equations for the
two unknowns FBC and FCD yields:
FBC = 100.4 lb
FCD = 112.0 lb
EXAMPLE (continued)
y
FBC =100.4 lb
FBA
45
15
x
Now move on to ring B.
A FBD for B should look
like the one to the left.
WF
The scalar E-of-E are:
   Fx = FBA cos 45 – 100.4 cos 15 = 0
   Fy = FBA sin 45 + 100.4 sin 15 – WF = 0
Solving the first equation and then the second yields
FBA = 137 lb
and WF = 123 lb
CONCEPT QUESTIONS
1000 lb
1000 lb
1000 lb
(A)
(B)
(C)
1) Assuming you know the geometry of the ropes, you cannot
determine the forces in the cables in which system above?
2) Why?
A) The weight is too heavy.
B) The cables are too thin.
C) There are more unknowns than equations.
D) There are too few cables for a 1000 lb
weight.
GROUP PROBLEM SOLVING
Given: The box weighs 550 lb 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.
GROUP PROBLEM SOLVING
(continued)
y
FB
FC
5
30˚
FBD at point A
3
4
A
x
FD = 550 lb
Applying the scalar E-of-E at A, we get;
+   F x = FB cos 30° – FC (4/5) = 0
+   F y = FB sin 30° + FC (3/5) - 550 lb = 0
Solving the above equations, we get;
FB = 478 lb and FC = 518 lb
ATTENTION QUIZ
1. Select the correct FBD of particle A.
30
A 40
100 lb
F1
A)
A
F2
B)
30
40°
100 lb
A
F
C)
30°
D)
F2
40°
F1
30°
A
A
100 lb
100 lb
ATTENTION QUIZ
2. Using this FBD of Point C, the sum of
forces in the x-direction ( FX) is ___ .
Use a sign convention of +  .
A) F2 sin 50° – 20 = 0
B) F2 cos 50° – 20 = 0
C) F2 sin 50° – F1 = 0
D) F2 cos 50° + 20 = 0
F2
20 lb
50°
C
F1
THREE-DIMENSIONAL FORCE SYSTEMS
Today’s Objectives:
Students will be able to solve 3-D particle equilibrium problems by
a) Drawing a 3-D free body diagram, and,
b) Applying the three scalar equations (based on one vector
equation) of equilibrium.
In-class Activities:
• Check Homework
• Applications
• Equations of Equilibrium
• Concept Questions
• Group Problem Solving
• Attention Quiz
1. Particle P is in equilibrium with five (5) forces acting on it in
3-D space. How many scalar equations of equilibrium can be
written for point P?
A) 2
B) 3
D) 5
E) 6
C) 4
2. In 3-D, when a particle is in equilibrium, which of the
following equations apply?
A) ( Fx) i + ( Fy) j + ( Fz) k = 0
B)  F = 0
C)  Fx =  Fy =  Fz = 0
D) All of the above.
E) None of the above.
APPLICATIONS
You know the weights of
the electromagnet and its
know the forces in the
chains to see if it is a safe
assembly. How would you
do this?
APPLICATIONS
(continued)
Offset distance
This shear leg derrick
is to be designed to lift
a maximum of 200 kg
of fish.
How would you find
the effect of different
offset distances on the
forces in the cable and
derrick legs?
THE EQUATIONS OF 3-D EQUILIBRIUM
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 #1
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.
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 #1
(continued)
F1 = {300(4/5) j + 300 (3/5) k} N
F1 = {240 j + 180 k} N
F2 = {– 600 i} N
F3 = {– 900 k} N
F4 = F4 (rB/ rB)
= 200 N [(3i – 4 j + 6 k)/(32 + 42 + 62)½]
= {76.8 i – 102.4 j + 153.6 k} N
F5 = { Fx i – Fy j + Fz k} N
EXAMPLE #1
(continued)
Equating the respective i, j, k components to zero, we have
Fx = 76.8 – 600 + Fx
= 0;
Fy = 240 – 102.4 + Fy = 0 ;
solving gives Fx = 523.2 N
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.
EXAMPLE #2
Given: A 600 N load is supported
by three cords with the
geometry as shown.
Find: The tension in cords AB,
Plan:
1) Draw a free body diagram of Point A. Let the unknown force
magnitudes be FB, FC, FD .
2) Represent each force in the Cartesian vector form.
3) Apply equilibrium equations to solve for the three unknowns.
EXAMPLE #2 (continued)
FBD at A
FD
z
FC
2m
1m
2m
A
30˚
y
FB
x
600 N
FB = FB (sin 30 i + cos 30 j) N
= {0.5 FB i + 0.866 FB j} N
FC = – FC i N
= FD { (1 i – 2 j + 2 k) / (12 + 22 + 22)½ } N
= { 0.333 FD i – 0.667 FD j + 0.667 FD k } N
EXAMPLE #2 (continued)
Now equate the respective i , j , k
components to zero.
 Fx = 0.5 FB – FC + 0.333 FD = 0
 Fy = 0.866 FB – 0.667 FD = 0
 Fz = 0.667 FD – 600 = 0
FBD at A
FD
FD = 900 N
FB = 693 N
FC
2m
y
1m
2m
A
30˚
FB
x
Solving the three simultaneous equations yields
FC = 646 N
z
600 N
CONCEPT QUIZ
1. In 3-D, when you know the direction of a force but not its
magnitude, how many unknowns corresponding to that force
remain?
A) One
B) Two
C) Three
D) Four
2. If a particle has 3-D forces acting on it and is in static
equilibrium, the components of the resultant force ( Fx,  Fy,
and  Fz ) ___ .
A) have to sum to zero, e.g., -5 i + 3 j + 2 k
B) have to equal zero, e.g., 0 i + 0 j + 0 k
C) have to be positive, e.g., 5 i + 5 j + 5 k
D) have to be negative, e.g., -5 i - 5 j - 5 k
GROUP PROBLEM SOLVING
Given: A 3500 lb motor and
plate, as shown, are in
equilibrium and supported
by three cables and
d = 4 ft.
Find: Magnitude of the tension
in each of the cables.
Plan:
1) Draw a free body diagram of Point A. Let the unknown force
magnitudes be FB, FC, F D .
2) Represent each force in the Cartesian vector form.
3) Apply equilibrium equations to solve for the three unknowns.
GROUP PROBLEM SOLVING (continued)
FBD of Point A
z
W
y
x
FD
W = load or weight of unit = 3500 k lb
FB
FC
FB = FB(rAB/rAB) = FB {(4 i – 3 j – 10 k) / (11.2)} lb
FC = FC (rAC/rAC) = FC { (3 j – 10 k) / (10.4 ) }lb
FD = FD( rAD/rAD) = FD { (– 4 i + 1 j –10 k) / (10.8) }lb
GROUP PROBLEM SOLVING (continued)
The particle A is in equilibrium, hence
FB + FC + FD + W = 0
Now equate the respective i, j, k components to zero
(i.e., apply the three scalar equations of equilibrium).
 Fx = (4/ 11.2)FB – (4/ 10.8)FD = 0
 Fy = (– 3/ 11.2)FB + (3/ 10.4)FC + (1/ 10.8)FD = 0
 Fz = (– 10/ 11.2)FB – (10/ 10.4)FC – (10/ 10.8)FD + 3500 = 0
Solving the three simultaneous equations gives
FB = 1467 lb
FC = 914 lb
FD = 1420 lb
ATTENTION QUIZ
1. Four forces act at point A and point
A is in equilibrium. Select the correct
force vector P.
A) {-20 i + 10 j – 10 k}lb
z
F3 = 10 lb
P
F1 = 20 lb
F2 = 10 lb
A
y
B) {-10 i – 20 j – 10 k} lb
C) {+ 20 i – 10 j – 10 k}lb
x
D) None of the above.
2. In 3-D, when you don’t know the direction or the magnitude
of a force, how many unknowns do you have corresponding
to that force?
A) One
B) Two
C) Three D) Four
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