Announcements
Equilibrium of a Particle in 2-D
Today’s Objectives
• Draw a free body diagram (FBD)
• Apply equations of equilibrium to solve a
2-D problem
Class Activities
• Applications
• What, why, and how of a Free
Body Diagram
• Equations of equilibrium in 2-D
• Springs and pulleys
• Examples
Engr221 Chapter 3
1
Applications
For a spool of given
weight, what are the
forces in cables AB
and AC?
Applications - continued
For a given cable
strength, what is the
maximum weight
that can be lifted?
Engr221 Chapter 3
2
Applications - continued
For a given engine weight, what are the forces in the
cables? What size of cable should you use?
Equilibrium of a Particle in 2-D
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 engine, we need to
learn how to draw a Free
Body Diagram and apply
equations of equilibrium.
Engr221 Chapter 3
3
The What, Why, and How of a Free
Body Diagram
Free Body Diagrams are one of the most important things
for you to know how to draw, and how to use.
What is it? - A drawing that
shows all external forces acting
on a particle.
Why draw it? - It helps you write
the equations of equilibrium used
to solve for the unknowns (usually
forces or angles).
How?
• Imagine the particle to be isolated or cut free from its
surroundings
• Show all the forces that act on the particle
- Active forces: They tend to move the particle
- Reactive forces: They tend to resist the motion
• Identify each force and show all known magnitudes
and directions as variables
A
Engine mass = 250 Kg
Engr221 Chapter 3
FBD at A
4
Equations of 2-D Equilibrium
Since particle A is in equilibrium,
the net force at A is zero.
A
ΣF = 0 (vector equation)
A
or TB + TD + W = 0
FBD at A
In general, for a particle in equilibrium, ΣF = 0 or
ΣFx i + ΣFy j = 0 = 0 i + 0 j (A vector equation)
Or written in scalar form,
ΣFx = 0 and ΣFy = 0
These are two scalar equations of equilibrium (E of E).
They can be used to solve for up to two unknowns.
Example A
Engine mass = 250 Kg
FBD at A
Write the scalar Equations of Equilibrium (E of E):
+ → ΣFx = TB cos 30º – TD = 0
+ ↑ ΣFy = TB sin 30º – 2.452 kN = 0
Solving the second equation gives: TB = 4.90 kN
From the first equation, we get: TD = 4.25 kN
Engr221 Chapter 3
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Springs
Spring Force = spring constant × deformation
F=k ×s
Cables and Pulleys
With a frictionless pulley, tensions are equal
Engr221 Chapter 3
6
Example B
Given: Sack A weighs 20 lb with
geometry as shown
Find: Forces in the cables and
weight of sack B
Plan:
1. Draw a FBD at point E
2. Apply E of E at point E to solve
for the unknowns (TEG & TEC)
3. Repeat this process at point C
Example B - continued
A FBD at E should look like
the one to the left. Note the
assumed directions for the two
cable tensions.
The scalar E of E are:
+ → ΣFx = TEG sin 30º – TEC cos 45º = 0
+ ↑ ΣFy = TEG cos 30º – TEC sin 45º – 20 lbs = 0
Solving these two simultaneous equations for the
two unknowns yields:
TEC = 38.6 lb
TEG = 54.6 lb
Engr221 Chapter 3
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Example B - continued
Now, move on to ring C.
A FBD for C should look
like the one to the left.
The scalar E of E are:
+ → Σ Fx = 38.64 cos 45° – (4/5) TCD = 0
+ ↑ Σ Fy = (3/5) TCD + 38.64 sin 45° – WB = 0
Solving the first equation and then the second yields:
TCD = 34.2 lb and WB = 47.8 lb
Example C
Given: The car is towed at constant
speed by the 600 lb force
The angle θ is 25°
Find:
The forces in the ropes AB
and AC
Plan:
1. Draw a FBD at point A
2. Apply the E of E to solve for the forces
in ropes AB and AC
Engr221 Chapter 3
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Example C - continued
600 lb
FBD at point A
A
25°
FAB
30°
FAC
Applying the E of E at point A, we get:
+ → ∑ Fx = FAC cos 30° – FAB cos 25° = 0
+ ↑ ∑ Fy = -FAC sin 30° – FAB sin 25° + 600 = 0
Solving the above equations, we get:
FAB = 634 lb
FAC = 664 lb
Questions
1) When a particle is in equilibrium, the sum of forces acting
on it equals ___
A) a constant
B) a positive number C) zero
D) a negative number
E) an integer
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 θ
Engr221 Chapter 3
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Question
Select the correct FBD of particle A.
30°
A
40°
100 lb
F1 F2
A)
A
B)
40°
30°
100 lb
A
F
C)
30°
D)
F2
40°
F1
30°
A
A
100 lb
100 lb
Question
Using the FBD for point C, the sum of
forces in the x-direction (Σ FX) is ___ . Use
a sign convention of + →
A) F2 sin 50° – 20 = 0
20 lb
B) F2 cos 50° – 20 = 0
C) F2 sin 50° – F1 = 0
D) F2 cos 50° + 20 = 0
Engr221 Chapter 3
F2
50°
C
F1
10
Textbook Problem 3-10
The 500-lb crate is hoisted using the ropes AB and AC. Each
rope can withstand a maximum tension of 2500 lb before it
breaks. If AB always remains horizontal, determine the
smallest angle θ to which the crate can be hoisted.
For TAB = 2500 lb, θ = 11.31º
For TAC = 2500 lb, θ = 11.54º
Summary
• Draw a free body diagram (FBD)
• Apply equations of equilibrium to solve a 2-D problem
Engr221 Chapter 3
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Announcements
• F = mg
• F = ks
• Test Monday
Equations of 2-D Equilibrium - Review
Since particle A is in equilibrium,
the net force at A is zero.
A
ΣF = 0 (vector equation)
A
or TB + TD + W = 0
FBD at A
In general, for a particle in equilibrium, ΣF = 0 or
ΣFx i + ΣFy j = 0 = 0 i + 0 j (A vector equation)
Or written in scalar form,
ΣFx = 0 and ΣFy = 0
These are two scalar equations of equilibrium (E of E).
They can be used to solve for up to two unknowns.
Engr221 Chapter 3
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Free Body Diagrams - Review
• Imagine the particle to be isolated or cut free from its
surroundings
• Show all the forces that act on the particle
- Active forces: They tend to move the particle
- Reactive forces: They tend to resist the motion
• Identify each force and show all known magnitudes
and directions as variables
A
Engine mass = 250 Kg
FBD at A
Textbook Problem 3-14
The unstretched length of spring AB is 2 m. If the
block is held in the equilibrium position shown,
determine the mass of the block at D.
FAB
FAC
W
Mass
Engr221 Chapter 3
= 90 N
= 102 N
= 126 N
= 12.8 kg
13
Textbook Problem 3-32
Determine the un-stretched length of spring AC if a
force P = 80 lb causes the angle θ = 60º for
equilibrium. Cord AB is 2 ft long. Take k = 50 lb/ft.
Fs = 40 lb
Luns = 2.66 ft
Equations of 3-D Equilibrium
Since the particle is in equilibrium, the
net force at the origin is zero.
So F1 + F2 + F3 = 0
A
or ΣF = 0
FBD at A
In general, for a particle in equilibrium in 3-D, ΣF = 0 or
ΣFx i + ΣFy j + ΣFz k = 0 = 0 i + 0 j + 0 k
Or written in scalar form:
ΣFx = 0, ΣFy = 0, and ΣFz = 0
These are the three scalar equations of equilibrium. They
can be used to solve for up to three unknowns.
Engr221 Chapter 3
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Procedure for Analysis
Free-Body Diagram
• Establish the x, y, and z axes in any suitable orientation
• Label all known and unknown force magnitudes and directions
on the diagram
• Assume senses for unknown magnitudes
Equations of Equilibrium
• Use scalar equations of equilibrium when it is easy to resolve
each force into its x, y, and z components (not often)
• If the 3-D geometry is difficult, express the forces in Cartesian
form, use ΣF = 0, and equate the i, j, and k components to 0
• If the answer is negative, the assumed sense should be reversed
Example 3-D Problem
The 50-kg pot is supported from A by the three
cables. Determine the force acting in each cable
for equilibrium. Take d = 2.5 meters.
FAB = 580 N
FAC = 312 N
FAD = 312 N
Engr221 Chapter 3
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Announcements
• Homework notes
• Test Monday
– One crib sheet (given)
– Calculator
– Pencil
– Eraser
– FBD’s, units, process are grading considerations
Procedure for Analysis - Review
Free-Body Diagram
• Establish the x, y, and z axes in any suitable orientation.
• Label all known and unknown force magnitudes and directions
on the diagram.
• Assume senses for unknown magnitudes.
Equations of Equilibrium
• Use scalar equations of equilibrium when it is easy to resolve
each force into its x, y, and z components.
• If the 3-D geometry is difficult, express the forces in Cartesian
form, use ΣF = 0, and set the i, j, and k components to 0.
• If the answer is negative, the assumed sense should be reversed.
Engr221 Chapter 3
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Textbook Example 3.8
Textbook Example 3.8 - continued
Engr221 Chapter 3
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Example Problem
A small peg P rests on a spring that is contained inside the smooth
pipe. When the spring is compressed so that s = 0.15 m, the spring
exerts an upward force of 60 N on the peg. Determine the point of
attachment A(x,y,0) of cord PA so that the tension in cords PB and
PC equals 30 N and 50 N, respectively.
x = 0.1904 m
y = 0.0123 m
Example Problem
Determine the tension developed in cables OD and OB and the strut
OC, required to support the 50-kg crate. The spring OA has an
unstretched length of 0.8 m and a stiffness of kOA = 1.2 kN/m. The
force in the strut acts along the axis of the strut.
Answers TBD
Engr221 Chapter 3
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Chapter 3 Summary
• Draw a free body diagram (FBD)
• Apply equations of equilibrium to solve a 2-D problem
• Apply equations of equilibrium to solve a 3-D problem
Engr221 Chapter 3
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