Download Equilibrium of a Particle

ME 108 - Statics
Equilibrium of a Particle
Chapter 3
Applications
For a spool of given
weight, what are the
forces in cables AB
and AC ?
Applications
For a given weight of
the lights, what are
the forces in the
cables? What size of
cable must you use ?
Applications
Equilibrium of 2D Particle
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.
The What, Why, 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 helps you write the
equations of equilibrium used
to solve for the unknowns
(usually forces or angles).
The What, Why, How of a Free
Body Diagram (FBD)
Why ? 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.
FBD at A
Note : Engine mass = 250 Kg
Equations of 2D Equilibrium
Since particle A is in equilibrium,
the net force at A is zero.
Written in a scalar form,
SFx = 0 and S Fy = 0
These are two scalar equations of equilibrium (EofE).
They can be used to solve for up to two unknowns.
Write the scalar EofE:
+ S Fx = TB cos 30º – TD = 0
+ SFy = 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
Springs, Cables, and Pulleys
Spring Force = spring constant *
deformation, or
F=k*S
With a
frictionless
pulley, T1 = T2.
Example 1
Given: Sack A weighs 20 N
and geometry is as shown.
Find: Forces in the cables and
weight of sack B.
Plan:
1. Draw a FBD for Point E.
2. Apply EofE at Point E to solve
for the unknowns (TEG & TEC).
3. Repeat this process at C.
Solution – Example 2
A FBD at E should look like the one to
the left. Note the assumed directions
for the two cable tensions.
The scalar EofE are:
+  S Fx = TEG sin 30º – TEC cos 45º = 0
+ S Fy = TEGcos 30º – TEC sin 45º – 20 N = 0
Solving these two simultaneous equations
for the two unknowns yields:
TEC = 38.6 N
TEG = 54.6 N
Solution – Example 2
Now move on to ring C.
A FBD for C should look
like the one to the left.
The scalar EofE are:
 S Fx = 38.64 cos 45 – (4/5) TCD = 0
 S Fy = (3/5) TCD + 38.64 sin 45 – WB = 0
Solving the first equation and then the second yields
TCD = 34.2 N and WB = 47.8 N .
Equilibrium of a Rigid Body
• Applications
• Equilibrium in 2D
• Support reactions
• Free-body diagram
Applications
A 200 kg platform is suspended off an oil rig. How do we
determine the force reactions at the joints and the forces in
the cables?
How are the idealized model and the free body diagram used
to do this? Which diagram above is the idealized model?
Applications
A steel beam is used to support
roof joists. How can we
determine the support reactions
at A & B?
Again, how can we make use of an idealized model and a
free body diagram to answer this question?
Conditions for Rigid-Body
Equilibrium
Statics deals primarily with the description of the
force conditions necessary and sufficient to
maintain the equilibrium of engineering structures.
When a body is in equilibrium, the resultant of all
forces acting on it is zero. Thus, the resultant force
R and the resultant couple M are both zero, and
we have the equilibrium equations
R  F  0
M  M  0
Conditions for Rigid-Body
Equilibrium
Equilibrium of a rigid body requires
ΣF = 0 and ΣM = 0
or
ΣFx=0
ΣFy=0
ΣFz=0
ΣMx=0
ΣMy=0
ΣMz=0
So, in 3-D, we have six equilibrium equations.
In 2-D, we have three equilibrium equations.
Equilibrium in Two Dimensions
• Required and necessary condition is
ΣFx=0
ΣFy=0
ΣMz=0
Forces
In these summations, all of the following forces need to be
included:
• external forces: loads applied to the structure by the
environment (e.g., weight, service loads, etc.)
• internal forces: forces in structural members and
connections that are generated by straining of material.
• reactions:* forces that support the structure as a whole.
* Reactions are also internal forces, but to draw attention
to their importance, we list them separately.
Reactions in 2D
1. Flexible cable, belt, chain, or rope
Force exerted by a flexible cable
is always a tension away from the
body in the direction of the cable.
Reactions in 2D
2. Smooth surface:
Contact force is
compressive and is
normal to the surface.
3. Rough surface:
Rough surfaces are capable
of supporting a tangential
component F (frictional
force) as well as a normal
component N of the
resultant contact force R.
Reactions in 2D
4. Roller support
Roller, rocker, or ball support
transmits a compressive force
normal to the supporting
surface
Reactions in 2D
5. Freely sliding guide
Collar or slider free to move
along smooth guides; can
support force normal to guide
only.
Reactions in 2D
6. Pin connection
A freely hinged pin connection is capable of supporting a
force in any direction in the plane normal to the axis;
usually shown as two components Rx and Ry. A pin not
free to turn may also support a couple M.
Reactions in 2D
7. Built-in or fixed support
A built-in or fixed support is
capable of supporting an axial
force F, a transverse force V
(shear force), and a couple M
(bending moment) to prevent
rotation.
Reactions in 2D
8. Gravitational attraction
Reactions in 2D
9. Spring action
Spring force is tensile if spring is
stretched and compressive if
compressed. For a linearly elastic
spring the stiffness k is the force
required to deform the spring a unit
distance.
Reaction forces
• If a support prevents translation of a body
in a given direction, then a force is
developed on the body in the opposite
direction.
• Similarly, if rotation is prevented, a couple
moment is exerted on the body.
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 helps you write the equations of
equilibrium used to solve for the unknowns
(usually forces or angles).
The Process of Solving Rigid-Body
Equilibrium Problems
Step1: For analyzing an actual physical system, first we need
to create an idealized model.
Step2: Then we need to draw a free-body diagram showing all
the external (active and reactive) forces.
Step3: Finally, we need to apply the equations of equilibrium
to solve for any unknowns.
Procedure for Drawing a Free-Body
Diagram
1. Draw an outlined shape. Imagine the body to be
isolated or cut “free” from its constraints and draw its
outlined shape.
2. Show all the external forces and couple moments.
These typically include: a) applied loads, b) support
reactions, and, c) the weight of the body.
Procedure for Drawing a Free-Body
Diagram
3. Label loads and dimensions: All known forces and
couple moments should be labeled with their
magnitudes and directions. For the unknown forces and
couple moments, use letters like Ax, Ay, MA, etc.. Indicate
any necessary dimensions.
Procedure for drawing FBDs
(Book Version)
1. Decide on the rigid body (or portion of a rigid body) whose
equilibrium you want to analyze.
2. Imagine that this body is cut completely free (separated) from the
rest of the structure and/or its environment.
• in 2-D, think of a closed line that completely encircles the body.
• in 3-D, think of a closed surface that completely surrounds the body.
3. Sketch the body.
4. Sketch all external forces that are applied to the body.
5. Wherever the cut passes through a structural member, sketch the
internal forces that occur at that location.
6. Wherever the cut passes through a support, sketch the support
reactions that occur at that location.
Example
Example
Example
Example
Given: The link is pin-connected
at A and rests against a smooth
support at B.
Find: Horizontal and vertical
components of reactions at pin A.
Plan:
1. Put the x and y axes in the horizontal and vertical
directions, respectively.
2. Draw a complete FBD of the boom.
3. Apply the EofE to solve for the unknowns.
Solution
FBD
• Reaction N B is perpendicular to
the link at B
• Horizontal and vertical
components of reaction are
represented at A
Example
Given: Weight of the boom =
125 N, the center of mass is at
G, and the load = 600 N.
Find: Support reactions at A
and B.
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 EofE to solve for the unknowns.
Solution
Example
Draw the FBD for the simply supported I-beam shown.
The cross section is a W14x26 shape which has a weight
of 26 N/m.
y
6.67 m
y
z
2000 N
B
A
L = 10 m
x
y
6.67 m
y
z
2000 N
B
A
x
L = 10 m
260 N
y
Ax
2000 N
x
5m
Ay
6.67 m
By
y
Now apply
equilibrium
equations:
260 N
2000 N
Ax
x
5m
Ay
6.67 m
By
 Fx = 0 = Ax 
 Fy = 0 = Ay + By - 260 N - 2000 N
 MA = 0 = (260 N) 5 m + (2000 N) 6.67 m - By (10 m)
 By =14640 N.m/10 m = 1464 N
 Ay = 796 N
Note: we could have used the equations  Fx ,  MA and  MB to
obtain our solution.
Examples
A
B
y
C
x
W2
A
C
y
W1
B
Draw FBD for ACB
W
D
x
E
Draw FBD for ABCD
Example:
Neglecting friction,
determine the tension
in cable ABD and the
reaction at support C.
F
A
A:
Cx = 80 N 
Cy = 40 N 
D
C
100 mm 100 mm
F=120 N
250 mm
B
Example:
Rod AB is attached to
a frictionless collar at
A. Neglecting the
weight of AB,
determine angle .
P = 16 N
Q = 12 N
l = 20 cm
a = 5 cm
A:  = ??
Q
A

l
C
B
a
P
Static Indeterminacy
Statically indeterminate structures and mechanisms:
2-D
statically
indeterminate
example:
Find the
support
reactions in
terms of P.
P
y
x
L/2
L/2
Constraints - support conditions
For a structure to be in static equilibrium under general
loading, the number of support reactions must be equal to or
greater than the number of equilibrium equations (three in 2D and six in 3-D). This is a necessary but not sufficient
condition.
Redundant supports: When a structure has more supports
than is required for static equilibrium, it becomes statically
indeterminate.
Improper supports: For static equilibrium, the constraints
must be sufficient in number and arrangement so that the
structure has no rigid body motion capability. A structure
that has rigid body motion capability is called a mechanism.
example: For the systems shown below, determine the
degree of redundancy (i.e., the number of support
conditions beyond those required for equilibrium).
cable
(a)
(b)
(c)
cable
(d)
(e)
(a)
# constraints
redundancy
(b)
(c)
(f)
(d)
(e)
(f)
example: For the systems shown below, determine the
degree of redundancy (i.e., the number of support
conditions beyond those required for equilibrium).
cable
(a)
(b)
(c)
cable
(d)
(a)
# constraints 3
redundancy 0
(e)
(b)
(c)
2 or 1 4
-1 or -2 1
(f)
(d)
3
0
(e)
4 or 3
1 or 0
(f)
3
?
Two- and Three-Force Members
Simplify some equilibrium problems by
recognizing members that are subjected to
only 2 or 3 forces
Two-Force Members
When a member is subject to no couple
moments and forces are applied at only two
points on a member, the member is called
a two-force member
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.
Example of Two-Force Members
If we neglect the weight, the members can be treated as two-force
members. 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).
Two- and Three-Force Members
Three-Force Members
If a member is subjected to only three forces, it is necessary
that the forces be either concurrent or parallel for the
member to be in equilibrium
Remarks on pin-connected straight bars, cables and ropes:
Consider a pin-connected
bar. Observe that the
internal force supported by
the bar is purely axial. That
is, the internal force in the
bar is always directed along
the axis of the bar.
• Identical remarks apply to
cables and ropes.
• These are examples of
simple, but common, twoforce members.
Draw the free-body diagram of member ABC which is
supported by a smooth collar at A, roller at B, and
short link CD. Explain the significance of each force
acting on the diagram.
Determine the
horizontal and vertical
components of
reaction at the pin A
and the tension
developed in cable BC
used to support the
steel frame.
The mass of 700
kg is suspended
from a trolley
which moves
along the crane
rail from d = 1.7 m
to d = 3.5 m.
Determine the force along the pin connected knee strut BC
(short link) and the magnitude of force at pin A as a function
of positiion d. Plot these results of FBC and FA (vertical axis)
versus d (horizantal axis).
The jib crane is
supported by a pin at
C and rod AB. İf the
load has a mass of 2
Mg with its center of
mass located at G,
determine the
horizontal and vertical
components of
reaction at the pin C
and the force
developed in rod AB.
Example
Example
Example
Example
If the breaking strength of
cables AC and CE is 750
N and the breaking
strength of cables CD
and AB is 400 N, what is
the maximum mass M
that can be safely
supported by this system
of cables? (The cables
are clamped at joints A
and C.)