# Equilibrium of Concurrent, Coplanar Force Systems Powerpoint

```Equilibrium of
Concurrent,
Coplanar Force
Systems
EF 202 - Week 5
1
Equilibrium
•
Newton’s First Law - If, and only if, an
object’s “mass center” has zero
acceleration, then the sum of ALL of
the forces acting on the object (body)
is zero.
•
Where is a body’s mass center? We’ll
study that in Module 4. But in this
class, the entire body is at rest, so we
know that the mass center, wherever it
is, has zero acceleration.
2
Problem Solving
Strategy
1. Identify ALL forces acting on the
body by making a “free-body
diagram” (FBD).
2. Invoke Newton’s First Law. (Add all
the force vectors on the FBD and
set the result equal to zero.)
3. Solve.
3
Free-Body Diagram
1. Define the “body.”
2. “Free” the body from the rest of the
world by cutting through
EVERYTHING that connects the body
to anything else.
3. Sketch the freed body.
4. To the sketch, add EVERY force that is
(or may be) acting on the body.
4
Concurrent, Coplanar
Force System
•
Coplanar system
•
•
The lines of action of all forces lie in
a common plane.
Concurrent system
•
The lines of action of all forces
intersect at a common point.
5
Independent
Equations
•
In a given problem, we can find only as
many unknowns as we have
independent equations.
•
For a system of coplanar, concurrent
forces, Newton’s First Law yields only
two independent equations for a given
FBD.
6
Problem 3-4
Find magnitude and direction of
F.
1. Make FBD.
2. Newton’s First Law
3. Solve.
1. Figure is already an FBD.
2. F  0


7.5 kNsin 30 ˆi  cos 30 ˆj
2.25 kNcos60 ˆi  sin 60 ˆj 0
F cos ˆi  sin ˆj  4.5 kNˆi 
7
3. Collect and equate components.
For the ˆi components:
F cos  4.5 kN  7.5 kN sin 30  2.25 kN cos60  0
(1)
For the ˆj components:
F sin   7.5 kN cos 30  2.25 kNsin 60  0
(2)
Now, what?
COUNT EQUATIONS AND UNKNOWNS!!!!!!!
Two equations, (1) and (2), and two unknowns, F

and .
8
From (1) and (2),
F cos  7.125 kN
(3)
F sin   8.444 kN
(4)
Now, what?
For the angle, divide (4) by (3) to
get
tan  1.185    49.8
For the magnitude,
F F F 
2
x
2
y
F cos  F sin  
2
F 11.0 kN
9
2
Ideal Cable
•
•
Neglect weight (massless).
Neglect bending stiffness.
•
•
•
Force parallel to cable.
Force only tensile.
Neglect stretching (inextensible).
10
Problem 3-8
Find the forces in cables AB and
AC.
1. Make FBD.
2. Newton’s First Law
3. Solve.
12 kg
11
COUNT EQUATIONS AND UNKNOWNS!!!!!!!
12
13
14
Problem 3-40
Find the forces in all five cables.
Since there are 5 unknowns,
we’ll need at least 3 FBD’s.
30kg
15
16
17
Linear Spring - I
•
Like an ideal cable, but extensible and
resists either stretching or shortening.
•
•
Force is either tensile or compressive.
•
When the force is not zero, it is
proportional to the spring’s deformation
(stretch or contraction).
When the force is zero, the spring is
undeformed and its length is called its
free length or natural length.
18
Linear Spring - II
•
The constant of proportionality
between the force in a spring and its
deformation is called the spring’s
stiffness or its spring rate.
•
Stiffness is usually represented by the
letter k.
19

Equation for Spring
Tension
Compression
OR
F  k l0  l
F  k l  l0 
stretch
contraction
l0  natural/free/undeformed length
(length when F = 0)

20
Assumed Tension
F  k l  l0 
•
Directions of loads are reversed on
objects to which springs are attached
(Newton 
III).
21
Assumed
Compression
F  k l0  l
•
•
cannot be wrong!

Equation used for spring load can be
wrong!
22
Problem 3-14
Spring AB is stretched 3m.
What is the mass of the
suspended block?
23
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