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```									EGR 231 Engineering Statics:                                                             Fall 2011
Lecture 07: Equilibrium in 2-D

Today:
Homework Questions?
Newton’s 2nd Law
2-D equilibrium problems.
500N
Homework Problem Assignment 07:
Problem 2.44
Two cables are tied together at C and are loaded
as shown. Determine the tension
a) in cable AC, b) in cable BC.

y                       FB

Problem 2:51
A welded connection is in equilibrium under the action
of the four forces shown. Knowing that F A=8 kN and
FB = 16 kN, determine the magnitudes of the other two                                              FC
forces.                                                                                                 x

5      3
4
FA

FD

Problem 2.67:
A 600 lb crate is supported by several rope
and pulley arrangements as shown.
Determine for each arrangement the                 T                                                             T
tension in the rope.                                                                 T
T                                  T

(a              (b            (c                    (d            (e
)               )             )                     )             )
T               T             T                     T             T
Following Today’s Lecture you should be able to:
1) Write Newton’s 2nd law for the case of static equilibrium of particles.

2) Set up and find the forces needed to give static equilibrium of a particle.
Particle Equilibrium in 2-Dimensions:

A particle is in equilibrium if it is at rest, or if it has a constant
velocity. When in equilibrium, the sum of all forces acting on a
body is zero.
F  0
Free Body Diagram:
A diagram of a body which shows all forces that at upon it and
replaces all connections to the outside with other forces.

Step 1: Isolate the body from the surroundings.

Step 2: Show all forces. Include pure forces and all reactive
forces due to cut or removed constraints.

Step 3: Label know forces with values. Label unknown forces
with variables.

The equation    F  0 is a vector equation. In 2-dimensions this
can be broken down into two scalar equations...
one in the x- and one in the y- directions.
F  0
F1  F2  F3  0
ˆ                 ˆ                ˆ
F1x i  F1 y ˆ  F2 x i  F2 y ˆ  F3x i  F3 y ˆ  0
j                 j                j
ˆ
( F1x  F2 x  F3x )i  ( F1 y  F2 y  F3 y ) ˆ  0
j
F   x   i   Fy ˆ  0
ˆ        j
or breaking it down into two scalar equations:
x-dir:    F      x   0       and             y-dir:    F  y   0
y

Example Problem 1:                                                        F

Find the force F required
FA=150 lbf             θ
to provide static equilibrium
x
of the green particle.                                          O

30o
FB=200 lb

--------------------------------------------------------------------------------
Given: FA  150 ˆ + 0ˆ lb
i     j
F  200sin 30 ˆ  200cos30 ˆ lb
B                i               j
 100 ˆ  173.2ˆ lb
i        j
F  F cosθ ˆ  F sinθˆ
i         j

Find the force, F that makes the system in equilibrium:
F  0
FA  FB  F  0
(150 ˆ + 0ˆ)  (100 ˆ  173.2 ˆ)  ( F cos θ ˆ  F sin θˆ )  0
i   j         i         j              i          j
(150  100+ F cos θ) ˆ  ( 0  173.2  F sin θ)ˆ  0
i                          j

i:    0  150  100+ F cos θ         j:    173.2  F sin θ  0
F cosθ=50                              F sin θ  173.2

To solve divide the 2nd equation by the 1st.
F sin θ   173.2
tan θ                   =3.464                  θ  73.9o
F cos θ    50

Therefore
50     50
F=                   180.3 lb                  so F =180.3 lb at 73.9
o
cos θ cos 73.9
Example Problem 2
B                                  C
Find the maximum load value                                                          5       3
which does not exceed a cable                                        60o   A
4
tension of 780 lbf in either cable.
------------------------------------------------------------------

Given:                                                                     W
4      3                                                                 y       Draw FBD
FC  FC ˆ + FC ˆ
i      j
5      5                                                       FB                         5
FB   FB cos60 ˆ  FB sin 60 ˆ                                                      FC           3
i             j
4
W  0 ˆ  Wˆ
i    j                                                                    θ
60o
x
For Equilibrium:                                                                     A

F  0
FC  FB  W  0                                        W

4        3
( FC ˆ + FC ˆ )  (  FB cos 60 ˆ  FB sin 60 ˆ )  (  W ˆ )  0
i        j                 i            j          j
5        5
4                                3
i: 0  FC  FB cos 60             j: 0  FC  FB sin 60  W
5                                5
3
FC  0.625 FB             W  FC  FB sin 60
5
At this point we can see that FC < FB, so let FB = 780 lbf and then
calculate the other forces.
FB  780 lb      FC  0.625 FB  0.625(780)  487.5 lb f
3                   3
W  FC  FB sin 60  (487.5)  780sin 60  968lb f
5                   5
This clearly shows the rope holding the weight exceeds 780 lb, so
redo the calculations letting W = 780 lb.
3                   3
780  FC  FB sin 60  (0.625FB )  FB sin 60  1.241FB
5                   5
780
FB         628.5 lb and FC  0.625(628.5)  392.8 lb
1.241
Example 3
The 40 N collar A can slide on a frictionless vertical rod and is
attached as shown to a spring. The spring is unstretched when
h = 300 mm. Knowing that the constant of the spring is 560 N/m,
determine the value of h for which the system is in equilibrium.
-------------------------------------------------------------------------------
B

h
k=560 N/m

A

300 mm        C
Example 3 Solution:
The 40 N collar A can slide on a frictionless vertical rod and is
attached as shown to a spring. The spring is unstretched when
h = 300 mm. Knowing that the constant of the spring is 560 N/m,
determine the value of h for which the system is in equilibrium.
-------------------------------------------------------------------------------
B
Geometry                       FBD         Fs
300mm
h
θ                                                 k=560 N/m
300        L1
mm                         N                                      A
h                                   A

L2                                                    300 mm      C
W = 40 N
d     θ

Vector Forces:
FS  Fs sinθ ˆ + Fs cosθ ˆ
i           j
N   N ˆ  0ˆ
i    j
W  0 ˆ  40ˆ N
i     j

Spring Equation:             FS  k L  k ( L2  L1 )  560( L2  L1 )

Equilibrium:            F  0
FS  N  W  0
Fs sin θ ˆ + Fs cosθ ˆ  N ˆ  40ˆ  0
i           j     i     j
i: Fs sin θ  N  0                 j: Fs cos θ  40  0
N  Fs sin θ                          Fs cos θ  40
Geometry relationship:                                                        300mm

L1  0.3002  0.3002  0.4243m
300       L1
L2  0.300  (0.300  d )
2                 2
mm
h
0.3  d
cos θ                                                                       L2
L2
d    θ
0.3
sin θ 
L2

So now start combining the equations:

Fs cos θ  40                        FS  562( L2  L1 )
become
0.3  d
Fs            40                      FS  562( L2  0.4243)
L2
and combining these two give

40  ( L  0.4243)(0.3  d )  0.4243 
 2                      1      (0.3  d )
562            L2                 L1 

Since L2 and d are related by
L2  0.3002  (0.300  d )2

we finally get a single equation with a single variable, but it's not fun to solve being
nonlinear in form.
          0.4243                       
0.071174  1                                      (0.3  d )
    0.3002  (0.300  d )2             
                                       
Rewrite as a function equation to zero.
          0.4243                   
f (d )  0  1                                  (0.3  d )  0.071174
    0.3002  (0.300  d )2         
                                   
This can be solved by plotting the function and locating the root.
          0.4243                    
f (d )  0  1                                   (0.3  d )  0.071174
    0.3002  (0.300  d )2          
                                    

Matlab code used to generate graph:

L1=0.4243
k=562
s=0.3
W=40
d=linspace(0.11,0.12)
x=(1-(L1./sqrt(s.^2+(s+d).^2))).*(s+d)-W/k
h1=plot(d,x,'b')
set(h1,'linewidth',2.0)
xlabel('d')
ylabel('residual')
grid on

Therefore:        d = 0.115      or    h = d + 0.3 = 0.415 m

------------------------------------------------------------------------------------------------------

An alternative method is to use Mathematica (available at LANDesk Management
Example 4:                                                            y             FB
An irregularly shaped machine component
is held in the position shown by three clamps.
Knowing that Fa=940 N, determine the
magnitudes of three forces Fb and Fc exerted
by the other two clamps                               FA                      50o
x
70o

--------------------------------------------------------------------------------
Given: FA  940 ˆ + 0ˆ N
FC
i     j
F   F cos50 ˆ  F sin50 ˆ
B        B       i        B     j
  0.6428FB ˆ  0.7660FB ˆ
i             j
FC   FC cos70 ˆ  FC sin 70 ˆ
i             j
  0.3420F ˆ  0.9397F ˆ
i  C          j     C

Find the forces, FB and FC that give the system equilibrium:
F  0
FA  FB  FC  0
( 940 ˆ + 0ˆ )  (  0.6428FB ˆ  0.7660FB ˆ )  (  0.3420FC ˆ  0.9397 FC ˆ )  0
i    j                  i            j                  i             j
( 940  0.6428FB  0.3420FC ) ˆ +( 0  0.7660FB  0.9397FC )ˆ  0
i                             j

i:     0  940  0.6428FB  0.3420 FC            j: 0  0.7660 FB  0.9397 FC  0
940  0.6428FB                                      0.7660 FB
FC                                                 FC 
0.3420                                           0.9397
FC  2749  1.8795FB                              FC  0.8152 FB

To solve set the two expressions for FC equal:
2749
2749  1.8795FB  0.8152 FB                     FB                       1020 N
0.8152  1.8795
then
FC  0.8152 FB  0.8152(1020)  832 N
Workout Problem:
Two cables are tied together at C and are loaded
as shown. Determine the tension
a) in cable AC, b) in cable BC.

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