ME 108 - Statics
Equilibrium of a Particle
Chapter 3
Equilibrium in 3D
Introduction
•In the case of a 3-D system of forces & moments, up to 6
independent equilibrium equations can be obtained:
– The 3 components of the sum of the forces must
equal zero & the 3 components of the sum of the
moments about a point must equal to zero
– The procedure for determining the reactions on an
object subjected to a 3-D system of forces &
moments – is the same as in 2 dimensions
Free Body Diagram in 3D
 Consider types of reaction that can occur at the supports
Support Reactions
 Important to recognize the symbols used to represent
each of these supports and to clearly understand how
the forces and couple moments are developed by each
support
 As in 2D, a force that is developed by a support that
restricts the translation of the attached member
 A couple moment is developed when rotation of the
attached member is prevented
FBD – Support Reactions
FBD – Support Reactions
FBD – Support Reactions
FBD – Support Reactions
FBD – Support Reactions
FBD – Support Reactions
FBD – Support Reactions
– The Hinge:
• Familiar device used to support doors
• It permits the supported object to rotate freely
about a line, the hinge axis
• The z axis of the coordinate system is aligned with
the hinge axis in the figure:
FBD – Support Reactions
Pin connection; similar to hinge
z
y
x
FBD – Support Reactions
5.4 3-Dimensional Applications
Example
Example
Example
Example
The bar AB in figure is supported by the cables BC
& BD & a ball socket support at A. Cable BC is
parallel to the z axis & cable BD is parallel to the x
axis. The 200-N weight of the bar acts at its
midpoint. What are the tensions in the cables &
reactions at A?
Solution
Plan
1. Draw the free-body diagram of the bar AB by
isolating it from the support at A & the 2 cables.
2. Write the forces using cartesian vector notation.
3. Apply equilibrium equations to determine the
reactions at A & the tensions in the cables.
Solution
Solution
Draw the Free-Body Diagram:
Isolate the bar & show the reactions that may be
exerted on it.
The bar & socket support can exert 3 components
of force Ax, Ay & Az. The terms TBC & TBD represent
the tensions in the cables.
Solution
Solution
Apply the Equilibrium Equations:
The sums of the forces in each coordinate
direction equal to zero:
Σ Fx = Ax  TBD = 0
Σ Fy = Ay  200 N = 0
Σ Fz = Az  TBC = 0
Let rAB be the position vector from A to B.
(1)
Solution
Solution
The sum of the moments about A, with forces in N
& distances in m:
1
 M point A  rAB   TBC k   rAB   TBD i    rAB   200 j
2
i
1
0
j
0.6
0
k
0.4 
 TBC
i
1
j
0.6
 TBD
0
k
i
0.4  0.5
0
0
j
0.3

k
0.2
 200
0
  0.6TBC  40 i  TBC  0.4TBD j  0.6TBD  100 k
Solution
Solution
The components of this vector (the sums of the
moments about the 3 coordinate axis) each equal
zero:
Σ Mx = (0.6 m)TBC + 40 N-m= 0
Σ My = (1 m)TBC  (0.4 m) TBD = 0
Σ Mz = (0.6 m)TBD  100 N-m = 0
Solving these equations: TBC = 66.7 N, TBD = 166.7 N.
(Notice that we needed only 2 of the 3 equations to obtain
the 2 tensions. The 3rd equation is redundant)
Solution
Solution
Then from Eqs. (1):
Ax = 166.7 N, Ay = 200 N, Az = 66.7 N.
Example
The bar AC in figure is 4 m long & is supported by a
hinge at & the cable BD. The hinge axis is along the
z axis. The centerline of the bar lies in the x-y plane &
the cable attachment point B is the midpoint of the
bar. Determine the tension in the cable & the
reactions exerted on the bar by the hinge.
Example
Strategy
1. Draw the free-body diagram of bar AC by
isolating it from the cable & hinge.
2. Write cable force in vectorial form.
3. Apply equilibrium equation to determine the
reactions.
Example
Solution
Draw the Free-Body Diagram:
Isolate the bar from the hinge support & the cable
& show the reactions they exert.
The terms Ax, Ay & Az are the components of the
force exerted by the hinge & the terms MAx & MAy
are the couples exerted by the hinge about the x &
y axes. (Remember that the hinge cannot exert a
couple on the bar about the hinge axis.)
The term T is the tension in the cable.
Example
Solution
Example
Solution
Apply the Equilibrium Equations:
To write the equilibrium equations, first express the
cable force in terms of it components:
The coordinates of point B are (2 cos 30°, 2 sin
30°, 0) m, so the position vector from B to D is:
rBD = (2  2 cos 30°)i + [2  (2 sin 30°)]j + (1  0)k
= 0.268i + 3j  k (m)
Example
Solution
Divide this vector y its magnitude to obtain a unit
vector nBD that points from point B toward point D:
n BD
rBD

 0.084i  0.945 j  0.315k
rBD
Now we can write the cable force as the product
of its magnitude & nBD:
TnBD = T(0.084i + 0.945j  0.315k)
Example
Solution
The sum of the forces in each coordinate direction
must equal zero:
Σ Fx = Ax + 0.084T = 0
Σ Fy = Ay + 0.945T  100 N = 0
(1)
Σ Fz = Az  0.315T = 0
If we sum the moments about A, the resulting
equations do not contain the unknown reactions
Ax, Ay & Az.
Example
Solution
The position vectors from A to B & from A to C are:
rAB = 2 cos 30°i  2 sin 30°j (m)
rAC = 4 cos 30°i  4 sin 30°j (m)
The sum of the moments about A, with forces in N
& distances in m:
 M point A  M Ax i  M Ay j  rAB  Te BD   rAC   100 j
i
 M Ax i  M Ay j  1.732
0.084T
j
1
0.945T
k
i
j
k
0
 3.464  2 0
 0.315T
0
 100 0
 M Ax  0.315T i  M Ay  0.546T j  1.72T  346 k
Example
Solution
From the vector equation:
Σ Mx = MAx + (0.315 m)T = 0
Σ My = MAy + (0.546 m)T = 0
Σ Mz = (1.72 m)T  346 N-m = 0
Solving these equations:
T = 201 N, MAx = 63.4 N, MAy = 109.8 N.
Then from Eqs. (1), the forces exerted on the bar
by the hinge:
Ax = 17.0 N, Ay = 90.2 N, Mz = 63.4 N.
Example
The plate in figure is supported by hinges at A & B
& the cable CE. The properly aligned hinges do not
exert couples on the plate & the hinge at A does
not exert a force on the plate in the direction of the
hinge axis. Determine the reactions at the hinges &
the tension in the cable.
Example
Strategy
Draw the free-body diagram of the plate, suing the
given information about the reactions exerted by
the hinges at A & B. Before the equilibrium
equations can be applied, express the force
exerted on the plate by the cable in terms of its
components.
Example
Solution
Draw the Free-Body Diagram:
Isolate the plate & show the reactions at the hinges
& the force exerted by the cable.
The term T is the force exerted on the plate by the
cable CE.
Example
Solution
Apply the Equilibrium Equations:
Since we know the coordinates of points C & E, we
can express the cable force as the product of its
magnitude & a unit vector directed from C toward E:
T(0.842i + 0.337j + 0.421k)
The sum of the forces in each coordinate direction
equal zero:
Σ Fx = Ax + Bx  0.842T = 0
Σ Fy = Ay + By + 0.337T  400 = 0
(1)
Σ Fz = Bz + 0.421T = 0
Example
Solution
If we sum the moments about B, the resulting
equations will not contain the unknown reactions at
B.
The sum of the moments about B, with forces in N
& distances in m:
i
0.2
j
0
k
0
 0.842T
0.337T
0.421T
 M point A 
i
 0
j
0
Ax
Ay
k
i
0.2  0.2
0
0
j
0
k
0.2
 400
0
  0.2 Ay  80 i   0.0842T  0.2 Ax j  0.0674T  80 k
0
Example
Solution
The scalar equations are:
Σ Mx =  (0.2 m)Ay + 80 N-m = 0
Σ My =  (0.0842 m)T + (0.2 m)Ax = 0
Σ Mz = (0.0674 m m)T  80 N-m = 0
Solving these equations:
T = 1187 N, Ax = 500 N, Ay = 400 N.
Then from Eqs. (1), the reactions at B are:
Bx = 500 N, By = 400 N, Bz = 500 N.
Example
Given:The cable of the tower
crane is subjected to 840 N
force. A fixed base at A supports
the
crane.
Find: Reactions at the fixed
base A.
Solution
r BC = {12 i + 8 j − 24 k} m
F = F [nBC ] N
= 840 [12 i + 8 j − 24 k] / (122 + 82 + (– 242 ))½
= {360 i + 24 j − 720 k} N
FA = {AX i + AY j + AZ k } N
Solution
• From Equilibrium equation we get,
F + FA = 0
{(360 + AX) i + (240 + AY) j + (-720 + AZ ) k} = 0
• Solving each component equation yields
• AX = − 360 N ,
• AY = − 240 N , and AZ = 720 N.
Example
Hand Solution
example: A uniform pipe cover of radius 240 mm and mass 30 kg is
held in a horizontal position by the cable CD. The bearings at A and
B are self-aligning and bearing A resists thrust. Determine the cable
tension and the reactions at A and B.
T = 343.4 N
Ax = 49.1 N
Ay = 73.6 N
Az= 98.1 N
Bx = 245 N
By = 73.5 N
160 mm
y
C
240 mm
240 mm
B
x
A
z
D
r = 240 mm