When a rigid body is in equilibrium, both the resultant
force and the resultant couple must be zero.
0
0
0
0
kMjMiM
M
kRjRiR
FR
zyx
zyx
Forces and moments acting on a rigid body could be external
forces/moments or internal forces/moments.
Forces acting from one body to another by direct physical
contact or from the Earth are examples of external forces.
Fluid pressure acting to the wall of a water tank or a force
exerted by the tire of a truck to the road is all external forces.
The weight of a body is also an external force.
Internal forces, on the other hand, keep the particles which
constitute the body intact.
Since internal forces occur in pairs that are equal in magnitude
opposite in direction, they are not considered in the equilibrium
of rigid bodies.
The first step in the analysis of the equilibrium of rigid bodies
must be to draw the “free body diagram” of the body in
question.
Common Support / Connection Element Types in Two
Dimensional Analysis
In rigid bodies subjected to two dimensional force systems, the
forces exerted from supports and connection elements are shown
in the free body diagram as follows:
It should be kept in mind that reaction will occur along the
direction in which the motion of the body is restricted.
Equations of Equilibrium in Two Dimensional Case
If all the forces acting on the rigid body are planar and all the couples are
perpendicular to the plane of the body, equations of equilibrium become two
dimensional.
0
0 0 0
kMM
FRFRjRiRFR
z
yyxxyx
or in scalar form,
0 0 0 Oyx MFF
At most three unknowns can be determined.
Alternative Equations of Equilibrium
In two dimensional problems, in alternative to the above set of
equations, two more sets of equations can be employed in the
solution of problems.
Points A, B and C in the latter set cannot lie along the same line, if
they do, trivial equations will be obtained.
000
000
CBA
BAx
MMM
MMF
Two-Force Member
Members which are subjected to only two forces are named as “two force
members”. Forces acting on these members are equal in magnitude, opposite in
direction and are directed along the line joining the two points where the forces are
applied.
Weight is neglected. If weight
is considered, the member will
not be a two force member!
Hydraulic cylinder
P
P
P
P
P
P
Examples of two force members
P
By
Bx
Ax
Ay
FB
FA=FB FA
B
A
Three-Force Member
In rigid bodies acted on by only three forces, the lines of action of the forces must be
concurrent; otherwise the body will rotate about the intersection point of the two forces
due to the third force which is not concurrent. If the forces acting on the body are parallel,
then the point of concurrency is assumed to be in infinity.
A P B
FA FB P
Free Body Diagram (FBD)
The procedure for drawing a free body diagram which isolates a body or system
consists of the following steps:
1) If there exists, identify the two force members in the problem.
2) Decide which system to isolate.
3) Isolate the chosen system by drawing a diagram which represents its complete
external boundary.
4) If not given with the problem, select a coordinate system which appropriately
suits with the given forces and/or dimensions.
5) Identify all forces which act on the isolated system applied by removing the
contacting or attracting bodies, and represent them in their proper positions on
the diagram.
6) Write the equations of equilibrium and solve for the unknowns.
Ay
Ax
MO
Ox
Oy
Bx
Ay
Ax
Ff
Ff
Ax
By
Bx
Ay
MA
Ax
Ff
W=mg
T
N
W=mg
N
Ax
L
T
Ay
N
mg
mOg
Ax
Ay
N
Ff
mg T Ay By
Bx Ax
T
Ay mg
Ax
L
Ay
Ax
By Bx
T
1. Determine the magnitude P of the vertical force required to lift the
wheelbarrow free of the ground at point B. The combined mass of
the wheelbarrow and its load is 110 kg with center of mass at G.
2) A 35 N axial force at B is required to open the spring loaded plunger of the
water nozzle. Determine the required force F applied to the handle at A and the
magnitude of the pin reaction at O. Note that the plunger passes through a
vertically-elongated hole in the handle at B, so that negligible vertical force is
transmitted there.
3. The small crane is mounted on one side of the bed of a pickup truck. For
the position q=40o, determine magnitude of the force supported by the pin
at O and the oil pressure p against the 50-mm diameter piston of the
hydraulic cylinder BC.
4. The pin A, which connects the 200-kg steel beam with center of gravity
at G to the vertical column, is welded both to the beam and to the column.
To test the weld, the 80-kg man loads the beam by exerting a 300 N force
on the rope which passes through a hole in the beam as shown. Calculate
the torque (couple) M supported by the pin.
5. The pipe bender consists of two grooved pulleys mounted and free
to turn on a fixed frame. The pipe is bent into the shape shown by a
force P = 300 N. Calculate the forces supported by the bearings of the
pulleys.
6. The mass center of 15-N link OC is located at G, and the spring constant of
k=25 N/m is unstretched length when q=0. Calculate the tension T and the
reactions at O for q=45o.
7. Pulley A delivers a steady torque of 100 Nm to a pump through its shaft
at C. The tension in the lower side of the belt is 600 N. The driving motor
B has a mass of 100 kg and rotates clockwise. As a design consideration,
determine the magnitude of the force on the supporting pin at O.
8. Plate AB contains a smooth parabolic slot. Fixed pins B and C are
located at the positions shown in the figure. The equation of the
parabolic slot is given as y = x2/160 , where x and y are in mm. If it
is known that the force input P = 4 N, determine the forces applied
to the plate by the pins B and C and also the force output Q.
Q
P x
B
C y
A
140 mm 60 mm 40 mm
20 mm
46 mm
120 mm
Q
P=4 kN x
B
C y
A
140 mm 60 mm 40 mm
20 mm
46 mm
120 mm
By
C
b
a a
b
Tangent to the parabolic slot
o
x
.
tan
tanx
dx
dyy
mm.yx
y
6812
4060140
2046120
4
3
160
2
522160
60
160
60
22
b
b
a
N.ByN.QN.C
Q.B
.tansinQB.PM
Q.C.B
cosQcosCBF
Q.C.
sinQsinCPF
y
yC
y
yy
x
20729554944
9013760
05224046605220
0975080
00
4219060
00
bb
ba
ba