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

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