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

Nx Ny

Ry

Rx

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. A 54 kg crate rests on the 27 kg pickup tailgate. Calculate the

tension T in each of the two restraining cables, one of which is

shown. The centers of gravity are at G1 and G2. The crate is located

midway between the two cables.

2. 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.

3. A portion of the shifter mechanism for a manual car transmission is shown in the figure.

For the 8 N force exerted on the shift knob, determine the corresponding force P exerted by

the shift link BC on the transmission (not shown). Neglect friction in the ball and socket joint

at O, in the joint at B and in the slip tube near support D. Note that a soft rubber bushing at D

allows the slip tube to self-align with link BC.

4. A large symmetrical drum for drying sand is operated by the geared motor

drive shown. If the mass of the sand is 750 kg and an average gear-tooth force of

2.6 kN is supplied by the motor pinion A to the drum gear normal to the

contacting surfaces at B, calculate the average offset of the center of mass G

of the sand from the vertical centerline. Neglect all friction in the supporting

rollers.

x

5. It is desired that a person be able to begin closing the van hatch from the open

position shown with a 40-N vertical force P. As a design exercise, determine the

necessary force in each of the two hydraulic struts AB. The center of gravity of

the 40-kg door is 37.5 mm directly below point A. Treat the problem as two-

dimensional.

6. The bracket and pulley assembly has a mass of 40 kg with combined center of

gravity at G. Calculate the magnitude of the force supported by the pin at C and

roller at A when a tension of 400 N is applied in the vertical plane of the cable.

75 mm

D

G

B

400 N

30o

450 mm

375 mm

75 mm

100 mm

C

A

7. The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and

held in place by the spring which has an unstretched length of 200 mm. The cocking lever

rests against a smooth peg at B. Determine the magnitude of the support force at A and

the normal force on the peg at B when the lever is in the position shown.

k=50 N/m 300 mm

100 mm

300 mm 30o

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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