EQUILIBRIUM OF RIGID BODIES IN TWO DIMENSIONS. When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.

EQUILIBRIUM OF RIGID BODIES

IN TWO DIMENSIONS

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=FBFA

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 FBP

Free Body Diagram

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.

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 uniform beam has a mass of 50 kg per meter of length.

Compute the reactions at the support O. The force loads shown

lie in a vertical plane.

3. Calculate the magnitude of the force supported by the pin at C

under the action of the 900-N load applied to the bracket. Neglect

friction in the slot.

4. 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

Px

B

Cy

A

140 mm 60 mm 40 mm

20 mm

46 mm

120 mm

5. The member ABC and the pulley at C together have a mass of 30

kg, with center of mass at G. Calculate the magnitude of the force

supported by the pin at A. Dimensions are given in “mm”s.

6. The winch consists of a drum of

radius 30 cm, which is pin

connected at its center C. At its

outer rim is a ratchet gear having a

mean radius of 60 cm. The pawl

AB serves as a two force member

(short link) and holds the drum

from rotating. If the suspended

load is 500 N, determine the

horizontal and vertical components

reaction at the pin C.

60 cm

20 cm

70 cm

30 cm

7. 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

8. 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 =0. Calculate the tension T for static

equilibrium as a function of 090o). State the values of T and the

reactions at O for =arcsin(0.6).

9. 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.