Equilibrium of a Particle

Chapter 3

ME 108 - Statics

Applications

For a spool of given

weight, what are the

forces in cables AB

and AC ?

Applications

For a given weight of

the lights, what are

the forces in the

cables? What size of

cable must you use ?

Applications

Equilibrium of 2D Particle

This is an example of a 2-D

or coplanar force system. If

the whole assembly is in

equilibrium, then particle A is

also in equilibrium.

To determine the tensions in

the cables for a given weight

of the engine, we need to

learn how to draw a free

body diagram and apply

equations of equilibrium.

The What, Why, How of a Free

Body Diagram (FBD)

Free Body Diagrams are one of the most important

things for you to know how to draw and use.

What ? - It is a drawing that

shows all external forces

acting on the particle.

Why ? - It helps you write the

equations of equilibrium used

to solve for the unknowns

(usually forces or angles).

The What, Why, How of a Free

Body Diagram (FBD)

1. Imagine the particle to be isolated or cut free from

its surroundings.

2. Show all the forces that act on the particle.

Active forces: They want to move the particle.

Reactive forces: They tend to resist the motion.

3. Identify each force and show all known magnitudes

and directions. Show all unknown magnitudes and

/ or directions as variables.

Why ?

Note : Engine mass = 250 Kg

FBD at A

Equations of 2D Equilibrium

Since particle A is in equilibrium,

the net force at A is zero.

Written in a scalar form,

SFx = 0 and S Fy = 0

These are two scalar equations of equilibrium (EofE).

They can be used to solve for up to two unknowns.

Write the scalar EofE:

+ S Fx = TB cos 30º – TD = 0

+ SFy = TB sin 30º – 2.452 kN = 0

Solving the second equation gives: TB = 4.90 kN

From the first equation, we get: TD = 4.25 kN

Springs, Cables, and Pulleys

Spring Force = spring constant *

deformation, or

F = k * S

With a

frictionless

pulley, T1 = T2.

Example 1

Given: Sack A weighs 20 N

and geometry is as shown.

Find: Forces in the cables and

weight of sack B.

Plan:

1. Draw a FBD for Point E.

2. Apply EofE at Point E to solve

for the unknowns (TEG & TEC).

3. Repeat this process at C.

Solution – Example 2

A FBD at E should look like the one to

the left. Note the assumed directions

for the two cable tensions.

The scalar EofE are:

+ S Fx = TEG sin 30º – TEC cos 45º = 0

+ S Fy = TEGcos 30º – TEC sin 45º – 20 N = 0

Solving these two simultaneous equations

for the two unknowns yields:

TEC = 38.6 N

TEG = 54.6 N

Solution – Example 2

Now move on to ring C.

A FBD for C should look

like the one to the left.

The scalar EofE are:

S Fx = 38.64 cos 45 – (4/5) TCD = 0

S Fy = (3/5) TCD + 38.64 sin 45 – WB = 0

Solving the first equation and then the second yields

TCD = 34.2 N and WB = 47.8 N .

Equilibrium of a Rigid Body

• Applications

• Equilibrium in 2D

• Support reactions

• Free-body diagram

Applications

A 200 kg platform is suspended off an oil rig. How do we

determine the force reactions at the joints and the forces in

the cables?

How are the idealized model and the free body diagram used

to do this? Which diagram above is the idealized model?

Applications

A steel beam is used to support

roof joists. How can we

determine the support reactions

at A & B?

Again, how can we make use of an idealized model and a

free body diagram to answer this question?

Conditions for Rigid-Body

Equilibrium

Statics deals primarily with the description of the

force conditions necessary and sufficient to

maintain the equilibrium of engineering structures.

When a body is in equilibrium, the resultant of all

forces acting on it is zero. Thus, the resultant force

R and the resultant couple M are both zero, and

we have the equilibrium equations

0MM0FR

Conditions for Rigid-Body

Equilibrium

Equilibrium of a rigid body requires

ΣF = 0 and ΣM = 0

or

ΣFx=0 ΣFy=0 ΣFz=0

ΣMx=0 ΣMy=0 ΣMz=0

So, in 3-D, we have six equilibrium equations.

In 2-D, we have three equilibrium equations.

Equilibrium in Two Dimensions

• Required and necessary condition is

ΣFx=0 ΣFy=0

ΣMz=0

Forces

In these summations, all of the following forces need to be included:

• external forces: loads applied to the structure by the environment (e.g., weight, service loads, etc.)

• internal forces: forces in structural members and connections that are generated by straining of material.

• reactions:* forces that support the structure as a whole.

* Reactions are also internal forces, but to draw attention to their importance, we list them separately.

Reactions in 2D

1. Flexible cable, belt, chain, or rope

Force exerted by a flexible cable

is always a tension away from the

body in the direction of the cable.

Reactions in 2D

2. Smooth surface:

Contact force is

compressive and is

normal to the surface.

3. Rough surface:

Rough surfaces are capable

of supporting a tangential

component F (frictional

force) as well as a normal

component N of the

resultant contact force R.

Reactions in 2D

4. Roller support

Roller, rocker, or ball support

transmits a compressive force

normal to the supporting

surface

Reactions in 2D

5. Freely sliding guide

Collar or slider free to move

along smooth guides; can

support force normal to guide

only.

Reactions in 2D

6. Pin connection

A freely hinged pin connection is capable of supporting a

force in any direction in the plane normal to the axis;

usually shown as two components Rx and Ry. A pin not

free to turn may also support a couple M.

Reactions in 2D

7. Built-in or fixed support

A built-in or fixed support is

capable of supporting an axial

force F, a transverse force V

(shear force), and a couple M

(bending moment) to prevent

rotation.

Reactions in 2D

8. Gravitational attraction

Reactions in 2D

9. Spring action

Spring force is tensile if spring is

stretched and compressive if

compressed. For a linearly elastic

spring the stiffness k is the force

required to deform the spring a unit

distance.

Reaction forces

• If a support prevents translation of a body

in a given direction, then a force is

developed on the body in the opposite

direction.

• Similarly, if rotation is prevented, a couple

moment is exerted on the body.

Free Body Diagram (FBD)

Free Body Diagrams are one of the most important

things for you to know how to draw and use.

What ? - It is a drawing that shows all external

forces acting on the particle.

Why ? - It helps you write the equations of

equilibrium used to solve for the unknowns

(usually forces or angles).

The Process of Solving Rigid-Body

Equilibrium Problems

Step1: For analyzing an actual physical system, first we need

to create an idealized model.

Step2: Then we need to draw a free-body diagram showing all

the external (active and reactive) forces.

Step3: Finally, we need to apply the equations of equilibrium

to solve for any unknowns.

Procedure for Drawing a Free-Body

Diagram

1. Draw an outlined shape. Imagine the body to be

isolated or cut “free” from its constraints and draw its

outlined shape.

2. Show all the external forces and couple moments.

These typically include: a) applied loads, b) support

reactions, and, c) the weight of the body.

Procedure for Drawing a Free-Body

Diagram

3. Label loads and dimensions: All known forces and

couple moments should be labeled with their

magnitudes and directions. For the unknown forces and

couple moments, use letters like Ax, Ay, MA, etc.. Indicate

any necessary dimensions.

Procedure for drawing FBDs

(Book Version) 1. Decide on the rigid body (or portion of a rigid body) whose

equilibrium you want to analyze.

2. Imagine that this body is cut completely free (separated) from the

rest of the structure and/or its environment.

• in 2-D, think of a closed line that completely encircles the body.

• in 3-D, think of a closed surface that completely surrounds the body.

3. Sketch the body.

4. Sketch all external forces that are applied to the body.

5. Wherever the cut passes through a structural member, sketch the

internal forces that occur at that location.

6. Wherever the cut passes through a support, sketch the support

reactions that occur at that location.

Example

Example

Example

Example

Given: The link is pin-connected

at A and rests against a smooth

support at B.

Find: Horizontal and vertical

components of reactions at pin A.

Plan:

1. Put the x and y axes in the horizontal and vertical

directions, respectively.

2. Draw a complete FBD of the boom.

3. Apply the EofE to solve for the unknowns.

Solution

FBD

• Reaction N B is perpendicular to

the link at B

• Horizontal and vertical

components of reaction are

represented at A

Example

Given: Weight of the boom =

125 N, the center of mass is at

G, and the load = 600 N.

Find: Support reactions at A

and B.

Plan:

1. Put the x and y axes in the horizontal and vertical directions,

respectively.

2. Determine if there are any two-force members.

3. Draw a complete FBD of the boom.

4. Apply the EofE to solve for the unknowns.

Solution

Draw the FBD for the simply supported I-beam shown.

The cross section is a W14x26 shape which has a weight

of 26 N/m.

x

y

L = 10 m

y z A

B

2000 N 6.67 m

Example

x

y

L = 10 m

y z A

B

A x

A y B y

260 N 2000 N

5 m

6.67 m

x

y

2000 N 6.67 m

Fx = 0 = Ax

Fy = 0 = Ay + By - 260 N - 2000 N

MA = 0 = (260 N) 5 m + (2000 N) 6.67 m - By (10 m)

By =14640 N.m/10 m = 1464 N

Ay = 796 N

Note: we could have used the equations Fx , MA and MB to

obtain our solution.

Now apply

equilibrium

equations:

A x

A y B y

260 N 2000 N

5 m

6.67 m

x

y

Examples

A

B

C

W 1

W 2

x

y

x

y

W

A

B

C

D E

Draw FBD for ACB Draw FBD for ABCD

Example:

Neglecting friction,

determine the tension

in cable ABD and the

reaction at support C.

A

B D

C

F=120 N

100 mm 100 mm

250

mm

F

A:

Cx = 80 N

Cy = 40 N

Example:

Rod AB is attached to

a frictionless collar at

A. Neglecting the

weight of AB,

determine angle .

P = 16 N

Q = 12 N

l = 20 cm

a = 5 cm

A: = ??

l

a P

Q

A

B

C

Static Indeterminacy

2-D

statically

indeterminate

example:

Find the

support

reactions in

terms of P.

Statically indeterminate structures and mechanisms:

P

x

y

L /2 L /2

For a structure to be in static equilibrium under general

loading, the number of support reactions must be equal to or

greater than the number of equilibrium equations (three in 2-

D and six in 3-D). This is a necessary but not sufficient

condition.

Redundant supports: When a structure has more supports

than is required for static equilibrium, it becomes statically

indeterminate.

Improper supports: For static equilibrium, the constraints

must be sufficient in number and arrangement so that the

structure has no rigid body motion capability. A structure

that has rigid body motion capability is called a mechanism.

Constraints - support conditions

example: For the systems shown below, determine the

degree of redundancy (i.e., the number of support

conditions beyond those required for equilibrium).

cable

cable

(a) (c)(b)

(d) (e) (f)

(a) (b) (c) (d) (e) (f)

# constraints

redundancy

example: For the systems shown below, determine the

degree of redundancy (i.e., the number of support

conditions beyond those required for equilibrium).

cable

cable

(a) (c)(b)

(d) (e) (f)

(a) (b) (c) (d) (e) (f)

# constraints 3 2 or 1 4 3 4 or 3 3

redundancy 0 -1 or -2 1 0 1 or 0 ?

Two- and Three-Force Members

Simplify some equilibrium problems by

recognizing members that are subjected to

only 2 or 3 forces

Two-Force Members When a member is subject to no couple

moments and forces are applied at only two

points on a member, the member is called

a two-force member

If we apply the equations of equilibrium to such a

member, we can quickly determine that

the resultant forces at A and B must be equal in

magnitude and act in the opposite directions along

the line joining points A and B.

Example of Two-Force Members

If we neglect the weight, the members can be treated as two-force

members. This fact simplifies the equilibrium analysis of some rigid

bodies since the directions of the resultant forces at A and B are thus

known (along the line joining points A and B).

Three-Force Members If a member is subjected to only three forces, it is necessary

that the forces be either concurrent or parallel for the

member to be in equilibrium

Two- and Three-Force Members

Remarks on pin-connected straight bars, cables and ropes:

Consider a pin-connected

bar. Observe that the

internal force supported by

the bar is purely axial. That

is, the internal force in the

bar is always directed along

the axis of the bar.

• Identical remarks apply to

cables and ropes.

• These are examples of

simple, but common, two-

force members.

Draw the free-body diagram of member ABC which is

supported by a smooth collar at A, roller at B, and

short link CD. Explain the significance of each force

acting on the diagram.

Determine the

horizontal and vertical

components of

reaction at the pin A

and the tension

developed in cable BC

used to support the

steel frame.

Determine the force along the pin connected knee strut BC

(short link) and the magnitude of force at pin A as a function

of positiion d. Plot these results of FBC and FA (vertical axis)

versus d (horizantal axis).

The mass of 700

kg is suspended

from a trolley

which moves

along the crane

rail from d = 1.7 m

to d = 3.5 m.

The jib crane is

supported by a pin at

C and rod AB. İf the

load has a mass of 2

Mg with its center of

mass located at G,

determine the

horizontal and vertical

components of

reaction at the pin C

and the force

developed in rod AB.

Example

Example

Example

Example

If the breaking strength of

cables AC and CE is 750

N and the breaking

strength of cables CD

and AB is 400 N, what is

the maximum mass M

that can be safely

supported by this system

of cables? (The cables

are clamped at joints A

and C.)