EQUILIBRIUM OF A PARTICLE, THE FREE-BODY

DIAGRAM & COPLANAR FORCE SYSTEMS

In-Class Activities:

• Reading Quiz

• Applications

• What, Why and How of

a FBD

• Equations of Equilibrium

• Analysis of Spring and

Pulleys

• Concept Quiz

• Group Problem Solving

• Attention Quiz

Today’s Objectives:

Students will be able to :

a) Draw a free body diagram (FBD), and,

b) Apply equations of equilibrium to solve

a 2-D problem.

READING QUIZ

1) When a particle is in equilibrium, the sum of forces acting

on it equals ___ . (Choose the most appropriate answer)

A) A constant B) A positive number C) Zero

D) A negative number E) An integer

2) For a frictionless pulley and cable, tensions in the cable

(T1 and T2) are related as _____ .

A) T1 > T2

B) T1 = T2

C) T1 < T2

D) T1 = T2 sin

T1

T2

The crane is lifting a load. To

decide if the straps holding the

load to the crane hook will fail,

you need to know the force in the

straps. How could you find the

forces?

APPLICATIONS

Straps

For a spool of given

weight, how would you

find the forces in cables

AB and AC ? If designing

a spreader bar like this

one, you need to know the

forces to make sure the

rigging doesn’t fail.

APPLICATIONS (continued)

APPLICATIONS (continued)

For a given force exerted on the boat’s towing pendant, what are

the forces in the bridle cables? What size of cable must you use?

COPLANAR FORCE SYSTEMS

(Section 3.3)

To determine the tensions in

the cables for a given weight

of the cylinder, you need to

learn how to draw a free body

diagram and apply equations

of equilibrium.

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.

THE WHAT, WHY AND 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 is key to being able to write the equations of

equilibrium—which are used to solve for the unknowns

(usually forces or angles).

How ?

Active forces: They want to move the particle.

Reactive forces: They tend to resist the motion.

Note : Cylinder mass = 40 Kg

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

surroundings.

A

3. Identify each force and show all known magnitudes and

directions. Show all unknown magnitudes and / or directions

as variables .

FC = 392.4 N (What is this?)

FB

FD

30˚

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

FBD at A

A

y

x

EQUATIONS OF 2-D EQUILIBRIUM

Or, written in a scalar form,

Fx = 0 and Fy = 0

These are two scalar equations of equilibrium (E-of-E).

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

Since particle A is in equilibrium, the

net force at A is zero.

So FB + FC + FD = 0

or F = 0

FBD at A

A

In general, for a particle in equilibrium,

F = 0 or

Fx i + Fy j = 0 = 0 i + 0 j (a vector equation)

FBD at A

A

FB

FD A

FC = 392.4 N

y

x 30˚

EXAMPLE

Write the scalar E-of-E:

+ Fx = FB cos 30º – FD = 0

+ Fy = FB sin 30º – 392.4 N = 0

Solving the second equation gives: FB = 785 N →

From the first equation, we get: FD = 680 N ←

Note : Cylinder mass = 40 Kg

FBD at A

A

FB

FD A

FC = 392.4 N

y

x 30˚

SPRINGS, CABLES, AND PULLEYS

With a frictionless pulley,

T1 = T2.

Spring Force = spring constant *

deformation, or

F = k * s

T1

T2

EXAMPLE

1. Draw a FBD for Point C.

2. Apply E-of-E at Point C to solve for the unknowns (FCB &

FCD).

3. Knowing FCB , repeat this process at point B.

Given: Cylinder E weighs

30 lb and the

geometry is as

shown.

Find: Forces in the cables

and weight of

cylinder F.

Plan:

EXAMPLE

(continued)

The scalar E-of-E are:

+ Fx = FBC cos 15º – FCD cos 30º = 0

+ Fy = FCD sin 30º – FBC sin 15º – 30 = 0

A FBD at C should look like the one at

the left. Note the assumed directions for

the two cable tensions.

Solving these two simultaneous equations for the

two unknowns FBC and FCD yields:

FBC = 100.4 lb

FCD = 112.0 lb

FCD

FBC

30 lb

y

x 30

15

EXAMPLE (continued)

Fx = FBA cos 45 – 100.4 cos 15 = 0

Fy = FBA sin 45 + 100.4 sin 15 – WF = 0

The scalar E-of-E are:

Now move on to ring B.

A FBD for B should look

like the one to the left.

Solving the first equation and then the second yields

FBA = 137 lb and WF = 123 lb

FBC =100.4 lb FBA

WF

y

x

15 45

CONCEPT QUESTIONS

1000 lb 1000 lb

1000 lb

( A ) ( B ) ( C )

1) Assuming you know the geometry of the ropes, you cannot

determine the forces in the cables in which system above?

A) The weight is too heavy.

B) The cables are too thin.

C) There are more unknowns than equations.

D) There are too few cables for a 1000 lb

weight.

2) Why?

GROUP PROBLEM SOLVING

Plan:

1. Draw a FBD for point A.

2. Apply the E-of-E to solve for the

forces in ropes AB and AC.

Given: The box weighs 550 lb and

geometry is as shown.

Find: The forces in the ropes AB

and AC.

GROUP PROBLEM SOLVING

(continued)

Applying the scalar E-of-E at A, we get;

+ F x = FB cos 30° – FC (4/5) = 0

+ F y = FB sin 30° + FC (3/5) - 550 lb = 0

Solving the above equations, we get;

FB = 478 lb and FC = 518 lb

FBD at point A FC FB

A

FD = 550 lb

y

x 30˚

3

4

5

ATTENTION QUIZ

A

30

40

100 lb

1. Select the correct FBD of particle A.

A) A

100 lb

B)

30

40°

A

F1 F2

C) 30° A

F

100 lb

A

30° 40° F1

F2

100 lb

D)

ATTENTION QUIZ

F2

20 lb

F1

C

50°

2. Using this FBD of Point C, the sum of

forces in the x-direction ( FX) is ___ .

Use a sign convention of + .

A) F2 sin 50° – 20 = 0

B) F2 cos 50° – 20 = 0

C) F2 sin 50° – F1 = 0

D) F2 cos 50° + 20 = 0

THREE-DIMENSIONAL FORCE SYSTEMS

In-class Activities:

• Check Homework

• Reading Quiz

• Applications

• Equations of Equilibrium

• Concept Questions

• Group Problem Solving

• Attention Quiz

Today’s Objectives:

Students will be able to solve 3-D particle equilibrium problems by

a) Drawing a 3-D free body diagram, and,

b) Applying the three scalar equations (based on one vector

equation) of equilibrium.

READING QUIZ

1. Particle P is in equilibrium with five (5) forces acting on it in

3-D space. How many scalar equations of equilibrium can be

written for point P?

A) 2 B) 3 C) 4

D) 5 E) 6

2. In 3-D, when a particle is in equilibrium, which of the

following equations apply?

A) ( Fx) i + ( Fy) j + ( Fz) k = 0

B) F = 0

C) Fx = Fy = Fz = 0

D) All of the above.

E) None of the above.

APPLICATIONS

You know the weights of

the electromagnet and its

load. But, you need to

know the forces in the

chains to see if it is a safe

assembly. How would you

do this?

APPLICATIONS

(continued)

This shear leg derrick

is to be designed to lift

a maximum of 200 kg

of fish.

How would you find

the effect of different

offset distances on the

forces in the cable and

derrick legs?

Offset distance

THE EQUATIONS OF 3-D EQUILIBRIUM

This vector equation will be satisfied only when

Fx = 0

Fy = 0

Fz = 0

These equations are the three scalar equations of equilibrium.

They are valid for any point in equilibrium and allow you to

solve for up to three unknowns.

When a particle is in equilibrium, the vector

sum of all the forces acting on it must be

zero ( F = 0 ) .

This equation can be written in terms of its x,

y and z components. This form is written as

follows.

( Fx) i + ( Fy) j + ( Fz) k = 0

EXAMPLE #1

1) Draw a FBD of particle O.

2) Write the unknown force as

F5 = {Fx i + Fy j + Fz k} N

3) Write F1, F2 , F3 , F4 and F5 in Cartesian vector form.

4) Apply the three equilibrium equations to solve for the three

unknowns Fx, Fy, and Fz.

Given: The four forces and

geometry shown.

Find: The force F5 required to

keep particle O in

equilibrium.

Plan:

EXAMPLE #1

(continued)

F4 = F4 (rB/ rB)

= 200 N [(3i – 4 j + 6 k)/(32 + 42 + 62)½]

= {76.8 i – 102.4 j + 153.6 k} N

F1 = {300(4/5) j + 300 (3/5) k} N

F1 = {240 j + 180 k} N

F2 = {– 600 i} N

F3 = {– 900 k} N

F5 = { Fx i – Fy j + Fz k} N

EXAMPLE #1

(continued)

Equating the respective i, j, k components to zero, we have

Fx = 76.8 – 600 + Fx = 0 ; solving gives Fx = 523.2 N

Fy = 240 – 102.4 + Fy = 0 ; solving gives Fy = – 137.6 N

Fz = 180 – 900 + 153.6 + Fz = 0 ; solving gives Fz = 566.4 N

Thus, F5 = {523 i – 138 j + 566 k} N

Using this force vector, you can determine the force’s magnitude

and coordinate direction angles as needed.

EXAMPLE #2

1) Draw a free body diagram of Point A. Let the unknown force

magnitudes be FB, FC, FD .

2) Represent each force in the Cartesian vector form.

3) Apply equilibrium equations to solve for the three unknowns.

Given: A 600 N load is supported

by three cords with the

geometry as shown.

Find: The tension in cords AB,

AC and AD.

Plan:

EXAMPLE #2 (continued)

FB = FB (sin 30 i + cos 30 j) N

= {0.5 FB i + 0.866 FB j} N

FC = – FC i N

FD = FD (rAD /rAD)

= FD { (1 i – 2 j + 2 k) / (12 + 22 + 22)½ } N

= { 0.333 FD i – 0.667 FD j + 0.667 FD k } N

FBD at A FC FD

A

600 N

z

y 30˚

FB

x

1 m 2 m

2 m

EXAMPLE #2 (continued)

Solving the three simultaneous equations yields

FC = 646 N

FD = 900 N

FB = 693 N

y

Now equate the respective i , j , k

components to zero.

Fx = 0.5 FB – FC + 0.333 FD = 0

Fy = 0.866 FB – 0.667 FD = 0

Fz = 0.667 FD – 600 = 0

FBD at A FC FD

A

600 N

z

30˚

FB

x

1 m 2 m

2 m

CONCEPT QUIZ

1. In 3-D, when you know the direction of a force but not its

magnitude, how many unknowns corresponding to that force

remain?

A) One B) Two C) Three D) Four

2. If a particle has 3-D forces acting on it and is in static

equilibrium, the components of the resultant force ( Fx, Fy,

and Fz ) ___ .

A) have to sum to zero, e.g., -5 i + 3 j + 2 k

B) have to equal zero, e.g., 0 i + 0 j + 0 k

C) have to be positive, e.g., 5 i + 5 j + 5 k

D) have to be negative, e.g., -5 i - 5 j - 5 k

1) Draw a free body diagram of Point A. Let the unknown force

magnitudes be FB, FC, F D .

2) Represent each force in the Cartesian vector form.

3) Apply equilibrium equations to solve for the three unknowns.

GROUP PROBLEM SOLVING

Given: A 3500 lb motor and

plate, as shown, are in

equilibrium and supported

by three cables and

d = 4 ft.

Find: Magnitude of the tension

in each of the cables.

Plan:

W = load or weight of unit = 3500 k lb

FB = FB(rAB/rAB) = FB {(4 i – 3 j – 10 k) / (11.2)} lb

FC = FC (rAC/rAC) = FC { (3 j – 10 k) / (10.4 ) }lb

FD = FD( rAD/rAD) = FD { (– 4 i + 1 j –10 k) / (10.8) }lb

GROUP PROBLEM SOLVING (continued)

FBD of Point A

y

z

x

W

FB FC

FD

GROUP PROBLEM SOLVING (continued)

The particle A is in equilibrium, hence

FB + FC + FD + W = 0

Now equate the respective i, j, k components to zero

(i.e., apply the three scalar equations of equilibrium).

Fx = (4/ 11.2)FB – (4/ 10.8)FD = 0

Fy = (– 3/ 11.2)FB + (3/ 10.4)FC + (1/ 10.8)FD = 0

Fz = (– 10/ 11.2)FB – (10/ 10.4)FC – (10/ 10.8)FD + 3500 = 0

Solving the three simultaneous equations gives

FB = 1467 lb

FC = 914 lb

FD = 1420 lb

ATTENTION QUIZ

2. In 3-D, when you don’t know the direction or the magnitude

of a force, how many unknowns do you have corresponding

to that force?

A) One B) Two C) Three D) Four

1. Four forces act at point A and point

A is in equilibrium. Select the correct

force vector P.

A) {-20 i + 10 j – 10 k}lb

B) {-10 i – 20 j – 10 k} lb

C) {+ 20 i – 10 j – 10 k}lb

D) None of the above.

z

F3 = 10 lb

P

F1 = 20 lb

x

A

F2 = 10 lb

y