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

T1T2

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,SFx = 0 and S 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 S F = 0FBD at A

A

In general, for a particle in equilibrium,

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

FBD at A

A

FB

FDA

FC = 392.4 N

y

x30˚

EXAMPLE

Write the scalar E-of-E:

+ S Fx = FB cos 30º – FD = 0

+ S Fy = FB sin 30º – 392.4 N = 0Solving 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

FDA

FC = 392.4 N

y

x30˚

SPRINGS, CABLES, AND PULLEYS

With a frictionless pulley, T1 = T2.

Spring Force = spring constant * deformation, or

F = k * s

T1T2

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:

+ S Fx = FBC cos 15º – FCD cos 30º = 0

+ S 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

x30

15

EXAMPLE (continued)

S Fx = FBA cos 45 – 100.4 cos 15 = 0

S 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 lbFBA

WF

y

x15 45

CONCEPT QUESTIONS

1000 lb1000 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 = 0Solving the above equations, we get; FB = 478 lb and FC = 518 lb

FBD at point AFCFB

A

FD = 550 lb

y

x30˚

34

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 (S 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 QUIZ1. 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) (S Fx) i + (S Fy) j + (S Fz) k = 0

B) S F = 0C) S Fx = S Fy = S 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 SFx = 0 SFy = 0 SFz = 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 (S F = 0 ) .This equation can be written in terms of its x, y and z components. This form is written as follows. (S Fx) i + (S Fy) j + (S 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} N3) 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

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

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

SFz = 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 NFD = 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 AFCFD

A

600 N

z

y30˚

FBx

1 m2 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 AFCFD

A

600 N

z

30˚

FBx

1 m2 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 (S Fx, S Fy, and S 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, henceFB + 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 givesFB = 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.

zF3 = 10 lb

P

F1 = 20 lb

x

A

F2 = 10 lb

y