SRSD 2093: Engineering Mechanics

2SRRI SECTION 19ROOM 7, LEVEL 14, MENARA RAZAK

PLAN OF THE LECTURE

Today’s Objectives:

Students will be able to :

a) Draw a free-body diagram (FBD), 2-D and 3-D free-body diagrams

b) Apply equations of equilibrium to solve 2-D and 3-D problems.

In-Class Activities:

• Applications

• What, Why, and How

• Equations of Equilibrium

• Analysis of Spring and Pulleys

• Group Problem Solving

The crane is lifting a load. To decide if thestraps holding the load to the crane hookwill fail, you need to know the force inthe straps. How could you find theforces?

Straps

APPLICATION

For a spool of given weight,how would you find the forcesin cables AB and AC? Ifdesigning a spreader bar likethis one, you need to know theforces to make sure the riggingdoesn’t fail.

APPLICATION (CONTINUE)

For a given force exerted on the boat’s towing pendant, whatare the forces in the bridle cables? What size of cable must youuse?

APPLICATION (CONTINUE)

What? - It is a drawing that shows all external forces acting on the particle. 2 connections in FBD; ::Spring ::Cables and pulley

- A particle is at equilibrium if:

At rest

Moving at a constant velocity

- Newton’s First Law of Motion is applied: ΣF = 0 ΣF = vector sum of all forces acting on the particle

- Newton’s Second Law of MotionΣF = mama = 0 (When the force fulfills Newton’s First Law of Motion,

where a = 0)Thus, particle moves in: Constant velocity or At rest

WHAT? WHY? HOW? OF A FREE BODY DIAGRAM

Free-body diagrams are one of the most important things for you to know how to draw and use for statics and other subjects!

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).

WHAT? WHY? HOW? OF A FREE BODY DIAGRAM

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

FB

FD

30˚

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

FBD at A

A

y

x

HOW?

SAMPLE SPRINGS

Springs

– Linear elastic spring: change in length isproportional to forces acting on it

– Spring constant is represented by stiffness, k– elasticity of spring

– Magnitude of force when spring is elongatedor compressed:

Consider two types of connections often encountered in particle equilibrium problems

T1

T2

CABLES AND PULLEYS

Cables and pulley

– Assumption: Cables (or cords) has negligibleweight and cannot stretch

– Tension always acts in the direction of the cable

– Tension force must have a constant magnitude forequilibrium

– For any angle θ, the cable is subjected to aconstant tension T

– With a frictionless pulley and cable T1 = T2.

Procedure of drawing a FBD:

– Draw outlined shape

– Show all the forces:

>> Active forces: particle in motion>> Reactive forces: constraints that prevent

motion

– Identify each forces>> Known forces with proper magnitude

and direction

>> Later used to represent magnitude and

directions

FREE-BODY DIAGRAM (FBD)

The bucket is held in equilibrium by thecable, and instinctively we know that theforce in the cable must equal the weightof the bucket. By drawing a free-bodydiagram of the bucket we can understandwhy this is so. This diagram shows thatthere are only two forces acting on thebucket , namely, its weight W and theforce T of the cable. For equilibrium, theresultant of these forces must be equal tozero, and so T = W.

• A particle is subjected tocoplanar forces in the x-y plane

• Resolve into i and j componentsfor equilibrium:

– ∑Fx = 0

– ∑Fy = 0

• Scalar equations of equilibriumrequire that the algebraic sumof the x and y components toequal to zero

COPLANAR FORCE SYSTEMS

A particle is subjected toA particle is subjected to coplanar forces in the x-y planeplane

Resolve into i and jcomponents for equilibrium:

Fx = 0

Fy = 0y

Scalar equations of equilibrium require that the algebraic sum of the and components toof the x and y components to equal to zero

• Procedures for analysis:

Free-Body Diagram• Establish the x, y axes

• Label all the unknown and known forces

Equations of Equilibrium • Apply the equations of equilibrium:

• Label all the unknown and known force

– ∑Fx = 0

– ∑Fy = 0

• For spring, apply F = k*s to find spring force. Mark negative result force as the reverse forces

COPLANAR FORCE SYSTEMS

To determine the tensions in thecables for a given weight ofcylinder, you need to learn how todraw a free-body diagram andapply the equations of equilibrium.

This is an example of a 2-D orcoplanar force system.

If the whole assembly is inequilibrium, then particle A is also inequilibrium.

COPLANAR FORCE SYSTEMS

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 = 0FBD at A

A

In general, for a particle in equilibrium,

F = 0 orFx i + Fy j = 0 = 0 i + 0 j (a vector equation)

A

FB

FDA

FC = 392.4 N

y

x30˚

FBD at A

EQUATION OF 2-D EQUILIBRIUM

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

FDA

FC = 392.4 N

y

x30˚

EQUATION OF 2-D EQUILIBRIUM

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 N and geometry is as shown.

Find: The forces in the ropes AB and AC.

EXAMPLE 1

FBD at point AFCFB

A

FD = 550 N

y

x30˚

3

4

5

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 N = 0

Solving the above equations, we get;

FB = 478 N and FC = 518 N

EXAMPLE 1: SOLUTION

EXAMPLE 2

EXAMPLE 2: SOLUTION

EXAMPLE 2: SOLUTION

1. Draw a FBD for Point D.

2. Apply E-of-E at Point D to solve for the unknowns (FCD & FDE).

3. Knowing FCD, repeat this process at point C.

Given: The mass of lamp is 20 kg and geometry is as shown.

Find: The force in each cable.

Plan:

GROUP PROBLEM SOLVING

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

+ Fy = FDE sin 30 – 20 (9.81) = 0

+ Fx = FDE cos 30 – FCD = 0

Solving the above equations, we get:

FDE = 392 N and FCD = 340 N

FBD at point D

FDE

FCD

D

W = 20 (9.81) N

y

x

30˚

GROUP PROBLEM SOLVING: SOLUTION

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

+ Fx = 340 – FBC sin 45 – FAC (3/5) = 0

+ Fy = FAC (4/5) – FBC cos 45 = 0

Solving the above equations, we get;

FBC = 275 N and FAC = 243 N

FBD at point C

FCD =340 N

FAC

C

y

x

45˚

FBC

3

45

GROUP PROBLEM SOLVING: SOLUTION

You know the weight of theelectromagnet and its load. But,you need to know the forces inthe chains to see if it is a safeassembly. How would you dothis?

APPLICATION

EQUATION OF 3-D EQUILIBRIUM

This shear leg derrick is tobe designed to lift amaximum of 200 kg of fish.

How would you find theeffect of different offsetdistances on the forces inthe cable and derrick legs?

Offset distance

APPLICATION (CONTINUED)

This vector equation will be satisfied only whenFx = 0Fy = 0Fz = 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

EQUATION OF 3-D EQUILIBRIUM

EXAMPLE I

EXAMPLE I: SOLUTION

EXAMPLE I: SOLUTION

EXAMPLE 2

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 geometryshown.

Find: The force F5 required to keepparticle O in equilibrium.

Plan:

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 2: SOLUTION

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: SOLUTION

1) Draw a free-body diagram of Point A. Let the unknown force magnitudes be FB, FC, FD .

2) Represent each force in its 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 3

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 AFCFD

A

600 N

z

y30˚

FBx

1 m2 m

2 m

EXAMPLE 3: SOLUTION

Solving the three simultaneous equations yields

FC = 646 N (since it is positive, it is as assumed, e.g., in tension)

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

FCFD

A

600 N

z

30˚

FB

x

1 m2 m

2 m

EXAMPLE 3: SOLUTION

Given: A 17500-N (≈ 1750-kg) motor and plate, as shown, are in equilibrium and supported by three cables and d = 1.2 m

Find: Magnitude of the tension in each of the cables.

Plan:

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

W = load or weight of unit = 17500 k N

FB = FB (rAB/rAB) = FB {(1.2 i – 0.9 j – 3 k) / (3.354)} N

FC = FC (rAC/rAC) = FC { (0.9 j – 3 k) / (3.132) } N

FD = FD (rAD/rAD) = FD { (– 1.2 i + 0.3 j – 3 k) / (3.245) } N

FBD of Point A

y

z

x

W

FB FC

FD

GROUP PROBLEM SOLVING

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 = (1.2/ 3.354)FB – (1.2/ 3.245)FD = 0

Fy = (– 0.9/ 3.354)FB + (0.9/ 3.132)FC + (0.3/ 3.245)FD = 0

Fz = (– 3/ 3.354)FB – (3/ 3.132)FC – (3/ 3.245)FD + 17500 = 0

Solving the three simultaneous equations gives the tension forces

FB = 7337 N

FC = 4568 N

FD = 7098 N

GROUP PROBLEM SOLVING

The End