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Homework 1

SCI 220 Engineering Mechanics

Homework 1

1.

2.

A and B

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Homework 1

3.

If the particle is subjected to the three forces and is in equilibrium,

Deterimine F3

4.

Determine the three forces acting on the lamp in figure below and draw the forces acting

on the 8kg suspended lamp.

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Homework 1

5.

Hint: determine the unit vector of each tension.

6.

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SCI-220 Engineering Mechanics

Spring 2020

Free Body Diagram

Moment of a Force

Principle of Moments

SCI-220 Eng Mechanics

Announcements

•

Textbook

– “Engineering Mechanics,” Statics / Dynamics, R. C. Hibbeler, Prentice Hall

Spring 2020

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SCI-220 Eng Mechanics

•

•

•

•

•

•

The Free-Body Diagram

Particle Equilibrium

Three-Dimensional Force Systems

Moment of a Force – Scalar Formation

Moment of Force – Vector Formulation

Principle of Moments

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SCI-220 Eng Mechanics

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SCI-220 Eng Mechanics

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SCI-220 Eng Mechanics

WHY TREAT A BODY AS A PARTICLE

In a real world there is no such things as

particles, which occupy zero volume in

space. However, when the body doesn’t

rotate, we can treat the body as a particle.

Even, when the body rotates but its angular

velocity remain constant, we can still treat

the body as particle, since it rotational

quantities remain unchanged during the

motion.

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SCI-220 Eng Mechanics

Condition for the Equilibrium of a Particle

• Particle is at equilibrium if

– at rest

– moving at a constant velocity

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SCI-220 Eng Mechanics

Condition for the Equilibrium of a Particle

• Newton’s first law of motion

Equilibrium

∑F = 0

where ∑F is the vector sum of all the

forces acting on the particle

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SCI-220 Eng Mechanics

Condition for the Equilibrium of a Particle

• Newton’s second law of motion

∑F = ma

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The Free-Body Diagram FBD

Best representation of all the unknown forces (∑F)

which acts on a body

A sketch showing the particle “free” from the

surroundings with all the forces acting on it

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SCI-220 Eng Mechanics

The Free-Body Diagram FBD

Two common connections in particle

equilibrium

– Spring

– Cables and Pulleys

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SCI-220 Eng Mechanics

The Free-Body Diagram

Spring

• Spring constant or stiffness k,

defines the elasticity of the spring

• Magnitude of force F when spring

is elongated or compressed

F = ks

stretched

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spring’s deformed length l

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SCI-220 Eng Mechanics

3.2 The Free-Body Diagram

• Spring deformed distance, s

where s is determined from the difference in

spring’s deformed length l and its undeformed

length lo

s = l – lo

– If s is positive, F “pull”

onto the spring

– If s is negative, F “push”

onto the spring

stretched

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spring’s deformed length l

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SCI-220 Eng Mechanics

The Free-Body Diagram

Example

Given lo = 0.4m and k = 500N/m

Magnitude of force F

deformed

To stretch it until l = 0.6m, A force, F = ks

deformed

F=(500N/m)(0.6m – 0.4m) = 100N is needed

To compress it until l = 0.2m,

A force, F = ks

deformed

=(500N/m)(0.2m – 0.4m)

= -100N is needed

stretched

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spring’s deformed length l

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The Free-Body Diagram

Cables and Pulley (Assumptions)

• Negligible weight and cannot stretch.

• Only support tension or pulling force.

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SCI-220 Eng Mechanics

The Free-Body Diagram

Cables and Pulley (Assumptions)

• Tension always acts in the direction of

the cable

• Tension force in a continuous cable

must have a constant magnitude for

equilibrium

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The Free-Body Diagram

• Cables and Pulley

For any angle θ, the cable is subjected to

a constant tension T, throughout its length

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The Free-Body Diagram

Procedure for Drawing a FBD

1. Draw outlined shape

– Isolate particle from its surroundings

2. Show all the forces

– Indicate all the forces

– Active forces: set the particle in motion

– Reactive forces: result of constraints and

supports that tend to prevent motion

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The Free-Body Diagram

Procedure for Drawing a FBD

3. Identify each forces

– Known forces should be labeled with

proper magnitude and direction

– Letters are used to represent magnitude

and directions of unknown forces

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The Free-Body Diagram

Example 3.1

The sphere has a mass of 6kg and is

supported.

Draw a free-body diagram of

i)the sphere,

ii)the cord CE

iii)and the knot at C.

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The Free-Body Diagram

Solution

FBD at Sphere

• Two forces acting,

weight and the force

on cord CE.

• Weight of 6kg

x(9.81m/s2) = 58.9N

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The Free-Body Diagram

Solution

Cord CE

• Two forces acting, force

of the sphere and force

of the knot

• Newton’s Third Law: FCE

is equal but opposite

• FCE and FEC pull the cord

in tension

• For equilibrium, FCE =

FEC

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The Free-Body Diagram

Solution

FBD at Knot

• Three forces acting, force by cord CBA, cord CE

and spring CD

• Important to know that

the weight of the sphere

does not act directly on

the knot but subjected to

by the cord CE

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FBD & Coordinate System

In solving equilibrium equation, it is

good to resolve forces into a

component of their coordinate system.

The problems could be solved in either

1D, 2D or 3D depends on the

movement, required accuracies and

etc.

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3.3 Equilibrium in Coplanar (2D) Systems

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

Equilibrium

resolve forces into a component

equilibrium equations

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of their coordinate system

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3.3 Equilibrium in Coplanar (2D) Systems

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

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3.3 Equilibrium in Coplanar (2D) Systems

• Procedure for Analysis

1. Free-Body Diagram

– Establish the x, y axes in any suitable

orientation

– Label all the unknown and known forces

magnitudes and directions

– Sense of the unknown force can be

assumed

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3.3 Equilibrium in Coplanar (2D) Systems

• Procedure for Analysis

2) Equations of Equilibrium

– Apply the equations of equilibrium

∑Fx = 0

∑Fy = 0

– Components are positive if they are

directed along the positive axis and

negative axis are negative, if directed

along the negative axis

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3.3 Equilibrium in Coplanar (2D) Systems

Example 3.2

Determine the tension in

cables AB and AD for

equilibrium of the 250kg

engine.

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3.3 Equilibrium in Coplanar

(2D) Systems

To solve for tension in AB and AD, you need to

establish equilibrium equation. You need to

establish the Cartesian coordinate system.

Where to locate the origin of the coordinate

system?

Locating the origin of coordinate system at point

D, make it difficult for determining the point A

in later stage (distance of A from D were not 30

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given).

SCI-220 Eng Mechanics

3.3 Equilibrium in Coplanar (2D) Systems

The best is to locate origin of the coordinate

system at point A. By doing so, it is easy for

you to resolve the component of tension force

in AD and AB in x and y direction of 2D

Cartesian coordinate system.

Always try to solve the equilibrium AT A POINT

where all the forces of the system acts.

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3.3 Equilibrium in Coplanar

(2D) Systems

Solution

FBD at Point A

– Initially, two forces acting, forces

of cables AB and AD

– Engine Weight

= (250kg)(9.81m/s2)

= 2.452kN supported by cable CA

– Finally, three forces acting, forces

TB and TD and engine weight

on cable CA

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3.3 Equilibrium in Coplanar

(2D) Systems

resolve forces into a component

Solution equilibrium equations

+→ ∑Fx = 0;

TBcos30° – TD = 0

+↑ ∑Fy = 0;

TBsin30° – 2.452kN = 0

Solving,

TB = 4.90kN

TD = 4.25kN

*Note: Neglect the weights of the cables since they

are small compared to the weight of the engine 33

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3.4 Equilibrium in 3D Force Systems

• Make use of the three scalar equations to

solve for unknowns such as angles or

magnitudes of forces

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3.4 Equilibrium in 3D Force Systems

• Make use of the three scalar equations to

solve for unknowns such as angles or

magnitudes of forces

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3.4 Three-Dimensional Force Systems

• For particle equilibrium

∑F = 0

• Resolving into i, j, k components

∑Fxi + ∑Fyj + ∑Fzk = 0

• Three scalar equations representing algebraic sums of

the x, y, z forces

∑Fxi = 0

∑Fyj = 0

∑Fzk = 0

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3.4 Equilibrium in 3D Force Systems

Example 3.4

A 90N load is suspended from the hook. The

load is supported by two cables and a spring

having a stiffness k = 500N/m.

Determine the force in the

cables and the stretch of the

spring for equilibrium. Cable

AD lies in the x-y plane and

cable AC lies in the x-z plane.

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3.4 Equilibrium in 3D Force Systems

Establish the coordinate system where the origin is

located at point A.

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3.4 Equilibrium in 3D Force Systems

Solution

FBD at Point A

– Point A chosen as the forces are

concurrent at this point

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3.4 Equilibrium in 3D Force Systems

Solution

Equations of Equilibrium,

∑Fx = 0;

FDsin30° – (4/5)FC = 0

∑Fy = 0;

-FDcos30° + FB = 0

∑Fz = 0;

(3/5)FC – 90N = 0

Solving,

FC = 150N

FD = 240N

FB = 208N

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3.4 Equilibrium in 3D Force Systems

Solution

For the stretch of the spring,

FB = ksAB

208N = 500N/m(sAB)

sAB = 0.416m

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Example

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

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Example

5. Select the correct FBD of particle A.

30

A 40

100 kg

F1

A

A)

F2

B)

30

40°

100 kg

A

F

C)

30

°

D)

A

100 kg

F1

30

°

F2

40°

A

100 kg

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SCI-220 Eng Mechanics

Example

7. 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

8. 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.

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Moment of a Force

tendency of rotation

caused by Fx

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4.1 Moment of a Force – Scalar Formation

• Moment of a force (Torque) about a point

or axis – a measure of the tendency of

the force to cause a body to rotate

about an axis passing through a

point.

•

Torque – tendency of rotation caused by Fx

or simple moment (Mo) z

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4.1 Moment of a Force – Scalar Formation

Magnitude

•

tendency of rotation

caused by Fx

For magnitude of MO, Force vector F

MO = Fd (Nm)

Counter clock wise

where d = perpendicular distance

from O to its line of action of force

Direction

•

Direction using “right hand rule”

Sense of rotation

Clock wise or Counter clock wise.

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SCI-220 Eng Mechanics

4.1 Moment of a Force •

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

– Scalar Formation

tendency of rotation

caused by Fx

– Vector Formation

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4.1 Moment of a Force – Scalar Formation

Resultant Moment

• Resultant moment, MRo = moments of

all the forces

MRo = ∑Fd

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

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Example 4.1

For each case, determine the moment of the

force about an axis that pass thorough

point O.

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Solution

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

Line of action is extended as a dashed line to establish

moment arm d.

Tendency to rotate is indicated and the orbit is shown as

a colored curl.

(a)Mo = (100 N )( 2m) = 200 N.m(CW )

clock wise

tendency of rotation

caused by Fx

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SCI-220 Eng Mechanics

Solution

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

(b)Mo = (50N )(0.75m) = 37.5N.m(CW )

clock wise

Extend force

line of action

(c )Mo = (40N )(4m + 2 cos30 m) = 229N.m(CW )

clock wise

tendency of rotation

caused by Fx

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Extend force

line of action

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SCI-220 Eng Mechanics

Solution

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

(d )Mo = (60N )(1sin 45 m) = 42.4N.m(CCW )

tendency of rotation

caused by Fx

Extend force

line of action

(e)Mo = (7kN)( 4m − 1m) = 21.0kN.m(CCW )

Extend force

line of action

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

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SCI-220 Eng Mechanics

•

Example 4.5

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

Determine the moment of the force about point O.

tendency of rotation

caused by Fx

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Solution

•

For magnitude of MO, Force vector F

MO = Fd (Nm)

where d = perpendicular distance

The moment arm d can be found from trigonometry,

d = (3)sin 75 = 2.898 m

Thus,

M O = Fd = (5)(2.898) = 14.5 kN m

Since the force tends to rotate or orbit clockwise about

point O, the moment is directed into the page.

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tendency of rotation

caused by Fx

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4.4 Principles of Moments (This is good for

2D vector analysis)

• Also known as Varignon’s Theorem

“Moment of a force about a point is equal

to the sum of the moments of the

forces’ components about the point”

• Since F = F1 + F2,

MO = r X F

= r X (F1 + F2)

= r X F1 + r X F2

Principles of Moments

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4.2 Cross Product

Direction

• Vector C has a direction that is

perpendicular to the plane containing A and

B such that C is specified by the right hand

rule

• Expressing vector C when

magnitude and direction are known

C = A X B = (AB sinθ)uC

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4.2 Cross Product

Laws of Operations

1. Commutative law is not valid

AXB≠BXA

Rather,

AXB=-BXA

• Cross product A X B yields a

vector opposite in direction to C

B X A = -C

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4.4 Principles of Moments (This is good for

2D vector analysis)

Resolve the Force components

Force F2 line of action passing through the

moment is equal to zero

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4.4 Principles of Moments (This is good for

2D vector analysis)

Principles of Moments

•

MO = r X F

= r X (F1 + F2) Resolve the Force components

= r X F1 + r X F2

Add the moments

Spring 2020

SCI-220 Eng Mechanics

4.4 Principles of

Moments (This is

good for 2D vector

analysis)

Resolve the Force components

Add the moments

Force F line of action passing

through the moment is equal

to zero

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Chapter Review

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Chapter Review

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Chapter Review

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Next lecture

• More Examples

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