Vector lesson and problems

ENGT 215 - StaticsENGT 215 - Statics

Chapter 2 – Resultant of Coplanar Chapter 2 – Resultant of Coplanar Force SystemsForce Systems

Chapter GoalsChapter Goals

Force SystemForce System – any set of forces treated as a group. – any set of forces treated as a group. Equivalent Force SystemsEquivalent Force Systems – any two systems of forces which – any two systems of forces which

have the same mechanical effect on a body.have the same mechanical effect on a body. ResultantResultant – a single force that is equivalent to a given force – a single force that is equivalent to a given force

system (same mechanical affect on the body).system (same mechanical affect on the body).

Force SystemForce System ResultantResultant

The goal of this chapter is to learn about equivalent The goal of this chapter is to learn about equivalent force systems and how to calculate the resultants for force systems and how to calculate the resultants for planar force systems.planar force systems.

Vector RepresentationVector Representation

Anatomy of a Force Vector:Anatomy of a Force Vector: Represented by arrow Represented by arrow ABAB showing the showing the line of actionline of action.. Length of Length of ABAB represents force magnitude at some convenient scale. represents force magnitude at some convenient scale. Direction for Direction for line of actionline of action is indicated by angle is indicated by angle θθ counter-clockwise counter-clockwise

from positive “x” reference axis (from positive “x” reference axis (Standard PositionStandard Position).). Arrowhead indicates Arrowhead indicates sensesense of force. of force. Reference coordinate system (x-y axes) is established at the point Reference coordinate system (x-y axes) is established at the point

of application of application AA..

Point of applicationPoint of application

FF

AA

BB

TailTail

TipTip

Vector RepresentationVector Representation

Equal Vectors - Equal Vectors - Two vectors are equal if they have the same magnitude AND the same direction as in (a). The line of action may be different as shown.

Negative Vectors - Two vectors are negatives of each other if they have the same magnitude and opposite directions as in (b).

Resultant of Concurrent ForcesResultant of Concurrent Forces

Adding Vectors – Finding Resultants: The resultant of multiple vectors is the vector sum of those vectors.

Vectors cannot be added algebraically, they must be added geometrically!

Two common methods for adding vectors (finding the resultant); Parallelogram Rule

Parallelogram - a planar four sided figure with parallel opposite sides.

Triangle Rule


Parallelogram Rule - Method for Adding Vectors : Recreate the two vectors to be added such that their tails are coincident.

The direction and magnitude of each vector should not change! Create a parallelogram by drawing two lines, one from the tip of each

vector and parallel to the other vector. The intersection of these lines is the tip of the resultant vector R = v + w. Create the resultant by drawing a new vector R.

Tail of R should correspond with tail of the other vectors v and w. Tip of R will be where construction lines intersect.

**NoteNote – – magnitude of magnitude of RR is not the sum of is not the sum of the magnitudes the magnitudes of of vv and and ww..

Parallelogram Rule:Parallelogram Rule: Properties of parallelograms useful for calculating force resultants:

Sum of the interior angles (2B + 2C) is 360Sum of the interior angles (2B + 2C) is 360o.o.

Opposite sides are equal in length.Opposite sides are equal in length. A + A + B = 180B = 180ºº A = A = CC


AA BB

CC

CC

BB


Triangle Rule - Method for Adding Vectors:Triangle Rule - Method for Adding Vectors: Recreate the two vectors such that the tail of second vector w is

coincident with the tip of the first vector v. The direction and magnitude of each vector should not change!

Create the resultant by drawing a new vector R. The tail of R should correspond with tail of first vector v. The tip of R should correspond with tip of second vector w.

Force TriangleForce Triangle – triangle formed by forces. – triangle formed by forces.


Polygon Rule: Polygon Rule: (Polygon – any closed figure with straight sides) An extension of the Triangle Rule.An extension of the Triangle Rule. Sum of 3 or more coplanar vectors can be accomplished by Sum of 3 or more coplanar vectors can be accomplished by

adding 2 vectors successively, forming a polygon.adding 2 vectors successively, forming a polygon.

Example: Example: RR = = aa + + bb + + cc RR = ( = (aa + + bb) + ) + cc

Force PolygonForce Polygon – polygon formed by forces. – polygon formed by forces.

a + b

R

ca

b

RR = ( = (aa + + bb) + ) + cc

Resultant of Concurrent ForcesResultant of Concurrent ForcesPolygon Rule: Polygon Rule: A more general form of Polygon Rule for adding vectors:

Recreate all summed vectors tip-to-tail. Create the resultant vector R.

The tail of R should correspond with the tail of the first vector v1.

The tip of R should correspond with the tip of the last vector v6.


Vector Sums are Commutative: Get the same result using (v + w) or (w + v). Applies to all methods for adding vectors.

ww

vv

RR

R R == w w ++ v v


Subtracting Vectors: A vector can be subtracted by adding its negative. Remember, the negative of a vector has the same

magnitude but the opposite direction.

Subtracting VectorsSubtracting Vectors

R = v + (-w)


Resultant:Resultant: a system of concurrent coplanar forces acting on a a system of concurrent coplanar forces acting on a rigid body may be replaced by a single rigid body may be replaced by a single ResultantResultant force which force which equals the vector sum of the given forces.equals the vector sum of the given forces.

The vector sums can be determined by two methods:The vector sums can be determined by two methods: Graphical Method – Graphical Method –

Use linear scale and protractor to lay out vectors as Use linear scale and protractor to lay out vectors as previously described. previously described.

Measure length and angle of resultant. Measure length and angle of resultant. Too Inaccurate! Too Inaccurate! Don’tDon’t UseUse!!

Trigonometric Method –Trigonometric Method – Lay out vectors as described but not necessarily to scale.Lay out vectors as described but not necessarily to scale. Label the known angles and lengths of sides (magnitudes).Label the known angles and lengths of sides (magnitudes). Compute the resultant using trigonometry.Compute the resultant using trigonometry.

Law of SinesLaw of Sines Law of CosinesLaw of Cosines

This is the method to be used in this class!This is the method to be used in this class!

Example 1:Example 1:

Determine the resultant of the two forces F1 and F2 acting on the hook.

F2 = 60N

60o

F1 = 54N

o

R

F

RF

NR

R

FFFFR

259.28

120sinsin

120sinsin

8.9877.98

120cos546025460

cos2

1

1

222

212

22

12

Example 1: SolutionExample 1: Solution Determine the resultant of the two forces F1 and

F2 acting on the hook.

F2 = 60N

60o

F1 = 54N

120o

F2 = 60N 60o

F1 = 54N R

Law of CosinesLaw of Cosines

Law of SinesLaw of Sines

RR = 98.8 N = 98.8 N 28.3 28.3°°

1.1. Draw the vector diagram.Draw the vector diagram.2.2. Use the Law of Cosines to find magnitude R.Use the Law of Cosines to find magnitude R.3.3. Use the Law of Sines to find direction of R.Use the Law of Sines to find direction of R.


Ropes support an I-beam Ropes support an I-beam of weight 1 kN. Thus of weight 1 kN. Thus resultant is 1 kN vertically resultant is 1 kN vertically downward.downward.

GivenGiven: when : when = 30 = 30º, tº, the resultant of Fhe resultant of F11 and F and F22 is 1 kN in the is 1 kN in the vertical downward direction. vertical downward direction.

FindFind: the value of F: the value of F11 and F and F22..

Example 2: SolutionExample 2: Solution

Solution:

1. Draw the vector diagram

2. Use the sine law to find

1 2 and F F

11

22

1000653

30 1301000

44620 130

FF N

Sin SinF

F NSin Sin

GivenGiven: when : when = 30 = 30º, tº, the resultant of Fhe resultant of F11 and F and F22 is is 1 kN in the vertical downward direction.1 kN in the vertical downward direction.

FindFind: the value of F: the value of F11 and F and F22..

290290ºº

240240ºº

Exercise 1:Exercise 1:

Determine the magnitude and direction of the resultant of the two forces acting on the eye hook.

Hint: The law of cosines is handy.

Exercise 1: SolutionExercise 1: Solution


Magnitude by Law of Cosines: Magnitude by Law of Cosines:

RR22 = F = F1122 + F + F22

22 – 2 F – 2 F11FF22cos 30cos 30RR22 = 85 = 8522 + 150 + 15022 – 2 (85)(150)cos 30 – 2 (85)(150)cos 30RR22 = 7641.4 N = 7641.4 N22

R= 87.4 NR= 87.4 N

FF1 1 = 85 N

FF22 = 150 N

3030ººRR

XX

YY

Angle by Law of Cosines: Angle by Law of Cosines:

FF2222 = F = F11

22 + R + R22 – 2 F – 2 F11RcosRcoscoscos = (F = (F11

22 + R + R22 – F – F222 2 ) / (2F1R)

coscos = (85 = (8522 + 87.4 + 87.422 – 150 – 1502 2 ) / (2*85*87.4)coscos = -0.5139 = -0.5139 = 120.9= 120.9ºº

*Note: to determine *Note: to determine an angle > 90an angle > 90º such º such as as , use the law of , use the law of Cosines instead of Cosines instead of the law of Sines.the law of Sines.R = 87.4 N R = 87.4 N 120.9 120.9





Magnitude by Law of Cosines: Magnitude by Law of Cosines:

RR22 = F = F1122 + F + F22

22 – 2 F – 2 F11FF22cos 35cos 35RR22 = 80 = 8022 + 65 + 6522 – 2 (80)(65)cos 35 – 2 (80)(65)cos 35RR22 = 2105.82 N = 2105.82 N22

R= R= 45.889245.8892 = 45.9 N = 45.9 N

3030ººAngle by Law of Sines: Angle by Law of Sines: (sin (sin / 65 N) = (sin 35 / 65 N) = (sin 35 / 45.8892 N) / 45.8892 N)sin sin = (65)(sin 35 = (65)(sin 35) / 45.8892 N) / 45.8892 N = 54.3355= 54.3355 = = + 10 + 10ºº = 64.3º= 64.3º

*Note: to determine *Note: to determine an angle > 90an angle > 90º such º such as as , use the law of , use the law of Cosines instead of Cosines instead of the law of Sines.the law of Sines.

FF1 1 = 80 N

FF22 = 65 N

RR

XX

YY

10101010

2525

R = 45.9 N R = 45.9 N 64.3 64.3

End Lesson 5End Lesson 5

Force Component: A resultant is a single force that is equivalent to the sum of a group of forces. In other words it can replace all those forces and give the same effect. The individual forces that the resultant replaces are called the components of the resultant.

Rectangular Components: 2 mutually perpendicular components. Typically selected along horizontal x and vertical y axes. Think of Fx and Fy as the projection of F onto the X and Y axes.

Rectangular ComponentsRectangular Components

F2

F1 F Example: Example: FF11 and and FF22 are are components of resultant components of resultant FF

Fx

Fy

F

x

y

0θ

Example: Example: FFxx and and FFyy are are rectangular components of rectangular components of FF

FFxx + + FFyy = = FF

Rectangular Components: Rectangular components are useful because they isolate the effect

of the resultant force in each direction. The effect of the force in the X direction acts independently of the

force effect in the Y direction (and vice versa). Using Rectangular Components, the effect of the resultant force in

each direction can be treated independently.


F2

F1 FExample: Non-rectangular componentsExample: Non-rectangular components

FF11 and F and F22 both have a force effect in both have a force effect in the X direction.the X direction.

Fx

Fy

F

x

y

0θ

Example: Rectangular componentsExample: Rectangular components

Only FOnly Fyy has an effect in Y direction and has an effect in Y direction and only Fonly Fxx has an effect in the X direction. has an effect in the X direction. They can be treated independently. They can be treated independently.


Rectangular Components Magnitude of rectangular components: If the magnitude and

direction of F is known, the magnitudes of the rectangular components can be calculated from right triangle trig.

Fx

FyF

x

y

0θ

FFxx = F cos = F cos θθ

FFyy = F sin = F sin θθ

MagnitudeMagnitude

y

Fx

Fy

F

x0θ

FFxx is called the is called the x-componentx-component of F. of F.FFyy is called the is called the y-componenty-component of F. of F.

Rectangular ComponentsRectangular ComponentsRectangular Components Magnitude of Rectangular Components:

What if F is directed as shown?

Assume Standard Position: Use Reference Angle ():

Fx

FyF

x

y

0

θ

Angle θ measured from +x axis. Equations yield direction sign of

component.

Example: if θ = 150°Fx = F cos 150 = -0.866 F

Fy = F sin 150 = 0.500 F

Use (positive acute angle from +x or –x axis).

Direction sign found by inspection.

Example: if θ = 150°, = 30°Fx = F cos 30 = 0.866 F -0.866 F

Fy = F sin 30 = 0.500 F

Fx

Fy

F

x

y

0 θ

Fx

FyF

x

y

0θ

Rectangular Components – Reference AngleRectangular Components – Reference Angle

For any angle θ in standard position, the Reference Angle α (pronounced “alpha”) is defined as the acute positive angle between the vector and the x-axis.

•Reference Angle α is acute (0º < α < 90º).•Reference Angle α is always positive.

I 0º < θ < 90º, then α = θII 90º < θ < 180º, then α = 180º - θIII 180º < θ < 270º, then α = θ - 180ºIV 270º < θ < 360º, then α = 360º - θ

To Calculate Reference Angle α II

IIIIII IVIV

IIII


Resolve the 60 N force on the ring into its horizontal (x) and vertical (y) components.

F = 60N

60°

Example 3: SolutionExample 3: Solution Resolve the 60 N force on the ring into its

horizontal and vertical components.

x

F = 60N

y

60°

Method 1: Standard PositionMethod 1: Standard PositionGiven:Given: F = 60N, F = 60N, = 60° = 60°

Find:Find: F Fxx, F, Fyy

Solution:Solution:

FFxx = F cos = F cos θθθθ = 360 = 360°° - - θθ = 360 = 360°° – 60– 60°° θθ = 300 = 300°°

FFxx = 60 cos 300 = 60 cos 300° ° NN

FFxx = 30.0 N = 30.0 N

FFyy = F sin = F sin θθ

FFyy = 60 sin 300 = 60 sin 300°°NN

FFyy = -51.96 = -51.96

FFyy = -52.0 N = -52.0 N

x

F = 60N

y

60°

Fy

Fx

Method 2: Reference AngleMethod 2: Reference AngleGiven:Given: F = 60N, F = 60N, = 60° = 60°


Solution:Solution:

FFxx = F cos = F cos →→

FFxx = 60 cos 60 = 60 cos 60°° N N

FFxx = 30.0 N = 30.0 N →→

FFyy = F sin = F sin ↓↓

FFyy = 60 sin 60 = 60 sin 60° ° NN

FFyy = 51.96 N = 51.96 N ↓↓

FFyy = 52.0 N = 52.0 N ↓↓


Resolve the weight of the 150 lb skater into components along the rail and normal to the rail.

W = 150 lb

1 ft

2 ft


Resolve the weight of the 150 lb skater into components along the rail and normal to the rail.

Given:Given: W = 150 lb, rail slope= 1:2. W = 150 lb, rail slope= 1:2.

Find:Find: W Wxx, W, Wyy

Solution:Solution:tan tan = 2 ft / 1 ft = 2 ft / 1 ft = tan= tan-1-1(2 / 1) (2 / 1) = 63.43= 63.43°°

WWxx = W cos = W cos

WWxx = 150 cos = 150 cos 63.434 lb 63.434 lb

WWxx = 67.1 = 67.1 lb lb oror WWxx = -67.1 = -67.1 lb lb

WWyy = W sin = W sin WWyy = 150 sin = 150 sin 63.43463.434°° lb lb WWyy = 134.2 lb = 134.2 lb or or WWyy = -134.2 = -134.2 lb lb

W = 150 lb

1 ft

2 ft

x

W = 150 lb

1 ft

2 ft

y

Wy

Wx


Resolve the 75 N force on the ring into its horizontal and vertical components.

F = 75N

48°


Resolve the 75 N force on the ring into its horizontal and vertical components.

Method 2: Reference AngleMethod 2: Reference AngleGiven:Given: F = 75 N, F = 75 N, = 48° = 48°


Solution:Solution:

FFxx = F cos = F cos ←←

FFxx = 75 cos 48 = 75 cos 48°° N N

FFxx = 50.2 N = 50.2 N ← ← oror -50.2 -50.2

FFyy = F sin = F sin ↓↓

FFyy = 75 sin 48 = 75 sin 48° ° NN

FFyy = 55.7 N = 55.7 N ↓ ↓ oror -55.7 -55.7

x

y

F = 75N

48°

Fy

Fx

F = 75N

48°

Rectangular ComponentsRectangular ComponentsRectangular Components - If the magnitudes of Fx and Fy are known,

the magnitude and direction of F can be calculated.

Fx

Fy

F

x

y

0 θ

Magnitude of F:

This is the Pythagorean Theorem

Direction of F:

is the angle of vector F with respect to x axis (pos. or neg.).

To find the direction angle θ, must know which quadrant vector lies in.

= tan-1Fy / Fx

F = (Fx2 + Fy

2)½

I 0º < θ < 90º, then θ = α II 90º < θ < 180º, then θ = 180º - αIII 180º < θ < 270º, then θ = 180º + αIV 270º < θ < 360º, then θ = 360º - α or θ = - α

Given:Given: F Fxx = -20 N, F = -20 N, Fyy = 5N = 5NFind: Find: FFMagnitude F = Magnitude F = (Fx2 + Fy2)½

F = (400 + 25)½ = 20.6 NDirection = tan-1Fy / Fx = tan-15 / 20= tan-1(0.25) = 14.04°2nd quadrant since Fx is neg.θ = 180 -14.04 = 166.0°θ = 166.0°F = 20.6 N 166.0°

F22 = Fx2 + Fy

2


The rectangular components of a force are given as Fx = -450 N and Fy = -300 N. Find the magnitude and direction of the force.

Fx= -450 N x

y

0

Fy= -300 N

Example 5: SolutionExample 5: Solution The rectangular components of a force are given

as Fx = -450 N and Fy = -300 N. Find the magnitude and direction of the force.

Given:Given: F Fxx = -450 N, F = -450 N, Fyy = -300 N = -300 NFind: Find: FFMagnitude – Magnitude – F = F = (Fx2 + Fy2)½

F = [(-450)2 + (-300)2]½ F = 540.83 NF = 541 NDirection- = tan-1Fy / Fx = tan-1300 / 450= tan-1(0.667) = 33.7°3rd quadrant since Fx and Fy are neg.θ = 180 + θ = 180 + 33.7°θ = 214°F = 541 N 214°

Fx= -450 N

F

x

y

0θ

Fy= -300 N

Fx= -450 N x

y

0

Fy= -300 N



The rectangular components of a force are given as Fx = 125 N and Fy = -288 N. Find the magnitude and direction of the force.

Fx= 125 N

x

y

Fy= -288 N

Exercise 4: SolutionExercise 4: Solution The rectangular components of a force are given

as Fx = 125 N and Fy = -288 N. Find the magnitude and direction of the force.

Given:Given: F Fxx = 125 N, F = 125 N, Fyy = -288 N = -288 NFind: Find: FFMagnitude – Magnitude – F = F = (Fx2 + Fy2)½

F = [(125)2 + (-288)2]½ F = 313.96 NF = 314 NDirection- = tan-1Fy / Fx = tan-1288 / 125= tan-1(2.304) = 66.5°2nd quadrant since Fy is neg.θ = - θ = -66.5°F = 314 N -66.5° or 314 N 293.5°


Fx= 125 N

F

x

y

0 θ

Fy= -288 N

Fx= 125 N

x

y

Fy= -288 N


Resultants by Rectangular ComponentsResultants by Rectangular Components

Resultants by rectangular components: The resultant of any number of concurrent coplanar forces can be calculated by resolving each into its rectangular components. Step 1: Resolve each force into its

rectangular components.

Step 2: All x components are directed horizontally and can be added algebraically to get x component of resultant (Rx)..

Rx = Fx = (F1)x+ (F2)x + (F3)x +… Step 3: All y components are directed vertically and can be added

algebraically to get y component of resultant (Ry)..

Ry = Fy = (F1)y+ (F2)y + (F3)y +…

Step 4: Using x and y components of resultant calculate magnitude and direction of resultant R.

R = R = (Rx2 + Ry

2)½ = tan--11Ry / Rx

F2

F1

x

y

0F3


Determine the resultant of the two forces using rectangular components; F1 = 3 kN 32° and F2 = 1.8 kN 105°.

F2 F1

x

y

0

Example 6: SolutionExample 6: Solution Determine the resultant of the two forces

F1 = 3 kN 32° and F2 = 1.8 kN 105°. Given:Given: F1 = 3 kN 32° and F2 = 1.8 kN 105°. Find: Find: ResultantResultant R R

F2

F1

x

y

0

X Components FX Components F11 – –

Fx1 = F1 cos θ1

Fx1 = 3000 cos 32° N

Fx1 = 2544.1 N

X Components FX Components F22 – –

Fx2 = F2 cos θ2

Fx2 = 1800 cos 105° N

Fx2 = -465.9 N

Y Components FY Components F11 – –

Fy1 = F1 sin θ1

Fy1 = 3000 sin 32° N

Fy1 = 1589.8 N

Y Components FY Components F22 – –

Fy2 = F2 sin θ2

Fy2 = 1800 sin 105° N

Fy2 = 1738.7 NX Components R – X Components R – Rx = Fx1 + Fx2

Rx = 2544.1 – 465.9 N

Rx = 2078.2 N

Y Components R – Y Components R – Ry = Fy1 + Fy2

Ry = 1589.8 + 1738.7 N

Ry = 3328.5 NMagnitude R – Magnitude R –

R = R = (Rx2 + Ry

2)½

R = [R = [(2078.2)2 + (3328.5)2]½ = 3923.9 N = 3920 N

Direction R – Direction R – = tan-1Ry / Rx = tan-13328.5 / 2078.2 = tan-1(1.602) = 58.0°Rx & Ry positive so 1st quadrantθ = = 58.0°

R = R = 3920 kN 58.0°

θ2F2 F1

x

y

0

θ1

R

Exercise 5A:Exercise 5A:

Determine the resultant of the two forces using rectangular components; F1 = 150 N 28° and F2 = 100 N 99°.

F2 F1

x

y

0

Exercise 5A - SolutionExercise 5A - Solution Determine the resultant of the two forces

F1 = 150 N 28° and F2 = 100 N 99°.Given:Given: F1 = 150 N 28° and F2 = 100 kN 99°.Find:Find: ResultantResultant R R

F2

F1

x

y

0

Y Components R – Y Components R – Ry = Fy1 + Fy2

Ry = F1 sin θ1+ F2 sin θ2

Ry = 150 sin 28° + 100 sin 99° N

Ry = 169.2 N

Magnitude R – Magnitude R –

R = R = (Rx2 + Ry

2)½

R = [R = [(116.8)2 + (169.2)2]½ = 206 N

Direction R – Direction R – = tan-1Ry / Rx = tan-1169.2 / 116.8 = 55.381°Rx & Ry positive so 1st quadrantθ = = 55.4°

R = R = 206 N 55.4°

θ2F2 F1

x

y

0

θ1

RX Components R – X Components R – Rx = Fx1 + Fx2

Rx = F1 cos θ1+ F2 cos θ2

Rx = 150 cos 28° + 100 cos 99° N

Rx = 116.8 N

Exercise 5B:Exercise 5B:

Determine the resultant of the three forces using rectangular components; F1 = 100 N, F2 = 200 N, and F3 = 300N.

F2

F1x

y

0

F3

20

40

60


Moment of a ForceMoment of a Force

What is the Moment of a Force?What is the Moment of a Force? A force can have two effects on a rigid body;A force can have two effects on a rigid body;

Translation - tends to move it linearly along its line of action.Translation - tends to move it linearly along its line of action. Rotation - tends to rotate it about an axis.Rotation - tends to rotate it about an axis.

Moment of a Force:Moment of a Force: the measure of a force’s tendency to rotate the measure of a force’s tendency to rotate the body about an axis. the body about an axis.

The moment of a force (moment) is also called torque (Ex: torque wrench).


What is the Moment of a Force? Moment of a Force in Action: Whenever there is a tendency for

rotational motion to occur, a Moment is at work.

What factors would What factors would increase the tendency increase the tendency for rotation to occur in for rotation to occur in the following examples?the following examples?

Moment of a ForceMoment of a Force Moment of a force : The tendency of a force to cause rotation

(moment of the force) depends on two factors; The magnitude of the force. The perpendicular distance (d) from the center of rotation (O)

to the line of action of the force. An important distinction:

Distance (d) - perpendicular distance from the center of rotation (O) to the line of action.

Distance (L) - distance from the center of rotation (O) to the point of force application.

LL

LL


Mathematical Definition of a Moment: The moment Mo of a force F about a point O is equal to the magnitude of the force multiplied by the perpendicular distance d from O to the line of action of the force.

Mo = F·d

Point O is referred to as the Moment Center Distance d is referred to as the Moment Arm.

Units - in terms of force & distance. SI Units: N·m or kN·m English Units: ft·lb or in·lb

LL

O


Let’s test the Let’s test the mathematical definition mathematical definition of a moment and see of a moment and see its affects using your its affects using your text book!text book!

AA

BB


Direction of a Moment: A moment is a vector just like a force is a vector. The direction of a moment vector is determined by the right hand rule.

Positive Moment - the force tends to cause a counter-clockwise rotation about the moment center .

Negative Moment - the force tends to cause a clockwise rotation about the moment center .

Positive MomentPositive Moment


Summation of Moments: In the 2D case (planar forces), moments can be added algebraically just like forces with the same line of action (Rx = F1x + F2x+ …).

Counter-clockwise moments positive. Clockwise moments negative.

oo xx

yyFF11

FF22

FF33

dd11

dd33

dd22

Mo = Mo1 + (-Mo2) + (-Mo3) +…

Moment of a ForceMoment of a Force Two ways to calculate the moment of a force about a point;

d=L

oo

x

y

FFxx

FF

Rectangular Component Method: Find the component of the force perpendicular to rod L. Calculate the moment using; Mo = Fx·d where

Fx is the normal component of Fd is the moment arm = L

Transmissibility Method: Extend the line of action to

find the moment arm length d ( distance from center to line of action).

Calculate the moment using; Mo = F·d where

d is the moment arm L

d

oo

L

FF


Determine the moment of a 500 N force about point o, if θ = 30°, 60°, 90°, and 120°.

θ

60°

200

mm

oo

F

Example 7: SolutionExample 7: SolutionDetermine the moment of a

500 N force about point o, if θ = 30°, 60°, 90°, and 120°.

θ

d

oo

200

mm

FF

θθ = 30 ° = 30 °sin sin θθ = d / .200 m = d / .200 md = .200 sin 30° md = .200 sin 30° md = .100 md = .100 m

MMoo= Fd= Fd

MMoo= - (500 N)(0.1 m)= - (500 N)(0.1 m)

MMoo= -50 Nm= -50 Nm or or

MMoo= 50 Nm = 50 Nm

θθ = 60 ° = 60 °sin sin θθ = d / .200 m = d / .200 md = .200 sin 60° mmd = .200 sin 60° mmd = 0.1732 md = 0.1732 m

MMoo= Fd= Fd

MMoo= - (500 N)(0.1732 m)= - (500 N)(0.1732 m)

MMoo= -86.6 Nm = -86.6 Nm oror

MMoo= 86.6 Nm = 86.6 Nm θθ = 90 ° = 90 °sin sin θθ = d / 200 mm = d / 200 mmd = 200 sin 90° mmd = 200 sin 90° mmd = 200 mm = 0.2 md = 200 mm = 0.2 m

MMoo= Fd= Fd

MMoo= - (500 N)(0.2 m)= - (500 N)(0.2 m)

MMoo= - 100 Nm= - 100 Nm or or

MMoo= 100 Nm= 100 Nm

θ

60°20

0 m

m

oo

F

Example 7: SolutionExample 7: SolutionDetermine the moment of a 500 N force

about point o, if θ = 30°, 60°, 90°, and 120°.

θθ = 120 ° = 120 ° = 180° - = 180° - θ θ = 180° - 120° = 60°= 180° - 120° = 60°sin sin = d / 200 mm = d / 200 mmd = 200 sin 60° mmd = 200 sin 60° mmd = 173.2 mm = 0.1732 md = 173.2 mm = 0.1732 m

MMoo= Fd= Fd

MMoo= - (500 N)(0.1732 m)= - (500 N)(0.1732 m)

MMoo= - 86.6 Nm= - 86.6 Nm or or

MMoo= 86.6 Nm= 86.6 Nm

θ

60°

200

mm

oo

F

θ

d

oo

200

mm

FF

Note:Note: Same value Same value as for as for = 60 = 60ºº


Determine the moment of force F= 150 lb about point o given the rod length (L) is 16 inches.

θ=128°

oo

F= 150 lb

L = 16 inL = 16 in


Determine the moment of force F= 150 lb about point O given the rod length (L) is 16 inches.

θ=128°

oo

F= 150 lb

ddL = 16 inL = 16 in

Given:Given: F = 150 lb F = 150 lbθθ = 128° = 128°L = 16 inL = 16 in

Find:Find: M Moo

Solution: = 180° - 128° = 52°° - 128° = 52°d = L sin d = L sin = 16 sin 52°° ind = 12.6082 ind = 12.6082 in

MMoo = Fd = (150 lb)(12.61 in) = Fd = (150 lb)(12.61 in)

MMoo = 1891 in lb = 1891 in lb

MMoo = (1891 in lb)(1 ft / 12 in) = (1891 in lb)(1 ft / 12 in)

MMoo = 157.6 ft lb = 157.6 ft lb


Varignon’s Theorem: The moment of a force about any point is equal to the sum of the moments produced by the components of the force about the same point.

This is the principle used in this method for moment calculation. In this case, the moment due to Fy is zero.d=

L

oo

x

y

FFxx

FF

FFyy

*For Moment calculations its important to *For Moment calculations its important to remember Transmissibility Law – Force remember Transmissibility Law – Force can be moved along line of action only. can be moved along line of action only. Don’t move FDon’t move Fyy tip to tail with F tip to tail with Fxx..


By Varignon’s Theorem, force F can be broken into By Varignon’s Theorem, force F can be broken into its components acting at point A. Using this approach, its components acting at point A. Using this approach, find the moment about O due to the 250 N force.find the moment about O due to the 250 N force.

oo

135°

200 mm

100 mm

AA

BB

FF

xx

yy

Exercise 7: Solution 1Exercise 7: Solution 1

Determine the moment of the 250 N force about point O.

Given:Given: F = 250 N; F = 250 N; = 135° = 135°

LLABAB= 100 mm; L= 100 mm; LBOBO= 200 mm= 200 mm

Find:Find: Moment M Moment Moo about point O. about point O.Solution:Solution: = 135°-90° = 45°Fx = Fcos = 250N (cos45°)

Fx = 176.7767N

Fy = Fsin = 250N (sin45°)

Fy = 176.7767N

MMoo = (0.2m)F = (0.2m)Fyy – (0.1m)F0.1m)Fxx

MMoo = (0.1m) = (0.1m) 176.7767N

MMoo= 17.68 Nm = 17.68 Nm

oo

135°

200 mm

100 mm

AA

BB

FF

x

y

FFxx

FFyy


Exercise 7: Solution 2Exercise 7: Solution 2

Determine the moment of the 250 N force about point O.

Given:Given: F = 250 N; F = 250 N; θθ = 135°; = 135°;


Find:Find: Moment M Moment Moo about point O. about point O.Solution:Solution:Tan (45° +) = LBO / LAB = 0.2 / 0.1 Tan (45° +) = 2.0(45° +) = 63.4349° = 18.4349°Fy = Fsin = 250 sin 250 sin 18.4349° N

Fy = 79.0567 N

Let d = L = LOA d = (0.22 + 0.12)½ = 0.22361 m

MMoo = F = Fyyd = (d = (79.0567 N)(0.2236 m)

MMoo = 17.68 Nm = 17.68 Nm

oo

135°

200 mm

100 mm

AA

BB

FF

LLOAOA

x

y

45°

FFxxFFyy


Transmissibility (review): The point of application of a force acting on a rigid body may be placed anywhere along its line of action.

Example below shows how this is useful.


Determine the moment of the 250 N force about point O.Use the geometry method in which d (moment arm

perpendicular to line of action) is determined.

oo

135°

200 mm

100 mm

AA

BB

FF

Example 8: SolutionExample 8: SolutionDetermine the moment

of the 250 N force about point O.

Given:Given: F = 250 N; F = 250 N; θθ = 135° = 135°


Find:Find: Moment M Moment Moo about point O. about point O.Solution:Solution: = 180° - 135° = 45°° - 135° = 45°tan = (LBC / 100) mm

LBC = (tan / 100) mm

LBC = tan 45°° / 100 = 100 mm

LCO = 200 – LBC = 100 mm = 90= 90°° - - = 45°°

d = d = LCO sin sin = 100 sin 45°° mmd = 70.71 mm = 0.07071 md = 70.71 mm = 0.07071 m

MMoo = Fd = (250 N)(0.07071 m) = Fd = (250 N)(0.07071 m)

MMoo = 17.68 Nm = 17.68 Nm

d

oo

135°

200 mm

100 mm

AA

BB CC

FF


Considering Varignon’s Theorem Considering Varignon’s Theorem andand Transmissibility, find the moment about O Transmissibility, find the moment about O due to the 250 N force.due to the 250 N force.

oo

135°

200 mm

100 mm

AA

BB

FF

xx

yy


Force CouplesForce Couples

Force Couple: A couple is defined as two parallel forces with equal magnitudes but opposite sense having different lines of having different lines of action separated by a distance (action separated by a distance (dd).

Net force of a couple of a couple is zero, but rotates in specified direction. Moment of a couple is not zero.Moment of a couple is not zero. Causes rotation about an axis perpendicular to plane of its Causes rotation about an axis perpendicular to plane of its

forces. forces. A force couple is a way to produce a moment without a net force.A force couple is a way to produce a moment without a net force.

EquivalentEquivalent

M = FdM = Fd


Moment of a Couple: The moment of a couple is given by;

M = Fd M = Fd

The moment of a couple is independent of The moment of a couple is independent of the choice of the moment center.the choice of the moment center.

The moment of a couple about any point, The moment of a couple about any point, anywhere, is the same.anywhere, is the same.

A couple may be transferred to any location A couple may be transferred to any location in its plane and still have the same effect.in its plane and still have the same effect.

OO

aa

MMOO = F (a+d) – Fa = F (a+d) – Fa

MMOO = Fa + Fd – Fa = Fa + Fd – Fa

MMOO = Fd = Fd thus M thus MO O is independent of distance a.

Force CouplesForce Couples Equivalent Couples: Two couples acting in the same plane are

equivalent if they have the same moment acting in the same direction produce the same mechanical effect.

If M = M’ then these areIf M = M’ then these are Equivalent Couples Equivalent Couples, and , and 0.4F0.4F = 0.3F’ = 0.3F’ F = (¾)F’F = (¾)F’


Addition of Couples: Addition of Couples: The moments about a point of two or The moments about a point of two or more couples acting in a plane may be added algebraically. more couples acting in a plane may be added algebraically. Using our convention Using our convention ccw ccw = positive = positive; ; cw cw = negative = negative..

dd11

dd22FF22

FF22

FF11

FF11

PP

Example:Example: Find the total Find the total moment about point P.moment about point P.

MMpp = -(F = -(F11dd11) + -(F + -(F22dd22)

What direction does the moment act?

Exercise 9:Exercise 9:Captain Bligh encounters rough seas. Determine the

moment of the couple required to keep his ship on track if F= 50 lb and the wheel diameter d = 30 inches.

-F-F FFdd

Exercise 9: SolutionExercise 9: SolutionCaptain Bligh encounters rough seas. Determine the

moment of the couple required to keep his ship on track if F= 50 lb and the wheel diameter d = 30 inches.

Given:Given: F = 50 lb F = 50 lbd = 30 ind = 30 in

Find:Find: M MSolution: M = F(d/2) M = F(d/2) + F(d/2) + F(d/2) = Fd = Fd M = (50 lb)(30 in) M = (50 lb)(30 in) M = 1500 in lb M = 1500 in lb

-F-F FFdd


Determine the moment of the force couple acting Determine the moment of the force couple acting on the block. on the block.

Then determine the force value for an equivalent Then determine the force value for an equivalent couple having opposing forces acting horizontally couple having opposing forces acting horizontally at C & D.at C & D.

AA

BBCC

DD

h=90 mmh=90 mm

w=120 mmw=120 mm

FFabab = 40 N = 40 N


Exercise 10: SolutionExercise 10: SolutionDetermine the moment of the Determine the moment of the

force couple acting on the force couple acting on the block. Then replace with an block. Then replace with an equivalent couple acting equivalent couple acting horizontally at C & D.horizontally at C & D.

Given: Given: FFabab = 40 N; w = 40 N; w = 120 mm; = 120 mm;h = 90 mmh = 90 mm

Find:Find: M and equivalent couple at M and equivalent couple at C & D.C & D.

Solution:Solution:

M = FM = Fababw w M = (40N)(120 mm) M = (40N)(120 mm) M = 4.80 Nm M = 4.80 Nm FFcdcdh = M = 4.80 Nm h = M = 4.80 Nm

FFcdcd = M/h = (4.80 Nm)/(0.900 m) = M/h = (4.80 Nm)/(0.900 m)

FFcdcd = 53.3 N (directions as shown) = 53.3 N (directions as shown)

AA

BBCC

DD

h=90 mmh=90 mm

w=120 mmw=120 mm



AA

BBCC

DD

h=90 mmh=90 mm

w=120 mmw=120 mm

FFcdcd

FFcdcd

Equivalent Equivalent CouplesCouples

Non-Concurrent Force SystemsNon-Concurrent Force Systems

In previous sections we;In previous sections we; Found the resultant of concurrent force systems.Found the resultant of concurrent force systems. Learned to calculate the moment of a force about a Learned to calculate the moment of a force about a

point (axis).point (axis). Learned about force couples and the moments they Learned about force couples and the moments they

produce.produce. In this section we focus on;In this section we focus on;

Non-Concurrent Force SystemsNon-Concurrent Force Systems.. Use what we learned about moments to find the Use what we learned about moments to find the

resultant of a resultant of a Non-Concurrent Force SystemNon-Concurrent Force System..

Force - Couple SystemsForce - Couple Systems A force system can be replaced by an equivalent system

consisting of a combination of forces and couples (moments). The net effect must be identical for each system – i.e. they

must both produce the same mechanical effect.

Equivalent Force SystemsEquivalent Force Systems – systems of forces are equivalent if; – systems of forces are equivalent if;they have the same resultant force, they have the same resultant force, andandthe same resultant moment about the same resultant moment about anyany selected point. selected point.

Are these EquivalentAre these Equivalent force force systems? (transmissibility) systems? (transmissibility)

Are these EquivalentAre these Equivalent force systems? force systems?

(parallel force displacement)(parallel force displacement)

Force - Couple SystemsForce - Couple Systems Force-Couple SystemForce-Couple System – any force may be moved to another – any force may be moved to another

point without changing its mechanical effect, provided that an point without changing its mechanical effect, provided that an appropriate couple (moment) is added.appropriate couple (moment) is added.

The value of the added couple (moment) is equivalent to the The value of the added couple (moment) is equivalent to the moment of the force at its original location about its new location.moment of the force at its original location about its new location.

FF FFNotNot

EquivalentEquivalent

FF FF FF

-F-F

FF FF

-F-F

FF

M=FdM=Fd

dd dd dd

Step 1:Step 1: Step 2:Step 2: Step 3:Step 3:Proper Method:Proper Method:


Replace the 2 KN force F with a force-couple system at point B.

F = 2 KNF = 2 KNAA

w=1.4 mw=1.4 m

h=390 mmh=390 mm

BB

Example 9: SolutionExample 9: Solution Replace the 2 KN force F with a force-couple system at point B.

Given:Given: F = 2KNF = 2KNh = 390 mmh = 390 mmw = 1.4 mw = 1.4 m

Find:Find: Equivalent force-couple system Equivalent force-couple system at B.at B.

Solution:Solution: Step 1:Step 1: Move the force to point B. Move the force to point B.Step 2:Step 2: Calculate moment about B Calculate moment about B due to F at A.due to F at A.M = F(0.390 m) M = F(0.390 m)

M = (2000 N)(0.390 m) M = (2000 N)(0.390 m) M = 780 Nm M = 780 Nm or -780 Nm or -780 NmStep 3:Step 3: Place moment at B. Place moment at B.

F= 2 KNF= 2 KNAA

w=1.4 mw=1.4 m

h=390 mmh=390 mm

BB

F=2 KNF=2 KN

AA

F=2 KNF=2 KN

AA

BB

BB

M=780 NmM=780 Nm

Step 1:Step 1:

Step 3:Step 3:


Replace the three forces shown with an Replace the three forces shown with an equivalent force-couple system at A.equivalent force-couple system at A.

FF11

FF22

FF33


To find the equivalent set of forces at A.

x x

o o o400 N cos 180 750 N cos 36.87 100 N cos 90

200 N

R F

1 o3tan 36.87

4

y y

o o o400 Nsin 180 750 Nsin 36.87 100 Nsin 90

550 N

R F


Find the moments about point A.

Using the line of action for the force at B. The force can be moved along the line of action until it reaches perpendicular distance from A

1 B

100 N 360 mm

36000 N-mm

M F d

��


Find the moments about point A.

The force at O can be broken up into its two components in the x and y direction

ox

oy

750 N cos 36.87

600 N

750 Nsin 36.87

450 N

F

F

Using the line of action for each component, their moment contribution can be determined.


Find the moments about point A.Using the line of action for Fx component d is 160 mm.

Fy component is 0 since in line with A.

2 Ox

600 N 160 mm

96000 N-mm

M F d

��

B i

1 2 3

36000 N-mm 96000 N-mm 0 N-mm

132000 N-mm

M M

M M M

k k k

k

��

��


The final result is

R = 585 N at 70.0o

M = 132 Nm

M = 132 Nm


Resultant of Nonconcurrent Resultant of Nonconcurrent Coplanar Force SystemsCoplanar Force Systems

Previously we found the resultant (single force equivalent to the Previously we found the resultant (single force equivalent to the given force system) for a given force system) for a ConcurrentConcurrent, , CoplanarCoplanar force system – all force system – all the forces passed through a single point.the forces passed through a single point.

In a In a NononcurrentNononcurrent ColanarColanar force system; force system; No concurrent point exists.No concurrent point exists. Line of action for resultant is not immediately known.Line of action for resultant is not immediately known.

Q:Q: If the forces are Nonconcurrent, how do we find the resultant?If the forces are Nonconcurrent, how do we find the resultant?A:A: We need to introduce moments into our resultant formulation.We need to introduce moments into our resultant formulation.

Remember! Equivalent Force SystemsRemember! Equivalent Force Systems – systems of forces are – systems of forces are equivalent if;equivalent if;

they have the same resultant force, they have the same resultant force, andandthe same resultant moment about the same resultant moment about anyany selected point. selected point.


Steps to find the resultant of a Steps to find the resultant of a Nonconcurrent CoplanarNonconcurrent Coplanar force system: force system: Step 1:Step 1: Find the magnitude and direction of the resultant Force. Find the magnitude and direction of the resultant Force.

Same as before. Same as before. Choose convenient x-y coordinate system.Choose convenient x-y coordinate system.

Resolve each force into x and y components.Resolve each force into x and y components.

Components of the reaction force are the algebraic sums of the Components of the reaction force are the algebraic sums of the individual force components;individual force components;

RRxx = = FFxx R Ryy = = FFyy

Find magnitude of the resultant Find magnitude of the resultant R = (RR = (Rxx22+ R+ Ryy

22))½½

Find the direction of the resultant Find the direction of the resultant αα = tan = tan-1-1RRy y / R/ Rxx then find then find


Steps to find the resultant of a Steps to find the resultant of a Nonconcurrent CoplanarNonconcurrent Coplanar force system: force system:

Step 2:Step 2: Find the location for the line of action of the resultant.Find the location for the line of action of the resultant. The moment of the original force system and the moment of the The moment of the original force system and the moment of the

resultant force about an arbitrary point must be equal.resultant force about an arbitrary point must be equal.

Calculate moment of the original force system about Calculate moment of the original force system about anyany convenient point.convenient point.

Knowing this and the resultant direction & magnitude, the Knowing this and the resultant direction & magnitude, the resultant location can be calculated.resultant location can be calculated.


Find the resultant and its location for the 2 force system shown.

FF11= 9.32 KN= 9.32 KNAA

w=1.4 mw=1.4 m

h=300 mmh=300 mm

BB CC

FF22= 10 KN= 10 KN4040ºº

Example 11: SolutionExample 11: Solution Find the resultant and its location for the 2 force system shown.

Solution:Solution: Step 1:Step 1: Magnitude & direction of Magnitude & direction of resultant.resultant.

Break 10KN force into components;Break 10KN force into components;

FFxx = 10KN cos 40 = 10KN cos 40º Fº Fyy = 10KN sin 40= 10KN sin 40ºº

FFxx = 7.66 KN = 7.66 KN F Fyy = 6.43 KN= 6.43 KN

Find RFind Rxx = = FFxx

RRxx = (9.32 – 7.66) KN = (9.32 – 7.66) KN

RRxx = 1.66 KN or 1.66 KN = 1.66 KN or 1.66 KN

Find RFind Ryy = = FFyy

RRyy = 0 + 6.43 KN = 0 + 6.43 KN

RRyy = 6.43 KN or 6.43 KN = 6.43 KN or 6.43 KN

F= 9.32 KNF= 9.32 KNAA

w=1.4 mw=1.4 m

h=300 mmh=300 mm

BBCC

F= 10 KNF= 10 KN4040ºº

AA

w=1.4 mw=1.4 m

h=300 mmh=300 mm

BBCC

FFxx= 7.66 KN= 7.66 KN

FFyy= 6.43 KN= 6.43 KN

F= 9.32 KNF= 9.32 KN


Solution:Solution: Find magnitude of R:Find magnitude of R:

R = R = (R(Rxx22+ R+ Ryy

22))½½ R = [R = [(1.66(1.6622+ 6.43+ 6.4322)])]½½ KNKNR = 6.64 KNR = 6.64 KN

Find direction of R:Find direction of R:αα = tan = tan-1-16.43/1.666.43/1.66αα = 75.5 = 75.5º = º = θθ

RR = 6.64 KN = 6.64 KN 75.5 75.5ºº

R= 6.64 KNR= 6.64 KN75.575.5ºº

AAF= 9.32 KNF= 9.32 KN

w=1.4 mw=1.4 m

h=300 mmh=300 mm

BBCC

FFxx= 7.66 KN= 7.66 KN

FFyy= 6.43 KN= 6.43 KN

75.575.5ºº

RRxx= 1.66 KN= 1.66 KN

RRyy= 6.43 KN= 6.43 KN

RR


Solution:Solution: Step 2: Find location of resultant.Step 2: Find location of resultant.Find moment of force system about C:Find moment of force system about C:

MMCC = = MMFF

MMCC = (9.32KN)(300mm) + 0 + 0 = (9.32KN)(300mm) + 0 + 0

MMCC = 2.796 Nm = 2.796 Nm or -2.796 Nm or -2.796 Nm

R must produce this same moment, so R must produce this same moment, so locate R accordingly. By inspection, R locate R accordingly. By inspection, R must be to the left of C to produce a must be to the left of C to produce a negative moment negative moment . Draw line of . Draw line of action and solve for daction and solve for dxx and d and dyy which which give Mgive MCC = 2.796 Nm = 2.796 Nm or -2.796 Nm or -2.796 Nm

F= 9.32 KNF= 9.32 KNAA

w=1.4 mw=1.4 m

h=300 mmh=300 mm

BB

CC

FFxx= 6.13 KN= 6.13 KN

FFyy= 5.14 KN= 5.14 KN

AA

BBCC

RR75.575.5ºº

yy

xx

ddyy

ddxx


Review data: MReview data: MCC = 2.796 Nm = 2.796 Nm

RRxx = 1.66 KN = 1.66 KN R Ryy = 6.43 KN = 6.43 KNAt x intercept, RAt x intercept, Rxx produces no moment produces no momentAbout C.About C.

RRyyddxx = M = MCC

ddxx = M = MC C / RRyy= 2.796 Nm / 6.43 KN= 2.796 Nm / 6.43 KN

ddxx = 0.434 m = 0.434 m

At y intercept RAt y intercept Ryy produces no moment produces no moment

RRxxddyy = M = MCC

ddyy = M = MC C / RRxx= 2.796 Nm / 1.66 KN= 2.796 Nm / 1.66 KN

ddyy = 1.684 m = 1.684 m

Any location along line of action is Any location along line of action is acceptable. Choose x = -0.434 , y = 0 acceptable. Choose x = -0.434 , y = 0 since located on part.since located on part.

AA

BBCC

RR75.575.5ºº

yy

xx

ddyy

ddxx

Y interceptY intercept

X interceptX intercept

AA

BBCC

RRθθ==75.575.5ºº

yy

xx

ddy y = 1.684 m

ddx x = 0.434 m

θθ


Find the resultant for the 3 force system Find the resultant for the 3 force system shown and locate it with respect to point A.shown and locate it with respect to point A.

AABB CC

yy

xx1.5 m1.5 m

33

FF11=500 N=500 N

DD

44

FF22=200 N=200 N

FF33=450 N=450 N

1.5 m1.5 m 1m1m


Find the resultant for the 3 force system Find the resultant for the 3 force system shown and locate it with respect to point A.shown and locate it with respect to point A.

R = 716 N R = 716 N 245 245

AABB CC

yy

xx1.5 m1.5 m

33

500 N500 N

DD

44

200 N200 N

450 N450 N

1.5 m1.5 m 1m1m

X = 2.77 mX = 2.77 m

Rx = -300 NRx = -300 NRy = -650 NRy = -650 NR = 716 N R = 716 N = 65.2= 65.2°° = 245°= 245°R = 716 N R = 716 N 245 245MMAA= -1800 Nm= -1800 NmX = 2.77 mX = 2.77 m


Resultant of Distributed Line LoadsResultant of Distributed Line Loads

So far we’ve dealt with loads developed So far we’ve dealt with loads developed by individual forces concentrated at a by individual forces concentrated at a point called point called Point Loads Point Loads oror Concentrated Concentrated LoadsLoads..

But what if a load is continuous along But what if a load is continuous along the length of a beam, say due to sacks the length of a beam, say due to sacks of concrete.of concrete.

This continuous loading is referred to as a This continuous loading is referred to as a Distributed LoadDistributed Load. It . It may be exerted along a line, over an area, or throughout an may be exerted along a line, over an area, or throughout an entire body (volume – remember gravity?). entire body (volume – remember gravity?). We will only We will only consider the line loadconsider the line load..

Goal:Goal: determine how to represent a distributed line load as a determine how to represent a distributed line load as a point load resultant which we are more familiar with.point load resultant which we are more familiar with.


Load IntensityLoad Intensity ( (ww) – the magnitude of the load per unit of length ) – the magnitude of the load per unit of length over which it acts. (lb/ft) (N/m)over which it acts. (lb/ft) (N/m)

example:example: 300 lb/ft - each foot of load represents 300 lb of force. 300 lb/ft - each foot of load represents 300 lb of force.

Types of Distributed Loads:Types of Distributed Loads:

Uniform Load Uniform Load – distributed load with – distributed load with constant intensity (constant intensity (ww). ).

example:example: 400 N/m uniform load 400 N/m uniform load distributed over 2m equals 800 N total.distributed over 2m equals 800 N total.

Triangular Load Triangular Load – distributed load – distributed load whose intensity varies linearly from 0 to whose intensity varies linearly from 0 to some maximum value (some maximum value (wwoo). ).

example:example: 300 N/m triangular load 300 N/m triangular load distributed over 2m equals 300 N total.distributed over 2m equals 300 N total.


Types of Distributed Loads:Types of Distributed Loads:

Trapezoidal Load Trapezoidal Load – distributed load – distributed load whose intensity varies linearly from a whose intensity varies linearly from a non-zero value to some maximum non-zero value to some maximum value. Can be treated as a uniform value. Can be treated as a uniform load plus a triangular load.load plus a triangular load.

example:example: a trapezoidal load with a minimum intensity of 200 N/m a trapezoidal load with a minimum intensity of 200 N/m and a maximum intensity of 300 N/m distributed over 2m equals and a maximum intensity of 300 N/m distributed over 2m equals (200 N/m)(2 m) + (½)(100 N/m)(2m) = 500 N total.(200 N/m)(2 m) + (½)(100 N/m)(2m) = 500 N total.


Equivalent Concentrated Force:Equivalent Concentrated Force: to determine the resultant of a to determine the resultant of a force system involving distributed loads, each distributed load force system involving distributed loads, each distributed load may be replaced by its equivalent concentrated load as follows;may be replaced by its equivalent concentrated load as follows;

Magnitude – Magnitude – equal to area of loading diagram.equal to area of loading diagram. Direction – Direction – according to distributed load direction (typically according to distributed load direction (typically

vertically down).vertically down). Location – Location – line of action passes through centroid of loading line of action passes through centroid of loading

diagram.diagram.


For Trapezoidal Load – For Trapezoidal Load – break into Uniform and Triangular load break into Uniform and Triangular load first (treat them independently).first (treat them independently).

Exercises 12:Exercises 12: Convert each distributed load to point load(s) of appropriate Convert each distributed load to point load(s) of appropriate

magnitude and locate correctly with respect to left end of beam.magnitude and locate correctly with respect to left end of beam.

2 m 3 m 1 m

30 kN6 kN/m

A B

3 m2 m

A

1 m

B

30 kN

Ra Rb

Rw

Law

A

1 kip/ft

B

3 kip/ft

9 ft

9 ft

A B

P2P1

Rb

LbLa

Ra

B

1 kip/ft5 kips

A

10 kips

4 kip/ft

9 ft

5 ft

6 ft 10 ft

5 ft

9 ft

A

10 ft

5 kips

6 ft

10 kips

B

P1 P2

RbRa

L1 L2

Exercises 12: SolutionExercises 12: Solution Convert each distributed load to a point load of appropriate Convert each distributed load to a point load of appropriate

magnitude and locate correctly on the beam.magnitude and locate correctly on the beam.

2 m 3 m 1 m

30 kN6 kN/m

A B

3 m2 m

A

1 m

B

30 kN

Ra Rb

Rw

LawA

1 kip/ft

B

3 kip/ft

9 ft

9 ft

A B

P2P1

Rb

LbLa

Ra

B

1 kip/ft5 kips

A

10 kips

4 kip/ft

9 ft

5 ft

6 ft 10 ft

5 ft

9 ft

A

10 ft

5 kips

6 ft

10 kips

B

P1 P2

RbRa

L1 L2

(6 kN/m)(5m) = 30 kN at 2.5 m(6 kN/m)(5m) = 30 kN at 2.5 m

(1 kip/ft)(9ft) = (1 kip/ft)(9ft) = 9 kip at 4.5 ft9 kip at 4.5 ft (½)(2 kip/ft)(9ft) = (½)(2 kip/ft)(9ft) = 9 kip at 6 ft9 kip at 6 ft

(½)(4 kip/ft)(9ft) = (½)(4 kip/ft)(9ft) = 18 kip at 3 ft18 kip at 3 ft (1 kip/ft)(16ft) = (1 kip/ft)(16ft) = 16 kip at 22 ft16 kip at 22 ft


Example 12: Example 12:

SolutionSolution

Example 12: Example 12: SolutionSolution


Determine the resultant force of the loads acting Determine the resultant force of the loads acting on the beam shown, and specify its location on on the beam shown, and specify its location on the beam with respect to point A.the beam with respect to point A.


Determine the Determine the resultant force of resultant force of the loads acting the loads acting on the beam on the beam shown, and shown, and specify its location specify its location on the beam with on the beam with respect to point A.respect to point A.


CHAPTER TWO.. The END!CHAPTER TWO.. The END!

RETIRED MATERIAL RETIRED MATERIAL FOLLOWS:FOLLOWS:

Exercise:Exercise:Considering that the 2 force systems shown are

equivalent (same mechanical effect), determine the tension in each of the equal length cables (left diagram) supporting a 1250 lb weight.

Equivalent Equivalent Force Force

SystemsSystems

SKIPSKIP

Exercise: SolutionExercise: Solution Determine the tension in each of the equal length cables

supporting a 1250 lb weight.

70°

1250 lb1250 lb

For equilibrium, the resultant For equilibrium, the resultant RR from the tension ( from the tension (TT11, , TT22) in the two cables must be ) in the two cables must be equal and opposite force vector from the 1250 lb weight.equal and opposite force vector from the 1250 lb weight.

1250 lb1250 lb

RR

TT11

TT22

3535°°

3535°°

110110°°

Solve for force triangle using Solve for force triangle using law of Sines:law of Sines:

TT1 1 / (sin 35/ (sin 35°)°) = R / (sin 110 = R / (sin 110°)°)

TT11 = R = R sin 35sin 35° / ° / sin 110sin 110°°

TT11 = (1250)(0.5736)/(0.9397) = (1250)(0.5736)/(0.9397)

TT11 =762.98 lb =762.98 lb

TT11 =763 lb =763 lb TT22 =763 lb =763 lb

SKIPSKIP

Vector lesson and problems

Education

resultant vector r

force vector

resultant of multiple

tip of r

tail of r

vectors v

angle of resultant

createthe resultant