Finding The Resultant Of A Group Of Forces

Finding the Resultant of A Group of Forces?V

This is the porch of AntoniGaudí ’s Güell Chapel.The sloping masonry piers liealong the lines of action of theresultants of the forces from thebrick arches and vaults thatconverge from above. Thus theyexperience only axial forcesunder the loading for which theyare designed.In this lesson, you will learnthe graphical method forfinding the resultant of a groupof forces, varying inmagnitude and direction, thatlie in a single plane.

To proceed with this lesson, click on the Next buttonhere or at the top of any page.When you are done with this lesson, click on theContents button here or at the top of any page toreturn to the list of lessons.

?V Finding the Resultant of A Group of Forces

The Problem: The forcesrepresented by the vectors onthis page all vary in magnitudeand direction.We would like to find theirresultant, including:

the magnitude of theresultant force,its direction, andthe location of its line ofaction.

FEDCB

A

G

Diagram Of Force Vectors1 MN


Step 1: Apply intervalnotation to the Diagram ofForce Vectors.Working left to right, we placecapital letters in the intervalsbetween the forces.Each force is named by theletters on either side. Forinstance, the leftmost force isAB. The force acting upwardto the right is EF.

FEDCB

A

G

Force AB

Force EF



Step 2: Construct the LoadLine.Now we combine the vectorsinto a load line, using intervalnotation to keep track of them.Working again from left toright, we plot the forces tip-to-tail. The first force, AB, isplotted onto the Load Line assegment ab. Segment ab isparallel, and equal inmagnitude, to AB on theDiagram of Force Vectors.Because AB acts downward,point b on the Load Line isbelow point a.

FEDCB

A

G


a

Load Line

b

1 MN


Next we plot BC on theDiagram of Forces assegment bc on the Load Line.Again, bc is parallel, andequal in magnitude, to BC.

FEDCB

A

G


a

b

1 MNLoad Line

c


Next we plot cd onto the LoadLine.

FEDCB

A

G


a

c

b

d

1 MNLoad Line


Next we plot de.

FEDCB

A

G


1 MN

a

c

b

d

eLoad Line


Because EF acts upward onthe Diagram of Force Vectors,point f occurs above point eon the Load Line.

FEDCB

A

G


1 MN

a

c

d

b

e

f

Load Line


Plotting fg completes the LoadLine.Notice that this Load Line isnot vertical. This is becauseits forces have horizontal aswell as vertical components.FE

DCB

A

G


1 MN

a

c

d

e

b

f

gLoad Line


Step 3: Find the magnitudeand direction of theresultant.The magnitude and directionof the resultant, ag, is foundmerely by connecting theends of the Load Line,working from the tail of ab tothe tip of fg.However, we do not yet knowthe location of the resultant’sline of action on the Diagramof Force Vectors.

FEDCB

A

G


1 MN

a

c

d

e

b

f

gLoad Line

Resultant ag


Step 4: Construct a forcepolygon.To find the location of theresultant’s line of action, webegin by establishing a pole,o, at any convenient locationnear the Load Line.We draw rays from this poleto the nodes on the Load Line.The Load Line now becomespart of a Force Polygon.

FEDCB

A

G


a

f

g

b

e

c

d

o

eForce Polygon

b

1 MN

Pole


Step 5: Construct thefunicular polygon.Drawing lines parallel to therays, and using intervalnotation to maintain order, weconstruct a Funicular Polygonover the original Diagram ofForce Vectors.Beginning at the left, a lineparallel to ray oa is drawn ininterval A of the Diagram ofForce Vectors. This line, alsolabeled oa, is placed so thatit intersects with force ABclose to its right-hand end.

FEDCB

A

Goa

FunicularPolygon


1 MN

f

g

a

b

eForce Polygon

oa

o

b

c

d

e


Next, a line parallel to ray obis drawn in space B of theDiagram of Force Vectors. Theleft end of this line is drawnto coincide with theintersection of oa and AB.The line of action of force BCis extended as necessary tointersect with opposite end ofob.

FEDCB

A

Gob

oa

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

oaob

o

b

c

d

e


Continuing to work from leftto right, we next construct ocin space C of the Diagram ofForce Vectors.The left end of this linecoincides with the intersectionof ob and the line of action ofBC.

obFE

DCB

A

G

ocoa

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

oaob

oco

b

c

d

e


Next, we construct od inspace D of the Diagram ofForce Vectors.Once again, we extend theline of action of force DE asnecessary to intersect withod.

oboc

FEDCB

A

God

oa

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

oaob

oc

od

o

b

c

d

e


Next, we construct oe inspace E of the Diagram ofForce Vectors.

oboc od

FEDCB

A

Goa oe

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

oaob

oc

od

oe

o

b

c

d

e


Next, we construct of.

oboc od

oe FEDCB

A

G

of

oa

Funicular Polygon


1 MN

f

g

a

eForce Polygon

oaob

oc

od

oe

ofo c

d

bb

e


Lastly, we construct og.With all rays of the FunicularPolygon constructed, we cannow find the location of theline of action of the resultant.ob

oc od

of

oe FEDCB

A

Gog

oa

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

oaob

oc

od

oe

of

og

o

b

c

d

e


Step 6: Find the line ofaction of the resultant.When the first and lastsegments of the FunicularPolygon, oa and og, areextended, they intersect onthe line of action of force ag,the resultant.The direction and magnitudeof ag are transferred to theDiagram of Force Vectors fromthe Force Polygon.Our solution is complete.

og

oboc od

of

oe FEDCB

A

Goa

Resultant ag

Funicular Polygon


1 MN

f

g

a

b

eForce Polygon

o

b

c

d

e


This technique is especiallyuseful for finding the resultantof parallel or nearly parallelforces. It is a good way, forexample, to find the resultantof a complex set of loads ona beam or truss.

Click on the Contents buttonto begin a new lesson.Click on the image of theGüell Chapel to return to thebeginning of this lesson.

og

oboc od

of

oe FEDCB

A

Goa

Resultant ag


1 MN

f

g

a

b

eForce Polygon

o

b

c

d

e

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Finding The Resultant Of A Group Of Forces