Lecture # 2.pdf Addition of Vectors and Scalars

Engg Mechanics

1. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
Vector Addition of Forces
Experimentalevidencehasshownthataforceisavectorquantity
sinceithasaspecifiedmagnitude.direction,andsenseanditadds
accordingtotheparallelogramlaw.
Twocommonproblemsinstaticsinvolveeitherfindingthe
resultantforce,knowingitscomponents,orresolvingaknown
forceintotwocomponents.
The parallelogram law must be
used to determine the resultant of
the two forces acting on the hook.

2. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
APPLICATION OF VECTOR ADDITION
There are three concurrent forces
acting on the hook due to the
chains.
We need to decide if the hook will
fail (bend or break).
To do this, we need to know the
resultant or total force acting on
the hook as a result of the three
chains.
F
R

3. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
VECTOR ADDITION ( by Parallelogram law)
All vector quantities obey the Parallelogram law of addition.
Two forces A and B are added to form a Resultant Force R = A + B using the:
following procedure:
• First join the tails of the components at a point so that it makes them concurrent.
• From the: head of B, draw a line parallel to A. Draw another line from the head of
Athat is parallel to B. These two lines intersect at point P toform the adjacent sides
of a Parallelogram.
• The diagonal of this parallelogram that extends to Pforms R, which then
represents
the resultant vector R= A+ B.

4. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
VECTOR ADDITION (by triangle rule)
We can a1so add Bto Ausing the Triangle Rule which is a special case of the
parallelogram law, where by vector Bis added to vector Ain a "head-to-tail"
fashion. i.e., by connecting the head of Ato the tail of B.
The resultant Rextends from the tail of Ato the head of B. In a similar manner, R
can also be obtained by adding Ato B.
The vectors can be added in either order i.e. R = A + B = B + A

5. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
Finding the Components of a Force
Sometimesitisnecessarytoresolveaforceintotwocomponentsinorderto
studyitspullingorpushingeffectintwospecificdirections.
Forexample.InFig.2.8a,Fistoberesolvedintotwocomponentsalongthe
twomembers,definedbytheUandVaxes.Inordertodeterminethe
magnitudeofeachcomponent,aparallelogramisconstructedfirstby
drawinglinesstartingfromthetipofF,onelineparalleltoUandtheother
lineparalleltoV.TheselinesthenintersectwiththeUandVaxes,forminga
parallelogram.TheforcecomponentsFuandFvarethenestablishedby
simplyjoiningthetailofFtotheintersectionpointsontheUandVaxes.
Thisparallelogramcanthenbereducedtoatrianglewhichrepresentsthe
trianglerule.

6. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
ADDITION OF FORCE IN CASE OF RECTANGULAR
COMPONENTS
•Each component of the force is shown as a
magnitude and a direction.
The rectangular components of force Fshown in Fig. are found using the
parallelogram law, so that F = Fx+ Fy, Because these components form a
right triangle, their magnitude and direction can be determined from:
•When a force is resolved into two
components along the xand yaxes, the
components are then called rectangular
components.

7. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
ADDITION OF FORCE IN CASE OF RECTANGULAR
COMPONENTS
Here the y component is negative since Fyis directed along the negative
y-axis.
The positive and negative sign is to be used only for computational
purpose, not for graphical representations in figures.
•Instead of using θ, the direction of F can
also be defined using a small “slope”
triangle, as shown in figure:
or
and
or

8. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
ADDITION OF SEVERAL FORCES
•Step 3 is to find the magnitude and
angle of the resultant vector.
•Step 2 is to add all the x-components
together, followed by adding all the y-
components together. These two totals
are the x and y-components of the
resultant vector.
•Step 1 is to resolve each force into
its components.

9. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
Break the three vectors into components, then add them.
F
Rx= F
1x -F
2x+ F
3x
F
Ry= F
1y +F
2y-F
3y
An example of the process:

10. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
EXAMPLE
Plan:
a) Resolve the forces into their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Given:Three concurrent forces acting
on a tent post.
Find:The magnitude and angle of the
resultant force.

11. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
EXAMPLE(continued)
For F
1
F
1x = 0N & F
1y = 300N
For F
2
F
2x = –450 cos(45°)N& F
2y = 450 sin (45°)N
F
2x = –318.2N F
2y = 318.2 N
For F
3
F
3x = (3/5) 600N & F
3y = (4/5) 600N
F
3x= 360N F
3y = 480N

12. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
EXAMPLE(continued)
Summing up all the xand ycomponents respectively, we get,
F
Rx= (0 –318.2 + 360) N = 41.80 N
F
Ry= (300 + 318.2 + 480) N = 1098 N
x
y

F
R
Using magnitude and direction:
F
R= ((41.80)
2
+ (1098)
2
)
1/2
= 1099 N
= tan
-1
(1098/41.80) = 87.8°

13. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
GROUP PROBLEM SOLVING
Plan:
a) Resolve the forces into their x and y-components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Given:Three concurrent forces
acting on a bracket.
Find:The magnitude and angle
of the resultant force.

14. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
F
1= {800 cos(60°) + 800 sin (60°) } N
= { 400 + 692.8 } N
F
2= {-600 sin (45°) + 600 cos(45°)} N
= { -424.3 + 424.3} N
F
3= {(12/13) 650 −(5/13) 650} N
{ 600 −250 } N
GROUP PROBLEM SOLVING (continued)

15. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.
GROUP PROBLEM SOLVING (continued)
Summing up all the x andycomponents, respectively, we get,
F
Rx= { (400 –424.3 + 600) N= 575.7 N
F
Ry=(692.8 + 424.3 –250)N = 867.1 N
Now find the magnitude and angle,
F
R= ((575.7)
2
+ (867.1)
2
)
½
= 1041 N
= tan
–1
( 867.1 / 575.7 ) = 56.4°
From positive x-axis, = 56.4°
x
y

F
R

16. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.

17. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.

18. Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbelerand Kai BengYap
© Pearson Education South Asia PteLtd
2013. All rights reserved
.