# Mechanics

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Lecture No. 03                                        BASIC MECHANICAL ENGINEERING

Addition of a System of Coplanar Forces

Vector Notation
In many problems it will be necessary to resolve a force into 2
components that are perpendicular to each other.
y

ˆ
j

iˆ   x
O
2 vectors, iˆ and ˆ that have the direction shown and magnitude 1 - unit
j
vectors.

ˆ
The i and ˆ provide direction!
j

ˆ
-suppose I want a vector 4 units long in the x-direction V  4i

 
-a vector 5 units long in the negative y-direction V  5  ˆ  5 ˆ
j     j

                                        
ˆ
Is P  0.6i  0.8 ˆ a unit vector?
j                     Yes P  0.6 2  0.8 2  1

Suppose we have a force, with magnitude Fx , that lies on the x-axis.
                              
ˆ                               ˆ
Fx  Fx i and one on the y-axis Fy  Fy i .

What is the resultant, F ?
        
ˆ
F  Fx  Fy  Fx i  Fy ˆ
j
           
Remember: Fx , Fy and Fx are vectors
Fx , Fy and F are magnitude of vectors, which are scalars


What does F look like?
y                                                                 
What is the magnitude of F ?
2
F     Fx  Fy
2

                            
Fy                           F                What is ?
 Fy     
  tan 1        
 Fx     
        
ˆ
j            

iˆ             Fx             x

Instructor: Imran Gilani                                                                   1
Lecture No. 03                                           BASIC MECHANICAL ENGINEERING

               
Given F what is Fx and Fy

Fx  F cos iˆ

Fy  F sin  ˆ
j
FX  F cos
What is FX and F ?
F  F sin 

Coplanar force resultants
                                  
Given 3 forces, P , Q , and S their resultant is R  P  Q  S
y

 P
S
x

Q

                                   ˆ
Q  Q x (i )  Q y ( ˆ)
j       
ˆ
P  Px i  Py ˆj                                              ˆ
S  Sx i  S y ˆ
j
ˆ
Q   Qx i  Q y ˆ j

   
ˆ              ˆ            ˆ
R  P  Q  S  P x i  Py ˆ  Qx i  Q y ˆ  S x i  S y ˆ
j            j             j

ˆ     
R   Px  Qx  S x  i  Py  Q y  S y ˆj    

ˆ                 
R   Px   Qx   S x  i  Py   Q y  S y ˆ
j      
RX                            R


ˆ
R  Rx i  R y ˆ
j
So:
R x   Fx
Sign convention! You have either + or - components.
R y  F y

From now on, drop magnitude (| |) sign for all scalars. All vectors will
have arrows.

P Vector            P  Magnitude of P.

Once you have the components, the resultant vector can be sketched and
found using:

2
Ry
R  Rx  R y                tan  
2    2     2

Rx

Instructor: Imran Gilani                                                            2
Lecture No. 03                                                       BASIC MECHANICAL ENGINEERING

1). Given: Replace the 6 kN and 4 kN forces by a single force, expressed in vector                nota-
tion.

4 kN                   y

40

6 kN

30

x


            F4 x       y
F4

40o      F4 y         
F6
30o                  
                     F6 y
F6 x
x

       
R  F6  F4
            
F6    F6 x  F6 y

F6          ˆ
 F6 x i  F6 y ˆ
j

F6                ˆ
 F6 cos30  i  F6 sin 30  ˆ
j

F6                  ˆ
 6000 cos30 i  6000 sin 30  ˆ
j

F6          ˆ
 5200 i  3000 ˆ j
            
F4    F4 x  F4 y

F4          ˆ
 F4 x i  F4 y ˆ
j

F4                    ˆ
 F4 sin 40  ( i )  F4 cos 40  ˆ
j

F4                        ˆ
 4000 sin 40 ( i )  4000 cos 40  ˆ

j

F4             ˆ
 2570 i  3060 ˆ     j

ˆ                 ˆ
R  5200 i  3000 ˆ  2570 i  3060 ˆ
j                 j

ˆ
R  2630 i  6060 ˆ N
j

Instructor: Imran Gilani                                                                     3
Lecture No. 03                                         BASIC MECHANICAL ENGINEERING


2). Given: Previous problem. Find R using scalar notation.

4kN
y

40

6kN

30

x

Rx  Fx                                       R y  F y
Rx   F6 x  F4 x                             R y   F6 y  F4 y
Rx  6 cos30  4 sin 40                      R y  6 sin 30   4 cos 40 
Rx  2.63 kN                                   R y  6.06 kN


ˆ
R  Rx i  R y ˆ
j

ˆ
R  2.63 i  6.06 ˆ kN
j

Instructor: Imran Gilani                                                       4

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