Mechanics

							     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