Law of parallelogram of force // by Er. Akash Gupta

 Law of   parallelogram 

At a point if two forces acting simultaneously are represented by two adjecent side of parallelogram corresponding to their magnitude and direction ,then their resultant force ,volumn and direction that can be drawn from the erossion point of these force can be shown .

    

                        .... According to parallelogram law of vector ,their resultant vector will be represented by the diagonal of the parallelogram.....

     

Resultant of two forces acting at a point -

       

                         Let us ,if these two forces P and Q work at an angle at a point o , then the value of their resultant force R is found from the following formula , 

   Now we get , 

                                 R= √ P^2 + Q^2 + 2 PQ cos α

Cases :- 

          (a) when two vector acting on the same direction, then Ө=0 , cos Ө=1 and sinӨ=0

             Now , 

                          R = √A^2 +B^2 +2AB

                       Or , R = √ (A+B)^2

                        Or , R = A+B

Then ,            tanβ = B×0 / A+B = 0

                 Or,    β = 0 

          (b) when two vector acting on right angle to each other , then Ө= 90degree , sinθ=1 and cos Ө = 0 

              Now, 

                              R = √ A^2 +B^2 + 2AB

                          Or , R = √ (A+B)^2

   Then ,          tanβ = B×( 1 )/ A+ B×0

                         Or , β = tan^-1  B/A

           (C) when two vector acting on opposite direction ,thenӨ = 180degree , cosӨ= -1 and sinӨ= 0

                Now, 

                              R = √A^2 +B^2 +2AB(-1)

                          Or, R = A-B or B-A

   Then ,          tanβ = B×(0)/A+B(-1)

                               β = 0 or 180 degree

Exp.- Two forces of 3N and 4N are acting on at a point such as the angle b/w them in 60 degree , find the resultant force ?

Solution -       let us assume ,    the resultant force is R 

                                                     

                           Where ,   P = 3N , Q = 4N and  α= 60 degree 

        

                 Now,      

                               R = √p^2 + Q^2 + 2 PQ cos α

                           Or, R = √3^2 +4^2 + 2(3×4) cos 60

                           Or,  R = √9+16+12 = √37 = 6.08 N

 

               Finding the angle of Ө, 

                                             tanӨ = 3× sin60 / 4+3 cos60

                                          Or, Ө = tan^-1   0.472 or 25.3 degree 

       Thus , the magnitude of resultant force is 6.08N and direction is 25.3 degree.///////