Traingle law of force and Lami's theorem // by Er. akash gupta

 Traingle law of force =>

                           If two forces acting simultaneously an a point are represented by the two sides of a traingle taking an order corresponding to their magnitude and direction so their resulting force can be represented by the third arm of the traingle in the opposite order or three forces acting on a point if in magnitude and direction , represented by all the three arms of a traingle in order then these forces will be in equilibrium .

  Lami's theorem - 

                               According to Lami's theorem , if three forces acting in a point remain in equilibrium , then each forces will be proportional to the sin angle between two other forces.
                    Suppose, three forces P,Q and R are working at a point O, the force is represented by the p line OB , Q line OA and R line OC,
              Let, ㄥBOC =α, ㄥBOA = √ and ㄥAOC = β.                   ( where, √ = gama )
Then, By lami's theorem ,
Farmula;:-
                  A/sinα = B/sinβ = C/sin√


Exp. - A lamp weighing 50N is suspended from a ceiling .A horizontal force of magnitude 20N .Acts on the string that is used to suspend the lamp ,calculate the tension in the string from the angle of including of the string from the vertical direction .

 salution - According from Lami's theorem ,
                        α = 90+θ , β = 90 and √ = 180-θ
                                             (Where, √ = gama )
                        
                 And , A = 50N 
                            B = ? 
                            C = 20N
    We know that ,
                                 A/sinα = B/sinβ= C/sin√
  

Then,           50/sin(90+θ) = B/sin90 = 20/sin(180-θ)
            
                Or ,     50/cosθ = B/1 = 20/sinθ
   
Consider ,         50/cosθ = 20/sinθ
                Or,      sinθ/cosθ = 20/50
            
                Or,      tanθ = 0.4
               
                Or,      θ = tan^-1 (0.4)= 21.8

Now ,                 B/1 = 20/sinθ
    
                Or,       B/1 = 20/sin21.8
    
                Or,        B =  20/0.37
       
                Or,         B = 54.054

Thus , tension along the string = 54.054N
           And  inclination of string from the       vertical  direction = 21.8 degree.



                              

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