How to calculate the tension developed in ropes using Lami's theorem?/ Lami's theorem application for calculating forces.

 Eg:

Calculate the tension developed in the ropes AC & BC in the following drawing using Lami's theorem.

To find:

Tension developed in the ropes AC & BC

Calculation:

Let us draw a line DE, parallel to the support AB as shown below. The line DE should pass through intersection point C.

As ∠ A = 42°, the ∠DCA will be equal to 42°

Note: 

The alternate angles of a line cutting the two parallel lines are always equal to one another.

Similarly,  ∠ECB = ∠B = 35°

As an applied load is in the vertical axis, ∠DCF = ∠ECF = 90°

( As shown in the above drawing)

Now, 

∠ACF = [∠ACD + ∠DCF]

            = [ 42° + 90°]

             =  132° 

(As shown in the above drawing)

∠BCF = [∠BCE + ∠ECF]

            = [ 35° +  90°]

             =  125° 

∠ACB =  [360° - (∠ACF + ∠BCF)]

 (As total value of a angle = 360°)

∠ACB =  [360° - ( 132° +  125° )]

             =  103° 

By applying Lami's theorem,

T1 / sin125° = T2 / sin132° = 300N / sin103°

Where, T1 & T2 are the tensions in ropes.

Therefore,

T1 = [(300N ➗  sin103°) x sin125°]

     = [ (300N ➗ 0.9743 ) x 0.819 ]

     = [ 307.91N x 0.819 ]

  T1 = 252.18N

T2 = [(300N ➗  sin103°) x sin132°]

     = [ (300N ➗ 0.9743 ) x 0.743 ]

     = [ 307.91N x 0.743 ]

  T2 = 228.77N

Tension in rope AC = T1 = 252.18N

Tension in rope BC = T2 = 228.77N

     Thank you for going through these calculation steps❤. Have a good day 😄.

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