Question

A and B are two pegs separated by 13 cm. A body of 169 kgwt is suspended by the thread of 17 cm connecting to A & B,such that two segments of strings are perpendicular. Then tensions in shorter and longer parts of string are..

Answer 1

see diagram.

CD = L SinФ = (17 - L) CosФ        --- (1)
      =>  tan Ф = 17/L  -  1
      =>  L = 17/(1+tanФ)     -- (2)

AB = L CosФ + (17 - L) SinФ  = 13 cm    --- (3)
 
- (1) * cosФ + (3) * sinФ =>
          (17 -L) cos²Ф + (17-L) sin²Ф  = 13 sinФ
     =>      17 - L = 13 sinФ  cm   
     =>      17 - 17/(1+tanФ) = 13 sinФ
     =>      17 tanФ = 13 (1+tanФ) sinФ
     =>      17 = 13 (cosФ + sinФ)
    =>       17/13  = √2 Sin (π/4 + Ф) 
     =>        Ф = Sin⁻¹ (17/13√2)  - π/4 = 22.62°
     =>      sinФ = 0.384       cosФ = 0.923 
                          => L = 12.0 cm nearly

Writing the equations of static Equilibrium for concurrent forces:
     T1 / Sin(180°-Ф)  =  T2 / Sin(90°+Ф)    = 169 / Sin90°
 =>  T1 /sinФ = T2 / CosФ = 169

The tension forces in the smaller and larger parts of the string are :
=>   T1 = 64.896 kg wt.             T2 = 155.987 kg wt
respectively.

Answer 2

See diagram.

CD = L SinФ = (17 - L) CosФ        --- (1)
      =>  tan Ф = 17/L  -  1
      =>  L = 17/(1+tanФ)     -- (2)

AB = L CosФ + (17 - L) SinФ  = 13 cm    --- (3)
 
- (1) * cosФ + (3) * sinФ =>
          (17 -L) cos²Ф + (17-L) sin²Ф  = 13 sinФ
     =>      17 - L = 13 sinФ  cm   
     =>      17 - 17/(1+tanФ) = 13 sinФ
     =>      17 tanФ = 13 (1+tanФ) sinФ
     =>      17 = 13 (cosФ + sinФ)
    =>       17/13  = √2 Sin (π/4 + Ф) 
     =>        Ф = Sin⁻¹ (17/13√2)  - π/4 = 22.62°
     =>      sinФ = 0.384       cosФ = 0.923 
                          => L = 12.0 cm nearly

Writing the equations of static Equilibrium for concurrent forces:
     T1 / Sin(180°-Ф)  =  T2 / Sin(90°+Ф)    = 169 / Sin90°
 =>  T1 /sinФ = T2 / CosФ = 169

The tension forces in the smaller and larger parts of the string are :
=>   T1 = 64.896 kg wt. T2 = 155.987 kg wt
= 65kg wt. =156kg wrt