Question

answer fast quick please​

Answer 1

Answer:

T₁ = 25 (√3 - 1 )

T₂ =25 (√6 - √2 ) / 2

Explanation:

To find ------> Tensions in the strings

Solution------> We know that ,

Sin ( A + B ) = SinA CosB + CosA SinB

Putting A = 45° and B = 30° , we get ,

Sin ( 45° + 30° ) = Sin45° Cos30° + Cos45° Sin30°

= ( 1 / √2 ) ( √3 / 2 ) + ( 1/√2 ) ( 1/2 )

= √3 / 2√2 + 1 / 2√2

= ( √3 + 1 ) / 2√2

Sin135° = Sin ( 180° - 45° )

= Sin 45°

= 1 /√2

Sin150° = Sin ( 180° - 30 ° )

= Sin ( 30° )

= 1/2

ATQ, Tension in the strings are T₁ and T₂ and object is in rest , so by Lami's theroem ,

T₁ / Sin135° = T₂ / Sin150° = 25 / Sin75°

T₁/( 1/√2 ) = T₂/ ( 1/2 ) = 25 / (√3 + 1 ) / 2√2

=> √2 T₁ = 2 T₂ = 50√2 / ( √3 + 1 )

Now taking first and third equal we get,

√2 T₁ = 50√2 / ( √3 + 1 )

=> T₁ = 50 / ( √3 + 1 )

=> T₁ = 50 ( √3 - 1 ) / ( √3 + 1 ) ( √3 - 1 )

=> T₁ = 50 ( √3 - 1 ) / (√3 )² - ( 1 )²

=> T₁ = 50 ( √3 - 1 ) / ( 3 - 1 )

=> T₁ = 50 ( √3 - 1 ) / 2

=> T₁ = 25 ( √3 - 1 ) N

Now taking second and third equal , we get,

2 T₂ = 50√2 / ( √3 + 1 )

=> T₂ = 25√2 / ( √3 + 1 )

=> T₂ = 25√2 ( √3 - 1 ) / ( √3 + 1 ) ( √3 - 1 )

=> T₂ = 25√2 ( √3 - 1 ) / ( √3 )² - ( 1 )²

=> T₂ = 25√2 (√3 - 1 ) / ( 3 - 1 )

=> T₂ = 25 ( √6 - √2 ) / 2 N