Question

A body of mass 5 kg is suspended by the string making angle 60 degree and 30 degree with the horizontal then t1 and t2

Answer 1

Given

Weight =5kg

Angles made with horizontal =60° and 30°

(1st diagram is configurations diagram and 2nd one is free body diagram)

Let tension in 1st string =t1

Tension in 2nd string =t2

Now using lami's theorem

$\frac{f1}{ \sin \alpha } = \frac{f2}{ \sin\beta } = \frac{f3}{ \sin\gamma }$
Here we are taking

F1=t1

F2=t2

F3=w=5kg

Now angles


$\alpha = 60 + 90 = 150$
$\beta = 30 + 90 = 120$

$\gamma = 90$


(let make is little easier - how we get alpha beta and gamma

In lami's theorem we take opposite sin angle to the applied load so we can see the free body diagram the opposite angle to the t1=150, opposite angle to t2=120 and opposite angle to w=90)

W=5kg

W=5x9.81 N

W=49.0333 N

Taking approximately =50N (or you can take g as 10)


$\frac{f1}{ \sin150 } = \frac{50}{ \sin90 }$

F1=sin150x50 (sin90=1)

F1=1/2 x50. (sin180=sin30)

F1 = t1=25N


$\frac{f2}{ \ \sin120 } = \frac{50}{ \sin90 }$


F2=50xsin120


$f2 = 50 \times \frac{ \sqrt{3} }{2}$

$= 25 \sqrt{3}$

=43.30N


Hence

T1=25N

T2=43.30N

This is ur ans hope it will help you in case of any doubt comment below

Answer 2

T2=43.30 this is your answer