Question

please solve this

stank you :)))​

Answer 1

Answer:

Balance the torque around hinge

I am assuming that angle at which rod is steady is alpha .

So length of rod will be

$\sin( \alpha ) = \frac{3}{l} \\ \\ l= 3 \cosec( \alpha )$

Now

$mg \frac{l}{2} \cos( \alpha ) = tl \sin( \alpha ) \\ \\ t = \frac{mg}{2} . \cot( \alpha ) = \frac{150}{2} \cot( \alpha ) \\ \\ t= 75 \cot( \alpha )$

So tension depends on what angle the rid is steady