Question

How to do this through lami theorem?

Answer 1

From Lami's theorem we know that if three coplanar, concurrent and non-collinear forces, having magnitudes as `A,B and C`, keep an object in static equilibrium, then

`A /sinα = B/ sinβ = C/ sinγ` ......(1)
where `α, β and γ` are the angles directly opposite to three forces `A, B and C` respectively.

Three coplanar, concurrent and non-collinear forces are tensions in two strings and weight `mg` of of the particle acting downwards.
Taking `g=10ms^-2`, `mg=21\ N`
`angle ABP=sin^-1(40/104)=22.62^@`
`angle PAB=sin^-1(40/50)=53.13^@`
`=>angle APB=180-53.13-22.62=104.25^@`

`angle` between tension in `PA` and weight `=90+22.62=112.62^@`
`angle` between tension in `PB` and weight `=90+53.13=143.13^@`

Using (1) we get

`21/(sin104.25)=T_(PA)/(sin112.62)=T_(PB)/(sin143.13)`

Using first equality we get

`T_(PA)=20\ N`

Similarly

`T_(PB)=13\ N`