How to do this through lami theorem?

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How to do this through lami theorem?

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Answer 1 · 2018-04-28T12:44:27

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

$T_(PB)=13\ N$

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How to do this through lami theorem?

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