Question

If 'O' is in equilibrium then tensions T1 and t2 are (P+20) Newton and (q+60) Newton. Then
the value of P+q is (tan37=3/4)

I gave a pic

Pls tell fast

Answer 1

$\Large{\bf{\underline{\underline{\green{Solution}}}}}$

$\sf{\green{Given}}\begin{cases}</p><p>\sf{T_{1} = (P + 20) \: N}\\</p><p>\sf{T_{2} = (Q + 60) \: N}\\</p><p>\end{cases}$

The system is in equilibrium, So all the component of forces acting on it are equal

$\boxed{\sf{\blue{Horizontal\:component}}}</p><p>\\</p><p>\sf{\implies\: T_{1}cos37\degree = T_{2}cos53\degree}\\ \\</p><p>\sf{\implies\: \frac{4}{\cancel{5}}T_{1} = \frac{3}{\cancel{5}}T_{2}}\\ \\</p><p>\sf{\implies\: T_{1} = \frac{3}{4}T_{2}\: \: \: \: \green{\longrightarrow\: (1)}}$

$\boxed{\sf{\blue{Vertical\:component}}}</p><p>\\</p><p>\sf{\implies\: T_{1}sin37\degree + T_{2}sin53\degree = 100}\\ \\</p><p>\sf{\implies\: \frac{3}{5}T_{1} + \frac{4}{5}T_{2} = 100}\\ \\</p><p>\sf{\implies\: \frac{1}{5}(3T_{1} + 4T_{2}) = 100}\\ \\</p><p>\sf{\implies\: 3T_{1} + 4T_{2} = 500}\\ \\</p><p>\sf{\implies\: 3(\frac{3}{4}T_{2}) + 4T_{2} = 500 \:\:\: \purple{[Using \: (1)]}}\\ \\</p><p>\sf{\implies\: \frac{9}{4}T_{2} + 4T_{2} = 500}\\ \\</p><p>\sf{\implies\: \frac{9T_{2} + 16T_{2}}{4} = 500}\\ \\</p><p>\sf{\implies\: 9T_{2} + 16T_{2} = 2000}\\ \\</p><p>\sf{\implies\: 25T_{2} = 2000}\\ \\</p><p>\sf{\implies\: \underline{\underline{T_{2} = 80\:N}}}\\ \\</p><p>$

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$</p><p>\sf{\star\: T_{1} = \frac{3}{4}T_{2} = \frac{3}{4} × 80 }\\ \\</p><p>\sf{\implies\: \underline{\underline{T_{1} = 60\: N}}}\\</p><p>$

Now, given that

$</p><p>\sf{\red{\star\: T_{1} = (P + 20) \: N}}\\</p><p>\sf{\implies\: 60 \: N = (P + 20) \: N}\\</p><p>\sf{\implies\: P = 40 \: N}\\ \\</p><p>\sf{\red{\star\: T_{2} = (Q + 60) \: N}}\\</p><p>\sf{\implies\: 80\:N = (Q + 60) \: N}\\</p><p>\sf{\implies\: Q = 20 \: N}\\ \\</p><p>\Large{\boxed{\green{\sf{\therefore\: P + Q = 60\: N }}}}</p><p>$