First slide

Mechanical equilibrium force cases

Question

If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ are x, y, if 20 N is vertically down. Then x, y are

Moderate

A

20 N, 30 N
B

20√3N, 20 N
C

20√3N , 20√3N
D

10 N, 30 N

Solution

$\large \frac{{{T_1}}}{{\sin {{60}^0}}} = \frac{{{T_2}}}{{\sin {{150}^0}}} = \frac{{20}}{{\sin {{150}^0}}}$

$\large \therefore \frac{{{T_2}}}{{\sin {{150}^0}}} = \frac{{20}}{{\sin {{150}^0}}}$

$\Rightarrow T_{2} = 20 N$

$\large \frac{{{T_1}}}{{\sin {{60}^0}}} = \frac{{20}}{{\sin (180 - {{30}^0})}}$

$\large \frac{{{T_1}}}{{\sin {{60}^0}}} = \frac{{20}}{{\sin {{30}^0}}} \Rightarrow {T_1} = \frac{{20\sqrt 3 /2}}{{1/2}}$

$\large x = {T_1} = 20\sqrt 3 N$
$\large x,y = 20\sqrt 3 N,20N$

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