First slide

Mechanical equilibrium force cases

Question

If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ respectively.

Moderate

A

20N, 30N
B

20√3N, 20N
C

20√3N, 20√3N
D

10N, 30N

Solution

$\large \frac{{20}}{{\sin 150}} = \frac{{{T_2}}}{{\sin 150}} = \frac{{{T_1}}}{{\sin 60}}$
$\large \Rightarrow {T_2} = 20N$
$\large {T_1} = 20\,\,\tan 60 = 20\sqrt 3N$

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