Question

11. If forces of magnitude 12 kg-wt, 5 kg-wt and 13 kg-wt act at a point are in equilibrium, then the angle between the first two forces is
(a) 30°
(b) 90°
(C) 60°
(d) 45°​

Answer 1

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$\alpha = 90(degree)$

Given:

The resultant of the first two should be equal and opposite to the third force,

$13 {}^{2} = 5 {}^{2} + 12 {}^{2} + 2 \times 5 \times 12 \: \cos( \alpha ) \\ 169 = 169 + 120 \cos( \alpha ) \\ \cos( \alpha ) = 0 \\ \alpha = 90(degree)$

silentlover45.❤️

Answer 2

Answer:

α=90(degree)

Given:

The resultant of the first two should be equal and opposite to the third force,

\begin{lgathered}13 {}^{2} = 5 {}^{2} + 12 {}^{2} + 2 \times 5 \times 12 \: \cos( \alpha ) \\ 169 = 169 + 120 \cos( \alpha ) \\ \cos( \alpha ) = 0 \\ \alpha = 90(degree)\end{lgathered}

13

2

=5

2

+12

2

+2×5×12cos(α)

169=169+120cos(α)

cos(α)=0

α=90(degree)