Three forces $F_{1} = 2 ¯ i + ¯ j - ¯ ¯ ¯ k N, F_{2} = 2 ¯ i + 3 ¯ j - 3 ¯ ¯ ¯ k$ N, $F_{3} = a (¯ i + ¯ j - ¯ ¯ ¯ k)$ N act simultaneously on a particle. The value of $^{'} a^{'}$ so that the particle may be in equilibrium is:

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Solution

The correct option is B -4
For equilibrium $F_{1} + F_{2} + F_{3} = 0$
$F_{1} + F_{2} = 4 ¯ i + 4 ¯ j - 4 ¯ ¯ ¯ k$
Hence, $F_{3}$ should be $- 4 (¯ i + ¯ j - ¯ ¯ ¯ k)$ for $F_{1} + F_{2} + F_{3} = 0$ and hence $a = - 4$

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Three forces F1=2¯i+¯j−¯¯¯kN,F2=2¯i+3¯j−3¯¯¯k N, F3=a(¯i+¯j−¯¯¯k) N act simultaneously on a particle. The value of ′a′ so that the particle may be in equilibrium is:

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Three forces $F_{1} = 2 ¯ i + ¯ j - ¯ ¯ ¯ k N, F_{2} = 2 ¯ i + 3 ¯ j - 3 ¯ ¯ ¯ k$ N, $F_{3} = a (¯ i + ¯ j - ¯ ¯ ¯ k)$ N act simultaneously on a particle. The value of $^{'} a^{'}$ so that the particle may be in equilibrium is:

The correct option is B -4
For equilibrium $F_{1} + F_{2} + F_{3} = 0$
$F_{1} + F_{2} = 4 ¯ i + 4 ¯ j - 4 ¯ ¯ ¯ k$
Hence, $F_{3}$ should be $- 4 (¯ i + ¯ j - ¯ ¯ ¯ k)$ for $F_{1} + F_{2} + F_{3} = 0$ and hence $a = - 4$