First slide

Scalars and Vectors

Question

Four forces are acting on a particle. one force of magnitude 3 N is directed upward, another is directed 37^o East of North having magnitude 5 N, third is directed in south-west direction is of magnitude $4 \sqrt{2} N$ and fourth force is $\sqrt{5 n} N .$ If the particle is in equilibrium. The value on n is _______.

Moderate

Solution

$\begin{array}{l} Here, {\vec{F}}_{1} = 3 \hat{k} \\ {\vec{F}}_{2} = 5 \cos 37^{\circ} + 5 \sin 37^{\circ} \hat{i} = 4 \hat{i} + 3 \hat{j} \\ and {\vec{F}}_{3} = - 4 \sqrt{2} \cos 45^{\circ} \hat{i} - 4 \sqrt{2} \sin 45^{\circ} \hat{j} \\ = - 4 \hat{i} - 4 \hat{j} \end{array}$
For equilibrium of the particle
$\begin{array}{l} {\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} + {\vec{F}}_{4} = 0 \\ or 3 \hat{k} + 3 \hat{j} + 4 \hat{i} - 4 \hat{i} - 4 \hat{j} + {\vec{F}}_{4} = 0 \\ ∴ {\vec{F}}_{4} = \hat{j} - 3 \hat{k} \\ ∴ {\vec{F}}_{4} = \sqrt{1 + 9} = \sqrt{10} \\ ∴ n = 2 \end{array}$

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