A particle is at the origin at t = 0. Two forces $\vec{F_{1}} = 70 \hat{j} N and \vec{F_{2}}$ act upon it and it experiences a constant acceleration $\vec{a},$ in the direction as shown in the figure. To make the acceleration of the particle zero, another force $\vec{F_{3}}$ is applied on it. Find $\vec{F_{3}}$ . (Given $sinθ = 4 / 5, cosθ = 3 / 5, tanθ = 4 / 3 and \vec{F_{2}}$ unknown. Neglect gravity)

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$- (30 \hat{i} + 40 \hat{j})$

$40 \hat{i} + 30 \hat{j}$

$- (40 \hat{i} + 30 \hat{j})$

$30 \hat{i} + 40 \hat{j}$

answer is B.

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Detailed Solution

The sum of forces

${\vec{F}}_{1} + {\vec{F}}_{2} = m \vec{a}$

$= 5 (10 cosθ \hat{i} + 10 sinθ \hat{j})$

$= 30 \hat{i} + 40 \hat{j}$

According to question,

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

$∴ {\vec{F}}_{3} = - ({\vec{F}}_{1} + {\vec{F}}_{2}) =$

$- 30 \hat{i} - 40 \hat{j}$

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