For the free body diagram shown in the figure | vedclass

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Question

(A) For the net acceleration of the body to be zero,the net force acting on the body must be zero.Let the additional force required be $\overrightarro

Vedclass · Accepted Answer

$\sqrt{2} 	ext{ N}, 45^{\circ}$

Vedclass · Answer

(A) For the net acceleration of the body to be zero,the net force acting on the body must be zero.Let the additional force required be $\overrightarrow{F} = F_x \hat{i} + F_y \hat{j}$.The forces acting on the body are $5 \hat{i}$ (along positive $x$),$-6 \hat{i}$ (along negative $x$),$7 \hat{j}$ (along positive $y$),and $-8 \hat{j}$ (along negative $y$).The condition for equilibrium is $\sum \overrightarrow{F} = 0$.$\overrightarrow{F} + (5 \hat{i} - 6 \hat{i}) + (7 \hat{j} - 8 \hat{j}) = 0$$\overrightarrow{F} - 1 \hat{i} - 1 \hat{j} = 0$$\overrightarrow{F} = 1 \hat{i} + 1 \hat{j}$The magnitude of the force is $|\overrightarrow{F}| = \sqrt{1^2 + 1^2} = \sqrt{2} 	ext{ N}$.The angle $	heta$ with the positive $x$-axis is given by $	an 	heta = \frac{F_y}{F_x} = \frac{1}{1} = 1$.Therefore,$	heta = 45^{\circ}$.

For the free body diagram shown in the figure,four forces are applied in the $x$ and $y$ directions. What additional force must be applied and at what angle with the positive $x$-axis so that the net acceleration of the body is zero?

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