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(A) Since point $P$ is at rest,the net force acting on it is zero. We resolve the forces into horizontal $(x)$ and vertical $(y)$ components.For the h

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	ext{ N}, \frac{3}{\sqrt{2}} 	ext{ N}$

Vedclass · Answer

(A) Since point $P$ is at rest,the net force acting on it is zero. We resolve the forces into horizontal $(x)$ and vertical $(y)$ components.For the horizontal direction ($x$-axis),taking the right direction as positive:$\Sigma F_x = 0 \Rightarrow F_1 + (1 	ext{ N}) \sin 45^{\circ} - (2 	ext{ N}) \sin 45^{\circ} = 0$$F_1 = (2 	ext{ N}) \sin 45^{\circ} - (1 	ext{ N}) \sin 45^{\circ} = (1 	ext{ N}) \sin 45^{\circ} = 1 	imes \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} 	ext{ N}$.For the vertical direction ($y$-axis),taking the upward direction as positive:$\Sigma F_y = 0 \Rightarrow (1 	ext{ N}) \cos 45^{\circ} + (2 	ext{ N}) \cos 45^{\circ} - F_2 = 0$$F_2 = (1 	ext{ N}) \cos 45^{\circ} + (2 	ext{ N}) \cos 45^{\circ} = (3 	ext{ N}) \cos 45^{\circ} = 3 	imes \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} 	ext{ N}$.Thus,$F_1 = \frac{1}{\sqrt{2}} 	ext{ N}$ and $F_2 = \frac{3}{\sqrt{2}} 	ext{ N}$.

There are four forces acting at a point $P$ produced by strings as shown in the figure. Point $P$ is at rest. The forces $F_1$ and $F_2$ are respectively:

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