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(A) Since point $P$ is at rest, the net force acting on it is zero $(\sum F = 0)$.We resolve the forces into horizontal $(x)$ and vertical $(y)$ compo

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 $(\sum F = 0)$.We resolve the forces into horizontal $(x)$ and vertical $(y)$ components.For the $x$-direction: $\sum F_x = 0$$F_1 + 1 \sin 45^{\circ} - 2 \sin 45^{\circ} = 0$$F_1 = 2 \sin 45^{\circ} - 1 \sin 45^{\circ} = (2 - 1) \sin 45^{\circ} = 1 	imes \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} 	ext{ N}$.For the $y$-direction: $\sum F_y = 0$Taking upward as positive and downward as negative:$1 \cos 45^{\circ} + 2 \cos 45^{\circ} - F_2 = 0$$F_2 = 1 \cos 45^{\circ} + 2 \cos 45^{\circ} = (1 + 2) \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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