Question
There are four forces acting at a point P produced by strings as shown in which is at rest. Find the forces $F_1$ and $F_2$.
✓
Answer
As the particle is rest or a = 0. So resultant force due to all forces will be zero.$\therefore$ Net components along X and Y-axis will be zero.
Resolving all forces along X-axis$\text{F}_\text{x}=0$
$\text{F}_1+1\cos45^\circ-2\cos45^\circ=0$ or $\text{F}_1-1\cos45^\circ=0$
$\text{F}_1=\cos45^\circ=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.414}{2}=0.707\text{N}$
Resolving all forces along Y-axis$\text{F}_\text{y}=0$
$-\text{F}_2+1\cos45^\circ+2\cos45%\circ=0$
$-\text{F}_2=-3\cos45^\circ$
$\text{F}_2=3.\frac{1}{\sqrt{2}}=\frac{3\sqrt{2}}{2}=\frac{3\times1.414}{2}=3\times0.707=2.121\text{N.}$
Resolving all forces along X-axis$\text{F}_\text{x}=0$
$\text{F}_1+1\cos45^\circ-2\cos45^\circ=0$ or $\text{F}_1-1\cos45^\circ=0$
$\text{F}_1=\cos45^\circ=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.414}{2}=0.707\text{N}$
Resolving all forces along Y-axis$\text{F}_\text{y}=0$
$-\text{F}_2+1\cos45^\circ+2\cos45%\circ=0$
$-\text{F}_2=-3\cos45^\circ$
$\text{F}_2=3.\frac{1}{\sqrt{2}}=\frac{3\sqrt{2}}{2}=\frac{3\times1.414}{2}=3\times0.707=2.121\text{N.}$
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