- Home
- Standard 11
- Physics
3-1.Vectors
hard
Two forces acting on point $A$ along their side and having magnitude reciprocal to length of side then resultant of these forces will be proportional to
A
$\frac {1}{BC}$
B
$\frac {1}{AD}$
C
$\frac {1}{BD}$
D
$\frac {1}{CD}$
Solution
$\overrightarrow{\mathrm{F}}_{1}=\frac{1}{|\mathrm{AC}|}=\frac{1}{\mathrm{b}}$
$F_{2}=\frac{1}{|A B|}=\frac{1}{c}$
$F_{Net}=\sqrt{F_{1}^{2}+F_{2}^{2}}=\sqrt{\frac{1}{h^{2}}+\frac{1}{c^{2}}}=\frac{\sqrt{b^{2}+c^{2}}}{\sqrt{b^{2} c^{2}}}=\sqrt{\frac{a^{2}}{b^{2} c^{2}}}$
$F_{\text {net }}=\frac{a}{b c}$
Area of $\Delta=\frac{1}{2} \times \mathrm{bc}=\frac{1}{2} \times \mathrm{a} \times \mathrm{x}$
$\mathrm{x}=\frac{\mathrm{b} \mathrm{c}}{\mathrm{a}}$
$F_{\text {net }}=\frac{a}{b c}=\frac{1}{x}$
$F_{net}=\frac{1}{|A D|}$
Standard 11
Physics
