Gujarati
Hindi
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

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