Two forces acting on point A along their side | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\overrightarrow{\mathrm{F}}_{1}=\frac{1}{|\mathrm{AC}|}=\frac{1}{\mathrm{b}}$

$F_{2}=\frac{1}{|A B|}=\frac{1}{c}$

$F_{Net}=\sqrt{F_{1}^{2}+F_{2

Vedclass · Accepted Answer

$\frac {1}{AD}$

Vedclass · Answer

$\overrightarrow{\mathrm{F}}_{1}=\frac{1}{|\mathrm{AC}|}=\frac{1}{\mathrm{b}}$

$F_{2}=\frac{1}{|A B|}=\frac{1}{c}$

$F_{Net}=\sqrt{F_{1}^{2}+F_{2}^{2}}=\sqrt{\frac{1}{h^{2}}+\frac{1}{c^{2}}}=\frac{\sqrt{b^{2}+c^{2}}}{\sqrt{b^{2} c^{2}}}=\sqrt{\frac{a^{2}}{b^{2} c^{2}}}$

$F_{	ext {net }}=\frac{a}{b c}$

Area of $\Delta=\frac{1}{2} 	imes \mathrm{bc}=\frac{1}{2} 	imes \mathrm{a} 	imes \mathrm{x}$

$\mathrm{x}=\frac{\mathrm{b} \mathrm{c}}{\mathrm{a}}$

$F_{	ext {net }}=\frac{a}{b c}=\frac{1}{x}$

$F_{net}=\frac{1}{|A D|}$

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

Similar Questions

The vectors $\vec{A}$ and $\vec{B}$ are such that

$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$

The angle between the two vectors is

A particle has displacement of $12 \,m$ towards east and $5 \,m$ towards north then $6 \,m $ vertically upward. The sum of these displacements is........$m$

Two forces $3\,N$ and $2\, N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $ 6\,N$ and the resultant become $2R$. The value of is ....... $^o$

If $A$ and $B$ are two non-zero vectors having equal magnitude, the angle between the vectors $A$ and $A - B$ is

When $n$ vectors of different magnitudes are added, we get a null vector. Then the value of $n$ cannot be