The vector sum of two forces is perpendicular | vedclass

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

Question

$\overrightarrow{\mathrm{P}}=$ vector $\mathrm{sum}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$

$\overrightarrow{\mathrm{Q}}=$ vector

Vedclass · Accepted Answer

are equal to each other in magnitude

Vedclass · Answer

$\overrightarrow{\mathrm{P}}=$ vector $\mathrm{sum}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$

$\overrightarrow{\mathrm{Q}}=$ vector differences $=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$

Since $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ are perpendicular

$	herefore \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0$

$\Rightarrow(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})=0 \Rightarrow \mathrm{A}^{2}=\mathrm{B}^{2}$

$\Rightarrow|\mathrm{A}|=|\mathrm{B}|$

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

Similar Questions

In projectile motion, the modulus of rate of change of velocity

The co-ordinates of a moving particle at a time $t$, are give by, $x = 5 sin 10 t, y = 5 cos 10t$. The speed of the particle is :

A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)

A body of mass $m\, kg$ is rotating in a vertical circle at the end of a string of length $r$ metre. The difference in the kinetic energy at the top and the bottom of the circle is

A body of mass $m$ is suspended from a string of length $l$. What is minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in the vertical plane with the point of suspension as the centre of the circle