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

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3-2.Motion in Plane

normal

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

cannot be predicted

are equal to each other

are equal to each other in magnitude

are not equal to each other in magnitude

Solution

$\overrightarrow{\mathrm{P}}=$ 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

$\therefore \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0$

$\Rightarrow(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}) \cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})=0 \Rightarrow \mathrm{A}^{2}=\mathrm{B}^{2}$

$\Rightarrow|\mathrm{A}|=|\mathrm{B}|$

Standard 11

Physics

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normal

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

Solution

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