Question

Two forces ${\overrightarrow{F}}_1=500N$ due east and ${\overrightarrow{F}}_2=250N$ due north have their common initial point. ${\overrightarrow{F}}_2-{\overrightarrow{F}}_1$ is

• A. $250\sqrt{5}N,{\tan }^{-1}\left(2\right)W$ of $N$
• B. $250N,{\tan }^{-1}\left(2\right)W$ of $N$
• C. Zero
• D. $750N,{\tan }^{-1}\left(3/4\right)N$ of $W$

Answer 1

The correct option is A $250\sqrt{5}N,{\tan }^{-1}\left(2\right)W$ of $N$

${\overrightarrow{F}}_1=500N$ due east ${\overrightarrow{F}}_2=250N$ both forces are perpendicular to each other.

$|{\overrightarrow{F}}_2-{\overrightarrow{F}}_1|=\sqrt{{\left({\overrightarrow{F}}_1\right)}^2+{\left({\overrightarrow{F}}_2\right)}^2-2{\overrightarrow{F}}_1{\overrightarrow{F}}_2\cos {90}^0}$

$=\sqrt{{\left(250\right)}^2+{\left(500\right)}^2}$

$=250\sqrt{5}N$

$\tan \theta =\frac{500}{250}$

$\theta ={\tan }^{-1}\left(2\right)wofNorth$