Question

Three forces of magnitude $6N$,$6N$ and $\sqrt{72}N$ act at a corner of a cube along three sides as shown in the figures.The resultant of these forces is:

• A. $12N$ along OB
• B. $18N$ along OA
• C. $18N$ along OC
• D. $12N$ along OE

Answer 1

Answer: (D)

$12N$ along OE

Given : $\overrightarrow{F_1}=\sqrt{72}N$ along OG, $\overrightarrow{F_2}=6N$ along OA and $\overrightarrow{F_3}=6N$ along OB

Since the three force are mutually perpendicular to each other, so magnitude resultant force is given by

$|F_R|=\sqrt{F_1^2+F_2^2+F_3^2}$

$\therefore$ $|F_R|=\sqrt{(\sqrt{72}{)}^2+6^2+6^2}=12N$ along OE