MHT CET · Physics · Mathematics in Physics
Figure shows three forces \(\overrightarrow{\mathrm{F}}_{1}, \overrightarrow{\mathrm{F}}_{2}\) and \(\overrightarrow{\mathrm{F}}_{3}\) acting along the sides of an equilateral
triangle. If the total torque acting at point ' 0 ' (centre of the triangle) is zero then the
magnitude of \(\overrightarrow{\mathrm{F}}_{3}\) is
- A \(\frac{\mathrm{F}_{1}-\mathrm{F}_{2}}{2}\)
- B \(\mathrm{~F}_{1}-\mathrm{F}_{2}\)
- C \(\mathrm{~F}_{1}+\mathrm{F}_{2}\)
- D \(\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{~F}_{1}+\mathrm{F}_{2}\)
Step-by-step Solution
Detailed explanation
The perpendicular distance of point \(\mathrm{O}\) from the sides \(\mathrm{AB}\) or \(\mathrm{BC}\) or \(\mathrm{AC}\) will be same (say r). Torque will be zero if total moment of force about point \(\mathrm{O}\) is zero Hence
\(\mathrm{rF}_{1}+\mathrm{rF}_{2}-\mathrm{rF}_{3}=0\)
\(\mathrm{F}_{3}=\mathrm{F}_{1}+\mathrm{F}_{2}=(4+2) \mathrm{N}-6 \mathrm{~N}\)
\(\mathrm{rF}_{1}+\mathrm{rF}_{2}-\mathrm{rF}_{3}=0\)
\(\mathrm{F}_{3}=\mathrm{F}_{1}+\mathrm{F}_{2}=(4+2) \mathrm{N}-6 \mathrm{~N}\)
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