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6.System of Particles and Rotational Motion
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The resultant of the system in the figure is a force of $8N$ parallel to the given force through $R$. The value of $PR$ equals to

A
${1/4}\,\,RQ$
B
${3/8}\,\,RQ$
C
${3/5}\,\,RQ$
D
${2/5}\,\,RQ$
Solution
By taking moment of forces about point R, $5 \times PR – 3 \times RQ = 0$
$⇒$ $PR = \frac{3}{5}RQ$.
Standard 11
Physics
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