6.System of Particles and Rotational Motion
medium

Two like parallel forces $20 \,N$ and $30 \,N$ act at the ends $A$ and $B$ of a rod $1.5 \,m$ long. The resultant of the forces will act at the point ........

A

$90 \,cm$ from $A$

B

$75 \,cm$ from $B$

C

$20 \,cm$ from $B$

D

$85 \,cm$ from $A$

Solution

(a)

Net torque should be same for the new point $20(0.75)+30(0.75)=50(x)$

Solve for $x$

Standard 11
Physics

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