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6.System of Particles and Rotational Motion
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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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