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Two persons of equal height are carrying a long wooden beam of length l. They are at a distance $\frac{l}{4}$ and $\frac{l}{6}$ from the nearest end of the rod. The ratio of the normal reactions at their heads i s

\[\begin {array} {1 1} (a)\;2:3 & \quad (b)\;1:3 \\ (c)\;4:3 & \quad  (d)\;1:2 \end {array}\]
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$N_1+N_2=mg$
Torque about the centre
$N_1 \times L/4= N_2 (L/2 -L/6)$
$\large\frac{N_1}{4}$$ =N_2 \large\frac{2}{6}$
$N_2= \large\frac{4}{3} N_1$
$\large\frac{N_1}{N_2}=\frac{4}{3}$
hence c is the correct answer
answered Nov 28, 2013 by meena.p
edited Jun 21, 2014 by lmohan717
 

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