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A thin uniform rod A B of mass $m = 1 kg$ moves translationally with acceleration $a = 2 m / s^{2}$ due to two antiparallel forces $F_{1}$ and $F_{2}$ . The distance between the points at which these forces are applied is equal to $l = 20 cm$ . Besides, it is known that $F_{2} = 5 N$ . Find the length of the rod.

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(1) $1.0 m$

(2) $5.5 m$

(3) $8.0 m$

(4) $2.0 m$

answer is A.

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Detailed Solution

Let x be the distance of centre point C of rod from D. Then,

A thin uniform rod AB of mass m = 1 kg moves translationally with acceleration a = 2 m/s^2 due to two anti parallel forces F1 and F2 The distance between the

$F_{2} - F_{1} = ma$

or $F_{1} = 3 N$

Further, $τ_{c} = 0$

$∴ F_{2} x = F_{1} (0.2 + x) 5 x = F_{1} (0.2 + x) 5 x = 3 (0.2 + x) x = 0.3 m$

$∴$ Length of rod $= 2 (x + 0.2) = 1.0 m$

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