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A thin uniform rod AB of mass m=1 kg moves translationally with acceleration a = 2 ms-2 due to two anti-parallel forces F1 and F2. The distance between the points at which these forces are applied is equal to l = 20 cm. Besides, it is known that F2 = 5 N. Find the length of the rod, in metre.

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Detailed Solution

Let x be the distance of centre point C of rod from D. Then F2−F1=ma⇒F1=3NFurther, τc=0⇒ F2x=F1(0.2+x)⇒ 5x=F1(0.2+x)⇒ 5x=3(0.2+x)⇒ x=0.3mSo, length of rod is L = 2(x + 0.2) = 1 m

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A thin uniform rod AB of mass m=1 kg moves translationally with acceleration a = 2 ms-2 due to two anti-parallel forces F1 and F2. The distance between the points at which these forces are applied is equal to l = 20 cm. Besides, it is known that F2 = 5 N. Find the length of the rod, in metre.