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 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. The length of the rod is .

Question

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 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. The length of the rod is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let x be the distance of centre point of rod from D. Then, F2-F1=m a or F1=3 N Further, τc=0 ∴ F2 x=F1(0.2+x) 5 x=F1(0.2+x) ∴ 5 x=3(0.2+x) or x=0.3 m ∴ Length of rod =2(x+0.2)=1.0 m

Q. A thin uniform rod AB of mass m=1kg moves translationally with acceleration a=2m/s2 due to two antiparallel forces F1​ and F2​. The distance between the points at which these forces are applied is equal to l=20cm. Besides, it is known that F2​=5N. The length of the rod is ____.

Let x be the distance of centre point of rod from D. Then, F2​−F1​=ma or F1​=3N Further, τc​=0 ∴F2​x=F1​(0.2+x) 5x=F1​(0.2+x) ∴5x=3(0.2+x) or x=0.3m ∴ Length of rod =2(x+0.2)=1.0m

Let $x$ be the distance of centre point of rod from $D$ .
Then, $F_{2} - F_{1} = ma$
or $F_{1} = 3 N$
Further, $τ_{c} = 0$
$∴ F_{2} x = F_{1} (0.2 + x)$
$5 x = F_{1} (0.2 + x)$
$∴ 5 x = 3 (0.2 + x)$
or $x = 0.3 m$
$∴$ Length of rod $= 2 (x + 0.2) = 1.0 m$