A thin uniform rod AB of mass 1.0 kg makes translationally motion with acceleration 2.0 m / s2 due to two anti-parallel force F1 and F2 . The distance between the point at which these forces are applied is equal to a = 20 cm. Besides, it is known that F2 = 5.0N . Find the length of the rod in m.

a=F2-F1m⇒2=5-F11F1=3NSum of moments about C.M is zeroF1(a+x)=F2×x⇒3(0.2+x)=5×x⇒x=0.3length of the rod is 2(a+x) = 1m

A thin uniform rod AB of mass 1.0 kg makes translationally motion with acceleration 2.0 m / s2 due to two anti-parallel force F1 and F2 . The distance between the point at which these forces are applied is equal to a = 20 cm. Besides, it is known that F2 = 5.0N . Find the length of the rod in m.

a=F2-F1m⇒2=5-F11F1=3NSum of moments about C.M is zeroF1(a+x)=F2×x⇒3(0.2+x)=5×x⇒x=0.3length of the rod is 2(a+x) = 1m

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A thin uniform rod AB of mass 1.0 kg makes translationally motion with acceleration 2.0 m / s² due to two anti-parallel force F₁ and F₂ . The distance between the point at which these forces are applied is equal to a = 20 cm. Besides, it is known that F₂ = 5.0N . Find the length of the rod in m.

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3 m

2 m

1 m

1 cm

answer is C.

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Detailed Solution

$a = \frac{F_{2} - F_{1}}{m} \Rightarrow 2 = \frac{5 - F_{1}}{1}$

$F_{1} = 3 N$

Sum of moments about C.M is zero

$F_{1} (a + x) = F_{2} \times x \Rightarrow 3 (0.2 + x) = 5 \times x$

$\Rightarrow x = 0.3$

length of the rod is 2(a+x) = 1m

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