A boy of mass 40 kg is hanging from a horizontal branch of a tree. The tension in his arms is minimum when the angle between the arms is:-

0^{o}

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90^{o}

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120^{0}

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180^{o}

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Solution

The correct option is A $0^{o}$
Resultant Tension should be equal to $m g = F$
$F^{2} = T^{2} + T^{2} + 2 T^{2} cos θ$

$F^{2} = 2 T^{2} (1 + cos θ)$

$T^{2} = \frac{F^{2}}{2 (1 + cos θ)}$

$T$ is min. when $(1 + cos θ)$ is max.

$\Rightarrow cos θ = 1$ or $θ = 0^{\circ}$

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A boy of mass 40 kg is hanging from a horizontal branch of a tree. The tension in his arms is minimum when the angle between the arms is:-

The correct option is A 0oResultant Tension should be equal to mg=F F2=T2+T2+2T2cosθ F2=2T2(1+cosθ) T2=F22(1+cosθ) T is min. when (1+cosθ) is max. ⇒cosθ=1 or θ=0∘

The correct option is A $0^{o}$
Resultant Tension should be equal to $m g = F$
$F^{2} = T^{2} + T^{2} + 2 T^{2} cos θ$

$F^{2} = 2 T^{2} (1 + cos θ)$

$T^{2} = \frac{F^{2}}{2 (1 + cos θ)}$

$T$ is min. when $(1 + cos θ)$ is max.

$\Rightarrow cos θ = 1$ or $θ = 0^{\circ}$