Question

an elephant is stopped by a rope wound twice round the rough trunk of a tree. if the elephant exerts 1000 kgf the minimum force required to stop the elephant is

Answer 1

Given info : an elephant is stopped by a rope wound twice round the rough trunk of a tree. [ coefficient of friction between rope and tree is 0.3 ]

To find : if the elephant exerts 1000 kgf, the minimum force required to stop the elephant is...

solution : in a belt and wrap friction relation,

$T_2=T_1e^{-\mu\theta}$

where T₁ = tension applied in the pulling side

T₂ = tension applied in resisting side

μ = coefficient of friction

and θ = angle of contact of rope

as rope wound twice round the rough trunk of tree, so θ = 2 × 360° = 720° = 4π

T₁ = 1000 kgf and μ = 0.3

so, T₂ = 1000e^{-0.3 × 4π}

= 1000 e^{-1.2π}

= 1000/e^(1.2π)

= 1000/43.3762

= 23.05 kgf

Therefore the minimum force required to stop the elephant is 23.05 kgf

Answer 2

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