Question

Q1.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

Step-by-step explanation:

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

Answer:

Step-by-step explanation:

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,

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