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CharlyH
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Report Thread starter 13 years ago
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A rope is coiled around 2 fixed bollards and one end is held with a force 50N. Find the greatest force which can be applied at the other end, P, without causing the rope to slip. The coefficient of friction between the rope and the bollard is 0.2.

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Jonny W
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First you have to prove that, in standard notation,

T_B = T_A exp[mu(\psi_B -\psi_A)]

See Question 4 on the attached exam paper and http://www.thestudentroom.co.uk/t81626.html

Let x be the angle shown in the attached diagram. Then cos(x) = 3a/6a = 1/2. So x = pi/3. So the angle of contact of the rope with each bollard is 5pi/3.

P
= (tension in middle part of rope)*exp(0.2*5pi/3)
= 50*exp(0.2*5pi/3)*exp(0.2*5pi/3)
= 406 N
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