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Mechanics 2 Couple Question

Hi all,

Got a question in MEI Mechanics 2 Ex3B Q12 (page38) which I keep getting wrong! Here it is:

Four Seamen are using a light capstan to pull in their ship's anchor at a steady rate. The diameter of the capstans drum is 1m and the spokes on which the men are pushing each project 2m from the centre of the capstan. Each man is pushing with a force of 300N horizontally and at right angles to his spoke. The anchor cable is taut; it passes over a frictionless pulley and then makes an angle of 20 with the horizontal.

Question: Find the tension in the cable.

Thanks for any help :smile:
Can you post your working so far so we can see where you've gone wrong? :smile:
Reply 2
Well firstly it seems to me that the moment at the centre of the capstan will be equal to the tension since theres no resistance in the system. So i calculate the moments from each force to the centre of the capstan. Using Moment = Force * Distance, So 300*2 = 600. Then 600*4= 2400. But the answer in the book is 4800 and i know im missing something very obvious or have misunderstood the question!

Thanks :smile:
Reply 3
Anyone?
You have correctly worked out that as the distance from the centre. To where the seamen push is 2m and each seaman pushes with 300N each seaman will give a momment of 300*2=600Nm this is then multiplied by four to give a total momment of 4*600=2400Nm.

Then the new part is that to convert from the momment at the end of the projections to. That of the rope setup an equation like this:

2400 = .5 * x

This is as whatever the momment is at the capstan will be half a meter from the centre and four times due to there being foursemen.

X is found to be 4800 N as required
(edited 10 years ago)
Reply 5
Original post by MechanicsMmm
blah


Why have you just resurrected a 6 year old thread?