Load when winching up a slope

Nov 29, 2007 9 Replies

I plan to winch a boat on it's trailer up a slight slope (1 in 12) an



into my shed. On the flat I only need a winch capable of overcoming th rolling resistance of the outfit. To hoist the whole lot into the air would need one able to take the entire weight (about 1 ton). M question is : does the load immediately go up to 1 ton (plus somethin for rolling resistance) however slight the slope, or is it proportional to the slope (rising to 1 ton as the slope approaches th vertical)? My intuition is that I would break my 1 ton chain block i the attempt ( I imagine hauling the load up just using a counterweight it would have to be at least a ton before the load would move surely


-- MikeP


Well my intuition is as follows - I have a car weighing about 1.5 tons or something, and it's not too much trouble for me to overcome its rolling resistance and push it along a flat road by myself. Add in a couple of mates and we could even push it up a reasonable slope.

Now consider how many blokes to lift the thing off the ground completely...?

David

Your intuition is wrong. Have a look at

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The force on the boat will be "in proportion" to the slope. By my reckoning, as the force to hold the boat still on the winch goes as the angle of the slope (in your case tan**-1 1/12 or 4.76deg) from zero at 0deg to 1tonne at

90deg and assuming it's linear without other effects (good enough), the strain on the winch when stationary will be 1000 x 4.76/90 kg, or about 53kg. To allow for overcoming friction and snatching etc., multiply this by five (say) and you need tackle with a breaking strain of 250kg. No liability accepted etc!

Mea culpa, Chris Dixon and Wikipedia are, of course, correct. I should have been using Sine not Tangent. Not much difference at small angles but the strain figure then comes out as 83kg, not 53kg - roughly the same outcome!

Rolling resistance + 1/12th of the weight for a 1 in 12 slope. (approximately.it actually the tan of the angle whereas 1 in 12 is usually the sine but (sine)~=(tan)~=(angle in radians) for small angles)

.....................................etc., multiply this by five (say

and you need tackle with a breaking strain of 250kg. No liabilit accepted etc! [/i][/color]

Okay - I've got it now (with the qualification that it's the sine o the angle not tan) So....... I'm hauling a big bucket of water out o the well ( I'd never got round to fitting a winder, just a singl pulley). I get it to the top and say to my wife standing behind me "Just hang on to this while I grab the bucket." She takes it - halvin the angle the rope makes with the horizontal - and disappears down th well

-- MikeP

Wives should look after the sick and leave the well alone.

I'll just get my coat. :-)

You be needin' to use them sines and cosines they went on about at school. You weren't one of those who whinged "what use is this in the real world" were you?

Up a 1 in 12 slope, a 1 ton(ne) boat would exert 1 / 12 of its weight on the rope/chain, i.e 0.08333ton(ne) plus the rolling resistance. You then need additional force (ie tension on the rope) to accelerate it up the slope - minimal at winching speeds.

Bob

To calculate the rolling resistance you need to identify the road surface first, if it is a good smooth surface you take 1/25 of the total weight as the rolling resistance (40kgs in the ton) To calculate the gradient resistance take the weight of vehicle /60 multiplied by the angle of the slope in degrees, this will give you the gradient resistance, add to rolling resistance to give the total winch power required.(if angle is greater than 45degrees take total weight) I am assuming that this is on a trailer if not you would have a third set of resistance called damage resistance to factor in

Dave

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