Force on winch to lift trailer ramp

[Mutaja]

Homework Statement

I was really unsure as to where to post this problem as it's got nothing to do with homework, yet it's a homework like problem. Please let me know if I've posted it in the wrong sub-forum.

It was a simple discussion at work where we debated how strong a winch we need to lift the trailer ramp.

I think all the details here are in the picture.

The question is how much force will be needed at the winch to pull up the ramp. The winch's "strength" is measured in kg. Will 1000kg be enough?

Homework Equations

The Attempt at a Solution

Personally, I think it'll be around 300-400kg, but I'm not sure how to explain my assumption.

My thinking is like this:

It weighs 100kg. If I were to pull it straight up at the middle, I assume it would be with a force under 100kg. Since there's a 45 degree or so for the wire, I'll multiply it by 2. + loss at the pulley. Then I just added some kg "to be on the safe side" and that's where my assumption is from.

I don't need any very in depth mathematical analysis although a rough "guide" would be ace.

Again, I'm sorry if I've posted on the wrong forum.

 

[verty]

Your hunch is mostly correct, I get a rough estimate of 349 kg, whereas 400kg would (barely) allow you to operate the ramp with one person on it. So actually 500kg would be better, I think.

 

[Mutaja]

Your hunch is mostly correct, I get a rough estimate of 349 kg, whereas 400kg would (barely) allow you to operate the ramp with one person on it. So actually 500kg would be better, I think.

Awesome, thanks. Do you have any rough formulas on how you did it?

No biggie, thanks a lot, I really appreciate it.

 

[verty]

Awesome, thanks. Do you have any rough formulas on how you did it?

No biggie, thanks a lot, I really appreciate it.

I do but because this is school math, basic trigonometry really, I don't think there is much good for going into it here. I mean, if you are interested to know more, why not get a book on trigonometry and learn more? Ask in the book section and you're sure to get replies.

That said, I will give you this bit of information. You thought the force on the wire would be less because it connects to the midpoint, actually that increases the force, it's more difficult to lift the ramp by the midpoint. Anyway, best of luck.

 

[haruspex]

I get a rough estimate of 349 kg.

Seems a bit high to me.
Taking moments about the pivot point, the torque from the ramp's weight is mg(0.8)cos(20) Nm. If the shortest distance from pivot to cable is h, tension in cable is mg(0.8)cos(20)/h N = 100(0.8)(0.94)/h kgwt = 75.2/h kgwt. I calculate h as 0.44m, giving 169 kgwt.
Am I doing something wrong? Are you taking the load as 1.6m from the pivot?

 

[verty]

Seems a bit high to me.
Taking moments about the pivot point, the torque from the ramp's weight is mg(0.8)cos(20) Nm. If the shortest distance from pivot to cable is h, tension in cable is mg(0.8)cos(20)/h N = 100(0.8)(0.94)/h kgwt = 75.2/h kgwt. I calculate h as 0.44m, giving 169 kgwt.
Am I doing something wrong? Are you taking the load as 1.6m from the pivot?

We don't know the weight distribution of the ramp. I was thinking those ramps tend to be heavier toward the end, so I assumed 100kg at the end of the ramp. Also it could happen that the ground falls away behind the truck in some locations, so perhaps the ramp will be steeper at times. So this seemed like a good assumption to make.

Then, I assumed the vertical force needed is twice what it would be by this very simple argument: the midpoint accelerates at half the speed of the end point (half the speed, twice the mass). And I assumed the triangle is isosceles.

Actually, now that you mention it, I do see a mistake. Not in the assumptions I made but there is an extra factor of cos(20) I think. So perhaps it should be 328kg. Anyway, this was a rough calculation.

Hopefully what I did is correct. Perhaps it was too rough but that was the whole idea (and it'll be off in the right direction, overestimating slightly).

 

[verty]

And just to fill this in, I've done an accurate calculation now which gives 338.7kg, so about halfway between 328 and 349. So my estimates weren't too rough, I don't think.

 

[haruspex]

Then, I assumed the vertical force needed is twice what it would be by this very simple argument: the midpoint accelerates at half the speed of the end point (half the speed, twice the mass). And I assumed the triangle is isosceles.

If you want to take a significant acceleration into account then it's the moment of inertia that needs to be considered. I doubt this needs anything like double the force.
Yes, the mass distribution need not be uniform, but all the mass at one end seems a bit extreme.
The main need for a safety margin comes from the "20 degree" angle. As that angle increases, the tension in the cable can increase quite rapidly.