What Load Weight is Needed to Suspend a Rope Over Water?

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Homework Statement


A 12 m long line, that streaches 5% of its length is streched in 0.5 m height over water.
The force that is used to keep the line streced is 5kN (kilo Newtons).
if an 90 kg man walks out the line and jumps up one meter and lands back on the line,
How heavy does the load, that the rope is tied to on each side need to be?

We assume that the height of the load is 50 cm, boxed shaped and the friction to the (flat) ground is 5 N
 
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I don't think you'll get any useful responses unless you make it much clearer.
What do you mean it "stretches 5% of its length"? It stretches up to that? It stretches that much for the 5kN tension, but will stretch more when the man walks on it?
Are the boxes square and in danger of tipping over, or long enough rectangles that we don't need to worry about that?
Do we need the boxes not to slip at all, or merely not so much that the man gets his feet wet?
If we know the static friction available from the boxes is 5N, why do we care what they weigh?
 
Thank you for the reply, I'll clear those thins up a bit.
The line stretches 5% under maximum tension, that it will hold, that is 15 kN, so it will stretch a bit more when the man walks on it.
By box I meant all sides equally long (50cm) (English is not my first language) so they are in the danger of tipping over and sliding, we want them to stay where they are.
And if the man has good enough ballance he should be able to get dry across the water.
 
Ok, let me play that back to you:
The rope is 12m when slack. It stretches linearly with tension up to a maximum of 5% at 15kN. It is set up across the water at a height of 0.5m under tension of 5kN (so 0.2m extension). The rope is weightless. The square-section boxes it is tied to at the ends are in danger of being pulled over by the rope and of sliding. The static friction available to the boxes is 5N.
It all makes sense except the last bit. Even if you make it 5kN they're already in danger of slipping before the man gets on the rope. Should it be stating a coefficient of friction?
 
I'm sorry, I am really bad in this friction business, how would you define the friction for this sort of setup?

The setup is supposed to be like this I think

http://www.slacklining.ca/wp-content/uploads/2012/02/Waterlining21.jpg

except that on ether side there is a box lifts the line up, holds is and that must not slide or tip over
 
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Normally you would specify a coefficient of static friction, often written μs. So if the box weighs B then the horizontal frictional force can reach μsB.
But you may have another problem. Haven't done the detailed calculation (involves a quartic) but the man might get his feet wet even if the boxes stay in place.
Seems like you are inventing a problem, not solving one you were given.
 
This is a problem I were given, but not for school.
We are going to set up a slackline like this and we were wondering if a tank with 1000 l of water on each side will hold the line or will it slide on wet anti slip tiles.
Is 1000 or each side enough or do we need more?

Thank you for you patience, it means a lot ;)
 
OK. But first let's look at what happens even if the boxes stay put. Is the load from the jumping man enough to dip the rope into the water? (At 5% extension it will be in the water, right?)
Any idea how you would calculate the max tension in the rope when the man jumps?
 
If we assume that the man stands in the middle when he jumps the length of the line needs to be
2[itex]\sqrt{6^2 + (0.5)2}[/itex]= 12.04 cm

So the line can only stretch 4 cm more so that the man won't touch the water.
Like you said we took 20 cm of the stretch away when we stretched the line up to 5 kN
and there are 40 cm more available until the line breaks, right, so how much force will be set extra on the line when the man jumps and lands back?
 
Think about potential energy. How much PE does the man have (relative to surface of water, say) when at peak of jump? You have to make some decision here about what you mean by jumping a given height. Is the height the vertical displacement of the mass centre relative to standing still in the rope? Seems reasonable. At peak of jump there is no KE, and no PE stored in the rope.
Next, think about the PE when at the lowest point after landing back on the rope. Again, no KE here, but PE both in terms of man's altitude and stretch in rope. You will have to assume something about the man's posture at this point. In practice, he will probably be somewhat crouched, giving him a lower PE than when standing erect.
You can then use conservation of energy to determine the PE stored in the rope.
(Until your last post, I was assuming the rope would not break, but simply cease to stretch. But I now see you are saying that it will stretch linearly up to max tension then snap. That means you can assume it stretches linearly throughout the calculation, then check at the end whether max tension was exceeded.)
Do you know the formula for the PE stored in a spring?
 
I'm thinking, is it reasonable to assume that the line stretches linearly? Isn't most of the stretch taken out in the beginning? Is it possible to define the stretch with a function that increases a lot in the beginning but less after some time, like ln(x)?

If we say that the mans feet are one meter above the line, and he is in average height, then the center of mass will be 0,57*180=102.6 cm higher than the mans feet.
http://hypertextbook.com/facts/2006/centerofmass.shtml
We can probably assume that the center of mass will be at 35% of his height, or what?
That means that the PE in the man is 90kg*9.8*2.526m=2227.932j

Isn't the formula for PE in a spring F=-kx?
 
Jonnigs said:
is it reasonable to assume that the line stretches linearly?
I'm no expert on the practicalities of real ropes and springs, but in school level mechanics it is always taught that the tension is proportional to the extension.
If we say that the mans feet are one meter above the line, and he is in average height, then the center of mass will be 0,57*180=102.6 cm higher than the mans feet.
http://hypertextbook.com/facts/2006/centerofmass.shtml
We can probably assume that the center of mass will be at 35% of his height, or what?
That means that the PE in the man is 90kg*9.8*2.526m=2227.932j
That's unreasonably precise. Let's just say for now that he can get his CoM 2.5m above the water. To achieve that in practice might involve bouncing up and down on the rope a bit, building up energy as on a trampoline.
Isn't the formula for PE in a spring F=-kx?
No, that's the force. It's analogous to KE: kx2/2.
 
But if we just assume for now that the line will hold and the mans feet won't get wet, will 1000 kg on each side hold the line and the man walking on it?
 
You'll still need to work out the max tension in the rope, and to calculate what that means for the boxes we also need the angle that tension makes to the horizontal, so might as well finish the calculation about extent of stretch first.