/ Physics/Engineering question

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Dave 88 - on 01 May 2014
Looking for some help to settle a discussion I had, please.

If you had a suspended load, for example 100kg weight, when it's hanging there with the rope vertical, there is 100kg on the rope and anchor (single anchor for arguments sake). Correct so far?

If you then attached a second rope to pull your load to the side, does this increase the force on the top anchor and rope?

My thinking is that it will increase (so effectively the 100kg load will "weigh more") and continue to do so the further the load is moved from hanging vertically.

Can anyone confirm this please?
Blue Straggler - on 01 May 2014
In reply to Dave 88:

Confirmed. Ignore the actual second rope, it is the deviation from vertical which increases the force. Picture a rope swing - when they snap the branch, it tends to be when they are at the apex of the swing i.e. farthest from vertical....
Jon Stewart - on 01 May 2014
In reply to Dave 88:

Is the second rope horizontal?
Dave 88 - on 01 May 2014
In reply to Blue Straggler:

Ok cheers. Do you know the calculation for this? Something like for every 10 degrees you move off vertical, the weight increase by X percent?
Henry Brown - on 01 May 2014
In reply to Dave 88:

Depending the angle between the two strands possibly. If its pulling directly sideways at around 90 degrees then yes it will incress the force
Dave 88 - on 01 May 2014
In reply to Jon Stewart:

For arguments sake let's say yes, but the second rope was actually slightly below so obviously a bit of downwards pull to add into the equation.
Dave 88 - on 01 May 2014
In reply to Henry Brown:

As per my reply to Jon, let's assume the second rope is at 90 to the vertical rope
jkarran - on 01 May 2014
In reply to Dave 88:

If the second rope is horizontal the answer is yes, the tension in the first rope and the associated load on its anchor increases as tension is increased in the second rope.

If the second rope is not horizontal then the answer is 'it depends'!

jk
psaunders - on 01 May 2014
In reply to Dave 88:
F = mg / cosine(angle).

So for the following angles, in degrees, from vertical:
10 = 1.5% greater force
20 = 6.5%
30 = 15%
40 = 31%
50 = 56%
60 = 100%
70 = 192%
80 = 476%

This is why equalised anchors should not be more than 120 degrees.
Post edited at 18:38
Jimbo C - on 01 May 2014
In reply to Dave 88:
> (In reply to Blue Straggler)
>
> Ok cheers. Do you know the calculation for this? Something like for every 10 degrees you move off vertical, the weight increase by X percent?

Simple trigonometry - the tension in the rope will be the weight divided by the cosine of the angle change from vertical

Jimbo C - on 01 May 2014
In reply to psaunders:

ha, just beat me to it :-)
Dave 88 - on 01 May 2014
In reply to jkarran:

Haha ok, it was more or less 90deg so I'm not too bothered about splitting hairs!

Is this pretty much the same as vector forces? And does anyone have an equation? (For this example, not standard vector forces).
Dave 88 - on 01 May 2014
In reply to psaunders:

Ah cracking cheers for that.

Thanks all, I can go to the pub satisfied that I was right!
Marmoteer - on 01 May 2014
In reply to Dave 88:

It is slightly challenging to answer this without the benefit of a diagram...

The answer to your question depends on the angle between the 2 ropes...anything less than 90 degrees will reduce the weight felt by the original anchor, angles greater than 90 degrees will increase the weight felt by the orginal anchor. This is the same problem as setting up a good belay.

So it depends how you are moving your load to the side...if the direction of pull makes the ropes into a "v" shape, the weight on the original anchor will decrease as more weight is taken by the new (sideward pulling) force as the load traverses.

However, if you start with the ropes making an "L" shape and you continue to open out the angle with the horizontal (sideward pulling) force remaining in that direction, the weight experienced on the orignal anchor will increase.

NB. For physicists/engineers/pedants (same thing?? :-) ), I have used "weight experienced" or "pull" interchangeably with force.
Dave 88 - on 01 May 2014
In reply to Marmoteer:

Yeah obtuse angle.

Haha I know, the OP was a minefield of incorrect terms! I knew I was never gonna nail it so I just had to try and keep it simple.
marsbar - on 01 May 2014
In reply to Dave 88:

That doesn't make sense to me.

Are you not exerting another force sideways with your rope?

Force doesn't just increase because of direction, the swing example is more likely to reflect a weakness in a particular orientation.

You can resolve your force into horizontal and vertical componants using basic trig if you like but this won't make it bigger.
marsbar - on 01 May 2014
In reply to Marmoteer:

I think I need a diagram.
Henry Brown - on 01 May 2014
In reply to marsbar:
The vertical force wont increase but a horizontal force will occour incressing the overall tension in the once vertical strand
Post edited at 18:53
Kevster - on 01 May 2014
In reply to marsbar:

surely the force decreases?

Weight just hanging on the rope, maximum force.

As the rope pulling to the side takes some of the weight, therefore there is less on the dangling rope?

The only way the force on the dangle rope and anchors is increased is is additional force is added through tension downwards. Or does the horizontal pulling rope remain in the same fixed anchor point so as the mass swings to the side, it is lifted slightly by the arc of the now not so vertical rope?

Confused on what is actually meant.
marsbar - on 01 May 2014
In reply to Henry Brown:

Ah that helps I had no idea the OP was talking about the tension.
Rick Graham on 01 May 2014
When I was doing my engineering degree we called it "the triangle of forces "

Where's Jim Titt when you need him :-)

puppythedog on 01 May 2014
In reply to Dave 88:

I'm calling troll.
crossdressingrodney - on 01 May 2014
In reply to Kevster:

> surely the force decreases?

> Weight just hanging on the rope, maximum force.

> As the rope pulling to the side takes some of the weight, therefore there is less on the dangling rope?

The second rope is pulling exactly sideways, so it doesn't take any of the weight. But now the first rope has to pull harder because, in addition to the upward force needed to hold up the weight, it must provide a sideways force to balance the pull of the second rope.

As was said above, if you draw a picture and resolve the forces you find that the tension in the first rope is equal to the weight divided by cos(theta), where theta is the angle between the first rope and the vertical. So, for example, if you pull the weight out to 60 degrees you double the tension.
Matt_b - on 01 May 2014
In reply to Marmoteer:
I think it is mostly engineers reading this. They can let a weight of 100 kg slip through. A physicist pedant wouldn't.
Post edited at 22:03
Rick Graham on 01 May 2014
In reply to Matt_b:

Us engineers live in the real world :-)
crossdressingrodney - on 01 May 2014
In reply to Rick Graham:

Do engineers not believe in the moon then? :)
Rick Graham on 01 May 2014
In reply to crossdressingrodney:

Blue ones maybe. Shit happens.
Bruce Hooker - on 01 May 2014
In reply to crossdressingrodney:

Yours is the right answer... if I remember correctly :-)
crossdressingrodney - on 01 May 2014
In reply to Blue Straggler:

Don't rope swings break at the lowest point rather than the highest?

Would have to get pen and paper out, but I think that the tension in the rope at the highest point (apex) is only weight times cos theta. That fits with that rather frightening thing that swings do when you go too high: you feel weightless for a moment and the rope goes a bit slack before snapping tight again. Or maybe it's that snapping tight that breaks the swing?
Rick Graham on 01 May 2014
In reply to crossdressingrodney:

Search on "triangle of forces", the horizontal and vertical components of all the forces need to balance out.
John Gillott - on 01 May 2014
In reply to Rick Graham:

> Search on "triangle of forces", the horizontal and vertical components of all the forces need to balance out.

Not when there is acceleration (which there is with circular motion, in the case of the swing).
crossdressingrodney - on 01 May 2014
In reply to Rick Graham:

Well that's only true if there's no acceleration, which is not the case for a swing.
Blue Straggler - on 01 May 2014
In reply to crossdressingrodney:

You may be right. I made a rushed reply because I felt sorry for the thread that had not had any answers within a minute of the OP. Might have got it wrong.
PeakDJ on 02 May 2014
In reply to Dave 88:

Draw a scale diagram showing the forces pulling on each rope. Downwards force before adding the second rope is about 980N. If you use a scale to show the size of each force, the third (and longest) side of the triangle = the tension in the rope...assuming that the anchored object isn't moving. If the second rope pulls horizontally then this will be greater than 980N.
Marmoteer - on 02 May 2014
In reply to Matt_b:

Now, the other thing we haven't considered is the mass of the ropes.....
:-)

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