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- Thread starter PEZenfuego
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The tension in a rope with significant mass and hanging under its own weight is given by

T = wy

Where w is the weight per unit length, y is the vertical distance above the bottom of the curve.

It is called a catenary.

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jbriggs444

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How could it be otherwise? If the forces were unequal then there would be a net force on the rope. For a massless rope, that would imply infinite acceleration!

If the rope has non-zero mass then the tension at either end of a small section of rope will vary based on the acceleration of or gravity upon that small segment. So if you have a 1 kg rope hanging vertically under 9.8 meter/sec^2 gravity then the top of the rope will be under 9.8 Newtons more tension than the bottom.

For acceleration or gravity aligned with the direction of the rope you get variable tension as above. For acceleration or gravity in the transverse direction you get curvature. Perhaps you have noticed that ropes sag.

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The tension in a rope with significant mass and hanging under its own weight is given by

T = wy

Where w is the weight per unit length, y is the vertical distance above the bottom of the curve.

It is called a catenary.

Oh, yes I've seen some of this. Doesn't it involve hyperbolic trig functions? Does this formula also apply to vertically hanging ropes? Could one, say, use this to calculate the tension on a chain link?

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How could it be otherwise? If the forces were unequal then there would be a net force on the rope. For a massless rope, that would imply infinite acceleration!

If the rope has non-zero mass then the tension at either end of a small section of rope will vary based on the acceleration of or gravity upon that small segment. So if you have a 1 kg rope hanging vertically under 9.8 meter/sec^2 gravity then the top of the rope will be under 9.8 Newtons more tension than the bottom.

For acceleration or gravity aligned with the direction of the rope you get variable tension as above. For acceleration or gravity in the transverse direction you get curvature. Perhaps you have noticed that ropes sag.

Sorry, I almost missed this post. I'm following you nicely I think. So, now follow my logic: Let's say that there are links in a chain and that every link can withstand the same amount of tension except for one link (the weak link). Now, let's assign values. Each strong link can withstand 15 N of tension and the one weak link can withstand 11 N of tension. We put the weak link on the bottom. If we are using a 1 kg chain and have a 1 kg mass hanging from said chain, would it be fair to say that the tension on the top link is 20 N (approximate gravity to be 10 m/s^2 for simplicity) and the tension on the bottom link is 10 N? If all of this checks out, then we can say that in this case, the chain broke because of its strongest link, rather than the weakest one.

And for a horizontal problem, could the same logic be applied if we put the weak link in the middle of the chain and the strong link on the end?

I realize that in almost every case these effects will be small compared to the relative strengths of the chain links (and also that finding a chain link that can only withstand 20 newtons would be difficult).

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jbriggs

How could it be otherwise?

Friction round a pulley. That is how machine belts work.

pezenfuego

And for a horizontal problem, could the same logic be applied if we put the strong link in the middle of the chain and the weak link on the end?

As jbriggs noted the cable or chain sags.

And if you think about it (and my formula) the minimum tension is at the bottom and the maximum at the supports (ends).

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SteamKing

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The catenary equations do not apply to a vertically hanging rope.

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steamking

The catenary equations do not apply to a vertically hanging rope.

Perhaps you'd like to explain where this fits in to the present discussion?

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Perhaps you'd like to explain where this fits in to the present discussion?

It answers a question that I asked.

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I asked because I am not convinced.

The equation I wrote is one of the five catenary equations and the tension in a vertical rope obeys it, when hanging under its own weight.

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AlephZero

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I asked because I am not convinced.

The equation I wrote is one of the five catenary equations and the tension in a vertical rope obeys it, when hanging under its own weight.

If you want to call the shape of a straiight vertical rope "a catenary", that's not very helpful IMO.

And

is wrong for a catenary, since the tension in the rope at the "bottom of the curve", where the rope is horizontal, is obviously non-zero (consider the free body diagram of one half of the rope), but the common sense meaning of "the vertical distance aboce the bottom of the curve" gives y = 0 and T = 0.The tension in a rope with significant mass and hanging under its own weight is given by

T = wy

Where w is the weight per unit length, y is the vertical distance above the bottom of the curve.

I think I know what you

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OK, the same equations apply to both chains and ropes.

Aleph Zero was correct in that for a complete catenary, suspended between two points we do not measure from the lowest point we measure from the directrix of the catenary which is at a distance c below this. So the tangent at the lowest point where dy/dx=0 is parallel to the x axis and c above it. Let us say this passes through the point A so that A is c above the origin as in my diagram. I will return to how we derive c later and include the condition for c to be zero.

Let us consider the equilibrium of the lowest point A and let this have a tension T_{0}, whatever that works out to be.

Because the slope of the chain or rope is zero at A T_{A} is purely horizontal.

If w the weight of rope or chain per unit length then Let T_{A} = wc for some distance c.

This is where we introduce c, but do not yet know that it is the distance shown on my diagram. That remains to be proved.

Now consider equilibrium of a length of rope or chain running from A to any point P(x,y) as shown.

This length is in equilibrium under the influence of three forces. T_{A}; The (different) tension T at P and the weight of the rope = w times the arc length s = ws

I said that there are 5 main equations relating to the catenary, this leads to the first.

[tex]\begin{array}{l}

T\cos \theta = wc........{\rm{ = }}\left( {{{\rm{T}}_{\rm{A}}}} \right).....{\rm{H}} \\

T\sin \theta = ws.......................{\rm{V}} \\

\tan \theta = \frac{s}{c}\quad s = c\tan \theta ....................{\rm{1}} \\

\end{array}[/tex]

Since [itex]\tan \theta = \frac{{dy}}{{dx}}[/itex] and

[tex]\frac{{ds}}{{dx}} = \sqrt {\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)} = \sqrt {\left( {1 + \frac{{{s^2}}}{{{c^2}}}} \right)} [/tex]

Then

[tex]\frac{{ds}}{{\sqrt {\left( {{c^2} + {s^2}} \right)} }} = \frac{{dx}}{c}[/tex]

Integrate

[tex]{\sinh ^{ - 1}}\left( {\frac{s}{c}} \right) = \frac{x}{c} + D[/tex]

Since s=0 when x=0, D=0

[tex]s = c\sinh \left( {\frac{x}{c}} \right)....................................2[/tex]

Also in 1 we have

[tex]\frac{{dy}}{{dx}} = \frac{s}{c} = \sinh \left( {\frac{x}{c}} \right)[/tex]

Integrate

[tex]y = c\cosh \left( {\frac{x}{c}} \right) + E[/tex]

Choose that when x=0, y=c, giving E=0

[tex]y = c\cosh \left( {\frac{x}{c}} \right).......................................3[/tex]

From 3 we have

[tex]{y^2} = {c^2}{\cosh ^2}\left( {\frac{x}{c}} \right) = {c^2}\left( {1 + {{\sinh }^2}\left( {\frac{x}{c}} \right)} \right) = {c^2}\left( {1 + \frac{{{s^2}}}{{{c^2}}}} \right)[/tex]

Which yields equation 4

[tex]{y^2} = {c^2} + {s^2}................................................4[/tex]

Finally if we square and add equations H and V from the beginning and substitute we obtain the fifth equation which I posted before.

[tex]\begin{array}{l}

{T^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = {w^2}\left( {{c^2} + {s^2}} \right) = {w^2}{y^2} \\

T = wy................................................5 \\

\end{array}[/tex]

This shows that the tension at any point P(x,y) is the same as the weight of a rope or chain hanging from that point to the directrix, where we have placed the x axis.

Finally for a rope or chain that is hanging straight down from a single support the origin and A coincide, as does the 'tangent' through A and the x axis because since there is no horizontal force T_{A}=0 in which case c = 0 since T_{A}wc and w≠0. Thus y is measured from the lowest point of the chain or rope.

Aleph Zero was correct in that for a complete catenary, suspended between two points we do not measure from the lowest point we measure from the directrix of the catenary which is at a distance c below this. So the tangent at the lowest point where dy/dx=0 is parallel to the x axis and c above it. Let us say this passes through the point A so that A is c above the origin as in my diagram. I will return to how we derive c later and include the condition for c to be zero.

Let us consider the equilibrium of the lowest point A and let this have a tension T

Because the slope of the chain or rope is zero at A T

If w the weight of rope or chain per unit length then Let T

This is where we introduce c, but do not yet know that it is the distance shown on my diagram. That remains to be proved.

Now consider equilibrium of a length of rope or chain running from A to any point P(x,y) as shown.

This length is in equilibrium under the influence of three forces. T

I said that there are 5 main equations relating to the catenary, this leads to the first.

[tex]\begin{array}{l}

T\cos \theta = wc........{\rm{ = }}\left( {{{\rm{T}}_{\rm{A}}}} \right).....{\rm{H}} \\

T\sin \theta = ws.......................{\rm{V}} \\

\tan \theta = \frac{s}{c}\quad s = c\tan \theta ....................{\rm{1}} \\

\end{array}[/tex]

Since [itex]\tan \theta = \frac{{dy}}{{dx}}[/itex] and

[tex]\frac{{ds}}{{dx}} = \sqrt {\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)} = \sqrt {\left( {1 + \frac{{{s^2}}}{{{c^2}}}} \right)} [/tex]

Then

[tex]\frac{{ds}}{{\sqrt {\left( {{c^2} + {s^2}} \right)} }} = \frac{{dx}}{c}[/tex]

Integrate

[tex]{\sinh ^{ - 1}}\left( {\frac{s}{c}} \right) = \frac{x}{c} + D[/tex]

Since s=0 when x=0, D=0

[tex]s = c\sinh \left( {\frac{x}{c}} \right)....................................2[/tex]

Also in 1 we have

[tex]\frac{{dy}}{{dx}} = \frac{s}{c} = \sinh \left( {\frac{x}{c}} \right)[/tex]

Integrate

[tex]y = c\cosh \left( {\frac{x}{c}} \right) + E[/tex]

Choose that when x=0, y=c, giving E=0

[tex]y = c\cosh \left( {\frac{x}{c}} \right).......................................3[/tex]

From 3 we have

[tex]{y^2} = {c^2}{\cosh ^2}\left( {\frac{x}{c}} \right) = {c^2}\left( {1 + {{\sinh }^2}\left( {\frac{x}{c}} \right)} \right) = {c^2}\left( {1 + \frac{{{s^2}}}{{{c^2}}}} \right)[/tex]

Which yields equation 4

[tex]{y^2} = {c^2} + {s^2}................................................4[/tex]

Finally if we square and add equations H and V from the beginning and substitute we obtain the fifth equation which I posted before.

[tex]\begin{array}{l}

{T^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = {w^2}\left( {{c^2} + {s^2}} \right) = {w^2}{y^2} \\

T = wy................................................5 \\

\end{array}[/tex]

This shows that the tension at any point P(x,y) is the same as the weight of a rope or chain hanging from that point to the directrix, where we have placed the x axis.

Finally for a rope or chain that is hanging straight down from a single support the origin and A coincide, as does the 'tangent' through A and the x axis because since there is no horizontal force T

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