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Composition Setup
 Excerpt
hiddentrue
The tension in ropes supporting an object can sometimes be much larger than the object's weight.
Image Added
Photo by Cheresa D. Theiral, courtesy 
...
 U.S. 
...
 Army.
Suppose that a rope bridge is constructed using a massless rope that does not stretch. Suppose further that a person is clipped to the rope by a carabiner, with all of their weight essentially supported at one point. Finally, suppose that the rope bridge is strung between anchor points that are separated by a horizontal distance of 12.0 m, and that the sag in the bridge when the person is at the midpoint is 1.20 meters. 
 Deck of Cards
idbigdeck
|

h4. Part A

Suppose that a rope bridge is constructed using a massless rope that does not stretch.  Suppose further that a person is clipped to the rope by a carabiner, with all of their weight essentially supported at one point.  Finally, suppose that the rope bridge is to be strung between anchor points that are separated by a horizontal distance of 12.0 m, and that the sag in the bridge when the person is at the midpoint is 1.20 meters.  If the person is at rest at the midpoint, what is the ratio of the tension in the rope to the person's weight?

System:  Person as [point particle] subject to external influences from the earth (gravity) and the rope (tension).

Model: [Point Particle Dynamics].

Approach:  We begin with a picture of the situation:

PICTURE

{note}Note that at the midpoint, the rope will be symmetric.  Thus, the angles on each side will be the same.{note}

We next construct a free body diagram for the person:

FBD

{note}Note that tension appears twice.  Again, the symmetry dictates that the tension on each side be the same.{note}

The person is at rest, so their acceleration is zero in both the _x_ and _y_ directions.  The _x_-direction equation is clearly satisfied since we have assumed that the tension and rope angles on each side of the person are the same.  [Newton's 2nd Law|Newton's Second Law] for the _y_ direction gives:

{latex}
 Card
labelPart A
Part A
If the person is at rest at the midpoint, what is the ratio of the tension in the rope to the person's weight?
Solution
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idsysa
 System: 
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Person as .
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idinta
 Interactions: 
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idinta
External influences from the earth (gravity) and the rope (tension).
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 Model: 
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.
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 Approach: 
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idappa
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iddiaga
  Diagrammatic Representation 
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iddiaga
We begin with a picture of the situation:
Image Added
 Note
Note that at the midpoint, the rope will be symmetric. Thus, the angles on each side will be the same.
We next construct a free body diagram for the person:
Image Added
 Note
Note that tension appears twice. Again, the symmetry dictates that the tension on each side be the same.
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diaga
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idmatha
  Mathematical Representation 
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idmatha
The person is at rest, so their acceleration is zero in both the x and y directions. The x-direction equation is clearly satisfied since we have assumed that the tension and rope angles on each side of the person are the same. Newton's 2nd Law for the y direction gives:
 Latex
\begin{large}\[ 2T\sin\theta - mg = 0 \]\end{large}
{latex}


which 
 
 is 
 
 solved 
 
 to 
 
 give:


{
 Latex
}
\begin{large}\[ \frac{T}{mg} = \frac{1}{2\sin\theta} \] \end{large}
{latex}


The 
 
 angle 
 θ 
θ is 
 
 determined 
 
 by 
 
 the 
 
 length 
 
 of 
 
 the 
 
 bridge 
 
 and 
 
 the 
 
 amount 
 
 of 
 
 sag 
 
 in 
 
 the 
 
 rope:


{
 Latex
}
\begin{large}\[ \theta = \tan^{-1}\left(\frac{h}{L/2}\right) \] \end{large}
{latex}


Thus:


{
 Latex
}
\begin{large}\[ \sin\theta = \frac{h}{\sqrt{h^{2}+(L/2)^{2}}}\]\end{large}
{latex}


and:


{
 Latex
}
\begin{large}\[ \frac{T}{mg} = \frac{\sqrt{h^{2}+(L/2)^{2}}}{2h}= 2.55 \]\end{large}
{latex}

{note}For a 180 lb person, this would correspond to a tension of about 460 lbs.  Note that to achieve less sag, more tension must be provided.{note}

h4. Part B

Consider the same rope bridge.  If the person is at rest exactly 1/6 of the way along the bridge (as measured horizontally) what is the tension in each side of the rope in terms of the person's weight?  Assume that the length of the rope is constant.

System and Model:  As in Part A.

The new picture is:  

PICTURE

which gives the free-body diagram:

FBD

{note}Note that now we _do not_ have symmetry about the location of the person and so both the angles and the tensions can be different on the two sides.{note}

The equations of Newton's Law give:

{latex}
 Note
For a 180 lb person, this would correspond to a tension of about 460 lbs. Note that to achieve less sag, more tension must be provided.
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 Card
labelPart B
Part B
Consider the same rope bridge. If the person is at rest exactly 1/6 of the way along the bridge (as measured horizontally) what is the tension in each side of the rope in terms of the person's weight? Assume that the length of the rope is constant.
Solution
System, Interactions and Model: As in Part A.
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idappb
 Approach: 
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iddiagb
  Diagrammatic Representation 
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The new picture is: 
Image Added
which gives the free-body diagram:
Image Added
 Note
Note that now we do not have symmetry about the location of the person and so both the angles and the tensions can be different on the two sides.
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  Mathematical Repesentation 
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The equations of Newton's Law give:
 Latex
\begin{large}\[ T_{1}\cos\theta_{1} - T_{2}\cos\theta_{2} = 0 \] 
\[ T_{1}\sin\theta_{1} + T_{2}\sin\theta_{2} - mg = 0\]\end{large}
{latex}


These 
 
 equations 
 
 cannot 
 
 be 
 
 solved 
 
 without 
 
 further 
 
 information. 
  
 The 
 
 extra 
 
 information 
 
 is 
 
 the 
 
 length 
 
 of 
 
 the 
 
 rope. 
  
 We 
 
 know 
 
 from 
 
 Part 
 
 A 
 
 that 
 
 the 
 
 rope's 
 
 length 
 
 is:


{
 Latex
}
\begin{large}\[ 2\sqrt{h_{A}^{2} + (L/2)^{2}} = \mbox{12.2 m} \]\end{large}
{latex}


We 
 
 can 
 
 also 
 
 write 
 
 the 
 
 length 
 
 of 
 
 the 
 
 rope 
 
 in 
 
 terms 
 
 of 
 
 the 
 
 sag 
 
 of 
 
 the 
 
 rope 
 
 at 
 
 the 
 
 1/6 
 
 L 
 
 point:


{
 Latex
}
\begin{large}\[ \mbox{12.2 m} = \sqrt{h_{B}^{2} + (L/6)^{2}} + \sqrt{h_{B}^{2} + (5L/6)^{2}} \]\end{large}
{latex}


which 
 
 can 
 
 be 
 
 solved 
 
 to 
 
 give:


{
 Latex
}
\begin{large}\[ h_{B} = \mbox{0.83 m}\]\end{large}
{latex}

Once _h_~B~ is known, the equations of 
Once hB is known, the equations of Newton's 
 
 2nd 
 
 Law 
 
 become:


{
 Latex
}
\begin{large}\[ T_{1} \frac{L/6}{\sqrt{h_{B}^{2} + (L/6)^{2}}} - T_{2}\frac{5L/6}{\sqrt{h_{B}^{2} + (5L/6)^{2}}} = 0\]\[ T_{1}\frac{h_{B}}{\sqrt{h_{B}^{2} + (L/6)^{2}}} + T_{2}\frac{h_{B}}{\sqrt{h_{B}^{2} + (5L/6)^{2}}} - mg = 0\]\end{large}
{latex}

Eliminating _T_~2~ then gives:

{latex}
Eliminating T2 then gives:
 Latex
\begin{large}\[ \frac{T_{1}}{mg} = \frac{5\sqrt{h_{B}^{2} + (L/6)^{2}}}{6h_{B}} = 2.17 \]\end{large}
{latex}


and 
 
 solving 
 
 for 
 _
T
_~2~ 
2 gives:


{
 Latex
}
\begin{large}\[ \frac{T_{2}}{mg} = \frac{\sqrt{h_{B}^{2} + (5L/6)^{2}}}{6h_{B}}= 2.01 \]\end{large}
{latex}

{tip}These results are messy, and so it pays to put the answers back into the equations of Newton's 2nd Law to check them.{tip}







 Tip
These results are messy, and so it pays to put the answers back into the equations of Newton's 2nd Law to check them.
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