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Deck of Cards
idbigdeck
h3.

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?

h4. Solution {

Solution

Card
labelPart A
Wiki Markup
Toggle Cloak

:

id

=

sysa

} *

System:
Cloak
idsysa

Person as .

Toggle Cloak
idinta
Interactions:
Cloak
idinta

External influences from the earth (gravity) and the rope (tension).

Toggle Cloak
idmoda
Model:
Cloak
idmoda

.

Toggle Cloak
idappa
Approach:

Cloak
idappa

Toggle Cloak
iddiaga
Diagrammatic Representation

Cloak
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.

Cloak
diaga
diaga

Toggle Cloak
idmatha
Mathematical Representation

Cloak
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
*  {cloak:id=sysa}Person as [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the earth (gravity) and the rope (tension).{cloak}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*  

{cloak:id=appa}

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diaga}

We begin with a picture of the situation:

!ropebridge1.jpg!

{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:

!ropebridge2.jpg!

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

{cloak:diaga}
{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}

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}\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}

{cloak:matha}
{cloak:appa}
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.

Cloak
matha
matha

Cloak
appa
appa

unmigrated-wiki-markup
Card
labelPart B
h3.

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.

h4. Solution *

Solution

System,

Interactions

and

Model:

*

As

in

Part

A.

{

Toggle Cloak

:

id

=

appb

} *

Approach:

* {cloak:id=appb} {

Cloak
idappb

Toggle Cloak

:

iddiagb
Diagrammatic Representation

Cloak
iddiagb

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.

Cloak
diagb
diagb

Toggle Cloak
idmathb
Mathematical Repesentation

Cloak
idmathb

The equations of Newton's Law give:

Latex
=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}

The new picture is:  

!ropebridge3.jpg!

which gives the free-body diagram:

!ropebridge4.jpg!

{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}

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Repesentation* {color}

{cloak:id=mathb}

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}

Eliminating T2 then gives:

Latex
{latex}

Eliminating _T_~2~ 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}

{cloak:mathb}
{cloak:appb}
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.

Cloak
mathb
mathb

Cloak
appb
appb