Project Help Outline by Keith Stroyan

Project 23:  A Falling Ladder

by A. Student and B. Pupil

with technical help from Da LabMonitor and Help Notebook

Introduction

Write this last.  Begin with the basic question and state your main finding clearly, but with few technical details.

Read the CD section 7.5 and the printed project.  The simple model of the falling ladder predicts that the vertical speed of the tip of the ladder approaches infinity, but that model assumes the tip of the ladder rests on the wall and fails to take the acceleration due to gravity into account.  Read Chapter 10 of the main text and the printed project for the law of gravity.

Make sure you remove all the Red text in this help NoteBook before you submit your project.

The Model for a Ladder Touching the Wall

Say briefly what this section is about bla, bla, bla

The Basic Set-up and Questions

Recall the setup for the basic model and write it up in your own words.

bla, bla, bla

Variables with units

t -

x -

y -

The fixed length parameter:

L -

The ladder is L feet long.  We let  x  denote the distance from the vertical wall along the horizontal floor to the base of the ladder and  y  denote the distance up the wall to the place where the ladder rests....

Using this figure, you can construct a simple mathematical expression defining the relationship between L, x, and y.  Derive this equation

RELATION BETWEEN THE VARIABLES WHEN THE LADDER RESTS AGAINST THE WALL... bla, bla, bla

x   ...   y     ...  .   L

Motion while the tip of the ladder contacts the wall

As the base of the ladder slides horizontally away from the wall, the tip on the wall slides down the wall.  Following is  an animation of the motion.  (Double click on the figure to see it move.)

Using the relationship between L, x, and y, that we found above, we can find a relationship between the horizontal and vertical velocities.  Derive this equation...  (See the main text CD Example 7.5 and Exercise CD 7.1.2.)  Following is some help.

Vertical velocity

Velocity of  x  or  y  is given by the ___  with respect to time.... this is the caclulus way of saying velocity...

By differentiating the equation that relates  x, y, and L in terms of  t, we obtain a formula for the velocity of the tip of the ladder as follows:

We compute (dy/dt) with implicit differentiation using the relationship between the variables  x, y, and the fixed length:

(x^(2 )) + (y^2) =  (L^2)

2 x (dx/dt) +  2 y (dy/dt)  =  0

(dy/dt) = -(x/y) (dx/dt)

In this project we model the situation where the horizontal speed, (dx/dt) = h  is constant, so the vertical velocity   v = (dy/dt)  is given by the formula

v = -(x/y) h

This model's prediction for the limiting velocity of the ladder in the vertical direction as the tip nears the ground is

(Underscript[Limit, y -> 0])[v] = ?? ??

because...

Velocity of the tip as a function of height  y  and constant horizontal speed  h.

Enter and explain the following sample computations and use the idea to work your project.

A more convenient expression for the velocity can be obtained by solving the ___ for  x ...

v := -(((Sqrt[(L^2) - (y^2)]) h)/y)
L = 20;
h = 1;
y = 1/12;
Print["Speed ", v, " at height ", y]

Enter and explain the next computation

Clear[y]
Solve[ v == -1100, y]

Enter and explain the next computation

N[(20/(Sqrt[1210001]))]

Enter and explain the next computation

Clear[y]
Solve[ v == -186000 5280/3600, y]

Give a numerical approximation for this solution

Height where Dad breaks the sound barrier

Does the model predict that the ladder will be able to break the sound barrier (have a velocity of 1100 ft/s)?  If so, at what point?

Height where Dad breaks the light barrier

Does the model predict that the ladder will be able to achieve the speed of light (186000 miles/s)?  If so, at what point?

Conclusion on the computations with the first model

Obviously, there's a physical error in our assumptions for the first model.  Falling ladders don't create a sonic boom.  We know that our model is valid initially, but it fails to take the physical law of gravity into account.  The wall counteracts gravity only until the downward acceleration is as much as gravity provides.  We describe how much that can be next.

Galelio's Law of Gravity

The only downward force acting on the ladder is gravity.  When the acceleration predicted by the first model is below the acceleration due to gravity, the wall can counteract part of it.  After this, the ladder will simply move with Galileo's law of gravity which says: "The acceleration due to gravity is constant for all objects."  In fps-units, this constant is  32 ft/(sec^2).  We describe this law in terms of calculus next.

Explain why Galileo's law can be expressed with calculus as

(dv/dt) = -g

g a constant depending only on the units, with  g = 32 in fps-units.

Acceleration while the tip stays in contact

Accleration is the time-rate-of change of velocity, (dv/dt), since  v = ((-x)/y) h, for a constant  h, we can express acceleration of the tip of the ladder in terms of the variables x, y, and the parameters of the model - assuming it remains in contact with the wall.

(dv/dt) = (d/dt) (((-x)/y) 
              h) = -h (d/dt) (x (y^(-1)))

now use the product rule and facts about the wall-contact-model to show

(dv/dt) = -(h^2) ((L^2)/(y^3))

Be careful with the Chain Rule.  For example, the derivative of  (y^5)  would be  ((d((y^5)))/dt) = ((dy^5)/dy)·(dy/dt) = 
          5 (y^4) (dy/dt).....

Where the tip leaves the wall

The acceleration for the model with the ladder touching the wall is: a = -(h^2) ((L^2)/(y^3)) or just  a = - ((L^2)/(y^3))  if  h = 1, so the y-value where the acceleration reaches  32  downward is given by:

Enter and explain the next computation

Clear[a, y, L];
a = ((-(L^2))/(y^3));
L = 20;
Solve[a == -32, y]
N[Solve[a == -32, y]]

When  y < 2.32, the model acceleration ___  _  so the model no longer applies.  Instead, we have the ladder in free fall with initial height  2.32... and initial velocity ___velocity at height 2.32 ...  _

Perform the next computation

Vertical Motion of the Tip After it Leaves the Wall

If we let  τ  be the time in seconds with τ = 0 at the time when the ladder reaches the acceleration -32 ft/(sec^2), then the equation for the position of the tip after that time is goverened by the differential equation

(dv/d)τ = -32

The velocity for these times is given by the derivative,

v  =  v[τ]  =  
          v0   -  32  τ

The formula for  y  is:

y  =  y[τ]  =  
                y0  +  v0 τ  -  16  (τ^2)

This is explained in greater detail in the text and written project.  Look that up and put the explanation for why these are the formulas for v[τ] and y[τ]  here...

Time and speed of the real crash

Once the ladder has achieved the acceleration due to gravity, how much longer before it strikes the ground?

Enter and explain the next computation (we are using exact solutions to make you think.  Try the approximate solutions you found above.)

Clear[y, τ, y0, v0];
y[τ_] = 
        y0 + v0 τ  -  16 (τ^2);
y0 = ((5^(2/3))/(2^(1/3)));
v0 = -(((2^(1/3)) (Sqrt[400 - ((5 (5^(1/3)))/(2^(2/3)))]))/(5^(2/3)))
Solve[y[τ] == 0, τ]
N[Solve[y[τ] == 0, τ]]

What is it's velocity in the vertical direction when it strikes the ground?

Enter and explain the next 3 computations

τCrash = 
      (1/160) (-(10^(1/3)) (Sqrt[400 - ((5 (5^(1/3)))/(2^(2/3)))]) + (Sqrt[5 (-5 + 240 (10^(2/3)))]))
N[τCrash]
Simplify[y[τCrash]]

y'[τCrash]
N[y'[τCrash]]

y'[0.197916]

Horizontal Motion

We have been primarily concerned with the vertical status of the ladder - now let's take a look back at it's horizontal status.  Does the new model change our expression for displacement in the x direction?  If so, how?  What is the x coordinate of the base of the ladder when the tip strikes the ground?  What is the x coordinate when you are halfway through the second model (that is, if the time found above were tCrash, what is the horizontal coordinate at tCrash/2)?

Enter and explain the next computation

Clear[x, y, L, t];
x[t_] = h t;
h = 1;
L = 20;
Solve[(Sqrt[(L^2) - x[t]^2]) == ((5^(2/3))/(2^(1/3))), t]

Enter and explain the next computation

tAccel = (Sqrt[(5/2) (160 - (10^(1/3)))]);

y0 = ((5^(2/3))/(2^(1/3)));
v0 = -(((2^(1/3)) (Sqrt[400 - ((5 (5^(1/3)))/(2^(2/3)))]))/(5^(2/3)));
y[t_] = 
      If[t < tAccel, (Sqrt[(L^2) - x[t]^2]), 
        y0 + v0 (t - tAccel ) -  
          16 (((t - tAccel))^2)]
tCrash = 
        (1/160) (-(10^(1/3)) (Sqrt[400 - ((5 (5^(1/3)))/(2^(2/3)))]) + (Sqrt[5 (-5 + 240 (10^(2/3)))]));
Plot[y[t], {t, 0, tCrash + tAccel}, 
        PlotStyle -> {RGBColor[0,  .4, 0]}, 
        AxesLabel -> {"t", "y"}];

Enter and explain the next computation

x[tAccel + tCrash] - 20
N[x[tAccel + tCrash] - 20]

More suggestions are contained in the printed project