Figure 23.1: Ladder & Wall
Emily's Dad decided to paint the garage. He put a ladder against the wall and climbed to the top. Then the base of the ladder slid out.... How fast was he going when he hit the ground?
In the section on linked variables, or related rates, in Chapter 7 of the core text, we discussed what happens as a ladder slides down a wall. That model is oversimplified in a way that we want you to discover.
The simple-minded model suggests that Dad will strike the ground at warp 9, well above the speed of light. This is nonsense. He certainly cannot exceed the speed of light and it seems highly dubious that he would even approach the speed of sound.
Recall the setup for the basic model. 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.
Since x2+y2=L2, we can solve for y and compute
. By the Chain Rule,
and the speed at which we pull the base away from the wall is constant,
. This predicts that the speed of the tip tends to infinity as x tends to L or y tends to zero, because the denominator of
tends to zero while the numerator tends to L. (See Section CD7.5 of the main text for more details.)
in several ways
Suppose the ladder is 20 feet long and we pull the base at the rate of 1 foot per second.
inches above the ground.
Write a computer program and check it by doing part 1 by hand.
Also use the computer to find roots in the next two parts.
It should be clear that Dad's speed never exceeds the speed of light. In fact, it seems doubtful that it could exceed the speed of sound. The model is certainly very reasonable when the tip of the ladder is high on the wall, so what goes wrong? This has a simple intuitive explanation, provided you have some intuition for what produces accelerations.
and have the three formulas above for this quantity.
Use
, your formula for
, and simplify your formula as much as posible.
Express your answer in terms of y and show that when x2+y2=L2 and
, 
The formula for Dad's acceleration tends to infinity as y tends to the ground, but the only thing to provide acceleration is gravity. (The garage wall can prevent it, but it can't provide it.)
The only force acting down on the ladder is gravity.
The wall counteracts gravity, but only until the downward acceleration is as much as gravity provides.
After that, the ladder simply falls under gravity.
Let us begin improving the model by using Galileo's law on Dad.
We will ignore the mass of the ladder and its motion, except for it keeping Dad in place while the ladder rests against the wall.
If we let t=0 at the time when the tip of the ladder reaches the acceleration -32 ft/sec2, the equation for the position of the tip after that time is
at t=0. This formula neglects air friction on Dad and the ladder.
. How much is g? How much is y[0]? How much is
?
A more extensive project would be to revise this model to take the effect of air friction into account. The corresponding computations for the previous exercise become more involved, but you should be able to write down the model in terms of a differential equation with some initial conditions after you work the air resistance project in the next section. We encourage you to explore this extension of the project if it interests you.
23.1
If we take t=0 at the time when the Dad's acceleration at the tip of the ladder is -32 and account for air resistance in the form of the next project, the differential equations describing his further descent are
A reasonable guess at b in these units is b=0.2. With this information, you can use the AccDEsol program to solve the initial value problem with your values of y0 and v0.
How much slower is Dad going because of air resistance?
Why does the air resistance have so little effect in this case?
where y[0]=y0 is the height on the wall where the acceleration of -32 is achieved, and v[0]=v0 is the velocity at this point.
The constant b depends on Dad's aerodynamics and would need to be measured.
with the position y0 and velocity v0 (be careful with signs, this should be negative) of Dad at the point where the ladder reaches an acceleration of -32. Run time to t=tfinal so that y[tfinal]=0, the ground.