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 Basic Thermodynamics ~ J. Pohl © ~ 2019 www.ThermoSpokenHere.com (A2070-012)

Imagine a retirement home is ablaze and flames have backed two elderly women to the opening of a rear, third story window. A ladder truck has arrived in a back alley but a building obstructs its reach.

To place the ladder at the window, it will be extended nearly vertical to the correct, laser determined, length L, then tipped downward to the angle θ. Lasers will obtain the dimensions (given in the sketch) and a computer will solve the geometry. To provide a test case, use the information of the sketch.

Calculate the length of the length, "L," and its angle of elevation, "θ."
♦  A first step is to label points on an initial sketch. At point A, the ladder touches the obstructing roof. Point B is at the center of the dome of that semi-circular (20ft dia) roof. Once programmed, laser controls of the truck can deal with the geometry.

Angle θ equals the sum of two angles:

θ = angle C0B + angle BOA.

By trigonometry of its triangle, angle C0B is determined in two steps:

0B = (322 + 8 2)1/2 = 33.0 ft

sin(C0B) = 8 / 33, hence angle C0B is 14.0°

Similarly, angle B0A is 17.6° and θ = 14.0° + 17.6° = 31.7°

What length of boom is required for the rescue?
♦  The distance from the base of the ladder to the wall is 52ft, so the length of the ladder is:

L cos( θ) = 52 ft or

L = 52 ft / cos(31.7°) = 61.1 ft.

First we write an implicit vector to the point P (where the women are at the bottom of the window) then we equate this to a second implicit vector to P.

0R + RP = 0Q + QP

We make these vectors as "explicit" as possible (distances are in in feet):

52 I + 30 K = 61.1 e0Q + d eQP

By inspection of the geometry, the unit vector e0Q can be written in terms of theta as:

e0Q = cos(31.7°) I + sin(31.7°) K.

Enter this information into the equation:

52 I + 30 K = 61.1[cos(31.7°) I + sin(31.7°) K ] + d eQP

Our next task is to obtain an explicit form for eQP, which is perpendicular to e0Q. In fact, at the point P, there are two unit vectors perpendicular to e0Q. We write them promptly as:

e1 = sin(31.7°)I - cos(31.7°)K and

e2 = - sin(31.7°)I + cos(31.7°) K

These vectors are written directly from e0Q by swapping the component magnitudes and changing one sign. That e0Q is perpendicular to 1) and 2) is easily proved by the conditions of perpendicularity:

e0Qe1 = 0  and e0Qe1 = 0

We seek ePQ which has a positive I component and a negative K component. When we enter this result into our equation we see that it has two unknowns which are d and D.

52 I + 30 K = D [ cos(31.7°)I + sin(31.7°) K ] + d[ sin(31.7°) I - cos(31.7°) K ]

Collect on I and K to obtain:

I:  52 = D cos(31.7°) + d sin(31.7°)

K: 30 = D sin(31.7°) - d cos(31.7°)

To solve these equations, multiply the first by sin(31.7°) and the second by ( -cos(31.7°)).

52 sin(31.7°) =  D cos(31.7°) sin(31.7°) + d sin2(31.7°)

- 30 cos(31.7°) = - D cos(31.7°) sin(31.7°) + d cos2(31.7°)