Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com () - () |
In competition, each climber will be timed for the performance of two climbs of an expert-level formation having an elevation of 10 meters. The climber will do the first climb wearing a backpack loaded with a mass equal to his or her weight. Once atop the first climb, each contestant will shed the backpack, slide down a "fire-man's pole" then climb the course again. The mass of the first climber is 50 kilograms.
Typically, persons would view this event as one that involved work. In the common meaning of the word, the event does involve "effort" or work. But let us look more closely.
Climb One, Backpack as System: modeled as a BODY. The mass of the backpack (50 kg) is constant; this is stated mathematically in two ways:
![]() | (1) |
Newton's Second Law applies to the system sketch shown.
![]() | (2) |
Regarding notation, the left-of-equality superscript is the vector space chosen: it is Cartesian, the 0XZ plane. The subscript is "BODY;" identifies the model to be applied to the backpack; we will follow its center of mass. Right of equality is ΣF, the sum of forces relevant to the system and its event.
We expand the derivative of momentum to become two terms. Since the mass is constant, the first term is zero. The forces acting on the backpack mass are those of gravity (often called a BODY-Force or force that acts "at a distance") and the external, "applied-by-something" force which we subscript as being of the packs "straps." It is clear now that, the mass, the enveloping pack and a section of the straps are the system - all of it taken to be a BODY.
![]() | (3) |
We assume the lifting of the backpack to proceed quite slowly, with near zero speed.
![]() | (4) |
Our understandings and assumptions regarding "lift of a backpack" (on the back of a climber) are reasonable. Newton's 2'nd law is valid. We conclude "for all instances of the event:"
![]() | (5) |
As our next step we write the forces explicitly.
intend to scalar multiply the entire equation by a differential displacement strap force (Fstrap(t) by its displacement and the force of gravity (Fbyand the force of gravity. forces by ext step is to address the fact that the forces of Eqn (5) are displaced and thereby become work terms. But first let us discuss some common comments about work.
The effect, work of a system, occurs when the integral of the vector product of the force through its displacement is evaluated. Above, regarding the backpack, two forces apply. Inspecting we see that the gravity force is for certain constant over any event. It can be written as a constant times the system mass.
A sketch of the man as a BODY, presented at the right, is greatly exaggerated in size so the forces to be shown. Newton postulated two classes of force. Those that acted "at a distance" and others that acted directly upon the BODY, upon its surface. We rewrite the sum of forces of equation (2) as a single "Force of Gravity" (acting downward) and the sum of forces of contact with surroundings (acting upward) which we label as "Force of Support."
The sketch depicts the man at some arbitrary location during his climb. The actual climb occurs quite slowly and the longer it takes the smaller is the term left-of-equality in Equation (2). The case of the slow climb also corresponds to a minimum of the work.![]() | (3) |
Our next step is to write Equation (3) with the forces expressed in magnitude-direction form (below left). The gravity force has magnitude, mg, and its direction is downward, in the "-K" direction. The magnitude of the "force of support" is |F|, its direction id "K." Also written (below right) is a differential vector displacement of the climb.
![]() | (4) |
Work is the scalar product of a force times its displacement. With the model "BODY (mass at a point)," the mass and all associated forces is the same, identical displacement. Consequently multiplication of the entirety of Equation (3) (above left) by the displacement (above right) is physically valid. That operation (in two steps) produces:
![]() | (5) |
Let's discuss this equation momentarily. Each term right-of-equality represents differential work, each term implies the potential of a force acting through its differential displacement consequent with an event of the BODY. The first term right-of-equality is known. It has a magnitude, "mass times gravity" and the direction of that force is downward, toward Earth. The unit vector, "minus K," quantifies the direction. The second differential of work is non-zero. That work occurs with the surroundings.
With a "climbing event" of finite displacement, upon integration ,that is, the differential terms of Equation (5) become finite. Simple limits of integration for an event are: z = z1 to z1 + H.
![]() | (7) |
In competition, each climber will be timed for the accomplishment of two climbs of an identical, expert-level formation having an elevation of 10 meters. Climb One: The climber will climb wearing a backpack loaded to the equal of his or her weight. Once atop the first climb, contestant will shed their backpacks, slide down a "fire-man's pole," then commence Climb Two: The same formation but without the backpack.
Model each climber and backpack as an Extended BODY. Apply Conservation of Mass and Newton's Second Law, and the Definition of Work. What do these principles tell us?