In three previous articles, I described how I deal with all problems of kinematics of the same three basic equations. I also said that this approach encourages students to discuss the options of physical problem in terms of fundamental principles. The purpose of this paper is to show how I continue to focus on basic-principle approach when the laws of Newton. Future articles will demonstrate how I turn to other topics such as dynamic rotation, mechanical energy conservation and static equilibrium.

Because my options are limited by the Equation Editor, I use the rather unusual notation. Much of it is summarized in an article entitled "teaching kinematics". Additional notation introduced for the first time in this article is: 1. an amount is represented by sum (the conditions to be summarized). 2 Pi is represented by FT. 3 Static and kinetic friction coefficients are represented by the United Kingdom and United States, respectively. A typical problem solution cinematic starts with the statement of one or more of the three fundamental equations like this:

...Position to time...Speed time.

...X = Xo + VoT + (AT**2)/2................................V = Vo + AT)

...Equation specific specific libellĂ©sĂ‰quation

Many students who accept my message comment how easily they can solve most of their duties kinematic problems. More important still, many of these students start actually solve the problems of Physics in terms of fundamental principles. Let me show you two examples.

Problem. A block of mass Mj is pushed rightward by a horizontal force **P**. A small block of mass Mi rests on it. Coefficients of static and kinetic friction on all surfaces are the United Kingdom and United States, respectively. If the block at the top of the page does slips on the bottom one, what is the acceleration of the blocks? (b) if the block at the top of the page does not bottom one slip out, what is the maximum **p** can be?

Analysis. We will use a Galilean repository at rest to Earth with horizontal and vertical x axis y axis. If the upper block is not dragging, two blocks have the same Aix = Ajx = acceleration. The upper block is receive (i) just lower block (normal force neither pointing upwards and friction force fi right-pointing). (ii) the weight of the upper block is MiG. The lower block is touch: (i) block higher (normal force nor pointing downwards and frictional force fi pointing Leftwards - what reaction forces to those of the lower block on the upper block); (ii) the floor (normal force Nj and strength friction fj left-pointing). (iii) the weight of this block is MjG. (iv) there is also a **P**, force pushing the lower block on the right. Newton second

................................................Newton's second law

...............................Block i..........................................................Block j

.Sum (Fix) = MiAix....Sum (FIY) = MiAiy...Sum (Fjx) = MjAjx...Sum (Fjy) = MjAjy

.....fi = MiA..................NI - MiG = 0... P - fi - fj = MA...New Jersey - Ni - MjG = 0

the first equation gives neither = MiG. When put us this into the equation y second and then solve for Nj, we find Nj = (Mi + Mj) g. Given that the lower block is rolling around the floor, fj = UkNj = Uk (Mi + Mj) g. After placing it with the x-fi first equation in the second equation x, we

...P-MiA - Uk (Mi + Mj) G = MDY

Therefore

...A = (P - Uk (Mi + Mj) G) /(Mi_+_Mj).

(b) the top block slides when it static friction force reaches fimax = UsNi = UsMiG. With this inserted in the first equation-x, we have Amax = fimax/Mi = UsMiG/Mi = UsG. Finally, after placing the result in the equation for acceleration found in (a) and problems for P, we find

...Pmax = (US + Uk)(Mi_+_Mj) g.

Problem. Coin is sitting on a rotating horizontal platform distance R = 20 cm from the Centre of the platform. If the platform is a T = 1.1 s rotation period, which is the minimum coefficient of friction required to maintain the piece in place?

Analysis. We study the motion of the room. We'll use radial (outward) and the axes of vertical reference (upwards). Since there is no tangential acceleration, so can be ignored. Room is affected just the platform (vertical upward normal inward force and n radial static force of friction f). The weight of the piece is MG... With the second law of Newton in radial and vertical directions, we have

.................................Newton's second law

...SUM(Fr) = MAr......................................................Sum (Fz) = MAz

...-f = M(-V**2)/R.....................................................N MG = 0

Since we want we have minimum coefficient of friction, f = UsminN, which, when combined with radial and vertical, equations yields UsMG = (MV * 2) / r. Thus the minimum coefficient of friction is

...Usmin = (V * 2) / RG.

Speed is related to the period by V = 2 (pi), R/t. insert this and given in equation Usmin values we find that

...Usmin = 0.67.

Two problems I just use examples, major emphasis has been on a basic principle, in this case Newton's second law. If you look at how I teach students to approach constant-acceleration kinematics, you'll see that my approach here is an obvious extension to what I was doing it. I try very hard to keep students from ever "formula hunting". You can find many more examples in the physics Web site help. In the next article, I demonstrate how I approach the problem of work-energy theorem solutions.

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.

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