In my last post, I translated (using Google Translate) the first page of a Japanese high school physics textbook (here). The translation isn’t great (I wouldn’t use it for teaching), but it is informative.
Here is the imperfect translation of the text above.
Problem Practice – Announcing the Touch Theorem
An object A of mass M is placed on a horizontal floor. An object of mass B is placed on top. Both are stationary. When the magnitude of gravitational acceleration is 1:
- What is the magnitude of the vertical force received by object A from object B?
- What is the magnitude of the vertical force received by object A from the floor?
Underline the word: “stationary”. I will explain why later. Although it is a simple problem in dynamics, by considering such a problem, the power of physics will follow me. The first point is which object I focus on now.
Let’s first focus on A (Figure 2-11 (a)). So, let’s put the power of physics to work on object A with arrows. From the theory of gravity and touch, first of all gravity Mg: the only thing that is touching object A is the floor and object B. The floor supports the object upward. Therefore, we describe the force on the face of A perpendicularly upward. Likewise, because it is receiving force to press down, it is good to go now.
We will explain in detail. Let’s formulate the equation of equilibrium of forces working on A. Please look at Figure 2-11. Since the block is stationary, the downward N1 and Mg are balanced,
N1 = N 2 + Mg ……… (1)
I want to find N 2, but it can not be obtained by ① expression alone (there are two unknowns N 1 and N 2 in one expression).
Therefore, focusing on object B (Fig. 2-11 (b)), focusing on all the objects appearing in the problem sentence in order is a trick to solving physics (in the same way as before Gravity and then the theorem of touch).
Draw the arrows of gravity and normal Forces. Let gravitational force mg and this normal force be N2. This N is the vertical drag force B receives upward from A. The expression of equilibrium of forces working on object B is:
N3 = mg ….. (2)
Furthermore, we use the “law of action and reaction” mentioned earlier to connect equation (1) and equation (2) (Figure 2-12)
A receives force from B (N2)
B receives force from A (N3)
N2 = N3 ……… ③
Therefore, we obtain (2) M of (1) by
N2 = N3 = mg from ②, ③ expression. Since M = M + 3 fg (from equation (1)).
N1 = mg + Mg = (m + M) g – – – – – – – – – – – – – – – – – – (2)
If you consider it, you will be able to improve your view very much. Let’s go to the next question.
Like I said, the translation isn’t good enough to use in class, but I find the following interesting:
- I’ve looked through my textbooks, and can’t find examples of students being treated with this level of intellectual respect until A-level (is it just my textbooks?)
- There is no attempt to capture the reader’s interest with pictures of rollercoasters. It’s straight in with the physics. (I find the chatty tone charming, but that might be an artifact of the translation).
- This is the first of several pages of model answers – each one appearing to build upon the previous one. As I flick through the book, I can’t see anything other than proper ‘old-school’ physics. I believe it’s for year 10 – just forces and motion. I will ask about this when (hopefully) I am introduced to a Japanese physics teacher (thank you Mary).
I am now searching through for the questions at the end of the chapter. There may be another blog to come!