Study Blog Part 5: Revision

I managed to catch some of the Ro8 games. Curious vs. Parting was pretty interesting to see (mainly due to the storyline!) and MC vesus Roro was pretty tense as well (especially G4!) but Taeja vesus SK was tense as fuck. I kinda suprised people have been resorting to ling/bling/muta play a lot in TvZ, which kinda suprises me, but it really creates far more interesting games. I don't think I was the only one giving out a cry of excitment when SK ALMOST blew up Taeja's PF. God, that must be a heart wrenching moment for SK

i) Topics covered

Just repetition this week, so I worked on the exercises which were to be handed in. They are due the end of next week, but I managed to get them all done last Friday! Yay!

ii) Time spent/what sessions?

3 x 2 hours of exercise working
1 x 2 hours of miscellanous work (repetition, final polish on exercises)

i) List of exercises
I thought that I would upload some of the exercises that I have been doing which are to be handed in. They are done in Swedish though, but most of the text is of course understandable regardless of language knowledge (assuming that you know how to follow mathematical text!).

ii) Solving problems

1)

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*The task at hand is to find the angle between the vector BA and the plane BCD. Let's draw a picture
*n is the normal vector to the plane BCD. It is defined as:
*Right hand rule gives us the direction
*Sarrus rule gives:

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*Let's now draw a picture with only n and BA:
*We want to find the angle theta. The scalar product gives:

2)

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* We'll calculate the center of mass of the subareas and then we add the center of mass for all of the subareas:
*Pappus rule gives:
*The figure can be seen as a circle sector

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*If we take a specific look at A3 we get this figure:
*As A3 = AT - A4 and m = (rho) x A, where (rho) is the same for all bodies in the diagram we can state the following relationship:
*We can find rGT and rG4 with the help of the center of mass for a circle sector

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*rG3 is then described as:
*The center of mass for the whole figure can now be found with the following relationship:

*We can see that:


3)

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*What's the value of P when N = 0?
*We don't have to find N2 or N3, because we are only interested in the rotation around point A

4)

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*Given:
*The truck is at a constant velocity, which means that it is in equilibrium, as the sum of F = sum of M = 0
*No friction is acting in A as the truck has an rear engine design
*What is the maximum value on P? => What is the greatest value of P while the truck is still in equilibrium?

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*1,2 and 4 into 3 gives:

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*We now plug in the values of our variables in our formula:
*If we use a calculator we get:
*We can also see that the dimension inside of the brackets must be 1, as [m]/[m] = 1.
*We can therefore conclude that if mg is a force than P must also be a force.

Hopefully, the pictures aren't to blurry. If you want me to take better pictures, I can reupload them again. At the moment, I ain't motivated enough to do so, as I do have the originals on paper

i) What topics?

Newton's laws have been scheduled for next week, so probably will mostly be about practical applications with Newton's laws.