#2021.1: Solutions that I Got Correctly But Didn't Trust My Guts- Friction on Friction (3/11/2021)

Hello everyone. I am creating a SERIES of note walls for my profile, where every solution I did to problems in Brilliant were correctly done but mismanaged to get the right answer. In other words, I DID NOT TRUST MY GUTS AND I F&@KED UP!!! (Sorry about the language.) Because of this, I am not be able to post my solutions to the problem and I can no longer share my approach but otherwise, do them here.

On this first episode, I want to share my solution to 3/11/2021 Daily Challenge—Friction on Friction.

So, to sum up the question from the daily challenge, what happens to the frictional force of the top book that is piled upon another book when both of their accelerations are doubled?

Here is my mathematical approach of the problem:

Newton's 2nd law of motion, Law of Acceleration, states that the acceleration of the books times their mass is equal to the sum of their net forces. So by plugging in the formula,

Fnet=maF_{net}=m\cdot a we will derived at the following solution:

From the set of equations above, it means that when the acceleration DOUBLES, the frictional force of the books also DOUBLES.

Here concludes one of those frustrating days that "Oh! It's so close yet so far!" moments. This alleviates my frustration and hoping to keep my sanity together (perhaps, I'm overexaggerating on this). Until then, hoping Brilliant.org will allow users who got the wrong answer still post their solutions on the comments section of the problems. Until then... John, signing off! :"D

PS: Please feel free to give a critical feedback regarding my explanation in the comments section below. Whether it be correcting some lacking to my solution/explanation, elaborating more the concepts behind the problem that I solved, or any other else, it will be really beneficial to learn from you guys. I'm willing to hear it from you so I can further improve my explanation skills. Thank you. :"D

PS 2: I thank @Jason Gomez for checking out my mistake on the previous solution. Your valuable response make me understand more about the derivation of formulas like this. If you are curious to the discussions we made in the comments section below, I will post a history note attachment to make all of you, readers, have a sense on what is going on. I highly suggest you read our discussions.

Historical Note 2021.1.1: Previous faulty solution:

You can click on the image to zoom in better. Thanks. ;)

Note by John Bryan Galiza
4 months, 2 weeks ago

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You may write your solution down below a top solution or second top solution if you have got it wrong and feel that your solution has a different ideology compared to other solutions, also

a=Fnetm2=Fnet2m2Fnetma=\dfrac{\frac{F_{net}}{m}}{2} = \frac{F_{net}}{2m} ≠ \frac{2F_{net}}{m}

Jason Gomez - 4 months, 2 weeks ago

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Thank you for the advise regarding posting solutions if I got the answer wrong.

Also, thank you for noticing my mistake in the complex fraction part. Now that you pointed this out, it would now mean that for this solution, if you DOUBLE the mass of the object, you DOUBLE the acceleration, wouldn't it? I do not get then now if how am I supposed to bring a 22 on the frictional force part of the equation. Can you please help me understand this so that I can re-fix my solution? I don't want to mislead anyone who will stop-by with the current information posted. Much appreciated from you. :)

John Bryan Galiza - 4 months, 2 weeks ago

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Oh, by the way. I managed to correct my solution by deriving a new one, which is the new one above. Is that how you derive the 2Fnet2 F_{net} thingy?

John Bryan Galiza - 4 months, 2 weeks ago

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Yes it’s right now :) it can have been done easily by taking Fnet=maF_{net}=ma therefore FnetaF_{net} ∝ a which directly gives the answer

Jason Gomez - 4 months, 2 weeks ago

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@Jason Gomez Ah yes. I could not agree more to that. But I intended to do the step-by-step method so that I could also understand how the solution was derived at the first place. Anyways, thank you for taking your time to look upon my solution. Best regards. :)

John Bryan Galiza - 4 months, 2 weeks ago

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