I just spent hours watching various videos that failed to explain this well, and you did it in 5 minutes. Im not gonna fail my physics exam because of you. Thanks man
He rounded the constant gravitational acceleration (9.8 m/s^2) to (10 m/s^2) for simplicity. And he used the mg*sin(theta) because the "opposite" side of his new triangle is parallel to the slope of the original incline. Hence is equal (parallel to) the force Fx. (the new triangle being the one with a hypotenuse equal to the line represented by -mg). Sin(theta) = opp/hyp. Hyp= -mg. -mg*sin(20) = opp. Opp=FxI hope this helps a little.
Been trying to figure out how to solve these type questions but ended up making all forces unit easy and clear now. Amazing and def recommending this video to friends and subscribing. Keep up the hard work boss!
Thank you, I have a test in about 1 hr, and the professor never really went over this problem, but it is on the practice test. I should be all set now!
this is beautiful. I had no idea how to approach this kind of problem and its going to be on my test tomorrow. After the video, I get it now! And I haven't even watched the other 12 videos. (yet) Thanks so much, I'm definitely subscribing. :)
Other than the fact that I'm pretty sure this video was recorded with a potato.. This is a fantastic demonstration of basic physics on an inclined plane! :)
I was stuck on this question for two days lol, not sure which value went where what was negative and wasn't and then i found this video and got the answer right on the first try without even using my brain because i was so sure I'd get it wrong again but didn't....wow...
he rounded off, FN= mgcos@ , mg for 9kg block is , 88.2N And he rounded off for the entire question for a more definitive answer I guess.. I tripped on the same thing! Haha Hope this helps , two years later...
Him not using 9.8 kept throwing my calculations off. I was wondering why my solutions manual wasn't matching up with my answers. but overall good explanation. cleared up alot of things
I think you should have taken the mass of the whole system (9+15=24) rather than just 9, in case if you want to take nine then you are considering the tension only on m=9 kg which is on the incline plane!!
Assume there was no friction and acceleration was 3.3m in 2.4s in the CW direction and we need to find the mass of the block on the incline? Probably really simple but I can't seem to get it.
that is basically given if it is not given then the reaction force and friction would be given and we will have to find it out by formula Friction=(mu)(reaction force)
Why did you change Fn to 90Ncos(20 degrees) when it should be 88Ncos(20 degrees)? It matters because for tension I got 87.6486N. Fn = Fgcos(20 degrees) = weight1Fgcos(20 degrees) = (9kg)(9.8m/s^2)cos(20 degrees) = 88Ncos(20 degrees) The weight of the first box is 88N.
9kg(9.8m/s^2)(cos20)=the component of the weight of the 9kg block perpendicular to the surface of the ramp which is also equals to the normal force but opposite direction.
Sw499erB0iZ That second triangle he makes can be considered a right triangle because fg is parallel to the far right side of his largest triangle. The hypotenuse of his largest triangle is a transversal that makes the new triangle a right one. 'theta' or however you spell it can be the value 20, so the horizontal, fx force, is equal to 90 (mass of box 9 *force of gravity 10m/s2) sin (20.) it's the sin because the fx is opposite to the angle 20 in the smaller triangle. Sin = op/hyp. Does this make more sense?
+Lollipop123Shana because both masses are connected my a rope/cable that we are assuming doesnt stretch. both accelerations are the same for each mass.
+Agboola Goodness this is probably too late but im pretty sure that you consider your y and x axis'. cos is your y-axis and sin is your x-axis, so finding your Fn= fgcos(angle) as its up and down.
The component adjacent to angle is always cosine Irrespective of direction where it is placed We always think that horizontal component is always cosine and vertical component is always sine Which is not always true!
Yeah, I hate it when people come off as scientific, yet are sloppy with their constants, in this case, the author using 10 m/s^2 instead of the more realistic 9.8 m/s^2. It's almost like they don't care to be precise. Authors should take some personal pride in being precise with science because precision is the backbone of scientific reasoning. I prefer to use 2-decimal place accuracy when writing the force of gravity, ie 9.81 m/s^2. But I will accept 9.8 m/s^2. But 10 m/s^2 is just lazy slop.
force = mass times acceleration. If we ignore air resistance all objects on earth fall with an acceleration of 9.8m/s^2. I think he rounded the value of acceleration up to 10
It is inaccurate and irresponsible to use 10.0 m/s^2 for the force due to gravity. You introduce a 2% error right from the start of your solution process. You should use 9.8 m/s^2 for the force due to gravity. This is very close to the actual value of 9.81m/s^2. Also, you lose credibility with thinking people who expect more accurate values for the force of gravity. You included 2 decimal places for your value for acceleration, why not include 2 decimal places for gravity, ie 9.81m/s^2.
holy shit are u making a big deal of him using 10 m/s^2? its not the end of the world, at the end of the day it made the calculations easier god damn calm ur ass down.
Well, if you don't care about what the scientific answer is, you are not doing science. Your statement is like saying alchemy is equal to chemistry. You had better be concerned about what the scientific answer is because the next time your drive your car, or cook your meal, or fly in a plane, or use anything built by scientists/engineers, you will clearly care. With attitudes like you have, you will guarantee errors in your designs which ultimately hurt human beings. You better care!
+MMM007 The tension is the same throughout the entire rope. if you find the tension in one spot, it will be the same for each spot on the rest of the rope.
+PressY2Satisfy That's not the case in real life, only if the exam paper specifies that the string is light and inextensible - you can make that assumption (like they do in A Level Mechanics).
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I just spent hours watching various videos that failed to explain this well, and you did it in 5 minutes. Im not gonna fail my physics exam because of you. Thanks man
2 years later, how did it go?
Your the man I find this method so easy we were never told we could make it linear !!!
He rounded the constant gravitational acceleration (9.8 m/s^2) to (10 m/s^2) for simplicity. And he used the mg*sin(theta) because the "opposite" side of his new triangle is parallel to the slope of the original incline. Hence is equal (parallel to) the force Fx. (the new triangle being the one with a hypotenuse equal to the line represented by -mg). Sin(theta) = opp/hyp. Hyp= -mg. -mg*sin(20) = opp. Opp=FxI hope this helps a little.
Went from having no idea how these problems worked to a pro in 5 minutes. Thank you Sir.
It's so relieving to see how people had got their problems solved freakinn 7 yrs ago..I warmly welcome those who are here in 2021!
Been trying to figure out how to solve these type questions but ended up making all forces unit easy and clear now. Amazing and def recommending this video to friends and subscribing. Keep up the hard work boss!
This is the most helpful thing I've come across! All the videos are so well explained! Thank you so much!
At first I didn't like how you made it linear, but then I found out how useful it is to find which way Ff is going. Thanks a lot! Great Video!
I say this with the relief of a desperate person with a physics test tomorrow: YOU ARE A LIFESAVER THANK YOU SO MUCH
Thank you, I have a test in about 1 hr, and the professor never really went over this problem, but it is on the practice test. I should be all set now!
wow, thanks for explaining it so clearly. You really clarified things for me!
Thanks for the great video. Much more clarity now.
Your explanation is so sooo much better than my professor. I aced this question on the exam
Thanks ! Clear and to the point!
this is beautiful. I had no idea how to approach this kind of problem and its going to be on my test tomorrow. After the video, I get it now! And I haven't even watched the other 12 videos. (yet) Thanks so much, I'm definitely subscribing. :)
I like the way you explain. Would you consider posting some torque problems? Thanks!
Great simple explanation. I could kinda do this, but this video really clarified things.
You are awesome! Thank you for posting this. :)
Thank you. This was a good review. :)
Great explanation thank you! You have helped me so much with my studies
Tomorrow is my exam in Dynamics and i dont know how would i deal with it but then i saw your vid. Thanks bro very helpful. :)
This Is Great, Really explained it well!
Different way than my professor did, but I like this way better. thanks so much!
Other than the fact that I'm pretty sure this video was recorded with a potato..
This is a fantastic demonstration of basic physics on an inclined plane! :)
This is assuming that physics and math work the same in Canada?
This was very helpful thank you!
Thank you. This helped a lot.
He made it look so easy. Thanks Sir
How that one become 88.6 N ?
I was stuck on this question for two days lol, not sure which value went where what was negative and wasn't and then i found this video and got the answer right on the first try without even using my brain because i was so sure I'd get it wrong again but didn't....wow...
"Linearizing" is innovative. However, why is Fx negative? Is it moving to the right and therefore positive?
How cn you find force fx by using impulse given tha velocity to the left and to the right
Thanks for your explanations,different way to solve it
Good job!
90 sin 20. Where did 90 come from? Do you multiply 9 by 10?
+Chad Hayag probably a little to late but it would seem so. im assuming he is using 10m/s^2 as gravity instead of 9.81m/s^2
he rounded off, FN= mgcos@ ,
mg for 9kg block is , 88.2N
And he rounded off for the entire question for a more definitive answer I guess..
I tripped on the same thing! Haha
Hope this helps , two years later...
How would we have approached this question if the masses were placed the other way around?? i.e (the 15Kg was the one on the incline)
did he just calculate sin in his head???
Tg
Good man.
Do you get paid if I skip the ads?
Very Nice Explanation Sir...
thank you i understand how to find tension now
Awesome.
Him not using 9.8 kept throwing my calculations off. I was wondering why my solutions manual wasn't matching up with my answers. but overall good explanation. cleared up alot of things
What pen are you using
where did you get 90sin from?
I think you should have taken the mass of the whole system (9+15=24) rather than just 9, in case if you want to take nine then you are considering the tension only on m=9 kg which is on the incline plane!!
Assume there was no friction and acceleration was 3.3m in 2.4s in the CW direction and we need to find the mass of the block on the incline? Probably really simple but I can't seem to get it.
Do you have tutorial vid bout impulse
Good one.
What about if you don't have normal force how will you find the acceleration
I solved this for practice. I was confused why my answer was wrong but close enough, turns out he used 10 as gravitational acceleration.
are we are not going to add the friction force?
Why do you use 10m/s^2 instead of 9.8m/s^2 for the acceleration of gravity?
IamArob0t We can use both. Just a .2 difference.
that's not proper acceleration. 9.81m/s^2 is the way to go, unless your on mt. Everest where it is 9.76m/s^2
Its just to make his calculations easier
I really want you to talk about how you got the mu. I really don't get why it's .25. Is it always .25?
that is basically given
if it is not given then the reaction force and friction would be given and we will have to find it out by formula Friction=(mu)(reaction force)
How did you work out that the angle was 20 degrees at 2:14
W mans
I dont understand why Fx and Ff are in the same direction, shouldn't it be opposite?
Why did you change Fn to 90Ncos(20 degrees) when it should be 88Ncos(20 degrees)? It matters because for tension I got 87.6486N.
Fn = Fgcos(20 degrees) = weight1Fgcos(20 degrees) = (9kg)(9.8m/s^2)cos(20 degrees) = 88Ncos(20 degrees)
The weight of the first box is 88N.
How do i find the normal force on the 9kg block?
9kg(9.8m/s^2)(cos20)=the component of the weight of the 9kg block perpendicular to the surface of the ramp which is also equals to the normal force but opposite direction.
thank you so much
Really don't understand how you got 90sin20 for Fx
Sw499erB0iZ That second triangle he makes can be considered a right triangle because fg is parallel to the far right side of his largest triangle. The hypotenuse of his largest triangle is a transversal that makes the new triangle a right one. 'theta' or however you spell it can be the value 20, so the horizontal, fx force, is equal to 90 (mass of box 9 *force of gravity 10m/s2) sin (20.) it's the sin because the fx is opposite to the angle 20 in the smaller triangle. Sin = op/hyp. Does this make more sense?
So I assume you were given coefficient of friction? Or did you assume it was 0.25?
its given
Josh Dougall brrruhhh
Why does m=9 kg for finding tension but m=9 kg + 15 kg when we solve for acceleration?
we don't care about what the scientific answer is, we care about how to do the problem so it doesn't matter
sir why you use mass of 9 kg only while calculating tension while tension is created due to both masses
Sabal Subedi acceleration is down to earth as 9kg is has greater weight than 6kg
How do you find the acceleration of only one mass
+Lollipop123Shana because both masses are connected my a rope/cable that we are assuming doesnt stretch. both accelerations are the same for each mass.
yeah !!!
I think you made a mistake 150- mgsintheta - frcitonal force = 9 a not 24 a
where did you get 90sin from?!!!!!!!!!!!!!!!!!!!!!!!!!1
Ben Feichtinger
hey can you check on your calculator 90sin20? 82.1650725655 is the answer right? but in the guys answer is 30.7N o.o
why did you use w*(90cos20) instead of (90 sin20)
+Agboola Goodness this is probably too late but im pretty sure that you consider your y and x axis'. cos is your y-axis and sin is your x-axis, so finding your Fn= fgcos(angle) as its up and down.
i keep getting 4.01 for the acceleration
TBE!
Thanks
Why is this video in comedy category
🤣🤣🤣🤣seriously.....
ugh this video confused me,... FN should be 90Sin20 I thought...wtf..
its 90cos20
The component adjacent to angle is always cosine
Irrespective of direction where it is placed
We always think that horizontal component is always cosine and vertical component is always sine
Which is not always true!
That is true, but I think he is referring to the weight being changed from 88N to 90N. He rounded the weight.
sin 20 degree*90=30.8 (3 SF)
@@sk851097 yes thats the right formula
Yeah, I hate it when people come off as scientific, yet are sloppy with their constants, in this case, the author using 10 m/s^2 instead of the more realistic 9.8 m/s^2. It's almost like they don't care to be precise. Authors should take some personal pride in being precise with science because precision is the backbone of scientific reasoning. I prefer to use 2-decimal place accuracy when writing the force of gravity, ie 9.81 m/s^2. But I will accept 9.8 m/s^2. But 10 m/s^2 is just lazy slop.
I got 4.0073 m/s/s as my acceleration
why did he multiply 15 kg to 10??? where did 10 come from?
it's g = gravity = 9.81m/s^2 he just rounded off to 10. If you actually use 9.81 (like on a test, the answer will be slightly different)
I ask because you deserve to be paid.
why is fg 15 * 10
force = mass times acceleration. If we ignore air resistance all objects on earth fall with an acceleration of 9.8m/s^2. I think he rounded the value of acceleration up to 10
dont really like how you put it linear
It is inaccurate and irresponsible to use 10.0 m/s^2 for the force due to gravity. You introduce a 2% error right from the start of your solution process. You should use
9.8 m/s^2 for the force due to gravity. This is very close to the actual value of 9.81m/s^2. Also, you lose credibility with thinking people who expect more accurate values for the force of gravity. You included 2 decimal places for your value for acceleration, why not include 2 decimal places for gravity, ie 9.81m/s^2.
holy shit are u making a big deal of him using 10 m/s^2? its not the end of the world, at the end of the day it made the calculations easier god damn calm ur ass down.
Well, if you don't care about what the scientific answer is, you are not doing science. Your statement is like saying alchemy is equal to chemistry. You had better be concerned about what the scientific answer is because the next time your drive your car, or cook your meal, or fly in a plane, or use anything built by scientists/engineers, you will clearly care. With attitudes like you have, you will guarantee errors in your designs which ultimately hurt human beings. You better care!
your saying low depth subject
slow down nobody is chasing you. ugh
What if you wanted to find tension for the other half?
+MMM007 The tension is the same throughout the entire rope. if you find the tension in one spot, it will be the same for each spot on the rest of the rope.
+PressY2Satisfy That's not the case in real life, only if the exam paper specifies that the string is light and inextensible - you can make that assumption (like they do in A Level Mechanics).
MMM007 then simply use free body diagram and calculate tension of each half