Physics - Pully on an Incline (1 of 2) Frictionless
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- เผยแพร่เมื่อ 20 ม.ค. 2013
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This lecture will cover Newton's Second Law: F = ma.
Problem Text:
A 5kg mass is placed on a frictionless incline making an angle of 30 degrees with the horizontal. A rope is attached and positioned over a pulley at the top of the incline. A 6kg mass is suspended from the free end. What is the acceleration of the system.
My left ear now understands physics.
Yes, our older videos had sound problems. We have fixed that for our newer videos.
I thought my headphones were damaged
Michel van Biezen thank you very much sir. I was able to get a full 100% in physics after I understood this video. Thank you sir.
hahaha
What about right
Luciano,
You only need to consider the tension in the rope if you are solving the problem using free body diagrams.
SO THE ANSWER WILL STILL BE THE SAME IF U USE FREE BODY DIAGRAMS
I would liket to thank you for the time that you put into these videos. I am attending MSU Bozeman and the instructor is no where as good as you, I have a lot better understanding of tension and these friction because of your videos. Thank you taking your time to make these.
Being able to solve these problems and the ecstatic feeling of when I am able to figure them out, with a little help from you and your videos, reminds me of why I am studying physics to begin with. Thank you so much sir
You have found the joy of learning and understanding. (It is like an explorer finding unexplored territory. ) 🙂
which grade?
This was absolutely amazing!!! Thank you so much. You are such an amazing teacher!
you're videos are great!! you explain things very clearly and you have helped me understand everything i've needed to know in my physics class :~) wish you were my class teacher
Sir ..you taught me perfectly..
Thanking you equally the perfection of this leecture
.god bless you sir.
I like when the teacher does the math at that very moment, because that is how the student will do... Good quality videos as ALWAYS Michel ! Please, never stop this amazing work and contribution to all physics lovers...
Thank you. We hope to keep going as long as we can. 🙂
Man you are too good, whatever i've been looking for you had each and everything thanks to you i'm doing good. Thanks you so much
6:36 this is a rare phenamenon called schrödinger's physics professor, when the professor actually reaches quantum state
haha lol
Hahaha
😂😂
i thought he was doing the doctor strange thing where he looks into multiple futures to find the answer
Very nicely done. Simple and clean cut. Great job!
-Thanks
It is important, that a teacher understands what he is talking about, so this guy does
Thanks so much. Explained way better than my text book. I feel so much better watching this.
i have a really bad univierity physiccs proffeser and this really clears up a lot of things, well done going over the formulas and how they interact with each part of the question, i will be explaining this to my friends shortly
Great! Glad you found our videos! 🙂
So glad I came across your video sir .After hours of searching and trying to make sense of things I'm finally able to solve such problems through this video.thank you so much sir
Glad you found us. We have thousands of videos on physics covering every topic.
This channel is a Holy grail!!! I love this channel
Good resource , really helped.
AP PHYSICS 1 EXAM TUESDAY, THIS IS SAVING ME!!! THANK YOU!!!
Still cant believe this lectures took place 10yrs ago
Im in 9th and thanks for helping me for IIT-JEE prep
Love from India❤❤
Your lectures are so good can understand literally every thing and i find a meaning in it and it is not at all boring.
Physics doesn't age. 🙂 Glad you found our videos and you find them helpful in preparing you for the JEE
Much clearer and more concise than my textbook. Thank you for uploading this.
You are welcome. Glad you found our videos.
excellent ! very well explained.
Thank you so much for making this video !!
Thank you I understand you more than my professor. Thank you!
Professor you are a life saver!
Thank you, now everything makes more sense!
Such a clean explanation..Thanks a lot!
very helpful, actually understood this
So helpful! Thank you so much
thanks for this video. really a big help.:)
My left ear enjoyed this. good vid though
+Giant Kannen weird, my right ear enjoyed this, but not my left :/
THUNGUNSPRO *X* you must have worn your left side of your earplug on your right ear lol 😂
Murathan,
Good luck on the test. Let me know how you did.
I know its so late but i deleted my old account and didnt see your reply.It was 3 years ago and i was going to take the test of physics in pullies e.t.c and i was so hopeless about the test then i fell into a rabbit hole and found these videos.And i was able to answer all the question right.I got A+ from the test thanks to your videos.Right now i am last grade in high school and in 70 days i am going to take university test in Math Physics Chemistry and Biology and once again i am watching your Torque and Harmonic Movement videos.Thank you sir.For making these videos.
Fantastic teacher!
Good explanation thank u sir
why is it only external forces on the system? why isn't the force of tension in the rope considered when calculating Fnet ?
This saved me a few hours, ima go back to work now :)
Glad you found our videos.
Your videos have been super helpful. Thank you!
Thank you. Glad you found our videos. 🙂
Okay, I am subscribing. Thank you.
I would like to thank you and my left ear for your efforts.
Our pleasure!
Thank you for all the help, appreciate it 👍
We are glad you found our videos and that you find them helpful! 🙂
This was so clear and made this kind of problem easy to understand. So Thank You! also cute bow btw ;)
Dania,
Thank you for the positive feedback. Much appreciated.
thanks a lot!!! you have made it very clear
Glad it helped!
Thank you!! Using this problem, how could I then find tension in the string?
The tension in the string can be calculated as follows:
T = m2g - m2a
You are a life saver man.
excellent sir
Thank u so much sir....Ur videos made me love physics.......😚😃😃😃
Thank you for writing. It is good to see that gaining understanding can lead to appreciation of the sciences.
Life saving video!
Oh thanks a lot it helped me in my exam
How would you solve for the hanging mass, if that is what the question asks for. I was given the Incline plane angle, mass and coefficient of friction, and the mass moved at a constant speed down the ramp?
You use the exact same equation and set the acceleration equal to zero. Then you'll solve for the hanging mass.
Sir awesome explanation
Hey sir u made my day
My left ear has learned very much
Splendid... magnetizing
Why can you subtract m1 g sin theta from m2g, even though they are not in the same direction?
When you add or subtract forces (which are vectors) you must subtract or add the x and y component of the vectors. (see the playlist on vectors) PHYSICS 1 VECTORS
sir because of u I toped in physics in my grade
what about the tension in the rope??
thank you for this awesome video. but still i have a question sir. when you got the Fnet how come did you
disregard Ft? why did you just use
m2g-m1g sin - m1g cos(mu) / Mtotal?
+Jay Cho Not sure what you mean with Ft. Are you referring to the tension in the string? Since the tension is an internal force, we don't have to consider it when we look at the whole system. Then we only need to consider external forces to the system. When you draw a free body diagram, then you do need to consider Ft.
thanks Dr :)
Two blocks with masses M1= 5 kg and M2= 15 kg are connected by a rope as shown. the horizontal surface has no friction and the pulley is weightless. If the block starts from rest , What is the speed of M2 if it has fallen 6 meters.
can someone answer this? please 😊
thanks teacher !!
What if you are not given initial angle?
You make it look so easy! I don't what happens to me when I start writing tests though... 😂
It comes down to recognizing the type of problem you have (what concepts are at play here), then to know the equations and concepts needed, and to know the steps to follow (as illustrated in the videos).
Michel van Biezen
Thanks for your reply! I'll do that 😄
Siryour explanations are excellent but please may you brighten oh bring the camera close to the written problem 😢
Yes, our older videos needed more light. We fixed that in the newer videos.
You saved me on my test. Thank you!!!!
Glad I could help!
I have a confusion that at 3:08 he breaks mg into its component then he breaks it in a format of x axis and y axis and I have studied that x axis is cos theta and y axis is sin theta then how come he wrote cos theta on y component because if u see then it will be y component of mg and should be sin but he wrote cos ,can anyone tell me why?
Don't automatically assume that the y-component is associated with sin and the x-component is associated with cos. It depends on the situation. USE: sin = opposite / hypotenuse cos = adjacent / hypotenuse
You are the best!
how do you know that the force that is perpendicular to the incline is cos and the one that is parallel is sin?
I always recommend that we use the definition: Cos (theta) = adjacent side / hypotenuse Sin (theta) = opposite / hypotenuse. Then substitute from the triangle you have drawn.
Thank You so much.
Thanks alot!!!!
shout out to this guy for teaching me in under 10 mins
Glad you found our videos and you found them helpful!
Thanks so much!
BRUH this guy is a hero
my left ear loved this, ty
Yes, our older videos were recorded in mono sound (before we figured out what we were doing). 🙂
What if you wanted to calculate the velocity of m2, how would you do that?
Since they are connected, the velocities are the same (magnitude) Just in the different direction.
so in all of these problems, its frictionless. how would you incorporate friction?
There are lots of examples in the playlists PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
you are an angel. thank you so much
You're welcome!
what would happen to the forces if the string wasnt parallel to the surface that the block is on?
Then you would have to decompose the tension into a parallel and vertical component
So this is what Gru does in his spare time.
Vinaka Vakalevu (thank you _in fijian)
life saver
Thank you so much!!!
You're welcome!
sir your videos are awesome
where is your Institue situated?
We live and teach in the Los Angeles area.
How would you solve for a, if the hanging mass is pulling the other mass down the slope instead of up the slope
Would the - sign in front of mgsintheta just turn into a + sign
Assuming m1 was large enough to make that happen, the numerator would become m1g sin(theta) - m2g
Thanks a lot sir
Most welcome
How will you get the moment of inertia of the pulley on this same problem?
We usually ignore friction (since it tends to be insignificant), but if the pulley has mass you must take into account its moment of inertia: see these two chapters: PHYSICS 13 MOMENT OF INERTIA APPLICATIONS and PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 starting at: th-cam.com/video/dEAtTpIG8AM/w-d-xo.html and th-cam.com/video/XHkDJwQ4Xng/w-d-xo.html
Thanks a lot!!!
You are welcome.
is the acceleration for each box the same as the entire system's acceleration? or is it different?
The magnitude of the acceleration of each box is the same as the acceleration of the system. (they are connected)
@@MichelvanBiezen thank you so much!! you are a lifesaver. Follow up question, if the angle was less than 0, would the boxes move in the other direction?
thank you sir. but i didn't get why m1g was not counted on m1,
this is me
Only the component parallel to the incline affects the acceleration of the system.
(m1g sin (theta))
Huh? What just happen?
👍 good job
do i need to worry about tension in the string on these types of questions?
There are 2 ways in which you can calculate the acceleration of the system. The way shown does not require that you know the tension. If you solve the problem using free body diagrams, you will need to include the tension.
right okay, thanks for the help!
My left ear found this really useful
Thanks.
I have a quick question, in this case how do we determine if g is negative or positive?
When used in the equations of kinematics g is a vector and the negative sign indicates that the direction of the vector (in this case the accleration due to gravity) is downward. In this example we use the MAGNITUDE of the vector g. Magnitudes of vectors are always positive and can never be negative.
I see. So now my question is why aren’t we using g as a vector and instead using its magnitude?
Does this same formula work on a flat surface? Would angle theta equal zero?
Yes the equations can be used with theta equal zero, but take a look at the other videos where the surface is flat to see how that is done.
Much thanks!
Ty!
You are welcome.
good man.
But how can we ask for personal help from you on some other areas
We try to answer questions when posed on these comments. But we don't have a lot of time to solve individual problems.
why is 9.8 m/s^2 positive and not negative?
g is a vector quantity and thus has a magnitude and a direction. The direction is given graphically, and here we are using just the magnitude of g in the equation and the magnitude of a vector is always positive.
Sir why you ignoring direction because m2 has force in -y direction while m1 has in -x direction ??? We add forces vectorially plzz answer ???
Notice the arrow drawn to indicate the direction of acceleration of the whole system. In this case (since the pulley redirects the direction of the forces, it makes it much more difficult to use the x-y coordinate system). Also note that the force on m1 is not in the x-direction.
@@MichelvanBiezen so you add these equation without use of coordinate system bcos according to coordinate system force on m2 is in negative y direction
They are 100% valid and using the principle of Newton's second law.
What happens to the tension forces on both blocks?
The tension on the string is internal to the system and does not affect the acceleration of the system. When you draw free body diagrams of each component separately, then you will need to consider the tension.
You great man
Thank you. Glad you like our videos. 🙂