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
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...
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
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
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.
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.
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.
Note how the force mg is perpendicular to the horizontal and the component mg cos(theta) is perpendicular to the incline. Thus the angle between those two forces must be equal to the angle of the incline.
@@Physicssimplifiedbyshivam To figure out which one should be sine, and which one should be cosine, one trick is to look at the limiting cases. Consider what happens when theta = 0, and when theta = 90 degrees. When theta = 0, sine = 0, and cosine equals its maximum value for real number inputs (i.e. 1). By inspection, determine whether the quantity you are trying to find should diminish to zero, or should equal its maximum value, when theta equals zero. If the former, then it is sine (or less likely, tangent). If the latter, then it is cosine. To determine if it should be sine or tangent, look at what happens when theta = 90 degrees. Does the quantity approach infinity (in which case it should be tangent), or does the quantity approach a maximum finite value (in which case it should be sine).
If you are referring to the acceleration calculation, the only two forces affecting the acceleration of the SYSTEM is the m2g force aiding the acceleration and mgsin(theta) opposing the acceleration. mgcos(theta) acts perpendicular to the motion and thus has no impact to the acceleration
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
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
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
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: ruclips.net/video/dEAtTpIG8AM/видео.html and ruclips.net/video/XHkDJwQ4Xng/видео.html
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?
@@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?
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.
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.
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.
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.
Do you have any tips for making the equations on the exam? Im often confused and not sure how to create the proper formula in fear that I possibly mix up some random values
The best way to solve this (and many other problems in physics) is to: 1) Make a good diagram and draw all the forces. 2) In this case start with Fnet = M total x a 3) Determine the assumed direction of the acceleration 4) F net = forces aiding the acceleration - forces opposing the acceleration. Hope that helps.
Draw a triangle where mg is the hypotenuse, and mg cos(theta) is the adjacent side and then move the mg sin(theta) to be the opposite side. Then use sin (theta) = opposite side / hypotenuse and cos (theta) = adjacent side / hypotenuse
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
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).
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.
@@MichelvanBiezen thank you much for answering.. So, if we could provide an amount of force that is equivalent to the weight of the hanging box (by pulling for example), would the one on the slope accelerate faster?? (only one mass to be devided by)
@@MichelvanBiezen thank you so much for answering.. So, if we could provide an amount of force that is equivalent to the weight of the hanging box (by pulling for example), would the one on the slope accelerate faster?? (only one mass to be devided by)
Yes, if a force equivalent to the weight of the hanging mass was applied to the mass on the slope only, the acceleration would be faster since the force would only accelerate the one mass. (A very good question)
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.
The mg vector is perpendicular to the horizontal plane and the mg cos (theta) vector is perpenedicular to the slope on the incline, therefore the two angles must be the same.
Tension is internal to the system and does not aid in the acceleration of the system. Only forces external to the system can make the system accelerate.
Good morning sir Here we can also use an alternative method using motion in connected bodies For the acc. of the mass on the inclined plane we can use equation T-mgsintheta=ma For the mass hanging we can use the equation mg-T=ma There after we can solve this Btw loved you explanation ❤️🩹❤️🩹
Historical reasons, and experimental measurements. The value of g happens to be approximately pi^2 in units of meters/second^2, because an early idea in defining the meter, is for the grandfather clock's pendulum to have a length of 1 meter and a swing period of 2 seconds. That way, every turning point of the pendulum will count the seconds. When you look at the theoretical equation for the swing period of a simple pendulum at a small angle amplitude, T=2*pi*sqrt(L/g), and if you set T=2 seconds, and L=1 meter, solving for g gives us g=pi^2 meters/second^2. The meter has evolved through several different standards to define it, but its approximate size has been preserved over its evolution. The next definition was based on the length of the longitude line from the equator to the north pole, such that it would equal exactly 10000 km in length. There was also a standard meter stick that was preserved under controlled conditions in a vault in France to reflect this definition, just as there had been a standard prototype kilogram stored in the same vault that (until recently) defined the kilogram. Now, the meter is defined from the speed of light in a vacuum, and the second's definition from the atomic clock.
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.
+Alan Ng No, I don't have an example and I can't think of where to look. However there is just a few ways in which this can be handled: 1) You are given the friction force of the pulley. Then you multiply that force times the distance traveled by the cable and you have the work done to overcome friction. 2) Sometimes they simply tell you how much energy is lost due to friction. 3) Or they have something pushing against the pulley causing friction. Then you must use F * mu * distance of the point on the pulley traveled.
In case of friction, tension in rope both sides the pulley is different. We need to calculate the tension ratio (higher/lower) by equating to e to the power (mu) (theta) and relate two tension values. Finslly using these tension values in FBDs can solve eqns simultaneously.
You would need to be given one more piece of information, such as the acceleration. Then you would use the exact same equation and same approach and solve the equation for m2
+jessica llain The two blocks are connected. They will have the same acceleration. (There are videos in the playlist that show just one mass on the incline)
This was posted 11 years ago and it is still top tier teaching. Thank you for the assistance.
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
Want to give props, this video gave me a second wind to deal with physics
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?
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
It is important, that a teacher understands what he is talking about, so this guy does
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. 🙂
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
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! 🙂
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
Much clearer and more concise than my textbook. Thank you for uploading this.
You are welcome. Glad you found our videos.
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.
Sir ..you taught me perfectly..
Thanking you equally the perfection of this leecture
.god bless you sir.
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.
Thanks so much. Explained way better than my text book. I feel so much better watching this.
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 😂
This was absolutely amazing!!! Thank you so much. You are such an amazing teacher!
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.
my left ear loved this, ty
Yes, our older videos were recorded in mono sound (before we figured out what we were doing). 🙂
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
This saved me a few hours, ima go back to work now :)
Glad you found our videos.
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.
I would like to thank you and my left ear for your efforts.
Our pleasure!
At 3:30, how are the cosine and sine assigned the way they are? What is he reason?
Note how the force mg is perpendicular to the horizontal and the component mg cos(theta) is perpendicular to the incline. Thus the angle between those two forces must be equal to the angle of the incline.
Thanks for this question shaheer....
@@MichelvanBiezen thanks for answering sir
@@Physicssimplifiedbyshivam To figure out which one should be sine, and which one should be cosine, one trick is to look at the limiting cases. Consider what happens when theta = 0, and when theta = 90 degrees.
When theta = 0, sine = 0, and cosine equals its maximum value for real number inputs (i.e. 1). By inspection, determine whether the quantity you are trying to find should diminish to zero, or should equal its maximum value, when theta equals zero. If the former, then it is sine (or less likely, tangent). If the latter, then it is cosine.
To determine if it should be sine or tangent, look at what happens when theta = 90 degrees. Does the quantity approach infinity (in which case it should be tangent), or does the quantity approach a maximum finite value (in which case it should be sine).
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
Good resource , really helped.
shout out to this guy for teaching me in under 10 mins
Glad you found our videos and you found them helpful!
5:12 what? why didnt you include m1gcos? and why did you include tension?
If you are referring to the acceleration calculation, the only two forces affecting the acceleration of the SYSTEM is the m2g force aiding the acceleration and mgsin(theta) opposing the acceleration. mgcos(theta) acts perpendicular to the motion and thus has no impact to the acceleration
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
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
AP PHYSICS 1 EXAM TUESDAY, THIS IS SAVING ME!!! THANK YOU!!!
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.
what about the tension in the rope??
You saved me on my test. Thank you!!!!
Glad I could help!
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
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: ruclips.net/video/dEAtTpIG8AM/видео.html and ruclips.net/video/XHkDJwQ4Xng/видео.html
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.
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?
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?
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
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 😊
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.
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.
Do you have any tips for making the equations on the exam? Im often confused and not sure how to create the proper formula in fear that I possibly mix up some random values
The best way to solve this (and many other problems in physics) is to: 1) Make a good diagram and draw all the forces. 2) In this case start with Fnet = M total x a 3) Determine the assumed direction of the acceleration 4) F net = forces aiding the acceleration - forces opposing the acceleration. Hope that helps.
why is it only external forces on the system? why isn't the force of tension in the rope considered when calculating Fnet ?
sir your videos are awesome
where is your Institue situated?
We live and teach in the Los Angeles area.
Hey sir,can you please help me understand how we got mgcostheta and mgsintheta ,thank you
Draw a triangle where mg is the hypotenuse, and mg cos(theta) is the adjacent side and then move the mg sin(theta) to be the opposite side. Then use sin (theta) = opposite side / hypotenuse and cos (theta) = adjacent side / hypotenuse
Also note that the angle of the incline equals the angle between mg and mg cos(theta)
Your videos have been super helpful. Thank you!
Thank you. Glad you found our videos. 🙂
My left ear has learned very much
Sir does the tension force apply here?
Since the tension force is internal to the system, we don't have to consider it when using this method to solve the problem.
Very nicely done. Simple and clean cut. Great job!
-Thanks
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
excellent ! very well explained.
So this is what Gru does in his spare time.
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.
Good explanation thank u sir
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.
How would you calculate the Tension by using the inclined side?
Please, I really need an answer. Is it the still the same regardless of the angle? Is it M1g+M1a?
T = m1gsin (theta) + m1a
Without friction it should be the same on both sides of the pulley
Michel van Biezen Thank you sir, You the real MVP. :)
Thank you very much! You really helped me in understanding how to approach a problem like this!
Glad you like our method.
sir because of u I toped in physics in my grade
Professor you are a life saver!
This channel is a Holy grail!!! I love this channel
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 😄
Where does the tension force goo??
The tension is internal to the system and therefore does not contribute to its acceleration.
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
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!
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.
My left ear found this really useful
Thanks.
Why do we add up the 2 masses??
Because both masses move together as a single system. Thus the net force acts on both masses.
@@MichelvanBiezen thank you much for answering..
So, if we could provide an amount of force that is equivalent to the weight of the hanging box (by pulling for example), would the one on the slope accelerate faster?? (only one mass to be devided by)
@@MichelvanBiezen thank you so much for answering..
So, if we could provide an amount of force that is equivalent to the weight of the hanging box (by pulling for example), would the one on the slope accelerate faster?? (only one mass to be devided by)
Yes, if a force equivalent to the weight of the hanging mass was applied to the mass on the slope only, the acceleration would be faster since the force would only accelerate the one mass. (A very good question)
@@MichelvanBiezen
Thank you so much
What about the tension force?
If you consider the whole system at once (as shown in the video), you don't have to consider internal forces such as the tension between masses.
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!
why is the angle 30° the same?
The mg vector is perpendicular to the horizontal plane and the mg cos (theta) vector is perpenedicular to the slope on the incline, therefore the two angles must be the same.
Awesome
Glad you liked it.
very helpful, actually understood this
why didn't you include tension?
Tension is internal to the system and does not aid in the acceleration of the system. Only forces external to the system can make the system accelerate.
BRUH this guy is a hero
Good morning sir
Here we can also use an alternative method using motion in connected bodies
For the acc. of the mass on the inclined plane we can use equation
T-mgsintheta=ma
For the mass hanging we can use the equation
mg-T=ma
There after we can solve this
Btw loved you explanation ❤️🩹❤️🩹
Did you try it and you got the same answer?
Thank you I understand you more than my professor. Thank you!
Can I get an explanation of why g equaled 9.8?
It is a measured value. You can learn all about gravity from the videos in this playlist: PHYSICS 18 GRAVITY
Historical reasons, and experimental measurements. The value of g happens to be approximately pi^2 in units of meters/second^2, because an early idea in defining the meter, is for the grandfather clock's pendulum to have a length of 1 meter and a swing period of 2 seconds. That way, every turning point of the pendulum will count the seconds. When you look at the theoretical equation for the swing period of a simple pendulum at a small angle amplitude, T=2*pi*sqrt(L/g), and if you set T=2 seconds, and L=1 meter, solving for g gives us g=pi^2 meters/second^2.
The meter has evolved through several different standards to define it, but its approximate size has been preserved over its evolution. The next definition was based on the length of the longitude line from the equator to the north pole, such that it would equal exactly 10000 km in length. There was also a standard meter stick that was preserved under controlled conditions in a vault in France to reflect this definition, just as there had been a standard prototype kilogram stored in the same vault that (until recently) defined the kilogram. Now, the meter is defined from the speed of light in a vacuum, and the second's definition from the atomic clock.
Why is Tension force not included in the system?
The tension is an internal force and has no impact on the acceleration. But when using free body diagrams to solve the problem, they will be needed.
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.
Such a clean explanation..Thanks a lot!
What if you are not given initial angle?
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.
hello. how to solve the tension?
Once you have calculated the acceleration, then use the hanging mass: T = m2g - m2a
thanks a lot!!! you have made it very clear
Glad it helped!
Thank you for all the help, appreciate it 👍
We are glad you found our videos and that you find them helpful! 🙂
Sir how to find the tension of the same problem
The tension = M2 g = M2 a where a is the acceleration of the system
@@MichelvanBiezen M2g - M2a
... Oki i get it thank you sir
Thank you so much for making this video !!
you are an angel. thank you so much
You're welcome!
What if the pulley have fricction?
+Michel van Biezen is there any example or document that i can read?
+Alan Ng No, I don't have an example and I can't think of where to look. However there is just a few ways in which this can be handled: 1) You are given the friction force of the pulley. Then you multiply that force times the distance traveled by the cable and you have the work done to overcome friction. 2) Sometimes they simply tell you how much energy is lost due to friction. 3) Or they have something pushing against the pulley causing friction. Then you must use F * mu * distance of the point on the pulley traveled.
In case of friction, tension in rope both sides the pulley is different. We need to calculate the tension ratio (higher/lower) by equating to e to the power (mu) (theta) and relate two tension values. Finslly using these tension values in FBDs can solve eqns simultaneously.
@@MichelvanBiezen sir, is the method to solve with friction suggested by me is correct or otherwise.
How do I solve for the m2 when only given theta=30° and m1 is 20kg?
You would need to be given one more piece of information, such as the acceleration. Then you would use the exact same equation and same approach and solve the equation for m2
Michel van Biezen that’s exactly what I was thinking but my professor only gave those 2 for the given. Thank you!
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))
Fantastic teacher!
excellent sir
No tension?????
We have lots of examples that also calcualte the tension in the string.
Thank you, now everything makes more sense!
Sir awesome explanation
So helpful! Thank you so much
What if you're only asked for the acceleration of mass_1 ?
+jessica llain
The two blocks are connected. They will have the same acceleration.
(There are videos in the playlist that show just one mass on the incline)
+jessica llain Well, in this problem it would still be the acceleration of the whole system. 3.12m/s^2