I am so happy I found your channel. I like how you explain the problems and the visual aspect of it, as I am a visual learner. Question: How would I determine the change in the kinetic energy of block A as it moves from the initial position to the final position(before the pulley) (for a given distance)? Thank you
The tension in the string is responsible for transferring force from one object to another. The tension acts in opposite directions at opposite ends of string segments. ie. if the tension acts up at the hanging block, it will act down at the pulley.
Let's say we were given a theta and the tension between the string, friction between m1 and the incline. M1 is accelerating up the incline and we are given m1. How would we solve for m2's mass?
sir , T1 and T2 you placed on the FBD of pulley , would be more approriate if called normal reaction ??, as T1 and T2 are components of the normal reaction due to the rope on pulley?
the vector sum of T1 and T2 should be exerted by the pulley bracket up and to the right, thus making the net translational force on the pulley zero. I left that out because it seemed like a digression in the lesson. additionally, note: T1 and T2 are not acting normal to the pulley, they are acting tangent to the pulley.
@@INTEGRALPHYSICS hi sir , the tension is developed on the rope only right, and the force b/w the pulley and rope is the normal reaction ??? , how do you directly take T1 and T2 forces on pulleys fbd , or do you draw the FBD of pulley ( including the rope which is in the groove of it )
Thank you so much that was so helpful, however the final acceleration that you found is the acceleration of the system pulley and blocks, if i need only the acceleration of the pulley i should multiply this acceleration by the radius of the pulley right? Secondly the acceleration will be directly equal to the acceleration of M1 and M2, if so , I can say M1 and M2 have the same acceleration. Please kindly correct me if I am wrong.
To find the angular acceleration of the pulley, divide the linear acceleration we calculated by the radius of the pulley. Both masses accelerate at the same rate.
I arbitrarily set up this problem so that the downward motion of M2 is positive. However be careful, one of the things that causes many people to get this problem wrong is not being consistent with sign conventions throughout the entire problem.
Im in university phyics and Im having a lot of trouble doing phyiscs problems. I am quite good at math including calculus and liniar algebra but can't wrap my head around physics questoins for whatever reason. Do you have any advice?
People often attribute the difficulty they have in physics to math; but I see LOTS of people who are VERY good at math struggle in physics. The hard part of physics is that each problem is a riddle in which has to be solved using a very short list of seemingly simple equations. Understanding the application and limitations of those equations (eg. F=ma) is tremendously important, but something many students and instructors overlook. Additionally I find many instructors take the 'logic' or 'intuition' necessary to solve a problem for granted. When you are just learning physics the difference between an obvious concept (eg. stuff speeds up as it falls downward) and a less obvious extension of that concept (eg. stuff has a vertical velocity of zero at its maximum height) can make students suddenly feel like they missed something.
I am so happy I found your channel. I like how you explain the problems and the visual aspect of it, as I am a visual learner. Question: How would I determine the change in the kinetic energy of block A as it moves from the initial position to the final position(before the pulley) (for a given distance)? Thank you
Thanks. The change in KE is going to be 1/2mv^2 provided you have solved for the final velocity.
Why is the toque caused by the tension and not the force of gravity on the masses and why is the tension pointing downwards
The tension in the string is responsible for transferring force from one object to another. The tension acts in opposite directions at opposite ends of string segments. ie. if the tension acts up at the hanging block, it will act down at the pulley.
Let's say we were given a theta and the tension between the string, friction between m1 and the incline. M1 is accelerating up the incline and we are given m1. How would we solve for m2's mass?
look just at m1 to find acceleration, then knowing 'a' and 't' look at m2 to solve for its mass.
If the system starts from rest, does it influence anything?
Nope.
sir , T1 and T2 you placed on the FBD of pulley , would be more approriate if called normal reaction ??, as T1 and T2 are components of the normal reaction due to the rope on pulley?
the vector sum of T1 and T2 should be exerted by the pulley bracket up and to the right, thus making the net translational force on the pulley zero. I left that out because it seemed like a digression in the lesson.
additionally, note: T1 and T2 are not acting normal to the pulley, they are acting tangent to the pulley.
@@INTEGRALPHYSICS hi sir , the tension is developed on the rope only right, and the force b/w the pulley and rope is the normal reaction ??? , how do you directly take T1 and T2 forces on pulleys fbd , or do you draw the FBD of pulley ( including the rope which is in the groove of it )
btw , do u have a telegram / discord chanel , where you discuss problems with subscribers ??
Thank you so much that was so helpful, however the final acceleration that you found is the acceleration of the system pulley and blocks, if i need only the acceleration of the pulley i should multiply this acceleration by the radius of the pulley right? Secondly the acceleration will be directly equal to the acceleration of M1 and M2, if so , I can say M1 and M2 have the same acceleration. Please kindly correct me if I am wrong.
To find the angular acceleration of the pulley, divide the linear acceleration we calculated by the radius of the pulley.
Both masses accelerate at the same rate.
Thank your sir for your quick response ❤️❤️❤️🙏! You saved my upcoming exam!
Great !!!
Thanks !!!
Glad you liked it!
Solve problems when m2 connected with 2nd pulley nd other end of 2nd pulley to fix support???
its in the works. =)
what is the reaction on the pulley?
The reaction by the support holding the pulley is equal (and opposite) to the vector sum of the two string tensions.
Hi, I think T2= m2a+m2g
I arbitrarily set up this problem so that the downward motion of M2 is positive. However be careful, one of the things that causes many people to get this problem wrong is not being consistent with sign conventions throughout the entire problem.
What happened when masses are equal
And when friction included
Im in university phyics and Im having a lot of trouble doing phyiscs problems. I am quite good at math including calculus and liniar algebra but can't wrap my head around physics questoins for whatever reason. Do you have any advice?
People often attribute the difficulty they have in physics to math; but I see LOTS of people who are VERY good at math struggle in physics.
The hard part of physics is that each problem is a riddle in which has to be solved using a very short list of seemingly simple equations. Understanding the application and limitations of those equations (eg. F=ma) is tremendously important, but something many students and instructors overlook.
Additionally I find many instructors take the 'logic' or 'intuition' necessary to solve a problem for granted. When you are just learning physics the difference between an obvious concept (eg. stuff speeds up as it falls downward) and a less obvious extension of that concept (eg. stuff has a vertical velocity of zero at its maximum height) can make students suddenly feel like they missed something.
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That's a lot of love for an Atwood Machine.