An incline, 2 masses, and a pulley. What could be more fun?

[@keldonchase4492]

At around the 12:44 mark in this video, the video seems to imply that since we got a negative acceleration for the system, that's how we know we had incorrectly determined the direction of kinetic friction. But after we change the sign of the kinetic friction, we still get a negative acceleration and then we can move on. I don't understand how an acceleration of -5.00390 was an indication that we needed to change the direction of the kinetic friction force yet an acceleration of -0.483741 allowed us to continue. Can someone clarify and help me understand? I'm not seeing it. I'd appreciate it!

[@FlippingPhysics]

Wow. I just looked at that and you are right. That's pretty poor on my part. Sometimes you don't notice something so obvious when you spend ~40 hours making a video like this. Oh well. A better statement would have been that, when we get an acceleration with a magnitude of ~5 m/s^2, which is half of free fall acceleration, that it cannot be correct. There is no way a situation like this would produce an acceleration which is half of free fall, therefore, we need to switch the direction of the force of kinetic friction. Hope that helps!

[@keldonchase4492]

@@FlippingPhysics Thanks for the quick reply, that helps!

[@tinimunson1242]

⁠@@FlippingPhysicswell why can’t it be half of the acceleration of free fall?? Is that acceleration too big or something??

[@fair-unfairfair9687]

@@tinimunson1242 To my understanding, after obtaining -5 m/s^2, it indicates that the block is descending down the incline. Consequently, the direction of friction is incorrect, and we need to reverse it to align with the opposing nature of friction to the motion.

[@DanielLovesPiano]

@@fair-unfairfair9687 wait that makes sense... even if the acceleration were a more reasonable amount, but in the same direction that we determined the force of friction to be in, wouldn't we still have to switch directions? Basically, I feel like the video explanation makes more sense then the 'half of free fall is unrealistic' explanation.

[@shaunworkenaour4250]

I really like the conceptual organization of this video. I've seen too many students erase so much valid work because they realized their force of friction was in the wrong direction. Great work!

[@FlippingPhysics]

I get so frustrated when students erase their work. It happened yesterday that a student erased their entire solution, tried again, showed me their work, and I am pretty sure their original work was much closer to what was correct, however, it was very difficult to tell because I could barely read it.

[@rithvikyagnamurthy6560]

20 minutes of flipping physics, what could be better? No, but seriously the demonstration worked really well, if only I had this video when learning pulleys and inclined planes, but this vid is really good.

[@FlippingPhysics]

Welcome to my Quality Control Team! I look forward to getting your perspective on my future videos.

[@dhritiyagnamurthy742]

you are a fat boi why exactly, funny? yes. little g broken into components hard work hard had.

[@PianosOfGreatness]

Studying for my physics midterm and this is amazing. This is amazing content

[@FlippingPhysics]

I am glad you appreciate it!

[@scienceseeker8377]

How the heck you have comment before a week and then video published before one hour

[@ayanchoudhary044]

True

[@maggiemerkle55]

This is outstanding. You've very nicely used the "holstered eqn" to eliminate background noise from the students' work. I liked the end where you went back to the 5 steps and emphasized the multiple ways forces can be summed. You make it very clear it is up to each student to decide what to do.

[@FlippingPhysics]

Thanks. I am particularly proud of this video.

[@joshuawilliams3221]

The physics works! :) Yes, it is a long video, but I think it is one of the best you have made! It would easily take double the time to do this example in class. This goes over and reviews so many concepts, it is fantastic! I love going back to compare the magnitude of the weights, glad you had the boys guess at first. Thank you!

[@FlippingPhysics]

Thanks. There is a lot in this video. I hope people find it useful!

[@joshuawilliams3221]

@@FlippingPhysics I am definitely going to be using this for AP 1, although I know you had it ear marked for AP C. Also on a side note, I really appreciate you putting the AP Exam FRQ videos on your list at the point students would know enough to do them, VERY helpful!

[@hammadtanoli684]

This approach to a problem is so helpful for students like me who are struggling with online learning. with Do you plan on completing the entire Physics C course material?

[@FlippingPhysics]

My plan is to be done by April of 2022. (I know that’s probably 1 year after what you were hoping for. Sorry.) I am quite a bit of the way through the curriculum at this point though. You can see my progress here. www.flippingphysics.com/ap-physics-c.html

[@hammadtanoli684]

@@FlippingPhysics I was thinking of self-studying Physics C this year after self-studying Physics 1 last year thanks to the enormous help of your Physics 1 videos. I think I will try this year using the APP1 material and your in-class videos. Thank you so much for helping me and my friends!

[@rohanr5150]

Couldn't you determine the direction the pulley would move by just summing the forces without friction? Then after seeing that the system moves in a negative direction, you could then sum them with friction opposing that direction (so friction would act in the positive direction) and get the answer that way? I just feel that it's easier.

[@risingstar9098]

Bro, you are doing a great work... Try out some IIT jee advanced questions, which is the engineering entrance exam in India

[@drfreeman6796]

Engineers be like. LoL.

[@Fuzzy100666]

Love your stuff and I recommend your videos to my students! To be honest, I don't recall ever seeing a problem with a pulley with mass AND friction (I do problems just with mass)....maybe you can make a video of that, if you haven't already done so? Thanks for all you do!

[@blueberrypeels6958]

10:37 I arrived at the equation at this segment however I have -m2gcostheta instead of -m2gsintheta. Is there something I did wrong?

[@arhantbagde8037]

Amazing video .... great efforts

[@RupamKumari-vo6wm]

Please make videos on CALCULUS Please please

[@FlippingPhysics]

Currently I am making videos about calculus based physics. So they have calculus in them....

[@Adumbb]

why do we assume the acceleration in a problem like this would be constant? is it because the kinetic friction is constant

[@eaglekraft687]

Hey I believe there is a mistake. In 19:50 we can see that the Net force in the + direction on both masses is incorrect. I agree with W₁-T. However, there is a mistake with the "-fₖ". If both force of tension and force of kinetic friction are pointing toward the positive side (as shown in the Free Body Diagram in 19:50), then it shouldn't be *T+fₖ*. What you have in the video would only be the case if force of kinetic friction was facing the negative direction of the system (parallel to the board) . However, the Free Body Diagram does support the equation for the Net force in the + direction for mass 2. Is what I am saying correct?

[@tinimunson1242]

Hello Flipping Physics, I really like your channel. But may I ask why do the tension forces in 8:09 cancel one another out? The masses are different, and I think the tension force is greater for a greater mass. The formula for force of tension is mg + ma, it seems like it would depend on acceleration, acceleration due to gravity, and MASS. Please correct me if I am wrong. Still, I love your channel!!

[@FlippingPhysics]

1) There is no equation for the force of tension. The only way to determine the force of tension is to draw a free body diagram and sum the forces. 2) When a string goes over a pulley, as long as the pulley is massless and has a frictionless axle, the force of tension on both sides of the pulley will be the same. 3) Here is an example where the forces of tension on either side of the pulley are _not_ the same: www.flippingphysics.com/2-mass-pulley-torque.html 4) Hope that helps and thanks for the love!

[@tinimunson1242]

Thank you that helps !!!!

[@ayanchoudhary044]

Sir , Please Can you teach Banking of Road concept. 🤔🤗🤔

[@carultch]

The banking angle is θ, above the horizontal. The weight of the car (m*g) acts directly downward. The normal force acts perpendicular to the roadway. N*sin(θ) is radially inward, N*cos(θ) is upward. The traction force (aka static friction) acts parallel to the roadway. Assume an inward friction force. F*sin(θ) acts downward, and F*cos(θ) acts radially inward. Balance forces in the vertical direction: N*cos(θ) - F*sin(θ) - m*g = 0 Balance forces in the horizontal direction, and equate to m*a: N*sin(θ) + F*cos(θ) = m*a Acceleration is centripetal acceleration, which is v^2/r Here are our equations so far, with the N and F terms on the left, and the remaining terms on the right. N*sin(θ) + F*cos(θ) = m*v^2/r N*cos(θ) - F*sin(θ) = m*g When solving for N and F, we get the following solutions: N = m*g*cos(θ) + (m*v^2*sin(θ))/r F = (m*v^2*cos(θ))/r - m*g*sin(θ) Note that F has a maximum magnitude of mu_s*N, which could occur both up along the slope or down along the slope, depending on the specific combination of v, g, r, and θ. In the special case that F = 0, the car is driving at the critical speed where there is zero lateral traction necessary, which is the safest possible speed to drive a banked curve. The equations in that occur in that case are: N = m*g/cos(θ) and v_crit = sqrt(r*g*tan(θ))

[@FlippingPhysics]

Carl, please know I very much appreciate that you have been cogently answering many questions on my videos. It is very helpful. Thank you!

[@CameoGuise]

So even if you get the direction of the force of friction wrong, you can always count on the fact that the final acceleration’s sign will correctly determine direction? Is this because the force of friction is lower than the force causing the acceleration? How could direction of friction be determined in different situations? Awesome videos, love the editing and layout

[@OussamaBezzad]

When it comes to friction, it always resists the motion. So if the object is moving to the right, it'll keep pulling it to the left until the object stops moving. When the objects is at rest, it stops pulling it, friction will never cause motion, it just resists it. Thus the friction force is always less than or equal to the sum of the motor forces applied, which means that both the net force (and acceleration) is always in the same direction of the movement. As a result, I believe we can always rely on the acceleration's sign to determine the right direction of friction. I hope this explanation makes sense.

[@carultch]

In this particular problem, you can check the direction of acceleration by temporarily assuming a frictionless incline and an ideal pulley. Check how m1*g compares to m2*g*sin(theta). If the former is greater, it accelerates up the incline, and if the latter is greater, it accelerates down the incline. With this knowledge, you then revisit the problem knowing which direction to assign friction and positive acceleration.

[@gagglz]

this is art

[@FlippingPhysics]

That is exactly how I feel about it.

[@Nitstar6174]

Answer to this video's title. Your terminal velocity videos is more fun.

[@scienceseeker8377]

Hi

[@broysthgaming3877]

why we cannot using Newton second to find acceleration by converting W₁ in to Fₐ on m₂ in the parallel direction. I mean why should devided by Total mass why not devided by m₁ like we did in our previous problem that has only 1 mass. I hope you understand what I wondered.