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!

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

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

@@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.

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.

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

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

I am glad you appreciate it!

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

True

Thanks. I am particularly proud of this video.

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

@@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!

Engineers be like. LoL.

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

@@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!

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

That is exactly how I feel about it.

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.

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.

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!

Thank you that helps !!!!

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))

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