How to Solve Inclined Plane Problems
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- เผยแพร่เมื่อ 28 มิ.ย. 2024
- Physics Ninja look at 3 inclined plane problems. 1) Determine the speed at the bottom of the ramp and the time is takes to get to the bottom. 2) Given an initial speed up the ramp find the max distance up the ramp (includes kinetic friction) 3) Find the tension in the rope if the block is in equilibrium.
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Great explanation, teacher I was struggling so much with this kind of problems. The result at the end is not correctly evaluated though, it should be T = 19,67 N
Thank you teacher. For more people like you in the world
Best video I've seen explain this. Should have so many more views.
I had to refresh before my final exams tomorrow and this was godsend. Thank you so much, I respect you sir deeply.
Your explanations are excellent thank you very much.
I've been struggling with this for years lol thank you so much! but really to understand this video, I have to self study about angles first 😅
You got this!
Third perspectives helps. Thanks.
Great explanation sir, keep it up.
Upload more video on physics.
thanks for your motive and go ahead.
can I ask.. how to get the angle if you have a mass 100kg , 5m long inclined plane and P.e of 2450J?
when you calculate acceleration in the first problem why do you use sin35 x 5.36? ive never learned it that way//used it like that with the sin component to find time...
Why is the angle for calculating components x and y 35degree and no 55degree. Still learning this.
When do you use sin vs cos? Thanks.
Using the relationship between Potential and Kinetic energy and solving for "h". "h" being the altitude necessary for block to gain the velocity 15 m/s. Why is equation h = v^2/2g incorrect when solving X = h/sin(35).
guys at 20.56 it was meant to be a + not -. Thats why you get 6.83
yes i thought so too
Fk and Wx are both -ve ,How is it meant to be +ve?
Is X at the front of the block or at the back of the block?
In problem 2 can you explain why the acceleration is down the incline when you have given an initial velocity upward?
slowing down, negative acceleration, or opposite the direction of the velocity. if acceleration is opposite the direction of motion, it indicates it's still moving in the original direction, but slowing down. equivalent to a negative acceleration.
Kinetic force should be opposite to Wx, right?
Bro ,
Plz do on thermodynamics and kinetics theory of gases
No, I didn't mean well. What force is applied to the inclined surface? It means that the force enters the inclined surface from behind and the inclined surface moves
@ 20:00 How did you go from mgsin(theta) "+" mu_kN = max, then substituted in N, mgcos(theta) to somehow get mgsin(theta) "-" mu_k(mgcos(theta)? How does substituting in N suddenly change the expression to "-"? And if I calculate (9.8)(sin35 "+" 0.15cos(35)) I get 6.83... If I use the negative expression you derived after subbing in the N, I get 4.41 m/s. Not sure what's going on here. Please elaborate?
I actually had the same question, but after solving, it was nothing but a typo
bless tysm
Great teaching, but Force is vector quantity and you should stay within the coordinate system you define. Tension should be negative.
Thank you at least I have understood a bit
Ben affleck, my hero!
at 20:01 you wrote (ma = mgsin - umgcos) but on top you have (mgsin + uN = ma), why did it change from addition to subtracting?
Because for this problem it’s sliding up the ramp and slowing down. Friction is in the opposite direction.
@@PhysicsNinja but shouldnt Wx be negative as well since its in the same direction as the friction?
@@rosalinechirishian5006 i think so too
@@rosalinechirishian5006 that's what im saying too, so would the acceleration be negative -6,8m/s^2 ??
@@rosalinechirishian5006 i was also thinking that Wx and Fk both pointing in the negative direction .Hence shud both be -ve
the shelf behind you is stressing me out
at 19:58, shouldn't it be plus, not minus? both forces acting downhill. also, it wouldn't it be more intuitive to make your positive x direction uphill? (direction of motion, and what they are asking for). i realize either way works.
I think I made an error on this problem. Sorry about that, was going too fast.
@@PhysicsNinja no prob, great vids! thanks a lot, really working for me and my son who is about to do the AP exam.
why is the force in the x component downward is it not sliding up
I believe it's because the initial velocity is 15m/s which means initially, the net force is 0 since it is in constant velocity. Although, once it starts to slide up, it accelerates downward, which means the forces in the X direction are pointing downward.
Hello, what force should be applied to the inclined surface so that the mass does not move
the block will not move if Fnet = 0N it is in static equillibrium (velocity and acceleration are 0)
You could of also calculate the time with Ep=Ek and a=v/t
The time to go down the ramp i mean
Since it is also frictionless so mechanical energy conserves
execuse me , but why in problem 2 the wx is going to the left not to the right
I may have made a mistake. Sorry about that
So when calculating time, we need to define a distance; here it ignores size of block 😂
Thank you but it is very very veryyyy veryyyyyy HARD!!
Why is g positive and not negative 9.8?
normally g is negative because it goes in the opposite direction of whatever force is being calculated, here we consider g as positive because we consider the left direction of the hypotenuse as the positive direction, which is why we get a negative for things like our final displacement because thats in the opposite direction of g, so the right direction of the hypotenuse. It doesnt matter though, the signs only mean direction, so you can have negative g and get positive answers., as long as ur directions are consistent g
Isn't Tension = 19.7N in the last question?
My thought to
Same here. He misread the answer and switched the positions of 6 and 9. It is 19.7 N in 3 s.f. (19.69374694 N)
yes even in Problem acceleration should be 4.42m/s2
*Problem 2
last question should be 19.67?
yes
He can't just go on and edit the video. He would have to create an entire new video and upload it to fix the answer. Hopefully, people will be able to realize that it's just a simple case of getting numbers mixed up in your head. I do this all the time, and according to everyone else that I talk to in the calculus/physics world, is something that almost everyone does.
How to make a video like this 😢
In the first question I'm almost certain you got the distance wrong. The distance is essentially the hypotenuse of the plane, and you used the height instead of the calculated distance to evaluate the time it takes to reach the final position.
And why are you calculating the acceleration twice in the kinematic equation? You have sin(35)•(5.62) but that isn't making sense.
Thanks, I’ll fix it
Just checked and first is correct. The sin35 comes from calculating the distance x=h/sin35
Okay, thanks for clarifying!
Does any one get tension 19.67
16:30
The time calculation is clearly wrong
21:00 accelaration is wrong
Expression is right, somehow I entered wrong in calculator.
@@PhysicsNinja We got the same answer sir, Its just that Wx and Ff must add up.
@@PhysicsNinja no
@@kevinbelza8294 I also get the same 6.83 if I enter (9.8)(sin35 "+" .15cos35) Not sure how subbing in N changed it from sin(theta) "+" mu_kN =max turned it into sin(theta) "-" mu_kN = max? How the hell does that happen? I get 4.41 if I use (9.8)(sin35 "-" .15cos35)
@@JosephAMuniz-hm4jh i also got 4.41 using his expression with a "-"
6:21 that isn't a right angle down by the Mg LMAO
There are more errors in this than a Tour de France drug test
wayyy complicated and confusing explanations...
It's understandable
it could be better...way better!@@Juxhats
@@Juxhatshe’s saying that cuz he’s probably in 5th grade
Wrong!
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