i have found in my many college classes that there are 2 types of teachers. teachers like you are the best, they actually want their students to learn and understand. and the other are there to flex their brains so to speak. making thngs way harder than they need to be to appear more impressive.
i just watched all 12 of your torque videos and i just want to say thank you so much!!! you help students like me who struggle so much with certain concepts. i feel so much better about my midterm now, thank you for taking the time to make these!
My goodness... I'm not so good at this critical thinking sometimes, so given this problem to solve even with all of the tools, I wasn't able to "put it together" like a puzzle. Thank you for the succinct walk-through on how to solve these ladder problems
I finished all my homework but on the quiz I finished so poorly because I couldn't understand which numbers to plug into the equation. At 5:15 you explained the equation perfectly. I love how you use "r * f * sin()" instead of " f*perpendicular and whatever". Both ways work but I'm glad I've learned about the other way
Thanks man, feels bad when you say that's alot to do and my physics department is probably laughing at the prospect of putting a problem so simple to them on the exam.
@Ben Fox, thanks for the "best" comment, I think they turned out pretty well but they (static equilibrium videos) have not gotten much traffic yet. Tell all your friends.
So in theory, if we solve for the friction force like we do in the video, but also find the magnitude of the normal force in by summing the forces in the y direction, we could find the value of mu correct?
I'll start off by saying thank you so much for these videos! I'm studying physics with the OU alongside my full-time job and having content like this really helps me understand concepts faster than using my textbooks! A quick question on this though, would there not be a coefficient of friction where the ladder meets the wall? Assuming the ladder is stationary there would be a coefficient of static friction and if the ladder were sliding there would be a coefficient of sliding friction.
Does the ladder not exert a normal force up on the person (equivalent in magnitude but opposite in direction to the force of gravity on the person) so that person doesn't fall straight through the ground? A little confused why this force is not in the free body diagram.
You said that the x distances given are the lever arms, but this doesn’t really make sense to me. With our torque equation where Tp + Tl = Tw, are we not talking about torques acted upon the ladder? Therefore, the ladder should be our lever arm right? However if we treat the x distances as lever arms, this ends up being the torques that the forces produce onto the floor, which doesn’t make sense
I found out how he got it. I would recommend some paper to visualise the setup. Take the force F_n 1.1 meters from our pivot point. Let L be the length of the ladder from the pivot point. The torque there is given by sin(theta)F_n*L. So, lets find L! Now with the paper draw a dotted line from the F_n to the ground. (paper coming in clutch now) Notice how this forms a triangle, where the angle theta is the same as the theta in sin(theta)F_n*L. In this new triangle, 1.1m is the Opposite, while L is the Hypotenuse (zamn I spelt it correct). Since Sin(theta)= Opp/Hyp, it is the same as Sin(theta)= 1.1/L. Heyyy, wait a minute, we now have to expressions with sin(theta). Lets plug in our new expression for Sin(theta) = 1.1/L into our torque equation Torque = sin(theta)F_n*L. That would mean Torque = sin(theta)F_n*L -(note: subbing in sin(theta) as 1.1/L ) Torque = (1.1/L) F_n*L. -(note: L denominator and L cancel) Torque = 1.1*F_n And there you have it! Thats just the perpendicular side times at the Force! Let me know if u didn't understand anything, or want me to draw something up. ps. F_n is F subscript n, hope that didn't confuse you.
At 3:39 when you said, “if these forces were to act by themselves, the would cause the ladder to rotate in the counter-clockwise direction” did you mean clockwise?
Well, i guess that just depends which normal force you get rid of to allow rotation. its only counterclockwise if you get rid of the wall. If yiu getvrid if the ground, it goes clockwise
Hey, actually I think I might have figured it out. When using the formula 'torque= r x F', we substitute the values 'torque= 5.5m x Fw' but if we choose to use the formula 'torque= r x F x sin θ', we substitute the values r= length of ladder, F=Fw , sin θ= sine of angle between the horizontal Fw and ladder or alternatively, sin θ= opposite/hypotenuse= 5.5m/length of ladder, all these into 'torque= r xF x sin θ' and the torques would be the same. Hope this helps!
Alas 6 years later I have the same question, that I just solved using your comment. Thanks. I just wish he had explained it as thoroughly as he usually does.
i have found in my many college classes that there are 2 types of teachers. teachers like you are the best, they actually want their students to learn and understand. and the other are there to flex their brains so to speak. making thngs way harder than they need to be to appear more impressive.
I appreciate your comment. Thanks very much and best wishes to you.
i just watched all 12 of your torque videos and i just want to say thank you so much!!! you help students like me who struggle so much with certain concepts. i feel so much better about my midterm now, thank you for taking the time to make these!
Great, thanks for commenting and I hope the midterm goes well.
wow my physics teacher makes this so much harder than it needs to be
best explanation ever. You helped solved my homework
+Chantel Jose Thanks for the positive comment. You can a listing of all my videos from my website, www.stepbystepscience.com
My goodness... I'm not so good at this critical thinking sometimes, so given this problem to solve even with all of the tools, I wasn't able to "put it together" like a puzzle. Thank you for the succinct walk-through on how to solve these ladder problems
My pleasure, glad to be helpful!
I was just using this to learn ladder examples before my test on momentum and torque and it’s very helpful.
Great to hear! Thanks for watching
I finished all my homework but on the quiz I finished so poorly because I couldn't understand which numbers to plug into the equation. At 5:15 you explained the equation perfectly. I love how you use "r * f * sin()" instead of " f*perpendicular and whatever". Both ways work but I'm glad I've learned about the other way
Great that you found the video explanation to be helpful, thanks for the comment!
I haven’t understood the lever arm thing until just now🤦♀️ I really appreciate your video😭🙏
So glad it was helpful, thanks for your comment
Thanks so much! I've got my Statics final in about 1.5 hours and this helped a bunch.
Great, hope it goes well!
thanks a lot for the clear and concise vid, really helped me through this torque class.
So glad it helped, thanks for the nice comment
Thank you my good sir. Really helpful for recapping for my AP physics test
+InfamousReaper Great, hope the test went well. You can see a listing of all my videos at my website, www.stepbystepscience.com
Amazing lesson. I have a physics final in 8 hours and i learned more from your videos than 3 months of lectures. God Bless!!
Hope the final went well, thanks for the great comment!
Yep, this was the video that made me understand after watching multiple lectures and other videos. Thanks!
You are welcome and thanks for letting me know.
Thanks man, feels bad when you say that's alot to do and my physics department is probably laughing at the prospect of putting a problem so simple to them on the exam.
No problem, thanks for watching.
my physics teacher suggested me to watch this really thank u for the brief explanation
You're very welcome! Thanks for watching!
Thank You Very Much! This was clear, and just what I needed
Thank you Sir for the time and patience to teach us!! So clearly explained! Thank you very much
Thank you so much for the great comment.
May the lord reward your hardwork
Thank you so much. Best wishes to you!
thanks!! i needed it soo much for my pre-medical entrance exam
You're welcome and best of luck!
That helped out a lot thanks! The best explanation of this problem I could find online.
@Ben Fox, thanks for the "best" comment, I think they turned out pretty well but they (static equilibrium videos) have not gotten much traffic yet. Tell all your friends.
Will do! I'll be checking out some more later as well.
Thank you for making me understand this concept...
It really helped
Glad to hear that, you're very welcome
This dude talks faster than anyone I've ever heard but still is easier to understand than my professor.
Glad it was clear and understandable!
you made this problem so easy to follow!
Thanks, it is a complicated one.
Thank you Sir!!! It's very helpfull
Most welcome!
Most Excellent!! thank you... great graphics... that's a cool program you're using.... so easy to read and follow...
Thanks for the positive comment. It is not really a program, just Apple Keynote with animations. Works pretty well.
Thanks for the detailed explanation 🙏😊
My pleasure 😊, thanks for watching
God bless you Sir!!! Thanks alot for your vidoes
So nice of you, you are most welcome.
Thank u for ur clear cut explanation sir...☺️
Most welcome, thanks for watching!
That was really awesome...... Keep it....!much appreciate
Really clear explanation. thanks a lot!
Great video, dude!
How do you decide which point is best to take the torque from?
The pivot point.
Thank you Sir you my best lecture ever may God bless
Very nice of you to say....thanks for watching and commenting.
Thank you! Your explanations were really helpful
Anytime, thanks for the comment.
So in theory, if we solve for the friction force like we do in the video, but also find the magnitude of the normal force in by summing the forces in the y direction, we could find the value of mu correct?
I'll start off by saying thank you so much for these videos! I'm studying physics with the OU alongside my full-time job and having content like this really helps me understand concepts faster than using my textbooks!
A quick question on this though, would there not be a coefficient of friction where the ladder meets the wall? Assuming the ladder is stationary there would be a coefficient of static friction and if the ladder were sliding there would be a coefficient of sliding friction.
We assume there is no friction between the wall and the ladder.
Does the ladder not exert a normal force up on the person (equivalent in magnitude but opposite in direction to the force of gravity on the person) so that person doesn't fall straight through the ground? A little confused why this force is not in the free body diagram.
Good question, the forces needed to solve the problem are the ones I included in the diagram. The normal force is at the bottom of the ladder.
Great video!
Thanks, glad you liked it!
Outstanding 🎉
Thanks so much!
You said that the x distances given are the lever arms, but this doesn’t really make sense to me. With our torque equation where Tp + Tl = Tw, are we not talking about torques acted upon the ladder? Therefore, the ladder should be our lever arm right? However if we treat the x distances as lever arms, this ends up being the torques that the forces produce onto the floor, which doesn’t make sense
Good question, but starting at about 5:00 minutes into the video I explain how we get around the fact that we do not know the lever arm.
I found out how he got it. I would recommend some paper to visualise the setup. Take the force F_n 1.1 meters from our pivot point. Let L be the length of the ladder from the pivot point. The torque there is given by sin(theta)F_n*L. So, lets find L!
Now with the paper draw a dotted line from the F_n to the ground. (paper coming in clutch now) Notice how this forms a triangle, where the angle theta is the same as the theta in sin(theta)F_n*L. In this new triangle, 1.1m is the Opposite, while L is the Hypotenuse (zamn I spelt it correct). Since Sin(theta)= Opp/Hyp, it is the same as Sin(theta)= 1.1/L.
Heyyy, wait a minute, we now have to expressions with sin(theta). Lets plug in our new expression for Sin(theta) = 1.1/L into our torque equation Torque = sin(theta)F_n*L. That would mean
Torque = sin(theta)F_n*L
-(note: subbing in sin(theta) as 1.1/L )
Torque = (1.1/L) F_n*L.
-(note: L denominator and L cancel)
Torque = 1.1*F_n
And there you have it! Thats just the perpendicular side times at the Force! Let me know if u didn't understand anything, or want me to draw something up.
ps. F_n is F subscript n, hope that didn't confuse you.
Thanks bro! With this video I subscribed to your channel! :)
Thanks for subscribing
clearly understood sir thanq this makes me good marks in my semister
Hope the semester ends well.
You can see a listing of all my videos at my website, www.stepbystepscience.com
Great explanation of the problem, thank you sir!
you are very welcome.
Why are we choosing the bottom end of the ladder to calculate the torque ?
Why can't we use the center of mass of the ladder ?
I have great thanks for you .keep it up 💯💯💯💯💯 percent
Thank you too!
Just a query can we take torque from the point on the wall? Btw nice lecture
Why aren't the weight forces broken down into components? Aren't they both at angles instead of being perpendicular to the ladder?
because they are acting in the y direction, not between the x and y directions.
very helpful ! Thank you very much!
do one on momentum
5.5m is the vertical distance from the pivot, aren't we supposed to use horizontal (perpendicular) distances ??
Shouldn't the Ff be (-247N)? It's in the opposite direction.
Thank you sir
Most welcome!
So good!
Nice, thanks!
nice explaination
Thanks and you are very welcome.
At 3:39 when you said, “if these forces were to act by themselves, the would cause the ladder to rotate in the counter-clockwise direction” did you mean clockwise?
Those downward forces are clockwise, aren’t they?
Well, i guess that just depends which normal force you get rid of to allow rotation. its only counterclockwise if you get rid of the wall. If yiu getvrid if the ground, it goes clockwise
Its because axis of rotation is at base of ladder
I love this but I still dont get that Xl and Xp, did u get those Xl and Xp from the question or did u just count them ? Thank you 😭
This is about 2 years too late, but those are given in the question
TY very well explained.
Thanks. You can also see a listing of all my videos by topic at www.stepbystepscience.com
Are we assuming that friction between ladder and wall = zero for this example?
Yes, You can see a listing of all my videos at my website, www.stepbystepscience.com
You're the man!!!
Ok, and thanks for the comment.
Super helpful
I appreciate this so much thanks!
You are very welcome. You can see a listing of all my videos at www.stepbystepscience.com
helpful, thank you!
You are very welcome.
Thank u sir
thnk you
You are very welcome, thanks for commenting!
How is the lever arm of the force from the wall (Fw) 5.5m when the pivot point is set at the base of the ladder touching the ground?
Hey, actually I think I might have figured it out. When using the formula 'torque= r x F', we substitute the values 'torque= 5.5m x Fw' but if we choose to use the formula 'torque= r x F x sin θ', we substitute the values r= length of ladder, F=Fw , sin θ= sine of angle between the horizontal Fw and ladder or alternatively, sin θ= opposite/hypotenuse= 5.5m/length of ladder, all these into 'torque= r xF x sin θ' and the torques would be the same. Hope this helps!
Alas 6 years later I have the same question, that I just solved using your comment. Thanks. I just wish he had explained it as thoroughly as he usually does.
BEST
thank you
You are very welcome, You can see a listing of all my videos at my website, www.stepbystepscience.com
wouldn't there be another up force at the another end of the ladder?
Yeah, if there was only one force then the ladder would be moving to the right.
Thank You!
+Raphael Banoub You are welcome!
You can see a listing of all my videos at my website,www.stepbystepscience.com
Great! They are awesome! Very helpful!
Why FW is directly horizontal ?
The ladder pushes into the wall and the wall pushes back. Newton's 3rd Law. Seems a bit odd but yes the forces are horizontal.
Step-by-Step Science Thanks
I damn sure appreciate cha!
❤
very good video but can you talk a bit slower next time
I am always trying to find a balance between too fast and too slow...thanks for watching and commenting.
I love you
Appreciated the video. Btw you sound like a congested Tom Hanks :D
Mmmm i get it now.thks amte
You are very welcome.
THANK U SHAHSJWDJOWJDWJDJW IM SCREAMING
I think that is good, You can see a complete listing of my videos at www.stepbystepscience.com
Excuse me sir! how did you get 1.1m and 1.6m? Is there any formula on how to get them?
Ur so fast when teaching but good also
Sorry for that, but thanks for the comment.
Why do u talk so fast I really don't get it especially with how fast u speak
Sorry, but you can try using the pause and replay features.
I reincarnated into this?!?
Awesome.
oops. solved for friction coeffiecient. It's .27
How did you get that?
Excellent! Thank you ❤️.
You are welcome!
Thank you sir
My pleasure, thanks for watching