Physics 15 Torque Example 2 (2 of 7) Mass on Rod and Cable
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- čas přidán 28. 08. 2024
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In this first of the seven part series I will show you how to find the tension of a horizontal cable attached to a wall and rod (attached to the same wall and at an angle) with a mass hanging from one-fifth from the end of the rod.
I've said it before and I'll say it again. *These are hands down the best physics lectures on youtube.*
I strongly agree.
@@jahson040 no one cares
Michel, you're the fucking man. Your method of assign the distances between their perpendicular forces is META AF.
I had no idea how to solve these problems untill i found your video- you are a real life saver. Thanks :)
Since every term in the equation has an ""L" in it you can divide both sides of the equation by L.
Absolutely awesome video. I appreciate you doing even the smallest steps of algebra to show your thought process. Thank you for all the help.
THANK YOU SO MUCH!!! My physics teacher simply uses powerpoints to teach our class and never goes through a problem and handwrites it on the board like you do.
Thanks, my left ear learned a lot from you.
Ivann Ruiz hahahaha
wish you were my physics teacher! very clear explanation. thank you
This is a really great question! Took me some time to figure it out on my own, and then watched what you did. Amazing, thank you!
Thank you so much for making these videos! I've been struggling with torque but your playlist has completely cleared things up for me :)
Professor Biezen, these lectures are incredible and helpful for all students in Statics.
We are glad these videos are helpful.
Wow, I never thought that I would actually understand this concept fully but after I looked at your example and tried it myself, I used what I learned from your videos to solve problems from my lectures and it made more sense than how my professor did it! Thanks
Great! I am glad these videos are helping.
i finally understand tension after watching your videos. so so valuable and helpful. thank you so much!
I spent 2 hours with my tutor trying to figure out how to solve this problem and still got it wrong. This video will save me in my upcoming test! thank you so much
Thank you for sharing, I learned a lot. I understand now why my teacher solves for L afterwards once writing down the equation first to see where d1, d2, and d3 is located. He also has a tendency to cancel out the L and then plug in the numbers once isolating L from each triangle. Very helpful ! : D
I actually spent some time discovering your website, it was fantastic and full of examples filled with different types of questions, perfectly suitable for students who are looking for a website to learn physics. Good job Sir! :)
Thank you for putting this up and walking us through problem. It was incredibly helpful.
Thank you sir. Your videos have been very useful for me in understanding the concepts. It has also built up my interest towards the subject. Thanks a lot.
This may have just saved my grade for one of my finals. Thank you for going through this problem, it was excellent.
Professor, I thought for clockwise the torque would be negative and counter-clockwise would be positive?
When taken as a vector you are correct.
Wow! You finally made that clear for me. Thanks!
Your videos are very helpful. So glad I found these videos regarding torque. Thank you!
Just started learning this and was wondering why you chose to us sin(45°) for d1 and d2 instead of cos? Wouldn't it have been simpler to factor out L and cos(45°) before shifting T?
Thanks man! you really helped me out.
I owe my life to you. Thank you so much!
Happy to help!
Thanks for making these videos, you really help me out.
Why no calculations for normal force acting on the wall?
Because r = 0
How come all of the videos online-and I mean all of them-cover either the wire angled down, or the wire connected to the end of the rod, while the questions on the exams and homework put the wire horizontal and connect it partway along the length of the beam?
I believe we covered many different situations, but we'll review it to see if we need to add more.
I appreciated with your method to describe physicsc sience system. And just recommend please the writing letter don't make too much bald, thank you.
i think the value of d1 should be 16/5 since 4/5 is just the value of the other end in the beam
So amazing ty it'll help for my test tomorrow
Good afternoon Professor, thank you very kindly!!!
thank u I really understand it now
Do we not consider the force exerted by the string holding the block?
Sir, may you instead take lever arms to be distances along the bar and only consider the component of each force normal to that line
what should we do, when we have 2 angles in the same triangle?
You can use either one of them.
sir please tell me why u didn't consider the tension in the string which is pulling the block of mass m upwards,isn't that a part of the problem
+Rahul Tiwari The tension is considered. It is the same as the weight of the block. That tension is multiplied by the perpendicular distance which give you the torque that tension produces.
but u haven't mentioned it when u were identifying all the forces
But why u haven't mentioned it while identifying all the forces
thank you thank you thank you!
Why are d1 and d2 horizontal distances and not the distance along the bar?
By definition, it has to be the perpendicular distance from the point of rotation to the line of action of the force.
Thank you sir for the great work
I have a question can we find the tension by sum of the force
Not in this case. This type of problem requires us to use the "sum of the torques" technique.
how would you find the direction of the force the hinge exerts on the wire?
Assume a point connection and then draw a triangle where the length of each of the sides of the triangle corresponds to the magnitude of the force of the each of the two cables and the beam.
why are we not considering the normal force exerted by the wall on the beam?
+Shameera Rafeek Good question. Since we picked to pivot point at the point where the beam is attached to the wall, the action of the force goes right through the pivot point, which means the moment arm is zero, and those forces don't contribute to the net torque.
Thank You soo much!
thanks very much sir
how would we set up an equation for the x and y components?
By using a graph, we have x and y components, you can try dividing forces which are not acting in the x and y components. ( Basically just dividing an angled force into 2 components- obviously x and y ). Therefore you'll be able to find equations when you changed all the forces into x and y components whereas Fx=0, Fy=0 when dealing with static equilibrium.
Note that when you divide forces which are not in components of x and y TO x and y components, you need to make sure that you've got the angles correct (sin,cos,tan). Ex- Wsin30=T1x
why do you use cos for the torque of the tension rather then sin theta
It depends on which angle you use. Always consider sin(theta) = opp side / hyp cos(theta) = adj side / hyp
omg, i think this is the first video ive seen with so many view and no dislikes.
very good
i paused the video and solved it before you added the mass of the beam, i assumed we was working with a "light beam" zero mass. my answer is 392 N. can any confirm this is correct? i used a different technique, by using the distance from the force acting on the beam to the pivet point
Yes, 392 N is correct.
@@MichelvanBiezen thank you. Really enjoying these videos 😃
why not just use L as the moment arm, rather than d?
The moment arm is the perpendicular distance from the line of action of the force to the pivot point.
Great video
I thought clockwise was supposed to be negative torque?
When taken as a vector, yes.
can you tell me why we don't count the torque of the tension that is lifting the weight ?
and thanks for the video
Superbb
I used,
cos(180-∅) = (d2) / (4/5L)
I know I am assuming that thickness of the rod is 0m but still my answer is very very different. I got 1.3N which obviously does not make sense but idk where I went wrong. Is it the trigo part or is it something else. What do you guys think?
I am often asked: "what did I do wrong?" by my students and even my own children and it is natural to want to know that. But each type of physics problem lends itself to a particular method for solving them. Using these specific methods will reduce errors and allow the gain of understanding especially in more difficult questions. Try to duplicate the method shown and see what happens.
@@MichelvanBiezen thank you so much. Honestly you are a physics god!!!
BEAUTIFUL
Great video! Did anyone get a slightly different answer? My answer varied slightly and they checked out. Just wanted to confirm. Thanks.
why are you cancelling the L on the equation?
If every term in the equation contains the same factor ( in this case L ), canceling them out will simplify the equation.
quick question;
So we can define direction of torque on our own? As we want?
If you are only calculating the magnitude, it doesn't matter. If you calculate the torque as a vector, then the convention is that counterclockwise is positive.
I didn't expect replay that fast :D hah
Thank You very much prof. van Biezen!
You are the best! :D
Cheers.
Should not be the counter clock wise direction is the positive direction ?
Yes, if taken as a vector quantity.
Clocks are negative
time does not exist, so, clockwise is an imaginary vector, You can use it any way you like..
can you introduce any good book for the basic Torque theory?
Hello, I have a question:
At d1 you did: (4/5L*sin45)
I did: (4/5L*cos45) and the answers did not match. Nevertheless sin45 and cos45 are the same.
Why is that?
Thank you! You are a helper!
How about the other terms in the equation. Did you get the same result for all of them?
Michel van Biezen I did. The other terms were like the ones in your lecture. So in the end the equation was divided by the cosø as shown in your equation. Then since i had 4/5L*cosø i just canceled them out (cosø). Thats how the results come out differently.
***** make sure you change the calculator mode to degree, it should not be radians.
why's d1= 4/5? and what happens if I don't cancel out the Ts
+James Mars
m is connected 1/5 from the end.
Sir I m talking about this problem how u considered 1/5L of d..
That is a "given" in the problem.
@@MichelvanBiezen Sir ok I m so sorry ,,,
why is the triangle d2, the length of the beam is L/2?
+MrDoYouWannaBeOnTop the centre of mass of the beam acts towards the midpoint so as the lengh of beam is L ,,its midpoint is L/2 :)
why he cancel " L"?
something i dont understand is why in the last video the distance that the tension force was acting through was from the middle of the cable to the pivot point, but here it isnt. can you clear that up? is it just because its a horizontal cable this time?
+footeythegreat Look for the :"perpendicular distance from the line of action of the force to the point of rotation"That means the distance line must be perpendicular to the line of action of the force (which in this case is the cable).
Yes I was confused by that too
Great video, although I was wondering how you were going to solve it without the mass of the beam haha.
Teacher How is the name of this method?
They call these "static problems", which means no motion. The technique used is: the sum of the forces add up to zero and the sum of the torques add up to zero.
YOU ROCK!!!!!!
Nice video! But you could have used a better lighting over the board.
Yes our older videos are not as good with the clarity and lighting. We had a lot to learn about making videos.
How do you know where to put the "pivot point"?
+Matthew Moran
You can choose the pivot point anywhere, however it is typically chosen to be at the bottom of the beam where it is connected to the post. You can try to choose a different pivot point to see what happens. (There is no wrong place, however at some locations you will not be able to find the solution)
how do you determine the line of action?
The line of action of a force is simply the direction of the force project forward and backward.
why is it at 1:28 the tension cable is pulling that way?
+MrDoYouWannaBeOnTop
Think about how the beam would move if the cable wasn't there.
+MrDoYouWannaBeOnTop if the cable had not been there the beam would have fallen clockwise ,,, but as tension is acting in Anti-clockwise direction it keeps the beam in equilibrium . :)
why you did the angle with sin n why not use cos?
+Muhammad Azamuddin
It doesn't matter if you use the sin or cos, as long as it is the correct trig function for the particular situation.
sin (theta) = opposite / hypotenuse
cos (theta) = adjacent / hypotenuse
+Michel van Biezen how about if we use length of beam as our d3 for tension could we?
This is so much easier if you use vectors.
WHICH COLLEGE ARE YOU A LECTURE AT???
i thought cw was negative?
If one looks at the torque as a vector, cw is the negative direction and ccw is the positive direction. If you are only calculating the magnitude (like in this example) it doesn't matter.
you respond faster than my teacher!!
i need Flashlight to see your equation.
You lost me round about 0.31. If I could understand this I might discover why my buildings keep falling down.
Beast!!
That fly though.
0 dislikes? wow!!!!!!!! And according to a calculator, 216/0 = infinity :V
216/0 is undefined lol but I see what you did there 😉
thank u-- pls teach me virtual work
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