Physics 15 Torque Example 1 (1 of 7) Mass on Rod and Cable
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- เผยแพร่เมื่อ 4 พ.ย. 2024
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In this first of the seven part series I will show you how to find the tension of a cable attached to a wall and rod with a mass hanging at the end of the rod.
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Because of your videos, I feel 100 times more confident going into my Physics exam today! Thanks so much!
my left ear learned a lot today
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Holy shit I laughed so hard
I was wondering why I couldn’t hear it. Thanks!
ooh，I thought there is something wrong with my airpods...
Hi Michael, been taking physics since grade 11 (I'm 25 now and am in first year engineering, sad I know) and for the first time I actually am feeling like I got what it takes to get good at this. I try to do these problems by myself before watching through and thanks to your excellent teaching methods I am finding at least half of the time I am on the right track... much better than before!
Don't worry about the "late" start on your engineering studies. (I didn't complete my bachelor degree until I was 27 years old, before I started graduate school). Life happens and we have to live it as it comes. We are glad the videos are helpful.
@@MichelvanBiezen yeah I wound up working trades for five-ish years before I decided I wanted to be an engineer, I didn't try my best in high school so I had to upgrade practically all my grade 12 credits to even qualify. But now that I'm here I don't wanna give up! Happy to report that I did pass my Physics Mechanics course, I imagine you're a big part of that! So thank you for your encouragement, here's hoping Waves and Fields will go well.
As a physics student, I just wanted to say thanks for all your videos! Even if they are not exactly what I'm looking for, they are quick and to the point which is perfect and helps me solve the problem at hand! Thanks again!
Anne
To find the horizontal force on the hinge, you must sum all the forces in the x-direction.
Therefore
Fx = T cos (30 degrees)
To find the vertical force at the hinge, you must use the sum of the torques with the pivot point at the right end of the horizontal beam.
Sir do you have lectures on calculus 3...multi variables
Pius Suh
Pius,
Only a few so far.
More will be coming later this year.
Bent jij toevallig Nederlands? Bij de naam van je kanaal is een Nederlands lijkende naam te zien.
Tijdens mijn jeugd woonde ik in Antwerpen
By using this relationship can we make an easy-to-use hand pumping system for hand pump
Now we have to make much effort to pullout the water from Borewell ?
Thank You for simplifying it and writing out the equations step by step. I've really been struggling with this but you are an excellent teacher and have helped me so much!
We are glad to be able to help. Thanks for writing.
You are born to teach. Thank you for the videos.
Stephy,
Thank you for your input. Much appreciated.
@@MichelvanBiezen haha. You reminded me of Sherlock's "yes Anderson, thanks for you input"
You taught me in 8 and a half minutes what I've been struggling for weeks to understand. Thank you, thank you, thank you!
Nick.
Great! Thanks for sharing and good luck on your exam!
I cannot thank you enough. You are by far the greatest teacher EVER
You are a dedicated teacher....
I have heard that you are the best teacher in the world.My grattitude toward you is immeasurable.Thank you very much😍😀🤗
Young forever 2022🙏🙏
There are many great teachers in the world. I hope to be one of them. Thank you for the comment
Shout out to this channel you do proper examples
Thank you. Glad you find these useful.
Thank you so much!! I was able to cram 12 chapters of physics in 3 days and get an A on the final by watching your videos.
Congraduation!!! You still had to do the work.
Bless your brain and your ability to teach. Literally, your videos are the only reason I'm passing physics 1 lol
WOW this is amazing!! Thank you so much! You made this so simple!!! I wish all physics professors taught like you everything makes sense now !! Subscribed :)
Wow!, thank you so much, your explanation is so straight forward and understandable, you're super cool. My prof never explain where he got anything that he writes, he just flows while we are not flowing with him.thank you so much!!!!!!
Nick,
Good luck with your exam.
why did i learn more from this 8 min video than a week of lectures
Static Equilibrium is frustrating because of the idea to relate Transnational Forces to Rotational Forces in an equilibrium system.
your are the best teacher ever!..thank you sir...
I have a physics test tomorrow and this helped me so much! Thanks :D
damn 6 year difference but im in the same boat
Im learning a lot from the videos
Thanks alot even tough you made this video long time ago.I was struggling to understand the concept and thanks to you i am able to solve my questions.
I'm studying for my AP physics torque quiz and this helped so much. You are a really great teacher
Same
Thank you very much for this video! I’m french so I’m training static and English at the same time ;) the perfect deal
C'est mon plaisir. Welcome to the channel.
Thanks Mr. Michel, I learned it much easier from you than my professor.
very good explanation
I GET IT! Finally, I see how it is nothing more than vector decomposition!
Thank you!
Thank you so sir.. this is so helpful ❤️ have a good life❤️
Thank you. Glad you find these videos helpful.
Legend. So Helpful
You did such an excellent job of explaining this
stay blessed ....u r such a wonderfull teacher!!!!
يسعد قلبك يا زلمة ❤️❤️
Big Respect ❤️❤️
kos
youre the best ever! i understood everything completely :D
That was an awesome example of how to find torque.
You just made the concept really simple, thanks.
I have my first year physics final tomorrow. Because of these videos I'm going to pass. Thank you
Good luck on your final
This is so useful and helpful for my final tomorrow! Thank you so much.
how did it go
@@gokublack9080 if i remember four years ago, it went decent lol. this material was on the test and i was able to get it right, but i suck at physics. thank god i dont have to use it for my job
u r a great teacher ..i wish to study with u for real
this video helped me a lot .thanx sir
What if the beam is not level and the big force is actually on a ramp connected to the end of the beam?
I have a bus with a foldimg rear door(5ft.) that is part of the ramp. The ramp extends another 7 ft to the ground. The pivot is 40" high. The door would need to be lowered to 25" to make the ramp accessible to the car, then lifted once the car is pulled into the ramp to straighten the door and ramp combination.
Your video has been very helpful to me, by the way, can you make a series that involves a the dynamic of a rotating rod?
Since we plan on filming today, we will add this if we get a chance. It will be some more days after that to make it through production.
i m frommorocco, my english is not good, but i learned that before applaying static equlibrium we must extact our system.
Thank You!!!
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I'm very appreciate for you time!!!
This video made everything so clear! Thank you!
OMG THANKYOU SO MUCH I ACTUALLY UNDERSTAND AHHHH
Great! Glad you found our videos. 🙂
There are 4 forces acting on the beam (not 3 as stated). Weight of beam, weight at end, tension and normal force at hinge.
What if (for the CCW moment), we multiplied the component of T (Tcos30) with the perp distance (Tsin30 OR 4tan30)? I tried but I got a different answer and I do not know whether it's due to a calculation error or a logical error. Could you help me verify the kind of the error by double-checking my method?
thanks for the concise explanation, very helpful
sir, did i mension you are awesome!!!
This video is so helpful! Thank you so much
I think to be clear, you chould say: We extract the beam and we will make static equlibrium of it. so we konw wha is interior to the system and what is exterior. what do you think about this.
Wow, you are such a great teacher! You put this wayyy better than my professor. Are you a Professor at a University? Because you should be!! Saw your Cal videos and your Physics tutorials are great as well, so far the best on TH-cam in breaking down these concepts. Very clear and straightforward. Do you do Chemistry tutorials as well?
shawawong
Thank you for the compliment. Yes we also have chemistry and astronomy videos.
Excellent.
WOW THIS NOW SEEMED EASY THANKS
Professor Biezen, these lecture are fantastic, however what point did you take the torque about?
The point about which we find the torque is where the horizontal beam is connected to the vertical post.
GOOD JOB PROFESSOR
anyone else only hear sound from left side?
+kslidzorous We had sound problems with our early videos.
That must be 2 Channel or more. U can set 1 channel therefore sounds will come out from any speaker
for correction
(+) counterclockwise
(-) clockwise
Yes, when used as vectors you are correct.
Thanks! You are a great teacher!!!!
Awesome Videos! You are a Physics lifesaver! THANK YOU SO MUCH
I wonder why you did not consider the horizontal force at the joint of the cable with the wall
thank god for you
First of all: thank you. But I have to point out Two important things.
The teacher drew (min. 0:50) the pivot...well. But he explained us that putting this pivot he eliminates the two unknown forces (reactions X and Y). Sorry it only eliminates another torque. The reactions stay or remain. Please just note that this force doesn't (or pair of forces Rx and Ry don't) produce moment or torque because its line of action passes through the fulcrum or pivot axis. . . Sorry about my English.
+walbani Very good point. I meant to say that it eliminated those two forces from CONTRIBUTING to the total torque.
O. K. thank you for your answer. I've been studied your videos. Thank you again.
This video was super helpful!!!!!!
Glad it was helpful!
Alright I'm one step closer to passing PHYS 106. Thanks!
Thanks I really struggle with these exact types of problems
in the beginning of the video, is it possible for the stick to stay in balence if there is no string to support it? and lets say that the stick is fixed to the wall
We can make the assumption that the vertical pole is anchored to the ground.
Thanks for a great video. Your explanation was easy to follow and replicate. Also using an example that is real helped as well We have a couple of swing arm tool balances at work, I am going to do some calcs tomorrow...
I believe that it should be positive when the direction is counterclockwise and negative when clockwise? I think what he is saying is the opposite
It doesn't matter unless you consider them to be vector quantities.
Sounds like I'm taking a physics class from Michael Scott... awesome!
Very helpful. Thank you.
You rock man! I never thought I would understand this :-). Thankyou!! a lot!
Why was the tension in the string to which m was hung not included in the equation?
Quick question: can we just draw the d3 along the bar? The only component of tension that causes torque is the y component anyway, and this component is normal to the bar. I find it easier to conceptualize this way.
There are multiple ways to approach this problem.
could you pls distinguish between the tow cases: calculating resultant and applaying static equilibruim.
At 3:41, you mention how turning clockwise gives you a positive torque, but using the Right hand rule, doesn't it state that a positive torque comes from Counterclockwise rotation giving you a negative value rather than positive?
If you consider them as vector quantities then yes.
This was great, thanks!
Learn a lot from this video. Thx for guiding! xD
Thank you, Thank you, Thank you, Thank you, Thank you sooo much
Glenda,
It is good to know these videos are helping people around the world. Thanks for taking the time to let me know.
Michel van Biezen Thank you, I've watched a couple of them and they are extremely helpful. I have my physics exam next week!
I solved this problem using the vertical projection of T multipied by d1 as the torque as the horizontal component of T is pointing along the beam and seems to not contribute to the torque of the beam and got a different answer? Could you please clear up how the tension force is acting at distance of d3? How would this make sense in terms of vectors?
What is material and diameter of cable to withstand the tension, also length and height of the wall and mounting of the cable is going to matter in the above example
Those concepts are explained in a different playlist. Here we are only considerting the concept of torque
@@MichelvanBiezen Will you please me the concerned link
Hello,
I wanted to know why we don't consider any y forces, X forces, or torque from the supporting cube touching the ground.
It seems to make sense to me that the whole structure would topple over if there are no forces?
Thanks
There are x and y forces where the post in embedded in the ground, but those forces would not add torque, since the distance from the point of rotation are considered "zero".
Wow man, I got an Exam and this is included, and I don't know how to solve it. You cleared it up big time for me. Thank you!
Small Q: If I converted the signs and made the clockwise -ve and the counter +ve, the answer will be the same; is it okay?
Great video, but shouldn't that should be Sin 60 or Cos 30, not sin 30? L/d3 = sin 60 and L/d3 = Cos 30, so T = 1697.4 N
It is probably a little confusing, but the angle theta = 30 degrees and therefore d3 = L sin(theta) = L sin(30).
I thought clockwise was negative and counterclockwise was positive. You can change the signs? In the fundamental videos, you said clockwise would be negative, not positive... Can you change signs? I'm confused. Besides that.... phenomenal job!
When calculating the magnitude of the torque, it doesn't matter what sign you use. However if you calculate the torque as a vector, then counter-clockwise is positive.
Excellent video!!
thank you sooo much for this video!!!! i was so lost in physics and now i finally know what's going on lol...
why did you draw d3 from the pivot to the center of the cable? Isnt it acting on the end of the beam? why wouldnt d3 just equal d1?
By definition, d3 is the perpendicular distance from the line of action of the tension to the point of rotation (pivot point). It is drawn correctly.
Awesome! thanks so much! this was very clear and I learned a lot.
Hello Michel,
I have put the figures in and i didn't get what you got. I got 2452.5N I didnt get 2940N. What am I doing wrong?
Thank you
,
Symon
Is it not Tcos(theta) since that would be the perpendicular force exerted.
It depends on the angle used. In this example, since we are using theta, we must use the sin,
Dear,
why the center of mass of rod is at its center? We also have mass at right side thus CM shall be to the right...
Each mass is taken separately. Thus the center of mass of the beam is at the center.
i will be wrong if i resolve the components of tension and then calculate torque of them but we that the x component will be zero?
There is no horizontal force acting, that's why he didn't use cos ø
Thank you very much.Your videos helped me a lot. :)
Tension is on the wrong place. its supposed to be on the other end so that it counteracts the two forces that act downwards.
I love all your videos, but can you tell me why "d3" was really necessary? Isn't everything just acting at the full radial length or at half the radial length? I understand the need for the angle (sin 30), but couldn't all the torques just be set up quicker/easier as +mgr +MGr/2 and -Trsinϴ from the FBD without d3 since the cable provides torque at the full radial distance anyway?
Yes, you can do it that way. The use of d1, d2, and d3 is to make it easier to initially get the concept and set up the initial equation correctly.
Thanks for the prompt reply! Keep up the good work!
I'm sorry. Great explaination, but it the direction of the Torques would matter. Because when I did it the textbook way. You get 980 N in the end.