Physics 15 Torque Example 3 (3 of 7) Mass on Rod and Cable
Vložit
- čas přidán 6. 04. 2013
- Visit ilectureonline.com for more math and science lectures!
In this third 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.
Anuj. As some students before, you found the error with the 1/2 factor. Very good!
sir Michel van Biezen, statics problems next please. Thanks.
Yes, the 1/2 was over-looked...but what really matters is knowing how to solve this problems. Great video!
Anyway, the correct final answer is supposed to be T=120483.4406N or just 1.2x10^5N
he got it wrong, I was confused for a bit but he saved me from my physics test
No, it's 12048.244
yes i agree, he made calcultion error
I'm still following right along with you MVB! You have provided lectures for EVERY single section we have covered so far in my phys-1 course! You are a God send! I truly thank you (again), with the utmost sincerity! You explain things in such a way that I can comprehend, and that takes talent! Well done, sir.
Felipe
You are correct. That is a mistake.
He has made my physics semesters easier and more clear. Appreciate your help!
Professor Biezen, thank you so much for these videos! :) Your explanations are succinct and sweet, just how physics should be :)
Your saving my life right now! God bless you immensely!
These videos are great. Very helpful in understanding multiple physics concepts
This video definitely simplifies things. Great teaching methods.
Tks
I hope you read this, your videos are f****** amazing! They helped me save valuable time and dramatic anxiety.
harold alvarez
Harold,
I do try to keep up to the comments. Much appreciated.
strategy works in all examples.thank you so much for the most productive videos..Love from South Africa
Thank you. Welcome to the channel! 🙂
Amazing and clear example, thank you!
Thank you so much, you saved my life! Literally
I'm glad that I scrolled once down to look at the comment section. I was just going to try that once again (after already working it up twice and getting the tension as 1.2*10^5.)
Do you know how to determine forces graphically, especially forces on the sheeleg crane
Fun problem. Thx for putting all of these together.
Thank you so much! You have no idea how helpful you have been to me! Much respect.
Everything static in here so summation of all forces in the y-direction equal to zero (2000*9.8+5000*9.8=Tsin(30) ) but the result comes different which is wrong in the equation?
Thank you. This helped a lot.
You are the best physic teacher ever👍
Glad you think so!
Sir v.good video for learning.
Sir how can I found Mg weight if we know the weight of L
Why is normal reaction from the point string makes angle of 30 is not considered?
great work prof
is their any app regarding this formula for physics of crane
it was a very useful video thank u so much. unfortunately missed that 1/2, anyway it was very helpfull
Good morning Mr. Michel van Biezen
Your lexplanation is very good .
I like your videos.
In this video d2=L/2Cos45°
Please look at the video 8:18 min. You have ignored denominator 2
Great video! Is another way to go about solving this problem? If you solved for all of the perpendicular forces and d3?
There are often multiple ways to solve these types of problems.
OML.You are awesome.subscribed!I just wish these videos were new so that I could thank you now..... :D
Glad you are enjoying them.
Thanks for the nice problem. You forgot to mention the pivot force on the beam. It got cancelled because you took the torque about the pivot point. Also you forget to divide Mg by 2. But anyway, it was nice all the way.
The answer is 1.2 X 10e5 N.
+Geo John I keep getting that too! The only thing I did differently to him here was not cancel out the Length.
+Geo John You are correct, he forgot to keep the 1/2 after canceling all the Ls from the equation.
120606.3829 ........
why can't you be my professor?? love your videos
Hey Michel, in this situation, how would one calculate the reaction force at the pivot point?
Would you split the tension into its two components?
Yes, you will have to find all the x and y components of each force including the tension.
Sometimes you can also find the reaction forces by picking other pivot points and calculating the sum of the torques.
Thanks for the response. I'm assuming that if I were to want to find the x and y components of the tension, I would use the tension as my hypotenuse but what would my angle be?
very well explained= thank u sir
Michel, Thank. You.
I’m definitely dedicating this engineering degree to you sir😹😹😹 you’re a life saver
Glad you found our videos! Keep it going! 🙂
for the final answer i got 12060N?
do you know about tachometer reading formula according to which it's reading increases
I see that it's reading difference hours meter time according to your wrist watch are different by a variety of different fraction to the real time according to your wrist watch???
please tell about that exact fraction according to which watch time of tachometer and wrist watch time are in ratio, forever right😌😌😌😌😌
thank you you helped me
Ap physics on the line!!!! Self studying sucks though no other kids to do labs with
Thank you Sir
you really helped me there sir.
Glad to be of help
thank you!
I have noticed that most people seem to prefer clockwise rotation as the positive rotation over anticlockwise rotation - I guess it's probably because the sun and the hands on a watch move clockwise, so it feels more natural and intuitive to use that movement as the positive rotation for those reasons.
I do as well. But when we use torque as a vector, the convention is that counterclockwise is positive.
Thank you
thank you
you are the best!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
the answer is 1.205*10^5
Thank you Abdullah.
how come we dont use the perpendicular components of force in this problem? arent we supposed to use mgcostheta?
There are two way to solve a problem like this. Using mgcostheta is one of them.
please sir i don't understand where you kept the half attached to Mgcos teta
That was an error. You found that error, good for you.
yes if u cancel L where is the 2... the other,, d sub 3 maybe its not only a sin of 15 deg.but instead it is sin15(10)....but never mind the explanation and effort are pretty good thank you Sir, its remind me the past..... but until now i'm still learning.....TY
If you cancel the L which is over 2 is it not going to be half the cos of theta
Wilhelm,
Yes you are correct. I forgot to carry the 1/2 (see below)
Yea I was confused about that as well :)
great videos only that i calculate my force vectors in terms of sines and cosines rather than changing the length, i end up with sines in my numerator and sin 60 in my denominator, im not sure if its better to go youre route
Jesus Perez
In the end everyone finds their favorite method. There are usually different ways to do a problem.
And you should pick the method that works best for you.
The methods that I use were chosen from experience with students. These methods seem to be the easiest understood by most.
When you were breaking the down the d3 triangle why did u not use the top right hand corner angle (15 degrees) and not the angle? (which ended up being 75 degrees?
Reagan,
You always want to break it down to a right triangle, where you know the hypotenuse or one of the sides as well as 1 angle.
@@MichelvanBiezen Please, I need your help in understanding, while that angle (the 15°) is the difference of the other two angles (45° and 30°). What rule or theorem states that? Please, I need to learn it, bcos I don't understand. Of course you can also give an explanation with your response. Thanks
A free-body-diagram with a coordinate system will make the problem mach easier to explain.
great video.
Michael van what type of triangle is d3 also isnt hyptonuse the longest side
d3 is a distance (not a triangle) It is the perpendicular distance from the line of action of the tension to the pivot point.
can i find distance d1=sin45*l (theta =45 and we have theta=90 the third theta 45)
for d2=sin45*l/2
for d3=sin 15 *l right ?
Jana,
sin(45) = cos (45)
However, based on the sketch, you should use cos(45)
sin 15 not = cos 15
jana yool
For d3, sin(15) L is correct.
Why isn't there a torque produced by the tension holding the mass? Is it because we are assuming the cable is massless?
We are only considering forces acting on the system, not forces within the system
@@MichelvanBiezen I see, thank you!
My mom was listening to me watch this as I study for the MCAT and her English is not that good, but she said "when teachers explain things this clearly, how can you not understand?." She then continued, "he sounds as charming as David Attenborough," who is considered a treasure of England, and a famous naturalist.
Binh,
Thank you for the compliment. I really like David Attenborough. I wish I could present material as well as he.
why didnt you take the tension in the other wire? the one between m and M?
That tension is equal to the weight of m, T = mg
@@MichelvanBiezen thank you!!
what about 1/2 on (L/2)......after cancellation..
+ishwor bista
Yes, there is an error. (see below)
You forgot a factor of 1/2 for the torque coming from the big M.
Yes, you found the error. Very good.
how come I got a different answer. I think Professor Biezen missing the L/2 for the 5000kg board.
Great problem, and you did forget the 1/2 for Mg*cos theta.
Thank you and yes, one of my students pointed that out already.
I think the 1/2 was overlooked, even the experts will make a mistake every now and then.
Yes, you found it as well. See the comments below.
I freaked when my answer was different but happy when I saw I caught the overlooked 1/2
Great job!
I tried doing it a couple of times but why am I still getting 120 606.4 instead of 120 483.4??? Anyone?
Look at the comments below.
oh! i think i know where went wrong. i used 9.81 for gravity instead of 9.8 (it made so much difference) but we can use 9.81 for gravity too right?
It doesn't matter. g = 9.81 m/sec^2 at the poles and g = 9.78 m/sec^2 at the equator.
i see. thank u professor! you've help so much and your videos are GREAT!
Good video, but does the force of friction act to the right of the beam? ( near the beam)
Also, is there a perpendicular normal force ? Or is at a 45 degree angle since the bream makes a 45 degree angle with the ground . I say this because in you previous video (example 2) , the beam made an angle with a vertical pole , so I assumed it applied a force to the beam at that angle.
Thank you
There are no friction forces involved here.
Problems like this one are solved by summing up all the torques. If there is no angular acceleration, the object is stationary, then the sum of all the torques add up to zero.
Sometimes you also need to sum up all the forces in the x direction and all the forces in the y direction and those sums will be zero as well.
In this problem, there are three forces, the weight of the beam, the weight of the hanging object, (both act directly downward) and the tension in the cable.
There are reactionary forces at the location where the beam touches the ground, but they can be ignored if you choose the pivot point at that location.
he is rock!!!
I thought the answer is dependent on the lowest no. of significant figures for multiplication and division. Shouldn't the answer be just 1 significant figure since there is the lowest sig fig is 1?
Chua,
The emphasis here is not significant figures but the understanding of the concepts.
The numbers in this problem were just randomly chosen with disregard to the significant figures.
But since you asked, the question now becomes: does the quantity "10" have one or two significant figures? unfortunately, it depends on who you ask. Technically speaking it has 2.
Frankly I am just clarifying so that I could make sense of it. Was the rule that I stated correct? Thanks for taking the time to reply as well, your videos do help:)
shouldn't d2 be mg l/2 cos theta1 and not theta2?
nvm, you corrected it.haha!
Good catch!
Thank You!
Near the pivot*
why isn't the normal force acting on the beam considered?
Any forces who's line of action goes through the pivot point can be ignored as they don't produce a torque
Do we account normal force too?
Depends on what part of the problem we are solving. We don't use it when we calculate the net torque.
I love you.
I would like to ask why the tension supporting the weight is not taken into account for this calculation,, hi professor...
The tension is equal to the weight.
@@MichelvanBiezen i see,,, so somehow the weight is already represented by the tension
Is it professor,,
That is correct. (we were on the road traveling so we couldn't respond to comments for a while).
@@MichelvanBiezen sorry for the haste, and enjoy yr trip 👻
you made an error sir.Just one.YOu forgot to use the 1/2.But you are the greatese sir
i didn't understand why did you cancel the length?
We divided both sides of the equation by L
i think at min 8;17 you forgot the 1\2 ?
Yes, as others before you, you found the error. I made some corrections. Thanks.
what about when they dont provide the length?
If the length of the beam is not given, you cannot calculate the torque.
sorry but you miss the half in Mgcos(45) *1/2
A Handsome guy (sir)!💝🇧🇩
SORRY,, Correct me if i'm wrong i'm lost of the cancellation.....
Nildo,
Are you referring to eliminating the "L"?
If so, the method used is dividing both sides of the equation by L.
Since the left side is equal to 0, 0 divided by L is still 0
Thank you sir,,,
What happens with the Tension of the hanging mass?
Michelle,
The tension in the cable holding the hanging mass is equal to the weight of the hanging mass = mg
then what is the difference between the tension we calculated in the problem and the tension which is equal to mg
how to find the pivot point?
+MrDoYouWannaBeOnTop
You can pick any point to pick the pivot point.
But typically you want to pick a point that will eliminate one unknown force.
oh okay. thanks!
Howdy Michel ... (is your first name pronounced like MICHAEL or like Michelle?) .. .. I was wondering what the FORCE is on the BOOM as it presses against the Pivot Point.. Would we just use Vector Addition to figure that out?.. Adding the Tension Vector to the (mg) Vector?.... yielding a Resultant Vector = to the Force on the Boom at the Pivot Point? by the way, I get a Kick out of your Wife editing your Videos when she adds comments. . like the HUNGRY Bear video .. you pointed out how your drawing of a bear was Skinny... so she changed the title to the STARVING Bear.. lol .. congrats on a Great series of VIDEOS!!.. I still refer to your Collection of Videos as the VanBiezen School of Engineering!!.. You cover everything... well mostly... Astronomy, Chemistry, Physics.. Algebra, Calculus.. Probablility.... on and on and ... ON!! THANK YOU!!.. like so many of your Fans here... I can agree that your Videos are the BEST!! They are Clear.. Organized... and use plenty of Examples to teach the Fundamentals... Thank you for all that you do.... .. Peace always.... -- Phil
Philip Y
Hi Philip. Still at it I see.
Most people call me "Mike", but when I grew up I was called Michel (pronounced like Michelle) (Like in the Beatle song: Michelle ma belle......
These videos have been an interesting journey. Where it will end up I don't know, but we are having fun doing it. It is however a lot of work (on top of the three jobs that I have).
Thanks for the comment.
To understand the forces on the boom better, look at the sum of all the forces in the x-direction and the sum of all the forces in the y-direction and they also add up to zero. (find the components of every force)
there is a error in video
+Nduduzo Mahlobo
Yes there is, thanks for letting me know.
Hello sir. I didn't heard clearly how you derived the sin 15° :(((
The 15 degrees = 45 degrees - 30 degrees in this case. To find the torque caused by the cable we take L cos (90 - 15) = L cos (75) = L sin (15)
@@MichelvanBiezen thankyou sir! 😭💕
Sir my ans comes out to be 124736N .......i guess u forgot to take that1/2....overall great video.....
Yes, you found the error.
After canceling the L's you forgot to put the remaining 1/2 that belongs in the Mgcos
A slightly toned down gru
Angle 15 for d3 , yu took needs little elaboration
Yes the angle should be 15 degrees
My answer is 120 483.44N, Did anyone get that?
+Zeino Madikizela That is correct. (although too many significant figures)
How many significant figures are recommended?
+Zeino Madikizela Typically, it the data is given with 3 significant figures, your answer should only have 3 significant figures. (It depends on the accuracy of the numbers given in the problem).
Thanks, totally forgot about taking the least accurate one into account as the number of significant figures.
T= (gSin(45)*(2500+200))/ sin(15)
Its 2000kg not 200kg
I got T= 120 483.44 N Sir . You made a mistake :) , you forgot 1/2
Yes, see the comments below. Thanks.
yup...answer is 120,615
i feel so bad bcos he did not include the 1/2 that should be multiplied with 5000 😂 😯✌
I THINK THE ANSWER IS 12048.3N
Yes, you are correct. (see comments below).
are you make an app over crane engineering and how much energy used by crane in running and lifting and efficiency of engine so we isolated the tachometer hours meter time and diesel consumption and various forms of energy in the same time and actual time or real time according to your wrist watch
this will help any crane engineering staff and operated machine
pollution solution become easy for millions of trillions of dollars world wide used in making pollution solution
It's never good to use 45 degrees in an example because you can't always be sure what angle in the triangle is being used. Also because sin 45 and cos 45 are the same.