Thank you so much. Great explanation. I like the way to explain things, you make complex material easy to understand. Can't wait for your vibrations videos :)
Matthew James dude can u plss explain me crank with counter clock wise motion I m little stuck in solving that.. I will be glad if u help on this as well..
Matthew James thnx bro so much for your revert.. I forwarded the same numerical to your given mail ID plss review and let me know the same way you did earlier one.. Waiting for your response ASAP brother..
Hi Matthew, Would the method be the same if there was an offset between A and C vertically ? Also wondering if there will be a video about free body diagrams and forces for this type of problem :) Thanks!
if Vcb=Vc-Vb then is it possible to put the value of Vc and Vb and get the value of Vcb anf if not then why again we have to look at the veloctiy triangle to get the value of Vcb ?
Hey mate, yep it's definitely true that *Vc/b* = *Vc* - *Vb*. Once you figure out *Vc* then you can easily plug in *Vb* and *Vc* back into that equation to find *Vc/b*. Be careful about solving this equation though! *Vc/b* and *Vc* and *Vb* are vectors, not scalars! Which means you can only solve them graphically (using velocity triangles) or simultaneously (using *i* *j* and *k* notation). Hope that makes sense, Cheers, Matt
Matthew James SIMILAR QUESTION IS GIVEN IN MY TEST BOOK .WHERE THE ANGULAR VELOCITY OF THE CRANK (A-B) IS 20 rad/sec .AND THE LENGTH OF THE CRANK RADIUS IS 10 CM. AND THE VELOCITY OF THE CONNECTING ROD (B-C) IN TERMS OF rad/sec IS . also given angle of A is 45 degree insted of 40 degree as per your diagram and the angle of C is given 30 degree .here angle c is known 1.100 rad/sec 2.50 rad/sec 3.10 rad/sec 4.non of these but in book it solved like this angular velocity of connecting rod W(bc)= (angular velocity of crank x cos 45*) /n where n= ratio of connecting rod and crank length= l/r n=l/r=sin 45*/ sin 30* n=root 2 so angular velocity of connecting rod W(bc)=( 20 x cos45*) / root 2 = 10 rad /sec.....(.ans) i understood solving graphically by your method but unable to apply in this question as i am getting different answer .please clarify my doubt .i have my exam very soon
Hi thank you for your great intuitive skills has really save me so many hours. how ever I have been trying to figure out how you were able o use the trignometry at 10:10 to get Vc and get the 13.08m per seconds. Can you use find a way to text it here or send me paper format of the solution. if you can use text , i will understand too. Thanks so much.
Hey mate, I use the sine rule: en.wikipedia.org/wiki/Law_of_sines Basically it says sin(a)/A = sin(b)/B where a is the angle opposite to side length A, and b is the angle opposite side length B. I apply the sine rule on the triangle in the bottom left of the screen. Hope that helps :)
Hey mate, sorry for the late reply. Not for at least a few months unfortunately. I'm creating a library of videos on Engineering Vibrations at the moment. Do you have any requests you want to make? Like do you want me to cover coriolis acceleration? Instantaneous centers? etc etc
+Raven Claw Hey mate, great question! This problem only deals with finding the angular velocity at an instant, but in order to find the maximum angular velocity of the rotating bar over the course of one complete cycle you'd have to follow the same process keeping angle BAC a variable (eg theta). Once you go though the mathematics you'll get w_BC as a function of theta. Then solve for dw/dtheta =0 to find the value of theta corresponding to the maximum and minimum values! Plug this value of theta back into the generalized equation to solve for wmax or wmin (whichever one it may be). Good luck!
Hey mate, if you changed the position of pin A to be vertically above where it is now then the problem only becomes a tad harder. The geometry would change which would adjust the angles, but the process of finding Vb then Vc then Vc/b then w_b/c is the exact same. Hope that helps :)
you just saved my semester man thank youuuu
Thank you so much. Great explanation. I like the way to explain things, you make complex material easy to understand. Can't wait for your vibrations videos :)
A brilliant video lesson! Thanks!
Great, cleared my doubts!!! Thanks a ton :)
Thanks a lot! Very clear explanation
Thank you so much, my mechanism was a little different but the way of the calculations are the same.
glad you found it helpful :)
Thank you very much your detailed explanation i like and it is helpful for me and all concern
hey! thanks bro, you're helpful. Comprehensive explanation
Thank you Mathew , that was good reminder .
glad you liked it :)
Thanks for the video!!!
superb,,,loved it..thanks
No worries :D
Thanx u so much brother for making it do simple to understand.. Thnx a lot..
I'm so glad you found this video helpful :)
Matthew James dude can u plss explain me crank with counter clock wise motion I m little stuck in solving that.. I will be glad if u help on this as well..
Hey mate can you email the problem to virtuallypassed@gmail.com and I'll try and answer that as soon as I can :)
Matthew James thnx bro so much for your revert.. I forwarded the same numerical to your given mail ID plss review and let me know the same way you did earlier one.. Waiting for your response ASAP brother..
Thanks a ton
great lesson. .
thank you. .
+Alistein 1 Thanks! :D
This made me to pass B.E exam.
Great to hear!!
Why is that Vc/b is not equal to the value of the difference of Vc and Vb?
Amazing. .keep it up mate
Thanks, mate :)
Hi Matthew, Would the method be the same if there was an offset between A and C vertically ?
Also wondering if there will be a video about free body diagrams and forces for this type of problem :)
Thanks!
Yes, the method will be the same.
i love you
I love him more than you do
Thank you
awesomee
thank you :)
now for acceleration please...
thank you sooooooooooooooooooo much
You're welcome :D
if Vcb=Vc-Vb then is it possible to put the value of Vc and Vb and get the value of Vcb anf if not then why again we have to look at the veloctiy triangle to get the value of Vcb ?
Hey mate, yep it's definitely true that *Vc/b* = *Vc* - *Vb*. Once you figure out *Vc* then you can easily plug in *Vb* and *Vc* back into that equation to find *Vc/b*. Be careful about solving this equation though! *Vc/b* and *Vc* and *Vb* are vectors, not scalars! Which means you can only solve them graphically (using velocity triangles) or simultaneously (using *i* *j* and *k* notation). Hope that makes sense,
Cheers,
Matt
Matthew James THANKS
Matthew James SIMILAR QUESTION IS GIVEN IN MY TEST BOOK .WHERE THE ANGULAR VELOCITY OF THE CRANK (A-B) IS 20 rad/sec .AND THE LENGTH OF THE CRANK RADIUS IS 10 CM. AND THE VELOCITY OF THE CONNECTING ROD (B-C) IN TERMS OF rad/sec IS . also given angle of A is 45 degree insted of 40 degree as per your diagram and the angle of C is given 30 degree .here angle c is known
1.100 rad/sec
2.50 rad/sec
3.10 rad/sec
4.non of these
but in book it solved like this
angular velocity of connecting rod W(bc)= (angular velocity of crank x cos 45*) /n
where n= ratio of connecting rod and crank length= l/r
n=l/r=sin 45*/ sin 30*
n=root 2
so
angular velocity of connecting rod W(bc)=( 20 x cos45*) / root 2 = 10 rad /sec.....(.ans)
i understood solving graphically by your method but unable to apply in this question as i am getting different answer .please clarify my doubt .i have my exam very soon
New universal linkage method and some examples:
www.mapleprimes.com/posts/204684-Lever-Mechanisms-
Hi thank you for your great intuitive skills has really save me so many hours. how ever I have been trying to figure out how you were able o use the trignometry at 10:10 to get Vc and get the 13.08m per seconds. Can you use find a way to text it here or send me paper format of the solution. if you can use text , i will understand too. Thanks so much.
Hey mate, I use the sine rule: en.wikipedia.org/wiki/Law_of_sines
Basically it says sin(a)/A = sin(b)/B where a is the angle opposite to side length A, and b is the angle opposite side length B.
I apply the sine rule on the triangle in the bottom left of the screen.
Hope that helps :)
are you going to upload relative acceleration videos anytime soon?
Hey mate, sorry for the late reply. Not for at least a few months unfortunately. I'm creating a library of videos on Engineering Vibrations at the moment. Do you have any requests you want to make? Like do you want me to cover coriolis acceleration? Instantaneous centers? etc etc
thanks. what is the rotational speed of link BC? can u calculate the corresponding angle BAC for the max rotational velocity of link BC?
+Raven Claw Hey mate, great question! This problem only deals with finding the angular velocity at an instant, but in order to find the maximum angular velocity of the rotating bar over the course of one complete cycle you'd have to follow the same process keeping angle BAC a variable (eg theta). Once you go though the mathematics you'll get w_BC as a function of theta. Then solve for dw/dtheta =0 to find the value of theta corresponding to the maximum and minimum values! Plug this value of theta back into the generalized equation to solve for wmax or wmin (whichever one it may be). Good luck!
2:43 you're running out of disk space is my brains
what happens if you offset A by an amount?
Hey mate! Do you mean if you move pin A to another place? Or do you mean if A becomes a slider (like C)?
say you offset A vertically?
Hey mate, if you changed the position of pin A to be vertically above where it is now then the problem only becomes a tad harder. The geometry would change which would adjust the angles, but the process of finding Vb then Vc then Vc/b then w_b/c is the exact same. Hope that helps :)
Matthew James I think it did, thanks!
you should have written the college textbooks. i hope that makes sense
Haha thanks mate :D
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