Mechanical Engineering: Equilibrium of Rigid Bodies (22 of 32) Ex. 6 Eq. of 3-Force Body
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- čas přidán 3. 10. 2015
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In this video I will find the angle of a 3-force body of a rod balanced in an inverted semi-sphere glass bowl.
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Thank you Michel. Now It's clear like water. now I can solve for L = 3R or 4R or anything. But I took a shortcut on solving the theta using calculator to solve for x . the I calculate the angle in first quadrant. I found that easy and time saver. Again Thanks a lot.
Glad it helped!
Nice explanation sir ❤ love from india
Thank you. Welcome to the channel!
there's something wrong with the distance from Fb to A when calculating the total moment. at 7:07 you took Rcostheta for the perpendicular distance(d) for the weight mg. that means R is equal to the distance from point A to the red dot at the center of the rod. But at 7:48 you take 2Rcostheta for the distance from Fb. 2R is the whole length of the rod but the Fb is not applied at the end of the rod.
If one fails to observe that FA is radial, and FB is normal to the rod, can he/she still proceed with this problem?
eg by assuming forces at A are FAX and FAY, and that the forces at B are FBX and FBY ?
I was searching the solve of this particular problem for a long time. Thanks for this one.
Also, can you suggest a book in which this/similar problems may be given. Or where did you find this one (book)?
THANKS again.
This is from a class I took (a long time ago).
A similar problem you can find in the book "Fundamentals of Physics" from Alonso and Finn, vol1.
4th chapter
exercise 4.40
Simply Wow
i think it would of been easier to use the AB and 2R therefore giving you a right angle triangle
Why isn’t FB towards the center of the sphere like FA?
If there is no friction, then the force MUST be perpendicular to the surface.
How can we assume the angles are the same from drawing the perpendicular line? Under the assumption that the drawing is not to scale, what would be the proof for this. I'm unable to grasp the basis of this.
Is it because the horizontal line segement has a magnitude equal to that of the radius. And, since all sides are equal, the angles would be equal as well?
If two lines are perpendiucular to one another, then the reference angles from those lines to the vertical and horizontal axis must be equatl (since they are perpendicular as well).
2:30 why is the angle of Fb off the vertical project also theta? I'm not sure how to derive that.
Oh, is this base off the parallelogram of lengths R? Since Fb mirrors the first perpendicular, you can reference it to derive theta?
Because the direction of Fb is also perpendicular to the rod.
How the angle between vertical and force B is theta? Explain Geometry
Thank you a lot, professor! Can I try to solve this problem in an alternative way? I assume that the center of mass in this equilibrium will be at the lowest point possible. Not sure if this assumption is correct , but if it is, I can try to write out the expression for the height of the center of mass, using theta. Then I only need to find the minimum of this expression. This way I don't need to analyze the forces.
I always recommend trying different approaches because that way you will understand the material better. Try it and see if it works and see if you get the correct answer at the end.
@@MichelvanBiezen Dear professor, thank you for your reply! I tried and the trigonometry got quite messy lol. I will keep working on it and see whether it works though.
That is part of the problem. Sometime you pick the method that makes solving the problem the easiest.
@@MichelvanBiezen Yeah, that's right!
Sir, why mgd1 is negative since mg acts at the center of the mass of the rod and it is inclined? Shouldn't it it be in counter clockwise direction?
Relative to the pivot point A, the mgd1 is negative (clockwise torque).
Doesn't mg act at the center of mass of the rod?
+82rah
That is correct. The drawing is a little off, but the problem was worked with the assumption that mg acts at the center of the rod.
for cosine 2theta,, wasn't it supposed to be (cos^2theta-sin^2theta)?
That is indeed the trig identity for cos(2 theta), but note that in the video we are referencing the force diagram, therefore the equation is not purely the trig identity you are referencing.
Lol I guess this is the lengthiest problem you have done