Resultant and Equivalent Force-Couple systems
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- čas přidán 10. 08. 2015
- This video screencast was created by Dr Terry Brown at the University of Technology, Sydney with Doceri on an iPad. Doceri is free in the iTunes app store. Learn more at www.doceri.com
I WAS SO USED TO RESOLVING INAPPROPRIATELY THAT 3:38 SEEMED COUNTER -INTUTIVE. THANKS FOR ENLIGHTMENT !
I just realized that I've been drawing the components wrong all the time! Thank you so much for showing me how to do it properly, this will make it much easier :))
thanks for the feedback. Glad to be able to help.
very usefull again ! thank you ver much!
great video, thank you so much.
you're welcome. glad you found it helpful
Thank you!
Thank you
thanks for the video, but how did u know that the moments are clockwise and anticlockwise?
I think my problem may be with the right hand rule.
Hi Hamz Gee, not sure exactly what the problem you are having is. Here's a bit of a long-winded video czcams.com/video/Vvlk0GUhVcQ/video.html of one way I demonstrate to students how to determine whether the moment of a force is clockwise or anti-clockwise. Once clockwise or anti-clockwise is determined, whether it's positive or negative depends on your assumption for what is positive or negative. In the original video above I have assumed clockwise as positive whereas in the video I've linked here, I have assumed anti-clockwise as positive. Hope this helps.
Ok I get you. Thanks so much
+Wilson Ambrose. Great. Happy to try to help. All the best.
Hello sir! and thank you for the video. If the question asked for the equivalent single force without specifying on what axis (supposed in your video that it wasn't specified to be on beam AB), will I use Mra/Fr to find the distance?
Hi, sorry I missed your comment. Did you still need help with this?
Tq u sir
thank you :)
You're welcome!
for this problem we only use external forces?
yes. In these problems we are finding force-couple systems that are equivalent to the original external loads (not finding the reactions). If you determine the reaction forces/moments for each of these equivalent systems you should get the same answers for the reactions. It's actually a good exercise to do this to help with understanding. I do this in this video czcams.com/video/iJpRW6dZDzY/video.html
To find the distance of the Resultant force, why do you divide the moment by the "y" component? why not the "x" or resultant?
hi Brandon, the x-component is not included because it passes through point A and thus has no moment about point A. Hope this clarifies things. cheers
Thank you good sir!
sir, how do you detect 35 cos30 in moment calculation and when it was sin
Only the vertical component of the 35N force causes moment about A. Vertical component is 35cos30
I'm a bit confused..How come for the 25N force, you used the perpendicular distance of 0.3m (till the axis of rotation) rather than to point A like you did with the other forces?
+Uthman Yunis I can try to explain,
the x-component of the 35N force does not create any moment about point A, so he only used the y-component. The distance between the y component and point A is along the line AB which is already perpendicular. Same for the 20N force.
+James Yi
Thanks for providing an explanation. You are correct in what you have said, but I think you are answering a different question to what Uthman asked. See my reply to Uthman.
+Uthman Yunis When calculating the moment you need the perpendicular distance between the line of action of the force and the point about which you are calculating moments. In the case of the 25 N force, its line of action is horizontal, so the perpendicular distance to point A is the vertical distance 0.3m. (Remember, you can move a force ALONG its line of action and not change its moment effect.)
What about the force 25N acting on 0.2m? Why does it act only on to 0.3m?
+Wilson Ambrose: Moment about point A = force x perpendicular distance between point A and line of action of force
So, for the 25N force, the perpendicular distance is 0.3m
What rule is the last rule you used to find the equivalent single force? (M_ra/F_ry)
not a 'rule' as such. Just a re-arrangement of the moment equation. Normally we have a force and a distance and want to know the moment. In this case we want to know the distance away the known force needs to be to create the known moment. We don't need to include F_rx component because it passes through the point A. The question could have asked: "how far above/below the beam (along the y-axis) do you need to place the equivalent single force?". In this case you would just use F_rx rather than F_ry. This is probably a worthwhile exercise to help your understanding. Hope this helps and hasn't confused you more.
Hey!
Really good video, but there's one thing that I can't wrap my head around and that's why you are using sin when calculating ΣFrx and cos when calculating ΣFry.
Shouldn't it be reversed since you're using the standard x/y coordinate system?
+Andre Alfonsson thanks for the feedback. sorry I didn't see your comment sooner. I use cos when calculating vertical component of the 35N force because the 30deg angle is defined relative to the vertical (y) direction. If you think about it in terms of right angle triangle and Pythagoras, then Frx is the "opposite" side and Fry is the "adjacent" side. Alternatively, I could redefine the angle as 60deg relative to horiz (x) direction and then I would use sin and cos as you suggest. Hope that helps. Cheers.
+Terry Brown Oh okay, yea that makes sense. Thanks ~
which is why you should not learn by rote memorization assuming cosine means x-component and sine means y-component. you must understand the free-body diagram
what about the distance from of force in X-axis
Nai, thanks for your question however I am unable to answer it because you have left out some information. Distance from where? and which force are you referring to?
how come you took the x and y components (of the resultant force ) on the tail of the arrow yet you took the x and y components (of 35 N ) towards the tip of the arrow
force is a sliding vector. therefore the 'arrow' representing the force can be drawn anywhere along the force's line of action. I could have drawn the resultant force with the tip of the arrow at A (and then draw the components in the same way as I did for the 35 N force) and it wouldn't change the solution. Hope this helps.
So before you find he distance d, do you always draw the resultant force at the axis of rotation (A in this example @ 7:09)?
you don't have to draw the resultant and moment at A (although in this question the equivalent force and moment at A were asked for), I included it in the solution because it helps some students to understand why the resultant single force is located to the right of A. Note that for the first part of the question, you could be asked to determine the equivalent force and moment about ANY point, e.g. B, C or anywhere else. The location of the single resultant force would still be the same, but the reference distance would be different, i.e. 0.39 to the left of B. Redo the problem using point B instead of point A and see for yourself :-)
Why is the resultant force facing downwards? Isn't is magnitude positive?
well not necessarily. F_resultant_x = +42.5N (leftward) and F_resultant_y = -50.3N (downward). so from this, it is obvious that the total resultant force is in between +x and -y plane which is pointing diagonally down at -49.8deg from the x-axis. look, resultant force is always gonna be yielded in positive value because you are using the hypotenuse formula, i.e. Total resultant force = sqrt(F_resultant_x^2 + F_resultant_y^2). so as you can see, regardless if the force is negative or positive, when you squared it, you will get positive magnitude in the end.
is the point of application the distance from the resultant force to the point
hi, it's been a long time since I did this video. At what time in the video is the bit where I talk about "point of application" that you don't understand?
point of application is just the location on the object where the force is applied
why do you use the vertical component of the 35 force for the moment??????
Why do you think I shouldn’t? The horizontal component does not cause a moment about point A because it’s line of action passes through point A (i.e. it’s perpendicular distance is zero, F x 0 = 0). Note that an alternative method to finding the moment caused by the 35N force would have been to find the perpendicular distance from the line of action of the force to point A. I.e. perpendicular distance is 0.2cos30. So then the moment calc would be 35 x 0.2cos30. The same answer as in my solution.
@@TerryBrownMechEng thx
Using the right hand rule counterclockwise should always be considered positive in my opinion but otherwise nice video
thanks for the comment. I agree, if positive x is to the right and and positive y is upwards. Note that in my solution y is downwards and so therefore using right hand rule, clockwise positive is actually correct :)
@@TerryBrownMechEng i stand corrected I actually just now got to resuming the video I commented before seeing the full solution 😂
I'm an MRA and calling me a clockwise is offensive.
You got the directions wrong
no, I didn't. Look at the sense I've assumed as positive. also, I'm calculating resultants and equivalents, NOT reactions. Perhaps you are expecting that I'm calculating reactions.