Mechanical Engineering: Ch 10: Forces on Cables (8 of 33) Concentrated Load: Example (Part 1)
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- เผยแพร่เมื่อ 5 ก.พ. 2025
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In this video I will calculate the various force acting at different points of a cable with concentrated load (Part 1).
Next video in this series can be seen at:
• Mechanical Engineering...
![EQUILIBRIUM: Cables - Problem #1 [1/2]](/img/n.gif)
Im a student from the Netherlands and ur videos help me alot
You're a world treasure. From Uganda
Thank you and welcome to the channel!
You are a very good educator. I can easily catch up to your discussions.Thank you for being born in this world. Please make lots and lots of videos like this. ❤
I love the way he says Newtons
Thank you so much. Your handwriting and the way you explain make easy for me to understand.
Fantastic
Love from india
Glad you liked it. Welcome to the channel! 🙂
I have a question, when calculating moment , we use negative sign when force is turning clockwise?
When using vector notation, yes.
Very good teacher, thank you very much
Thank you! 😃 Glad you liked it.
Deer Sr. I have a question, please, if there is a cable suspended between two pins say point A & E, and point E vertically little higher than A loaded with two unequal weights w1 > w2 at point B & D respectively and there is a pulley at point C between B & D and at a hight between A & E, and assume the cable is long enough, will the system have two sags at B & D or one sag only at point B due two the higher weight?. waiting for your reply. Thank you.
Sir appreciate for the video. You said that when considering moments at point C we can take one side if its cable, is it because they dont have bending moments like rigid structures? Kindly respond pls
Thank you for the wonderful videos! Could you please explain why we are able to ignore the other half of the cable?
@@elliotmartin8986 method of sections, in equilibrium we can cut a structure into parts because there won't be a change on the structure since it is at equilibria
why not calculating moment at D to the left so we can get Ay easily?
First i get the moment at A getting the Ex & Ey unknowns. Second i did a moment at D (to its right) leaving Ex & Ey as unknowns. Solving both Ex and Ey wont match with yours. Whyy is that? :(( Does it vary on where u get the moment? Any tips if so?
How comes when you try to take moments at point D, you get a different result?
Hi, upon using point D as my pivot, I found we can eliminate Ax and find Ay. But The Ay I calculated doesn't correlate with your answer. Any reasons?
stfu kid u poes dom
I think the reason behind it was the symmetry. If you cut the cables at Joint C, the horizontal distance from A-C is 3 and E-C is 3. I think that was the reason he cut it at Joint C.
So helpful
Glad you think so! 🙂
Thank you
I think the Ay in the first equation should be negative since it goes clockwise when getting Moment at E
The is correct and that is why the term was subtracted.
the direction of the moment of Ax at C still confuses me. how comes its different from that at point E
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Thank you for your endorsement. At least through the videos, we can reach more students. 🙂
Very helpful. Thank you!
I tried to calculate moment at point D, I got different result. I thought in cable we calculate moment at any point.
In which course do you take this topic? I am in my first year of civil engineering and taking this in static.. Although I am good at trusses but cables are a Lil bit different.. Due to solving an equation algebraically
This is exactly what I wanted to ask. Taking moment about point D is an easy way of getting Ay, but it gives different answer.
I had the same exact question
@@lapisbrainprint9607 me too why does point D five a different answer
Hi Michel!. i got your example but, i do not if can you performance an exercise with just 2 Forces with their respective masses, Greetings!
ótimo vídeo! ajudou muito.
Obrigado
I found Ey as 29333 N!
the 1m vertical distance is infact Not from point C to D but rather from point C to the origin! missed that at first.
Thanks for the lecture
I think clockwise is positive and counter is negative
The convention for torque is that clockwise is negative and counter-clockwise is positive (when used in vector format)
Why horizontal thrust forces are equal at both the ends
If they were not equal, then there would be a net force in the horizontal direction, and the whole system would accelerate to one side or the other.
This is driving me crazy, i'm given the sags and distances but i'm expected to find d. Can't find an example anywhere.
Actually, the simple fact that moments can be taken to left and right of any cable point was the key to solving it. My lecturer never mentioned that. Thankyou.
When moment about C=0..Why Ax is counter
Clockwise?? However Ax was clockwise when Moment about E =0
The line of action of Ax is above C and below E.
@@MichelvanBiezen Do you have an explanation why?
just by looking at the geometric of this cable Ey can't be greater than Ay
When taking the moments at point C, why do we get to calculate only the moments on the left side?
You can calculate the moment on either side. But there is a strategy that will get you the answers and this video shows you how to do that.
Thank you so much. Why does that strategy work? To solve for the reactions at A, we take the moments at E and C. But at C, we take only the moments at the left side to get the reactions. Why does that work? It seems like it wouldn't be equal or something
its because tension in cables is equal on either sides and at any point along the cable its as good as the contact point A and E
THANKS BRO
It's like method of section in trusses.
@@iconbylvergara6188 Unlike beams, cables can't take any inner moment (any moment will bend the cable until there's no moment) , so at every point along the cable the total moment must be equal to 0
can anyone show a solution for a multispan problem
How did that -2 come on 4:46?
Since we are doing the same thing to both sides of the equation, we could have picked any number.
In algebra i think its called Elimination? Correct me if I'm wrong.
When you start with Ex and Ey and take moment about C and A, respectively, you get different result. There's something wrong with this cable. This is so missed up!
I think the methodology youre doing makes sense but there is something wrong with the question. Are you sure that is a stable system ? Because when I choose a different pivot point it always gives a different answer for the x and y components of force at A. For example by taking D as a pivot point. Total moment about point D=0, A_x is gone because the perpendicular distance is 0. So we only have A_y and it turns out to be 14.666 N. Could you please check the question bec I have an exam tomorrow and Im pretty confused. Thanks :)
That is a good observation. Since this example problem is not a "real" problem, the vertical positions of the loads do not correspond to where they would actually be in a real situation and therefore you will get a different answer if you picked a different pivot point (D instead of C). But the methodology shown is correct and the one to follow. Your answer of 14,666 N is correct if you pick pivot point D.
Thanks a lot,we are grateful :)
A sub y is wrong
this is soooo wrong. sorry but not sorry. you take the cable as rigid body or you calculate it as 3 hinged arch. if you make MD=0 (right side) then Ey=48000