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M2 Moments Question

So I missed my mechanics lesson due to a university interview, and they started the moments topic. I have the M2 textbook and have even looked at the solutionbank but I simply do not understand this new way of moments. I'd be grateful if someone would kindly explain, I'm on q1a at the moment and am simply lost :confused:

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Reply 1
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Original post by KaylaB
So I missed my mechanics lesson due to a university interview, and they started the moments topic. I have the M2 textbook and have even looked at the solutionbank but I simply do not understand this new way of moments. I'd be grateful if someone would kindly explain, I'm on q1a at the moment and am simply lost :confused:

Spoiler



The moment of a force is equal to the magnitude into the perpendicular distance of the force from the pivot. Using this definition, most people have two approaches to moments - one is to resolve the distance to the force from the pivot, the other is to resolve the force. I prefer the former, and looks like you've tried that from your diagram. Note that if you take moments about A, the reaction force, R, isn't considered, giving you the force T.
Reply 2
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Original post by KaylaB
So I missed my mechanics lesson due to a university interview, and they started the moments topic. I have the M2 textbook and have even looked at the solutionbank but I simply do not understand this new way of moments. I'd be grateful if someone would kindly explain, I'm on q1a at the moment and am simply lost :confused:

Spoiler



nice pun :ahee:
Reply 3
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Original post by aymanzayedmannan
nice pun :ahee:

:teehee: I didn't event realise

Original post by aymanzayedmannan
The moment of a force is equal to the magnitude into the perpendicular distance of the force from the pivot. Using this definition, most people have two approaches to moments - one is to resolve the distance to the force from the pivot, the other is to resolve the force. I prefer the former, and looks like you've tried that from your diagram. Note that if you take moments about A, the reaction force, R, isn't considered, giving you the force T.


Yeah I do it the first way too, but if I take moments about A to find T, wouldn't I then be able to take moments about B to find the reaction force at A?
Reply 4
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Original post by KaylaB
:teehee: I didn't event realise



Yeah I do it the first way too, but if I take moments about A to find T, wouldn't I then be able to take moments about B to find the reaction force at A?


This would certainly be the case if R was not acting at an angle. However, we do not actually know the direction or R, hence it is simpler to break down R into a vertical and a horizontal component (which I like to label Y and X respectively). After doing this, resolving forces upwards or taking moments about B will give you the force Y and resolving forces sideways will give you the force X. To find the magnitude and direction of R, you would then form a vector triangle using X and Y and hey presto!
Reply 5
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Original post by aymanzayedmannan
This would certainly be the case if R was not acting at an angle. However, we do not actually know the direction or R, hence it is simpler to break down R into a vertical and a horizontal component (which I like to label Y and X respectively). After doing this, resolving forces upwards or taking moments about B will give you the force Y and resolving forces sideways will give you the force X. To find the magnitude and direction of R, you would then form a vector triangle using X and Y and hey presto!

So I'm currently here so far, I don't know what you mean by then resolving sideways as it's a beam and thus I don't imagine that it is flying sideways
Reply 6
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Original post by KaylaB
So I'm currently here so far, I don't know what you mean by then resolving sideways as it's a beam and thus I don't imagine that it is flying sideways


So far, so good!

Well, it's not flying upwards either, is it? :tongue: That's why Ry+Tsin⁡θ−6g=0\displaystyle R_{y} + T\sin\theta - 6g = 0 worked in the first place, right? Notice that RxR_{x} is being balanced by another force, Tcos⁡θT\cos\theta. Can you resolve forces horizontally using this information?
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Original post by aymanzayedmannan
So far, so good!

Well, it's not flying upwards either, is it? :tongue: That's why Ry+Tsin⁡θ−6g=0\displaystyle R_{y} + T\sin\theta - 6g = 0 worked in the first place, right? Notice that RxR_{x} is being balanced by another force, Tcos⁡θT\cos\theta. Can you resolve forces horizontally using this information?


Rx = Tcos(theta)

But then how can I find the value of T or even Rx from this?
Reply 8
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Original post by KaylaB
Rx = Tcos(theta)

But then how can I find the value of T or even Rx from this?


Did you not find the value of T by taking moments about A? If you know the value of T, you know cos(theta) and you can find Rx.
Reply 9
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Original post by aymanzayedmannan
Did you not find the value of T by taking moments about A? If you know the value of T, you know cos(theta) and you can find Rx.


By taking moments about A, I just get 3g = Tsin(theta) which I also found from previous working
Reply 10
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Original post by KaylaB
By taking moments about A, I just get 3g = Tsin(theta) which I also found from previous working


But here's the thing. You can find the angle theta. You have the length of the rod, and you know the distance of the force T from the pivot at A. It forms a right angled triangle so you can find it using trigonometry.
Reply 11
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Original post by aymanzayedmannan
But here's the thing. You can find the angle theta. You have the length of the rod, and you know the distance of the force T from the pivot at A. It forms a right angled triangle so you can find it using trigonometry.


Ahhh, so 1.5/4 = sin(theta) = 3/8
which could then be put in to 3g = Tsin(theta) to make
3g = T(3/8)
so T= 3g/(3/8) so T=8g?
Reply 12
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Original post by KaylaB
Ahhh, so 1.5/4 = sin(theta) = 3/8
which could then be put in to 3g = Tsin(theta) to make
3g = T(3/8)
so T= 3g/(3/8) so T=8g?


Does it match the book's answer? I think it's correct.
Reply 13
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Original post by aymanzayedmannan
Does it match the book's answer? I think it's correct.


Yeah it does. The solutions just work it out in like one or two steps and I'm here with a whole page of workings :colondollar:

Thank you sooo much for your help :h::h:
Reply 14
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Original post by KaylaB
Yeah it does. The solutions just work it out in like one or two steps and I'm here with a whole page of workings :colondollar:

Thank you sooo much for your help :h::h:


The solution bank is very unhelpful at times and it's difficult to understand what they're doing. I hope you understood the problem now!

No worries :colondollar:

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