# Can Someone Explain Moments to Me Please? (Physics)Watch

#1
So I really don't understand moments. Just any of it. Every time I feel like I understand it, I try and practice question and get it completely wrong. Please use an example question, thank you.
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1 year ago
#2
I recommend past paper questions for moments! Get one with the answers in the back and try to work through it like that. For me it just worked memorising the answers to the questions which are often repeated in the paper
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1 year ago
#3
(Original post by physconomics)
So I really don't understand moments. Just any of it. Every time I feel like I understand it, I try and practice question and get it completely wrong. Please use an example question, thank you.

Maybe watch this video? he really helped me, it's for GCSE but he really explains it well, I learnt it 2 years ago and can recite it to you by memory cuz of this guy!

Hope it helps
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#4
Maybe watch this video? he really helped me, it's for GCSE but he really explains it well, I learnt it 2 years ago and can recite it to you by memory cuz of this guy!

Hope it helps
See, I totally understand this but when theres a really complicated situation in a question it just flies over my head. I never know where to take the pivot from if there's multiple pivots or just ah! It's so infuriating haha, I'm so jealous of everyone who did it in their gcse!!
(Also is this guy a vampire?? He looks creepy as sh*t)
Thanks though!!
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#5
http://pmt.physicsandmathstutor.com/...%20Moments.pdf

Let's say Q1b, can anyone explain it for me? Like in words? Idk. This is driving me insane.
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1 year ago
#6
(Original post by physconomics)
http://pmt.physicsandmathstutor.com/...%20Moments.pdf

Let's say Q1b, can anyone explain it for me? Like in words? Idk. This is driving me insane.
We are given the tension in Q, and the weight of the rod. The only other force on the rod is the tension in P. By resolving forces vertically, we can deduce that the tension in P is 3.0N, as P+Q=W

To calculate d, we need to take moments about somewhere. As we have rotational equilibrium the moments about *any* point must be zero.

I'd chose to take moments about where P crosses the rod, but anywhere will work: The easiest places will be where the wires attach, as that eliminates one of the forces from the calculation.

(D=90cm, W=5,0N)

I'll let you take it from there.
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1 year ago
#7
(Original post by physconomics)
See, I totally understand this but when theres a really complicated situation in a question it just flies over my head. I never know where to take the pivot from if there's multiple pivots or just ah! It's so infuriating haha, I'm so jealous of everyone who did it in their gcse!!
(Also is this guy a vampire?? He looks creepy as sh*t)
Thanks though!!
Aha yeah i totally get what you mean - i did all this in my triple physics GCSE 2 years ago so most of it i cannot remember to save my life - sorry i wasnt much help, maybe just go on youtube and search for worked examples o0f harder questions - if you want i could find some for u but it wouldnt be today

i kinda get what you mean but i dont know how to explain it over a keyboard so im not much use there - sorry my physics knoledge only goes up to Triple AQA GCSE, i'd assume it would be the middle of both pivots then something idk sorry - he has more videos on pivots and moments if that helps

Aha yeah he does look pretty scary tbh - saved my GCSE'S though!!!

okay ive written way too much sorry
Good luck tho!
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#8
(Original post by RogerOxon)
We are given the tension in Q, and the weight of the rod. The only other force on the rod is the tension in P. By resolving forces vertically, we can deduce that the tension in P is 3.0N, as P+Q=W

To calculate d, we need to take moments about somewhere. As we have rotational equilibrium the moments about *any* point must be zero.

I'd chose to take moments about where P crosses the rod, but anywhere will work: The easiest places will be where the wires attach, as that eliminates one of the forces from the calculation.

(D=90cm, W=5,0N)

I'll let you take it from there.
So would it be:
5 x d = 90 x 2
d = 36cm?

(Thank you so much)
0
1 year ago
#9
(Original post by physconomics)
So would it be:
5 x d = 90 x 2
d = 36cm?

(Thank you so much)
Yes, that looks right.

It might be a good idea to convince yourself that you could take the moments at another point too - try where Q attaches.

Another common mistake with taking moments is getting the signs wrong. You need to have a convention for which direction (clockwise or anti-clockwise) is positive. In this question, I simplified it as there were only two forces, so said that the moments were equal and opposite. Some questions may require you to say that the sum of all moments is zero - getting the signs right there is key.
0
#10
(Original post by RogerOxon)
Yes, that looks right.

It might be a good idea to convince yourself that you could take the moments at another point too - try where Q attaches.

Another common mistake with taking moments is getting the signs wrong. You need to have a convention for which direction (clockwise or anti-clockwise) is positive. In this question, I simplified it as there were only two forces, so said that the moments were equal and opposite. Some questions may require you to say that the sum of all moments is zero - getting the signs right there is key.
3 x 90 = 5 x (90-d)
270 = 450 - 5d
d = 36cm
.. Well that's a good sign.
Thank you so much, I honestly think I'm beginning to understand it now. I'm going to be working at this all week until I've mastered it!! Is there anything I can do to repay you haha!!!
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