mechanics (non-uniform rods)

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<@&791435371564892232>

what have they done?? the M(A) bit

i get the 3R + R = 20
4R = 20, R = 5

but not the bit below

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Sum of clockwise moments

Equals sum of anti-clockwise moments

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yeah, please can u explain it, like in terms of the question

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It said find the distance from centre of mass from A

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we've found that R =5
where are we gonna use that

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So we let that distance be x

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yeah

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When we take moments we gonna substitute it

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centre of mass, thats the 20 right

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Yeh

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x is already here

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Alright,

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Yehh so the easiest way to find x is to take moments about A

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moment = F x d
so the F is the 5??
r= resultant force yeah?

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alright

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R is a reaction force

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yeah which is F

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Yh

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we use that in the equation moment = force x distance ?

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So the perpendicular distance from 3R is 1 right

To A

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Yeh

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lemme re send. so u and I dont have to scroll up all the time

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yeah

this ye

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the upwards forces about pivot A would make it rotate anticlockwise right

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yeah cuz 3R is greater than R? thats the reason innit

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Nah just imagine that only that 3R force was acting

It would wana rotate anticlockwise then

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is what i said right lol or irelevant

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Irrelevant

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oh

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whys that

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3R and R work together to rotate it anticlockwise

But have different torques

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yeah, theres no math to that right , u can just use the diagram cant u

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yh

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yeah im with u

whats a torque ?

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Moment

Same thingg

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oh alr

yeah then

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so what would be the anticlockwise moments about pivot A

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lemme re send cuz its annoying sccrolling up

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so add the distacnes and forces, and multiplty to get the meomoment ?

thats the bi tin the rectnagle right, which i drew

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so do it individually , for the 3R force its moment about pivot A is Force * perpendicular distance from A
3R * 1

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yeah

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and for R
Its R*2.5

And these work together so u must sdd them

Thats the total anticlockwise moment

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ahhh

2.5R and 3R

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And because its in equilibrium
Sum of anticlockwise is sum of clockwise which is 20x

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so 5.5R?

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Yh

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yeah = 20?

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20 is the force
We label the perpendicular distance from A to the weight as x

So 20x

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ye

ye 20x

we're tryna find x innit

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that would make it wana rotate clockwise so like downwrds

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"centre of mass of rod from A"

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Yhh cuz you know R

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R is 5

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5.5(5) =20x

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wdym by this again?

what would make it rotate clcokwise

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The weight

About pivot A

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y not anti clockwise, cuz its more towards the left, towards A

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imagine holding the rod at point A and letting it hang

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yeah it'd go clockwise

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If only the weight was acting it would

Yhh

But likewise

Imagine point B

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yeah but its not at A

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anticlockwise

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Yeh thats the jist of it

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yeah

so where were we again?

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5.5(5) =20x

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wait lemme re send, the q

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yeha

divide both sides by 20 to get x/

?

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Ye

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1.375

yeah?

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yup

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thank u so much my friend

what year are u actually? y13??

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U coulda solved it taking moments about B or the middle
But the algebra is more tedious

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ye

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Ye 😂

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haha fairs, what courses have u applied for?

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js maths
Wbu

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fairs, financial mathetmaics is my firm

which unis have u got offers from? u got 5/5 ?

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i see

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got 3 so far

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nice, from ?

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@ uni of sheffield

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durham bristol and Birmingham
But really wana go ww tbf

U get any offrrs yet?

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ww?

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ye bro 3/5

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Nice

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Warwick

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warick

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ye

just mathematics? or a joint

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js maths

Idk if im gonna regret it butt yhh

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alr

im guessin gu do maths fm phys?

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yeh

Its not that bad tbf

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yhh

what are ur predicteds

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surprisingly A* A* A*
But i feel like every teacher inflates it to let people apply

In the real thing id probs get 2 stars and an A

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ahh

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thats awesome dude

how do u revise maths btw?

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i just do qs thats it tbgf

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oh okay

imma try attempt it