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A level math help.

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im supposed to express w in term of W. myoo is a sixth root 3. AB is in equilibrium. i took moments about A but I keep getting W cancelling out please help.
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Reply 2
Original post by mrwonderful
im supposed to express w in term of W. myoo is a sixth root 3. AB is in equilibrium. i took moments about A but I keep getting W cancelling out please help.
image-bb2b328e-c3b3-47e6-86b5-d9b81c59f98a7406129060654171906-compressed.jpg.jpeg

Please post the full question along with all your working which shows the problem you’re having.
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Original post by Notnek
Please post the full question along with all your working which shows the problem you’re having.
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its question 9
(edited 4 years ago)
Reply 4
Original post by mrwonderful
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Attachment not found

its question 9

I can have a look in around 15 mins
Reply 5
Original post by mrwonderful
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its question 9

Can you please post your diagram again? I can't see it in your first post anymore.
Original post by Notnek
Can you please post your diagram again? I can't see it in your first post anymore.


Original post by Notnek
Can you please post your diagram again? I can't see it in your first post anymore.


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Reply 7
Original post by mrwonderful
x

There appear to be two w's in your working. One is the weight of the ladder but what's the other one?
Original post by Notnek
There appear to be two w's in your working. One is the weight of the ladder but what's the other one?

part c of the question says a load of weight w ia applied at B. The weight of the ladder is W. and i am to express w in terms of Big W
Original post by mrwonderful
part c of the question says a load of weight w ia applied at B. The weight of the ladder is W. and i am to express w in terms of Big W


If you're answering part (a) then don't use any information that isn't given until part (b)/(c).

The wall is smooth and in part (a) there is no weight hanging at B, so there is only one force acting at B. The force you're looking for is the normal reaction of the wall on the ladder, so you don't need to know what μ\mu is (just as well because you're never given a value for it).
Original post by Quirky Object
If you're answering part (a) then don't use any information that isn't given until part (b)/(c).

The wall is smooth and in part (a) there is no weight hanging at B, so there is only one force acting at B. The force you're looking for is the normal reaction of the wall on the ladder, so you don't need to know what μ\mu is (just as well because you're never given a value for it).


i got part A and B correct, i am just struggling a bit with part c.
Reply 11
Original post by mrwonderful
i got part A and B correct, i am just struggling a bit with part c.

Try taking moments about B instead. Post your working if you're still stuck.
Original post by mrwonderful
i got part A and B correct, i am just struggling a bit with part c.


Oh sorry, part c wasn't in the picture you posted so I assumed you were just overcomplicating part a :lol: can you post part c?
Original post by Quirky Object
Oh sorry, part c wasn't in the picture you posted so I assumed you were just overcomplicating part a :lol: can you post part c?


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Original post by Notnek
Try taking moments about B instead. Post your working if you're still stuck.

ok ill try that but why doesnt it work with moments about A?
Reply 15
Original post by mrwonderful
ok ill try that but why doesnt it work with moments about A?

EDIT: After actually trying the question I've realised that it's impossible wherever you take moments :smile:

You have to think carefully about what variables you have and what kind of equation you will arrive at if you take moments. You have

RB=36(W+w)R_B = \frac{\sqrt{3}}{6}(W+w)

So if you want an equation just involving WW and ww then you're going to need to eliminate RBR_B.

If you take moments amount A then that's going to give you an equation involving RBR_B, WW and ww but you already have an equation with those variables so this isn't going to help.

If instead you take moments about B then you'll get an equation involving WW and RBR_B (since RAR_A can be written in terms of RBR_B). So then you'll be able to write RBR_B in terms of WW and eliminate it from the above equation.
(edited 4 years ago)
Reply 16
@mrwonderful I recommend that you don't use the app when posting in the maths forum mainly because maths formatting doesn't show up properly (you may notice that in my post above) and attachments are very buggy. Use the mobile site instead if you're using your phone.
Original post by Notnek
Try taking moments about B instead. Post your working if you're still stuck.


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ok so i took moments about B and Big W cancelled out!
Reply 18
Original post by mrwonderful
X

After trying the question myself I've realised that it's impossible! Then I checked my (updated) version of the textbook and realised that they have corrected a typo. Here's the new version of the question:



μ\mu should be 35\frac{\sqrt{3}}{5}.

Thanks Pearson for wasting both of our times :grumble:

By the way for future questions please try to post everything in a single post. It was tricky looking through all your posts to find all the information as I looked through your working.
Original post by Notnek
After trying the question myself I've realised that it's impossible! Then I checked my (updated) version of the textbook and realised that they have corrected a typo. Here's the new version of the question:



μ\mu should be 35\frac{\sqrt{3}}{5}.

Thanks Pearson for wasting both of our times :grumble:

By the way for future questions please try to post everything in a single post. It was tricky looking through all your posts to find all the information as I looked through your working.


i knew it! it just wasnt working.with the myoo value given. I now got w is equal to a quarter big W

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