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Mathematics: Post your doubts here!

Ashique

Ashique

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Oh god, so it's not only me. Can someone please take a look that?> ^^

Edit:


user said:
assalamoalaikum wr wb!
i need help withQ:7 ii of the same paper..
Click to expand...
Part ii asks us to show that the maximum acceleration is 2, and not greater than that. For this to happen, there must be no sliding between the above and the below boxes.
So considering limiting equlibrium, F=uR, and F=ma
0.2*2000=200a
a=2,
hence a<= 2
 
S

SUMMERLOV35

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Ashique said:
Again, I request someone to help me out with the last part-
Can someone PLEASE have a look at questin number 7 iii part, please, I realllu don't get it-
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf
Click to expand...

this question is repeated , you will find it in 2003 or 2004 paper too...
anyways for part i
r= 4500
f= 3150 so coefficient is 0.7

for second part they want the acceleration as in
between the blocks so obviously top one would the one to slip away...
and i figured out that whenever they show that a would not be greater than a value for the forces you only
take frictional force
0.2x2000=200a
a=2

to find the max value of P
there is frictional force so
P-F=450X2

REMEMBER FOR NET FORCE=MA
you always take net force

hope this helps
 
mushoo

mushoo

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Ashique said:
Again, I request someone to help me out with the last part-
Can someone PLEASE have a look at questin number 7 iii part, please, I realllu don't get it-
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf
Click to expand...

i dont know whether its correct or not,
max friction will be 3150; force=fric (for the obj still not moving), [as we did in aprt i]
and in part ii, we have already taken out that the max possible a will be 2ms-2
so by applying newtns second law,
F-friction=ma
P-3150=450*2
P=4050N.

if its correct, the ans n the procedure, then lemme kno too :p :/
 
iKhaled

iKhaled

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Ashique said:
Follow this diagram, and hope you will understand, then use the sine and the cosine rule to calculate the angles and magnitude
Click to expand...
my question is why W1 and W2 has got such vectors like these? r they equal to their tensions in the string? :S
 
C

Chooi1993

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Can anyone explain this to me ? Between I lost the question but this is my mock exam question .
They want us to find out the least arg which satisfied both.
Please !
 

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mushoo

mushoo

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iKhaled said:
my question is why W1 and W2 has got such vectors like these? r they equal to their tensions in the string? :S
Click to expand...

yeah they are equal to thr tensions,
you solve the question logically, using horizontal components of the tensions, the ans will be the same, cox its in equilbrm!
so the tension in the strings r equal to the weight in opp dir!
 
2

2pac

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Hey can someone please help me with q2 of may 2003 paper 2.Its regarding polynomials and I only know the divisible method,although the question is asking for the coefficient method,which sadly I can't do.Would anyone please explain me the coefficient method or better just tell me how to do this using my preferred divisible method.
Thanks and please reply soon.
 
leadingguy

leadingguy

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every one please post all your queries with the links of question papers and mark schemes.

it wil enable others to answer more clearly and fastly. thanx
 
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