IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: mousa on December 04, 2009, 05:22:44 am
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well ppl i got a couple of Questions. see them attached below. hope you will help me as soon as possible., Thanks in advance.
Q 6 pg:57
Q 11 pg:58
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forgot how q6 pg57 is done :P
I have done it b4, mayb try
resolving 2g vertically downwards (opposite to P & Q)
So for P & Q
P = 2gcos40
Q = 2gcos20
Though i don't think its simple as this!
Then q11 pg.58
Just resolve T & 2T vertically & horizontally
So
VERTICALLY: Tcos 60 + 2Tcos
going upwards but 50N going downwards
Tcos 60 + 2Tcos
= 50 ----------eq.1
HORIZONTALLY: T & 2T go in opposite directions: Tsin60 = 2Tsin
---------eq.2
For
from eq. 2
Tsin60 = 2Tsin 
sin60/2 = sin
= sin^-1 (sin60/ 2)
u know
.....so find T from eq.1 !!
Hope this is right & clear
(PS: instead of a thanks plz give me +rep. power!! ;D)
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Thanks alot... thoughQ6 isnt tha much clear.
anymore replies ?? astar??
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In court today.
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res vertically Pcos 50+Q cos 20=2g
res horizontally Psin40-Qsin20=0 so P=Qsin20/sin40
P(cos50+sin20/sin40*cos20)=2g and P=2g/(cos50+sin20/sin40*cos20)
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res horizontally Tsin60 -2Tsinx=x so sinx=1/2 sin60=sqrt(3)/4
sub this into the equation below to find T
Res vertically Tcos 60+2Tcosx=50
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here
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here
Thanks alot sir, realy helpful, but how you assumed , all of the three forces, 2g ,p and Q act from the same point?? is that clearly ressembled by the diagram???
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Yes. I resolved vertically and horizontally as shown.
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Why are u being sued astar ??
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I sued barclays cos they cashed someone elses cheques from my account
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Dammit, and I thought England was heaven.