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Mechanics Questions

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mousa:
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

123:
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)

mousa:
Thanks alot... thoughQ6 isnt tha much clear.
anymore replies ?? astar??

astarmathsandphysics:
In court today.

astarmathsandphysics:
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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