Maths Mechanics help needed

[Mueez]

http://www.xtremepapers.me/CIE/Internat ... _qp_41.pdf
how 2 solve Q..4

 

[waleedsmz]

i) Since there are three forces acting on P, which are the two tensions and it's own weight, then you can apply Lami's rule. However, before you do that, you need to find the opposite angle of each force. And to do that, you'll need to get angle PAS and PSA first.

As you have a right angled triangle with three lengths given, you can apply the sine rule.
40/sinA = 50/sin90 = 30/sinS

PAS should give you 53.1' and PSA should give you 36.9'

You can now calculate all three angles at P... As the horizontal at P is parallel to AS ( also horizontal ) ..

Weight ----> opposite angle 90'
Long Tension ----> opposite angle 90+53.1
Short Tension ----> Opposite angle 90+36.9

By using Lami's theorem...
5/sin90 = T/sin143.1
Then T= 3N

 

[waleedsmz]

ii) As S is in equilibrium:

Force acting on the left = Force acting on the right

Therefore Friction = Tension X cos36.9
= 3cos36.9 = 2.4N

 

[Mueez]

can t we solve it by resolving forces into components

 

[waleedsmz]

iii) Again since S is in equilibrium, then force upwards = force downwards
This means that Normal Reaction force = W + Component of tension
R = W + 3sin36.9

Since Coefficient of friction X R = Friction
0.75 X R = 2.4 (given in ii )
R= 3.2N

Returning to the equation:
3.2= W + 3.2sin36.9
W= 1.4 N

I'd be thankful if you helped me in my question : viewtopic.php?f=26&t=6780&p=67053#p67053

 

[waleedsmz]

You can also do resolve forces into components... You'll need to get the angles too but I think Lami's rule is easier here...
I can answer it again by resolving forces if you want...