Mechanics again and again.... (f=ur)

Two lights strings are attached to a block of mass 20kg. The block is in equ on a horizontal surface ab with the strings taut. Diagram below
http://img230.imageshack.us/img230/9246/maths3.png


Given surface is smooth find value of T and magnitude of contact force..

2 the surface is rough and block is on the point of slipping. The frictional force has magnitude 25N and acts towards A. Find coeefficient of friction between block and the surface...

First part was easy.... I equated the x component of both side and it gave me the value of 130

and using that value, I worked the y component that gave me 150,
200 - 150 = 50 N for the contact force....

Second part, i am confused, f = uR

F would be 25 ?
R = the resultant aka 50 ?

but i cant get the answer right, its supposed to be 0.268N

P.s some explanation how to work these cases would help me greatly..

Thanks

Quote:

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Originally Posted by ramzel

First part was easy.... I equated the x component of both side and it gave me the value of 130

and using that value, I worked the y component that gave me 150,
200 - 150 = 50 N for the contact force....

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can you explain how you did this?

fx= 75cos30-Tcos60 = 0
fy= 75sin30+Tsin60-200=0

and I assume A is at the T end B is at the 75N end right?

Yes..... A is the side of T, and 75 N is on the side of B....

I resolved T, as it has to be in equilibrium, The x values should equate, and the answer was 130 (which is right according to my books)..

And the contact force would be 200 (downward force - the (130sin60 + 75 sin 60) = 50 acting downwards.... (again according to my answer sheet its right)...

For the second part, i am confused, Does T change ? usually for these type of question i use normal = perpendicular to mass aka here the y components.... and the F acting would be the x components....

Resolving that gives the 50 downwards and the 25 direction of A.... What am i missing ?