Problem of fluid -- Blocks, pulley, two fluids

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Queren Suriano
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Homework Statement


dynamic viscosity 1 = 1125 2 kgf-s / m2 , dynamic viscosity 2 = 62.5 2 kgf-s / m2 , Areas: A1 = 20 cm2 , A2 = 50 cm2
W1 = 15 Kgf, W2 = 40 Kgf.
The block (2) is displaced downwards with a
constant speed of 80 mm / sec, whereas
linear velocity distribution determined:
1) The weight of the block (3).
2) The speed of the block (1).
Neglect friction in the pulley
upload_2015-3-3_12-13-33.png

upload_2015-3-3_12-25-39.png


Homework Equations


the angle is 30º
(tangencial force) / area = (dynamic viscosity) (velocity/space betwteen the two surfaces)

sum forces=0 (Equilibrium)

The Attempt at a Solution



I did the problem of two ways, considering each block and considering the system, but I don't get the same answer.

What am I doing wrong??
 
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I don't understand why the force balance on block 2 is incorrect?? Because in the block 2 acts two tangetial forces due the fluid, and the component horizontal of the weight
 
Chestermiller said:
It should include only the component of the weight of block 2, not both blocks.

Chet
But the block 2 support the block 1, if I only consider the weight of block 2, it's just like the block 2 was alone, but it doesnt
 
Queren Suriano said:
But the block 2 support the block 1, if I only consider the weight of block 2, it's just like the block 2 was alone, but it doesnt
The weight of block 1 exerts a force on block 2 in the normal direction, but block 1 is not exerting a force on block 2 in the tangential direction. The fluid in the gap between block 1 and block 2 is exerting a force on block 2 in the tangential direction, but you've already included that in your force balance.

Chet