# Homework Help: Problem of fluid -- Blocks, pulley, two fluids

Tags:
1. Mar 3, 2015

### Queren Suriano

1. The problem statement, all variables and given/known data
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

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

sum forces=0 (Equilibrium)

3. The attempt at a solution

I did the problem of two ways, considering each block and considering the system, but I dont get the same answer.

What am I doing wrong??

Last edited: Mar 3, 2015
2. Mar 3, 2015

### Staff: Mentor

Your force balance on block 2 is incorrect, and your force balance on the overall system is incorrect. Your calculated value of F2 is correct. The force balance for the overall system should be the sum of the force balances on the individual blocks.

Chet

3. Mar 3, 2015

### Queren Suriano

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

4. Mar 3, 2015

### Staff: Mentor

It should include only the component of the weight of block 2, not both blocks.

Chet

5. Mar 3, 2015

### Queren Suriano

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

6. Mar 3, 2015

### Staff: Mentor

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