Fluid Dynamics Help (Force with a dynamic viscosity)

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Discussion Overview

The discussion revolves around a fluid dynamics problem involving the calculation of forces required to move a block on an inclined surface, considering the effects of dynamic viscosity and friction. Participants explore the application of fluid mechanics principles to determine the necessary force and the impact of an oil film on force reduction.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Colt 45 J presents a problem involving a block's movement on an incline and calculates the required force without lubrication to be 40.5 N.
  • Colt 45 J calculates the force with a 0.4 mm oil film to be approximately 2.554 N, suggesting a 94% reduction in force, which he finds unrealistic.
  • Another participant agrees with Colt 45 J's calculations but emphasizes the importance of adhering to international standards for unit notation.
  • A later reply proposes using shear force calculations, resulting in a 37% reduction in force, which the participant finds more realistic.
  • Colt 45 J questions whether the previously calculated force (2.40 N) is indeed the shear force, suggesting a need for clarification on the application of area in the calculations.

Areas of Agreement / Disagreement

Participants have differing views on the validity of the calculated force reductions, with some finding the initial 94% reduction unrealistic while others suggest alternative calculations leading to a 37% reduction. The discussion remains unresolved regarding the correct interpretation and application of shear force in this context.

Contextual Notes

There are unresolved assumptions regarding the definitions of shear force and the application of area in the calculations. Additionally, the impact of the incline angle on the calculations has not been fully explored.

Colt 45 J
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Problem Statement
A 50cm by 30cm by 20cm block weighing 150N is to be moved at a constant velocity of .8 m/s on an inclined surface with a friction coefficient of 0.27. (a) Determine the force f that needs to be applied in the horizontal direction. (b) if a .4mm thick oil film with a dynamic viscosity of .012Pa*s is applied between the block and inclined surface, determine the percent reduction in the required force.

Given: P=150N V=.8m/s a=0, mu(v)=.012Pa*s mu(f)=.27

Find: F(a), F(b), %r

Diagram:
http://img18.imageshack.us/img18/7915/fluids275.png The attempt at a solution

I solved the F(a) to be 40.5N using statics equations (since there is no acceleration F=ma=0)

When using the equations presented in my text I get:
F=mu(v)*A*V/l
F=.012*(.5*.2)*.8/.0004
F=2.4
F(b)=2.4/cos(20*)
F(b)=2.554N

Thus the percentage would be a ~94% reduction in force, which sounds highly unrealistic and leads me to think I did something wrong...

Some Help?

Thanks, Colt 45 J
 
Last edited by a moderator:
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Colt 45 J: Your answer currently looks correct. The oil reduces the required force by a large amount. By the way, please note the following international standard for writing units. Secondly, please do not post wide images directly to the forum page. Just post a text link to wide images. Or crop the image.

  1. Always leave a space between a numeric value and its following unit symbol. E.g., 50 cm, not 50cm.

  2. Numbers less than 1 must always have a zero before the decimal point. E.g., 0.5, not .5. See the international standard for writing units[/color] (ISO 31-0[/color]).
 
I was considering, would I end up using the shear force (calculated using the force already found). The result from doing that is about a 37% reduction, which seems much more realistic.

(V=F/A)

For the future I'll be sure to crop and use the ISO standards.
 
Colt 45 J: I think the force already found (F = 2.40 N) is already the shear force. Therefore, you would not divide it by A.
 

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