Fluid Dynamics Help (Force with a dynamic viscosity)

  • Thread starter Colt 45 J
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  • #1
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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 [Broken]





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:

Answers and Replies

  • #2
nvn
Science Advisor
Homework Helper
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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 (ISO 31-0).
 
  • #3
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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.
 
  • #4
nvn
Science Advisor
Homework Helper
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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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