Basic Question about Force for Fluid Mechanics

AI Thread Summary
To determine the force needed to pull a block of ice over a lubricated surface at a constant speed, the relevant equation is F = μ (V/h)A, where μ is the viscosity, V is the speed, h is the thickness of the water layer, and A is the contact area. The viscosity of water at 0°C is given as 1.79 × 10^–3 N∙s/m², with the area calculated as 0.80 m x 1.20 m. The velocity gradient is derived from the speed of the block divided by the thickness of the water layer, which is 0.10 mm. Substituting these values into the equation yields a force of approximately 8.6 N. This approach correctly applies Newton's law of viscosity to the problem.
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Homework Statement



The block of ice (temperature 0°C) shown in Figure P9.57 is drawn over a level surface lubricated by a layer of water 0.10 mm thick. Determine the magnitude of the force needed to pull the block with a constant speed of 0.50 m/s. At 0°C, the viscosity of water has the value η = 1.79 × 10^–3 N ∙ s / m^2.

I have attached the image. The surface is 0.80 x 1.20m, with force pointing to the right.


Homework Equations



Force = Pressure x Area
m = pV, p = 1.00 x 10^3 kg/m^3

not sure how these are helpful.


The Attempt at a Solution



Can I just multiply the viscosity by the speed by the area of the surface divided by the thickness of the water to get the force?

So F= ηv x A / thickness = 1.79 x 10^-3 Ns/m^2 x 0.5 m/s x 0.8 x 1.2 m /0.10 = 8.6 x 10^-2 N?

The units cancel out.. not sure if this is the right approach though. I don't really know where to begin. Can anyone please help / guide me?

EDIT: Sorry... it accidentally posted when I couldn't complete the subject title. I would want it entitled "Basic Question about Force for Fluid Mechanics
 

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I've just checked dynamic viscosity from Wikipedia. http://en.wikipedia.org/wiki/Viscosity

There's the equation ##F=\mu A \frac{u}{y}##. Maybe that can be helpful to check your answer. ##u## is the speed, ##A## is the area, and ##y## the separation.

Make sure though that you verify that the equation can be applied to your problem.
 
Seydlitz said:
I've just checked dynamic viscosity from Wikipedia. http://en.wikipedia.org/wiki/Viscosity

There's the equation ##F=\mu A \frac{u}{y}##. Maybe that can be helpful to check your answer. ##u## is the speed, ##A## is the area, and ##y## the separation.

Hmm so if my u = 1.79 × 10–3 N ∙ s / m2, A = 0.80m x 1.20m, v = 0.50 m/s, and y = .10 mm I should get...

F = uAv/y = 1.79x10-3 Ns/m2 * (0.80m x 1.20m) x 0.50 m/s / (0.10 mm x 10^-3 m/1mm) = 8.6 N

Can someone please check my math/logic and that my variables are correct? Is "0.10 mm" the right value to use for the separation?

Thanks for the formula.
 
physicswork said:
EDIT: Sorry... it accidentally posted when I couldn't complete the subject title. I would want it entitled "Basic Question about Force for Fluid Mechanics
fixed!
 
Thanks for fixing the title :)
 
The way this works is: the tangential force F exerted by the layer of water on the block (and by the ground on the layer of water) is equal to the shear stress τ times the contact area A. According to Newton's law of viscosity, the shear stress in the liquid layer is equal to the velocity gradient dv/dy within the fluid times the viscosity of the fluid μ. In this example, the velocity of the block is V, the velocity of the ground is zero, and thickness of the gap (0.1mm) is h, so the velocity gradient is V/h. So the shear stress on the block is τ = μ (V/h). So the tangential force is F = μ (V/h)A.
 
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