A 10-kg block slides down a smooth inclined surface. Determine the terminal velocity of the block if the 0.1-mm gap between the block and the surface contains SAE 30 oil at 60 °F. Assume the velocity distribution in the gap is linear, and the area of the block in contact with the oil is 0.1 m^2.

Question

(the diagram shows a slope downward 20 degrees from the horizontal)
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What I did:
T = 60 F = 15.56 C
Looked up viscosity of SAE 30 oil at 15.56 C:
mu = 0.8 N*s/m^2
h (height) = 0.1 mm = 1 * 10^-4 m

tau = F/A = mu * dv/dy
F = sin theta * mg

Rearranging with assumption that dv/dy is constant (linear velocity gradient mentioned in problem):

velocity = (tau * h)/mu = (sin (theta) * mgh)/(A * mu)
= (sin (20*pi/180)*10*9.8*10^-4)/(0.1*0.8) = 0.0419 m/s

The official textbook answer is 0.0883 m/s.

How do I get the correct answer?

Answer 1

First off, it needs to be Cos(20), you want the horizontal component not the vertical for you force.

Set that Fx = Tau*Area (Shear force)

Tau = mu * du/dy
The mu of SAE 30 is .38 N*s/m^2
du is what you're searching for
dy is .1mm (convert to meters)

Finally multiply Tau by the area and then set it equal to the Fx force, finally solve for du (basic algebra)

Final answer should be around .244 m/s

Answer 2

Simple high school physics.

First you have to take into account the Newtonian fluid shear stress:
tau=mu*(du/dy)
Since the velocity distribution in the layer is assumed to be linear, it follows that the velocity component u, parallel to the block motion, as a function of y, normal distance from the incline plane:
u(y)=(V*y)/h,
Finally we notice the shear stress at the block surface at each point is:
tau(block)=mu*(V/h)
Using some simple substitution we conclude with:
V=(W*cos70*h)/mu*A
Where W=weight, A=area of block.
Final answer is 0.0883 m/s