| du/dy (s-1) | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 |
| t (N m-2) | 0.0 | 1.0 | 1.9 | 3.1 | 4.0 |
Using Newton's law of viscocity
where m is the viscosity. So viscosity is the gradient of a graph of shear stress against vellocity gradient of the above data, or

Plot the data as a graph:

Calculate the gradient for each section of the line
| du/dy (s-1) | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 |
| t (N m-2) | 0.0 | 1.0 | 1.9 | 3.1 | 4.0 |
| Gradient | - | 5.0 | 4.75 | 5.17 | 5.0 |
Thus the mean gradient = viscosity = 4.98 N s / m2
r oil = 850 kg/m3
r water = 1000 kg/m3
g oil = 850 / 1000 = 0.85
Dynamic viscosity = m = 5´ 10-3 kg/ms
Kinematic viscosity = n = m / r

At the plate face y = 0m,
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Calculate the shear stress at the plate face
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At y = 0.34m,
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As the velocity gradient is zero at y=0.34 then the shear stress must also be zero.
Weight 46 800 = mg
Mass m = 46 800 / 9.81 = 4770.6 kg
Mass density r = Mass / volume = 4770.6 / 5.6 = 852 kg/m3
Relative density
m = 0.01008 poise
1 poise = 0.1 Pa s = 0.1 Ns/m2
m = 0.001008 Pa s = 0.001008 Ns/m2
g = 0.913