Given the density of 0.976 g/cm3.
1. Find the surface slope a. For a given point take the rise/ run or slope to get the angle.
In our case the run = 1 mm the rise = 0.29 mm.
a= Tan-1 (rise/run)= Tan-1 (0.29mm/ 1mm)= 16.2°
Thus the surface slope at the mid point is 16.2°
2. Get the hight at the midpoint = 3.18 mm.
3. Plug and chug.
Tau = r g h sin (a)
Tau = 0.976 g/cm3 X 980 cm/sec2 X 0.32 cm X sin (16.2)
Tau = 85.4 (g/cm3 cm/sec2 cm) = 85.4 g/cm sec2 - What's this?
Some dimensional analysis:
Force = MLT-2
Acceleration = LT-2
Area = L2
Stress = Force / area = MLT-2 / L2 = M/T2 L
Thus our units are OK for stress.
Last - lets get the exponent correct.
1m = 100 cm so our final answer is 8540 g/m sec2
Also have to convert g to kg
Thus -> 8.5 kPascals