The cylindrical tank containing water accelerates upward with az = 2 m/s². while it rotates about its vertical axis by n= 100 rpm. Determine the pressure at point A.
(there's a diagram, but I wish I could attach it).
pz⋅πR2=mg−mazp_z\cdot \pi R^2=mg-ma_zpz⋅πR2=mg−maz
pressure at point A in z-direction:
pz=m(g−az)πR2p_z=\frac{m(g-a_z)}{\pi R^2}pz=πR2m(g−az)
pressure at point A in radial direction:
pr=max2πRhp_r=\frac{ma_x}{2\pi Rh}pr=2πRhmax
ar=ω2R=2πn=2π⋅100/60=10.5a_r=\omega^2R=2\pi n=2\pi\cdot 100/60=10.5ar=ω2R=2πn=2π⋅100/60=10.5 s-1
m=2πR2hρ=2π⋅0.32⋅1.2⋅1000=678.6m=2\pi R^2h\rho=2\pi \cdot 0.3^2\cdot 1.2\cdot 1000=678.6m=2πR2hρ=2π⋅0.32⋅1.2⋅1000=678.6 kg
then:
pz=678.6(9.8−2)π⋅0.032=1872044p_z=\frac{678.6(9.8-2)}{\pi \cdot 0.03^2}=1872044pz=π⋅0.032678.6(9.8−2)=1872044 Pa
pr=678.6⋅10.52π⋅0.3⋅1.2=3150p_r=\frac{678.6\cdot 10.5}{2\pi \cdot 0.3\cdot 1.2}=3150pr=2π⋅0.3⋅1.2678.6⋅10.5=3150 Pa
pressure at point A:
p=pr+pz=31502+18720442=1872046.65p=\sqrt{p_r+p_z}=\sqrt{3150^2+1872044^2}=1872046.65p=pr+pz=31502+18720442=1872046.65 Pa
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