Answer:
a) 68.9 b/ ft²
b) 57.416
Explanation:
Volume of the cylindrical open container V = 5 gal
= (5)(231)
= 11155 in³
different diagrammatic expression has been shown in the attached file below for clear understanding.
Area A = 120 in²
V = hA
= [tex]\frac{1155}{120}[/tex]
= 9.625 in
Upward acceleration , [tex]a_z= 3ft/s^2[/tex]
[tex]\frac{dp}{dz} = \rho (g + a_z)[/tex]
[tex]dp= -\rho (g + a_z)dz[/tex]
[tex]\int\limits^P}_0 \,dp= -\rho (g + a_z) \int\limits^0_h \, dz[/tex]
[tex]P_b =[/tex] [tex]-\rho (g + a_z) h[/tex]
[tex]= (2.4)*(32.2+3) (\frac{9.625}{12})[/tex]
= 68.9 b/ ft²
b) The flowchart for the resultant force is shown in the diagram below:
with that:
[tex]F_f = P_bA[/tex]
[tex]F_f =\frac{(68.9)(120)}{(144)}[/tex]
[tex]F_f =57.416[/tex]
Thus, the resultant force that the container exert on the floor of the elevator during this acceleration = 57.416
