Question

A 5-gal, cylindrical open container with a bottom area of 120 square inches is filled with glycerin and rests on the floor of an elevator. a. Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft/s2 . b. What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (1 gal=231 in.3 )

Answer 1

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

    = $\frac{1155}{120}$

    =  9.625 in

Upward acceleration , $a_z= 3ft/s^2$

$\frac{dp}{dz} = \rho (g + a_z)$

$dp= -\rho (g + a_z)dz$

$\int\limits^P}_0 \,dp= -\rho (g + a_z) \int\limits^0_h \, dz$

$P_b =$ $-\rho (g + a_z) h$

$= (2.4)*(32.2+3) (\frac{9.625}{12})$

= 68.9 b/ ft²

b) The flowchart for the resultant force is shown in the  diagram below:

with that:

$F_f = P_bA$

$F_f =\frac{(68.9)(120)}{(144)}$

$F_f =57.416$

Thus,  the resultant force that the container exert on the floor of the elevator during this acceleration = 57.416