Answer :
The most nearly correct option is a. 291.12 lb.
To find the force acting on the bottom of the tank, we first need to calculate the weight of the water accelerated, which is 1.55 ft³.
Given that the density of water is approximately 62.4 lb/ft³, the weight of the water is [tex]\(1.55 \, \text{ft}³ \times 62.4 \, \text{lb/ft}³ = 96.72 \, \text{lb}\)[/tex].
Next, we consider the pressure exerted by this weight over the bottom area of the tank, which is circular with a diameter of 6 ft (since the tank is 3 ft deep and cylindrical).
Using the formula for pressure [tex](\(P = \frac{F}{A}\)),[/tex] where F is force and A is area, we calculate the force:
[tex]\[ F = \left(\frac{96.72 \, \text{lb}}{\pi \times (3 \, \text{ft})²/4}\right) \times (\pi \times (3 \, \text{ft})²/4) \][/tex]
[tex]\[ F \approx 291.12 \, \text{lb} \][/tex]
Therefore, the most nearly correct option is a. 291.12 lb.
The probable question may be:
An unbalanced vertical force of 60lb upward accelerates 1.55ft³ of water. If the water is 3ft deep in a cylindrical tank, which of the following most nearly gives the force acting on the bottom of the tank?
a. 291.12 lb
b. 40.74 lb
c. 183.60 lb
d. 528.96 lb
