A 13-ft-long, 6-ft-high rectangular tank open to the atmosphere is towed by a truck on a level road. The tank is filled with water to a depth of 5 ft. Determine the maximum acceleration or deceleration allowed if no water is to spill during towing.

a_x = 1.51 m/s^2 ... ( 3 sig fig )

Explanation:

Given:

- The Length of the tank L = 13 ft

- The height of the tank H = 6 ft

- The depth of water in tank d = 5 ft

- The angle the free surface while accelerating = θ

Find:

Determine the maximum acceleration or deceleration allowed if no water is to spill during towing.

Solution:

- Sketch a schematic of horizontally accelerating (a_x) tank. (See Attachment)

- The relation between the acceleration (a_x) and the angle (θ) is given by the following formula:

                                 tan ( θ ) = a_x / g

Where, g = 9.81 m/s^2 ... ( acceleration due to gravity )

- Now using Pythagoras Theorem and the schematic compute  tan ( θ ):

                                 tan ( θ ) = Δ h / (L/2)

- Δ h is the change in free surface depth when the tank accelerates. For maximum acceleration while avoiding spillage we have margin of:

                                 Δ h = H - d

- Now substitute back all the terms and evaluate the acceleration a_x:

                                 2*(H - d) / L = a_x / g

                                 a_x = 2*(H - d)*g / L

- Plug in the values:

                                 a_x = 2*(6-5)*(9.81) / 13

                                 a_x = 1.50923 m/s^2 = 1.51 m/s^2                            

can you make it more obvious. i can't see it properly