Respuesta :
Answer:
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              Â
