Abandoned Car   While cleaning trash from an estuary, students discovered an abandoned car exposed only at low tide. They decided to remove the eyesore. Their plan was at low tide to insert inverted steel washtubs inside the car with the bottoms of the tubs against the roof. The incoming tide would trap air in the tubs. Once floated, the car body would be dragged to much deeper water then sunk from sight. The car body (no motor or wheels ) weighs less than 600 pounds. The square wash tubs are one foot deep. Each holds 30 gallons. Will one tub be enough to float the car?

With not much thought, we realize we cannot determine if the plan will work. If the hulk had been there a long time, it might be full of mud and who knows what. Not knowing what the tubs must float makes this problem impossible. However, we can perform a calculations to establish if the plan will definitely fail. If we make some sweeping assumptions that favor the enent and the calculations say "no," then we can state that the plane will definitely fail."

♦  We assume there is no mud within the car body. A scenario sketch shows the car with one inverted washtub in plave. When the tide is out, the weight of the hulk is supported by the mud bottom. As the tide comes in, estuary water traps air in the tub. According to Archimedes, the lift force of the trapped air is its greatest when the greatest amount of water is displaced. So the greatest lift force will occur when the inverted tub is about submerged or when the tide is about level with the roof of the car. If the car does not lift then, the lift force will decrease as the tide gets higher. Steps of the solution are:

Free Body Diagram:  The hulk, immediately before it lifts (we hope) is in mechanical equilibrium. To apply the momentum equation requires a carefully drawn free-body-diagram (right).

Momentum Equation::  The Z-component of the momentum equation is:

Mass Equation :  The mass equation with the air trapped in the tub as system is:

The lifting process will last a few hours as the water level of the tide rises. Consequently the air will always have the same temperatuure as the water. We assume the water temperature to be constant. The process will be isothermal. Thus our mass equation with the ideal gas equation of state for air becomes:

Lastly, the hydrostatic equation (momentum applied to a column of sea water adjacent to the trapper air is written. This equation will tell us tyhe pressure of the water besude the air which is also the pressure of the air:

A method of solving these equations is to use 2) and 3) to solve for H
(ans: H = 0.97 ft) then return to 3) with H to determine p_airA_tubs = 8,716 lbf. Finally we put numbers into the momentum equation 1) to calculate the force of the mud F_mud. If the force of the mud is positive (upward), mud is still supporting the car so it will not lift. But if the force of the mud is negative, the car might lift... If it is not stuck in the mud.

Thus, for one tub the supporting force of the mud is 351 lbf. This means the car won't float, so we had better use more tubs. The calculation with three inverted tubs is:

If three tubs are used, either the mud maintains a downward force (pull) of 147 lbf, or the hulk will float.

The volume of the car will help it float but that volume is unknown and difficult to calculate so we assumed the worst case; a car volume is zero. The bottom of the car might be under mud and the tires stuck. So we have also assumed no problems with mud in the car, and no "sticking" effect.