A closed container of water contains a Styrofoam block attached to the bottom of the container by a massless string. When the system is accelerated upwards, what happens to the tension in the string?

Background: I am an MCAT Physics teacher and this question was on an MCAT. Assuming the water is compressible, Newton's first law would indicate that a higher than normal pressure would result at the bottom of the container and a lower than normal pressure would result at the top of the container. The net result would be an increase in buoyant force (force due to a pressure gradient); thus, increasing the tension. The question I have is, "If the water were considered to be incompressible, would the tension still increase?"

  • $\begingroup$ If $\mathbf{a}$ is the acceleration vector then effective gravity experienced by the accelerating container and all its contents is $\mathbf{g}_{eff}\equiv\mathbf{g}-\mathbf{a}$, in which $\mathbf{g}$ is the usual gravity vector. Just solve the problem with effective gravity acting on it. $\endgroup$
    – Deep
    Aug 28, 2017 at 5:13

2 Answers 2


Buoyancy does not depend on compressibility. The variation of pressure with depth is linked to the weight of a cylinder of liquid having cross-section $A$ extending from depth $y=-D$ to the surface $y=0$. The density is $\rho(y)$ and the pressure is $p(y)$.This cylinder has a weight $$W=gA\int_{-D}^0\rho(y)dy$$ that must be balanced by the difference between the pressure forces on its two ends $A(p(-D)-p(0))$. this gives $$p(-D)=p(0)+g\int_{-D}^0\rho(y)dy$$ and hence $$\frac{dp}{dy}=-g\rho(y)$$ For any real fluid, we can take $\rho$ to be a known function of $p$ and integrate this equation to find $p(y), \rho(y)$. A common model for the atmosphere follows from assuming an isothermal relationship.

However, to investigate a hypothetical fluid that is totally incompressible, we can just set $\rho$ to an arbitrary constant $\rho_0$ so that $dp/dy=-g\rho_0$ and the buoyancy of any object having volume $V$ and totally submerged at any depth is $\rho_0gV$.

You can check this question for information on the actual changes in ocean density from surface to deepest deeps. The change in pressure is a factor of about 1,000. The change in density is less than 5%. If the density actually changed as much as the pressure does, it would not be possible for ocean exploration vehicles to dive as they do. They could not overcome the buoyancy force.

ADDITION An answer that has now been deleted stated correctly that the effect of accelerating the system is merely to increase the effective value of gravitational acceleration. Therefore both mass and buoyancy forces increase in the same ratio.


First, how is it that Styrofoam can float?
Because when partially immersed in water the weight of the displaced water (upthrust) is equal to the weight of the Styrofoam.

If the Styrofoam is fully immersed why must it be tethered by a string?
Because the upthrust (weight of displaced water = mass of displaced water $\times g$) is now greater than the weight of the Styrofoam (mass of Styrofoam $\times g$) and the difference is equal to the tension in the string.

Tension = (mass of displaced water - mass of Styrofoam) $\times g$

If the container and its contents has an upward acceleration of $a$ then the effective value of $g$ is now higher ($= g+a$).

So the tension is now higher = (mass of displaced water - mass of Styrofoam) $\times (g+a)$.

This increase in tension is to be expected because if the experiment was performed in space or with the container in free fall ($g_{\rm effective} =0$) the tension in the string would be zero.

The upthrust is due to the pressure change from the bottom to the top of the Styrofoam or pressure gradient within the water.
If the effective value of $g$ increases then so does the upthrust and the weight of the Styrofoam in the same proportion.
So the tension in the string will also increase.


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