Question

In 1654 Otto von Guericke, inventor of the air pump, gave a demonstration before the noblemen of the Holy Roman Empire in which two teams of eight horses could not pull apart two evacuated brass hemispheres. (a) Assuming the hemispheres have (strong) thin walls, so that R in Figure may be considered both the inside and outside radius, show that the force required to pull apart the hemispheres has magnitude$\mathit{F}\mathbf{=}\mathit{\pi }{\mathit{R}}^{\mathbf{2}}\mathbf{}\mathbf{∆}\mathit{p}$, where$\mathbf{∆}\mathit{p}$is the difference between the pressures outside and inside the sphere. (b) Takingas, the inside pressure asrole="math" localid="1657253356406" $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$, and the outside pressure as$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$,find the force magnitude the teams of horses would have had to exert to pull apart the hemispheres. (c) Explain why one team of horses could have proved the point just as well if the hemispheres were attached to a sturdy wall.

Short answer

1. The derived expression for force is found to be$F=\pi R^2∆p$
2. Magnitude of the force is $2.6\times {10}^{4}N$.
3. The force of the wall on the sphere balances the force of the horses.

Step-by-step solution

The given data

1. From figure and data, radius= Rand pressure= p
2. Radius of hemisphere,$R=30cm$
3. Inside pressure,$p_i=0.10atm$
4. Outside pressure, $p_0=1.00atm$

Understanding the concept of pressure

We use the concept of pressure to find the force by integrating the formula of it concerning pressure and covered area. Also, we find the value of the force by using the values.

Formula:

Pressure on the body by force F on an area A,

$p=\frac{F}{A}$ (i)

a) Derivation of the general expression for force

Let’s assume that the team of horses is pulling the sphere towards the right. To pull the sphere apart, it must exert a force equal to the horizontal component of the total force. This can be found by the integration of force and component of each force.

Consider the force vector at angle$\theta$. Since pressure, $∆p=F/A$from equation (i), we can write the leftward component of force as $\left(∆p\right)\cos \theta \left(dA\right)$

Here, dA is the area at which the force is applied. We can consider the symmetry; so let dA be that of the ring of constant $\theta$on the surface.

Radius of the ring is $r=R\sin \theta$, R is the radius of the sphere; if the angular width of the ring is $d\theta$then its width is $Rd\theta$and its area is $dA=2\pi R^2\sin \theta d\theta$.

Thus, the net horizontal component of force of air is

$$\begin{gathered} F_k=2\pi R^2∆p{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{sin\theta cos\theta }d\mathrm{\theta } \\ ={\left|{\mathrm{\pi R}}^2∆{\mathrm{psin}}^2\mathrm{\theta }\right|}_0^{\frac{\mathrm{\pi }}{2}} \\ =\pi R^2∆p.............................\left(1\right) \end{gathered}$$

Hence, the derived expression for force is $\pi R^2∆p$

b) Calculation for the magnitude of the force

$$\begin{gathered} ∆p=p_o-p_i \\ =\left(1-0.10\right)atm \\ =0.90atm \\ =9.09\times {10}^{4}Pa\left(\because 1atm=1.01\times {10}^{5}Pa\right) \end{gathered}$$

Hence, the magnitude of force using equation ( 1) can be given as:

$$\begin{gathered} F_h=\mathrm{\pi }{\left(0.30\right)}^2\left(9.09\times {10}^4\right) \\ =2.6\times {10}^{4}N \end{gathered}$$

Hence, the magnitude of force is$2.6\times {10}^{4}N$

c) Explanation of the force of the horse team

One team of horses could be used if one half of the sphere was attached to a sturdy wall.

Hence, the force of the wall on the sphere balances the force of the horses.