An ornament for a courtyard at a World's Fair is to be made up of four large metal spheres

[netnomad]

Homework Statement

2.22 An ornament for a courtyard at a World's Fair is to be made up of four identical, frictionless metal spheres, each weighing 2√6 ton-weight. The spheres are to be arranged as shown in Fig. 2-20, with three resting on a horizontal surface and touching each other; the fourth is to rest freely on the other three. The bottom three are kept from separating by spot welds at the points of contact with each other. Allowing for a factor of safety of 3, how much tension T must the spot welds withstand?

Relevant Equations

$\theta$ is the angle between the horizontal plane the center of the top ball. From the geometry of the tetrahedron, $\tan \theta = \sqrt{2}$.

Relation between $T$ and the force acting through the centroid:
$$T = \frac{F_T}{\cos 30^\circ}$$

Factor of safety gives:

$$T_{\text{safe}} = 3T = \frac{3F_T}{\cos 30^\circ}$$

Relating $F_c$ and $F_T$:
$$F_c = 2F_T$$

The total acting force $F$:
$$F = 3F_c$$

Substituting $F_c = \frac{1}{3}F$:
$$F_T = \frac{1}{6}F$$

From equilibrium and using the principle of virtual work by applying a small displacement along the centroid (moving the ball to the center):
$$F = W \frac{dh}{dc}=W \tan \theta$$

Substituting $F = W \tan \theta$ into $F_T$:
$$F_T = \frac{1}{6}W \tan \theta$$

Substituting $F_T$ into $T_{\text{safe}}$:

$$\begin{aligned}
T_{\text{safe}} &= \frac{3F_T}{\cos 30^\circ}\\[6pt]
&= \frac{3}{\cos 30^\circ} \cdot \frac{1}{6}W \tan \theta \\[10pt]
&= \frac{W \tan \theta}{2 \cos 30^\circ}
\end{aligned}$$


Substituting $\tan \theta = \sqrt{2}$ and $W = 2\sqrt{6}$:

$$T_{\text{safe}} = \frac{2\sqrt{6} \cdot \sqrt{2}}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{2\sqrt{12}}{\sqrt{3}} = 4$$

After applying a small displacement along the axis between the center of the bottom ball and the projected center of the fourth ball, I got 4 as the answer, which is twice the correct value. I’m assuming I didn’t account for the fact that the tension force acts on both sides, so I ended up calculating it twice. Even though this makes sense to me intuitively, I’m not sure how to properly incorporate it into an equation or how to think about the tension force in this setup. Thank you!

 

[haruspex]

Your work is too hard to follow without a clear definition of $F, F_c, F_T$.

 

[netnomad]

You're right, sorry. I tried being more rigorous, but it turned out to be more confusing.

Spoiler: Force definitions

$F_T$ is the projection of the tension force onto the axis going from the center of a bottom ball to the projected center.

$F_c$ is the sum of two projected tension forces.

$F$ is the total force ($3 F_c$ as there are 3 balls) supporting the ball.

For simplicity, we can omit the safety factor and just focus on $F$. $F$ is the sum of all forces supporting the top ball in equilibrium. It acts through the centroid of the bottom triangle.

$$F = W \frac{dh}{dc}=W \tan \theta$$

$$F = 2\sqrt{6} \cdot \sqrt{2} = 4 \sqrt{3} $$

Now, this force should be split in 3, but considering the safety factor of 3, they cancel each other.

Then, the force per ball with a safety factor of 3:
$$F_{\text{single ball}} = 4 \sqrt{3}\,\text{ton-wt}$$

It consists of two (?) tension forces acting on the ball, so I divide this force by $2 \cos{\frac{\pi}{6}}$ and I get 4 as an answer.

 

[haruspex]

Still unclear.
Is F the weight of the top ball? If so, $F_c$ is the vertical component of the normal force between the top ball and a lower ball, right?
What plane are these projections onto?

 

[netnomad]

Here's a drawing. All the projections are done on the bottom plane. I calculated h as $c \tan\theta$, which allowed me to calculate dh/dc above.

So, moving bottom balls inward slightly as a result of these tension forces will move the top ball slightly up, changing its potential energy as:

$$F dc = W dh$$

 

[netnomad]

The difference between this and calculating using projections is that using projections, $F = \frac{W}{\tan\theta}$. I'm trying to understand what I did wrong with the virtual work principle

 

[netnomad]

I calculated the derivatives incorrectly, as the angle also changes. The correct approach is to either derive from $(2r)^2 = c^2 + h^2$ or use $\delta \theta$ to calculate $\delta c$ and $\delta h$. This leads to $F = -\frac{W}{\tan \theta}$, which resolves to 2 for the tension

 

[kuruman]

I took the straightforward approach and considered finding the tension on one of the balls (call it ball A) at the base of the tetrahedron by demanding that the sum of all the forces acting on it be zero. Since there is no friction, all forces are normal to the surface of the ball. Excepted is the force of gravity which acts at the center of ball A.

Additional forces are the normal force from the floor on which the ornament rests, two tensions from balls B and C also resting on the floor and the center-to-center compression from the top ball O. The centers of the four balls can be placed at four corners of a cube as shown in the figure on the right. The red triangle is in the plane of the surface supporting the ornament.

The Cartesian coordinate system has its origin at the top ball O. This makes writing unit vectors very easy. Note that gravity points along the $\{1,1,1\}$ direction (cube diagonal $d$.) Thus, in this system, the weight of any of the balls is $$\mathbf W=\frac{Mg}{\sqrt{3}}\{1,1,1\}.$$The procedure then would be to

1. Write all the forces on O and demand that the sum be zero. This will yield the magnitude of the compression force $f$ in terms of the weight $W$;
2. Write all the forces on A;
3. Demand that the net force on A be zero. Although in 3D this means three equations, only two equations are independent because of the symmetry of the situation. These two equations yield the tension $T$ and the normal force $N.$

It works and gives the expected answer. It is a good test of the method to solve for the normal force $N$ on one of the masses and verify that $3N=4Mg$, i.e. that the total normal force exerted by the floor on the four-mass system is equal to the weight of the system.

(Edited for typos and clarity. Also added the test in the last paragraph.)