# Problem on tension

1. Jan 27, 2015

### arpon

1. The problem statement, all variables and given/known data

An ornament for a courtyard at a World's Fair is to be made up of four identical, frictionless metal spheres, each weighing $2\sqrt{6}$ ton-wts.The spheres are to be arranged as shown, with three resting on 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 must the spot welds withstand?

2. Relevant equations

3. The attempt at a solution
At first, I calculated the angle between the vertical axis and the line segment which connects the centres of the top and bottom spheres. Let, the angle be $\theta$.;

In the picture, A, B , C and D are the centers of the spheres. D is the centre of the top sphere. Let the radius of a sphere be $R$ .
So, AB = BC = CA = AD = 2R ;
Applying pythagorean theorem for APC triangle, $AP = \sqrt{3} R$
As, G is the centroid, AG : GP = 2 : 1 ;
So, $AG = \frac{2R}{\sqrt{3}}$
From, the triangle AGD, $sin \theta = \frac{AG}{AD} = \frac{1}{\sqrt{3}}$
and, $tan \theta = \frac{1}{\sqrt{2}}$
The vertically downward (component) force of the top sphere on each of the bottom spheres will be
$= \frac{1}{3} \cdot 2 \sqrt{6} = \frac {2\sqrt{2}}{\sqrt{3}}$

Now, let's consider the force on the bottom sphere having centre A :

So the vertical component is $\frac {2\sqrt{2}}{\sqrt{3}}$
And so, horizontal component becomes $= \frac {2\sqrt{2}}{\sqrt{3}} \cdot tan \theta = \frac{2}{\sqrt{3}}$
Now, look at the forces on the bottom spheres in the horizontal plane.

This horizontal force on A, can be divided into two components, along CA and BA ;
To, get the tension in AC, I took the component along AC which is $= \frac{1}{2} \frac{2}{\sqrt{3}} cos 30 = \frac{1}{2}$
As, the safety factor is 3, the tension becomes $\frac{3}{2}$;
But it doesn't match the answer.

2. Jan 27, 2015

### TSny

Your work looks good to me all the way to the last part where you are considering the horizontal forces on the bottom spheres. Draw a free body diagram of one of the bottom spheres showing all of the horizontal force components acting on the sphere.

Last edited: Jan 27, 2015
3. Jan 27, 2015

### CWatters

Check the line that begins...

When you resolve a force into two orthogonal components the original force vector is the hypotenuse not the side of the triangle.

4. Jan 27, 2015

### Staff: Mentor

I confirm your result for the horizontal force. For the horizontal force balance, I get 2Tcos(30)=(horizontal force). This gives T = 2/3. If I multiply by the factor of safety, I get 2.

Chet

5. Jan 28, 2015