Solving Problem on Tension Homework Statement

[arpon]

Homework Statement

[/B]
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?

Homework Equations

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.

 

[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.

 

[CWatters]

Check the line that begins...

And so, horizontal component becomes...

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

 

[Chestermiller]

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

 

[arpon]

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.

Your suggestion was helpful to find out my mistake.