Solving Problem on Tension Homework Statement

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In summary: Thank you.In summary, the problem involves calculating the tension needed for the spot welds to withstand the weight of four identical, frictionless metal spheres arranged in a specific way at a World's Fair. The spheres are arranged with three on a horizontal surface and one resting freely on top of the others. After calculating the angle between the vertical axis and the line connecting the centers of the top and bottom spheres, the tension needed for the spot welds can be found by resolving the horizontal forces on the bottom spheres. After correcting a mistake in the calculation, the final answer is 2 ton-wts, taking into account a factor of safety of 3.
  • #1
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Homework Statement


upload_2015-1-27_19-17-45.png
[/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##.;
upload_2015-1-27_19-52-50.png

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 :
upload_2015-1-27_20-14-37.png

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.
upload_2015-1-27_20-12-51.png

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.
 
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  • #2
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:
  • #3
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.
 
  • #4
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
Chestermiller said:
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.
 

1. What is tension?

Tension is a force that is exerted on an object by a string, rope, or cable. It is directed along the length of the string and can either be pulling or stretching the object.

2. How do I calculate tension?

To calculate tension, you need to know the force acting on the object and the angle of the string with respect to the horizontal. You can then use the formula T = F*cos(theta) to find the tension.

3. What are some common problems involving tension?

Some common problems involving tension include finding the tension in a rope supporting a hanging object, calculating the tension in a pulley system, and determining the tension in a cable holding a bridge or building.

4. How do I approach solving problems on tension?

To solve problems on tension, it is important to draw a free body diagram and identify all the forces acting on the object. Then, use the equations of motion and/or Newton's laws to solve for the unknown tension.

5. What are some tips for solving tension problems effectively?

Some tips for solving tension problems effectively include breaking down the problem into smaller parts, labeling all forces and angles, double-checking your calculations, and using the correct units in your final answer. It is also helpful to practice solving different types of tension problems to become more familiar with the concepts and techniques.

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