Statics of bar under own weight.

In summary, the conversation discusses the weight of a vertical rod with constant area attached to the ceiling from one point. The formula for the weight of the rod is w(x)=ρ*x*g, where x is the distance from the free end. The conversation also raises the question of where the formula may be incorrect and if there are any other factors that need to be considered.
  • #1
Chacabucogod
56
0
Hi,


I was thinking about this and hit a "wall". If you have a vertical rod with constant area that is attached to the ceiling from one point its weight will be:

[itex]w(x)=ρ*x*g[\itex] (where 0 is the end of the bar)

now if you take a differential above x you will have that the downward force that it's experiencing is w(x), but above the dx is w(x+dx). Where did I go wrong here. Is there something I'm not taking into consideration?

Thank you.
 
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  • #2
w(x) represents the total weight of the rod only when x = L. Obviously, if x = 0 at the free end, w(0) = 0. As x runs from 0 to L, w(x), which represents the partial weight of the rod between the free end and the location x, according to your formula, must be increasing from zero to the total weight of the rod at the ceiling.
 

What is the definition of statics?

Statics is the branch of mechanics that deals with the study of objects at rest or in a state of constant motion.

What is the importance of studying the statics of a bar under its own weight?

The statics of a bar under its own weight is important because it helps us understand the distribution of forces acting on the bar and how those forces affect its equilibrium.

What are the key factors that affect the statics of a bar under its own weight?

The key factors that affect the statics of a bar under its own weight include the weight of the bar, the length of the bar, and the material properties of the bar.

How do you calculate the forces acting on a bar under its own weight?

To calculate the forces acting on a bar under its own weight, you can use the principles of static equilibrium and apply the equations of equilibrium to find the reactions at the supports of the bar.

What are some real-world applications of studying the statics of a bar under its own weight?

Studying the statics of a bar under its own weight has practical applications in construction, architecture, and engineering, where it is important to understand how different materials and structures will behave under their own weight.

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