What is the opposite of tension?

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The discussion centers on identifying the opposite of tension, which is compression force. When a material is pushed together, it experiences compression, contrasting with tension, which involves pulling. The relationship between applied forces and internal forces is explored, emphasizing that external forces lead to internal forces in materials, adhering to Newton's laws. The conversation also touches on how the distribution of forces affects deformation in materials, using examples like ropes and springs to illustrate these concepts. Understanding these forces is crucial for analyzing material behavior under different loading conditions.
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Homework Statement


Tension is the force that pulls. What is the force that pushes?

Homework Equations



F=T​

The Attempt at a Solution


Impact force? Pushing force? Something to do with Young's Modulus? F=stress*area?
 
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JDoolin said:

Homework Statement


Tension is the force that pulls. What is the force that pushes?

Homework Equations



F=T​

The Attempt at a Solution


Impact force? Pushing force? Something to do with Young's Modulus? F=stress*area?

When a bar is in tension, it stretches out, right. So when it is being pushed together, it is in __________
 
Think opposite of expansion.
 
compression force, perhaps.
 
that would be my best guess =]
 
Yes, compression force is correct. How much compression force can be applied is a function of the material. For example nice linear-elastic materials like steel are assumed to act the same in tension and compression; that is given some displacement (elongation or compression) the same magnitude of force must be applied. This is like a spring where F=kx only with a material it is stated as stress=(youngs modulus)(strain).
 
I realize now that I want to be able to describe two different types of forces; one describing intent (pushing force and pulling force) and the other describing the result. (compression force and tension force.)

(if I am not mistaken, the resulting compression or tension is less if the pushing or pulling results in acceleration.)
 
You are looking at the difference between external and internal forces. The external force is the applied force (sometimes called a boundary force). Because of the application of an external force you will generate internal forces. This is because everything must satisfy Newtons law F=ma. So if you make "fictitious cuts" as they are sometimes called in solid mechanics at any position along the material and sum forces on that piece, the forces must sum up correctly. If the material is static that means the external forces sum to zero, likewise if you study the sum of forces on any element of the material the forces must sum to zero as well. On the other hand if there is a net external force on material it will accelerate. For the same reason the sum of forces on any element must equal to the (elemental mass)(acceleration). That is the basic idea...
 
kjohnson said:
You are looking at the difference between external and internal forces. The external force is the applied force (sometimes called a boundary force). Because of the application of an external force you will generate internal forces. This is because everything must satisfy Newtons law F=ma. So if you make "fictitious cuts" as they are sometimes called in solid mechanics at any position along the material and sum forces on that piece, the forces must sum up correctly. If the material is static that means the external forces sum to zero, likewise if you study the sum of forces on any element of the material the forces must sum to zero as well. On the other hand if there is a net external force on material it will accelerate. For the same reason the sum of forces on any element must equal to the (elemental mass)(acceleration). That is the basic idea...

Hmmm. So if one pulls at 100 Newtons at one end of a rope, and another pulls at 100 Newton's at the other end of the rope, the internal force is 100 Newton's for the entire length of the rope. If the rope is ten meters long, does that mean there ore only 10 Newton's of tension in each one-meter section of the rope?
 
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In that case the idea is that the internal force is a constant = 100N. Think of a spring where hookes law F=kx applies. In such a case if a spring is loaded with 100N force at each end (tension or compression) than the resulting deformation will equal 100/k. Since the force function is a constant no calculus is needed. On the other hand if a 100N force is applied only on one side it will cause the spring to accelerate with acceleration 100/m. But the internal force function will not be uniform so to find the total deformation you would need to use hookes law on an elemental basis.
 
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