Conservation of Energy: Why are Normal Forces Zero?

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Discussion Overview

The discussion revolves around the concept of normal forces and their work in the context of conservation of energy. Participants explore why the work done by normal forces is considered to be zero, examining definitions and the relationship between force and displacement.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant asks for clarification on why the work done by normal forces is zero, referencing conservation of energy equations.
  • Another participant explains that the work done by a force is calculated using the equation ##W= Fd\cos\theta##, noting that the normal force is perpendicular to the displacement, leading to ##\cos(90^o)=0##.
  • A different participant states that the work is defined as ##dW=\vec{F}\cdot d\vec{r}##, asserting that normal forces fulfill the condition ##\vec{F}\cdot d\vec{r}=0##, thus making the work zero by definition.
  • One participant questions the direction of normal forces and the nature of displacement, suggesting that the angular velocity affects the direction of displacement.
  • A subsequent reply challenges the previous assertion about the direction of normal forces, stating that they remain the same before and after, and clarifies the nature of displacement in different scenarios.

Areas of Agreement / Disagreement

Participants express differing views on the implications of normal forces and displacement directions, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Some assumptions about the definitions of work and normal forces are implicit in the discussion, and the relationship between angular velocity and displacement is not fully explored.

t2r
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Hello everyone,
someone could explain me please, why the work of the normals forces are 0 ?



He used with conservation energy equations.
How should I refer to the displacement point ?

Thx everyone !
 
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Because the work done by a force is ##W= Fd\cos\theta##. The normal force is perpendicular to the displacement by definition in which case ##\theta =90^o## and ##\cos(90^o)=0##.
 
Well, the work is, by definition ##dW=\vec{F}\cdot d\vec{r}##, and a normal force is, again by definition a force which fulfils ##\vec{F}\cdot d\vec{r}=0##. So the work is zero almost by definition again.
 
The direction of the Normal forces is changed, and also the Δx is not downwards because the angular velocity.

Am I right ?
 

Attachments

t2r said:
The direction of the Normal forces is changed, and also the Δx is not downwards because the angular velocity.

Am I right ?
Nop, the direction of the Normal forces is the same before and after, and also the ##\Delta x## is downwards in one case and horizontal in the other. From where you can see that ##\vec{F}\cdot d\vec{r}=0## as we said before.
 
Last edited:

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