Resolving in the vertical direction and along the direction of the normal force

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Discussion Overview

The discussion revolves around resolving forces in a physics problem involving normal forces and weight components. Participants explore the implications of resolving forces in the vertical direction versus along the direction of the normal force, addressing potential contradictions in their results.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that resolving forces in the vertical direction leads to the equation R = W/cos(theta), suggesting that this approach is correct due to the absence of vertical acceleration.
  • Another participant argues that resolving forces along the direction of the normal force yields R = Wcos(theta), indicating a contradiction with the first result.
  • A later reply asserts that the first approach is correct, while the second is incorrect due to the presence of radial acceleration, which affects equilibrium in the normal direction.
  • One participant seeks clarification on the reference to radial acceleration, indicating a need for further understanding of the concept.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the two approaches to resolving forces, with some asserting that one method is correct while another participant questions the implications of radial acceleration on the second approach.

Contextual Notes

The discussion highlights the complexity of resolving forces in different directions and the impact of acceleration on equilibrium conditions, but does not resolve the contradictions presented.

Who May Find This Useful

Students and educators in physics, particularly those studying mechanics and force resolution, may find this discussion relevant.

influx
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Ok I know this is simple resolving but I'm a bit confused. If we resolve forces in the vertical y direction (as in the attached image), we obtain Rcos(theta)-W=0 and hence Rcos(theta) = W meaning R = W/cos(theta) [where theta = 18 degrees]

However, if we resolve forces along the direction of the normal force, R, we obtain R - Wcos(theta) = 0 and hence R = Wcos(theta) [again theta=18]. These 2 results for R appear to contradict so what have I done wrong?

Cheers
 
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influx said:
View attachment 82057
Ok I know this is simple resolving but I'm a bit confused. If we resolve forces in the vertical y direction (as in the attached image), we obtain Rcos(theta)-W=0 and hence Rcos(theta) = W meaning R = W/cos(theta) [where theta = 18 degrees]

However, if we resolve forces along the direction of the normal force, R, we obtain R - Wcos(theta) = 0 and hence R = Wcos(theta) [again theta=18]. These 2 results for R appear to contradict so what have I done wrong?

Cheers
The first approach as per the solution is correct, since there is no acceleration or component thereof in the vertical direction.

The second approach is incorrect, because if you resolve forces along the normal direction, the acceleration is not 0 in the normal direction; there is a component of the radial acceleration in that direction, so equilibrium does not apply in that direction normal to the plane.
 
PhanthomJay said:
The first approach as per the solution is correct, since there is no acceleration or component thereof in the vertical direction.

The second approach is incorrect, because if you resolve forces along the normal direction, the acceleration is not 0 in the normal direction; there is a component of the radial acceleration in that direction, so equilibrium does not apply in that direction normal to the plane.
Ah I see. With radial acceleration you are referring to an right?
 
influx said:
Ah I see. With radial acceleration you are referring to an right?
Yes, right.
 
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