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

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SUMMARY

The discussion centers on resolving forces in physics, specifically the normal force and weight components at an angle of 18 degrees. The correct resolution of forces in the vertical direction yields R = W/cos(18°), while the incorrect resolution along the normal force gives R = Wcos(18°). The confusion arises from the assumption of equilibrium in the normal direction, which is invalid due to the presence of radial acceleration. The first approach is validated as correct, emphasizing the importance of recognizing acceleration components in force resolution.

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  • Familiarity with trigonometric functions, particularly cosine.
  • Knowledge of radial acceleration and its implications in force analysis.
  • Ability to resolve forces in two dimensions.
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  • Study the principles of radial acceleration in circular motion.
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  • Review the concept of equilibrium and its limitations in dynamic systems.
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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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