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

In summary, the first approach of resolving forces in the vertical y direction yields the correct result of R = W/cos(theta) where theta = 18 degrees. However, the second approach of resolving forces along the normal direction is incorrect due to the presence of radial acceleration in that direction, making equilibrium not applicable. This explains the contradiction in the two results for R.
  • #1
influx
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ImageUploadedByPhysics Forums1429110373.315217.jpg

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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  • #2
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.
 
  • #3
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?
 
  • #4
influx said:
Ah I see. With radial acceleration you are referring to an right?
Yes, right.
 
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Related to Resolving in the vertical direction and along the direction of the normal force

1. What is the meaning of resolving in the vertical direction and along the direction of the normal force?

Resolving in the vertical direction and along the direction of the normal force means breaking down a vector or force into its components in the vertical direction and along the direction of the normal force, which is perpendicular to the surface. This is often done in physics and engineering problems to simplify calculations.

2. Why is it important to resolve forces in the vertical direction and along the direction of the normal force?

It is important to resolve forces in the vertical direction and along the direction of the normal force because it allows us to accurately determine the effects of these forces on an object. By breaking down a force into its components, we can better understand how it will affect the motion and stability of an object.

3. How do you resolve a force in the vertical direction and along the direction of the normal force?

To resolve a force in the vertical direction and along the direction of the normal force, you need to use trigonometry. The vertical component of a force is found by multiplying the magnitude of the force by the sine of the angle between the force and the vertical direction. The normal component is found by multiplying the magnitude of the force by the cosine of the angle between the force and the normal direction.

4. Can you resolve a force in any direction?

Yes, you can resolve a force in any direction. However, it is most common and useful to resolve forces in the vertical and normal directions, as these are often the directions in which forces act on objects.

5. What are some real-life examples of resolving forces in the vertical direction and along the direction of the normal force?

Resolving forces in the vertical direction and along the direction of the normal force is commonly used in construction, engineering, and sports. For example, when designing a bridge, engineers must resolve the forces acting on the structure to ensure its stability. In sports, such as basketball, players often use the backboard to resolve the force of their shot into both vertical and normal components to increase their chances of making a basket.

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