What is the Direction of Normal Force in Static Equilibrium on a Ramp?

[Coderhk]

Homework Statement

Part a of the image below. When I try to solve this question, I can't get the solution in the answer key. In the answer key the y component of the normal force is NCos(alpha) in the last line. Does the normal force not point perpendicular to the ramp?

Homework Equations

F=ma=0 since we are in static equilibrium
sum of torques = 0;

The Attempt at a Solution

In the picture below.

 

[haruspex]

Does the normal force not point perpendicular to the ramp?

It does indeed, but doesn't that mean that the vertical component is N cos(α) as stated?
In the attachment I see N_y evaluated as N sin(β-α). That would be taking N as acting along the bar, not normal to the ramp.

 

[Coderhk]

It does indeed, but doesn't that mean that the vertical component is N cos(α) as stated?

Shouldn't it be Nsin(beta-alpha)? I got to that answer based on my right most diagram. If the normal force was to be perpendicular to the surface it would be parallel to the bar right?

 

[haruspex]

. If the normal force was to be perpendicular to the surface it would be parallel to the bar right?

No, that would only be true if the bar were normal to the surface, which it is not.

 

[Coderhk]

No, that would only be true if the bar were normal to the surface, which it is not.

should the free body diagram look like this instead?

 

[haruspex]

should the free body diagram look like this instead?

It looks roughly right, but there seems to be (faintly) an angle labelled α between the normal and the horizontal. That would not be right.
The angle between the surface and the horizontal is α.