Finding what angle is the Normal force is.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 5K views
lion_
Messages
18
Reaction score
0

Homework Statement


The airplane traveling at a constant speed of 50m/s is executing a horizontal turn. If the plane is banked at ##\theta=15## when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature of the turn. Also what is the normal force of the seat on the pilot if he has a mass of 70kg.

Homework Equations


##\Sigma F_n = ma_n=N \sin \theta ##
##\Sigma F_b= 0 = N \cos \theta= W ##
##a_n=v^2/r##

The Attempt at a Solution


It is pretty much a plug and chug problem with the given equations. The issue I am having is that the problem uses cosine where I have sine and sine where I have cosine. Meaning that the normal force is 75 degrees from the vertical which is impossible or I must be interpreting the problem incorrectly. Please explain why this is or explain what is really happening in the picture where the plane is banking.
 

Attachments

  • plane problem.png
    plane problem.png
    1.9 KB · Views: 926
  • plane problem.png
    plane problem.png
    1.9 KB · Views: 848
Physics news on Phys.org
plane problem.png

Image didn't come out like I wanted too
 
I agree that 'banked at 15 degrees' should mean 15 degrees to the horizontal, and that leads to the equations you posted. On the other hand, banking at 75 degrees is not impossible.
The picture you attach shows the banking angle as being measured from the vertical. Did that picture accompany the question, or is it your illustration of what it would mean if it is at 15 degrees to the vertical?
 
I figured it out it was a little bit of common sense. First I pictured the plane flying horizontally the normal force and weight are parallel. If the plane tilts 90 degrees, then the normal force is perpendicular to the weight. So then some advanced geometry tells me if it shifts 15 degrees to the left it's going to move the normal force down 15 degrees which is 75. Picture was a bit confusing for me.
 
lion_ said:
it's going to move the normal force down 15 degrees which is 75
What do you mean "which is 75"? Do you mean 75 degrees to the horizontal?
lion_ said:
Picture was a bit confusing for me.
Are you saying the picture is part of the problem as provided to you? If so, it is clear that the given '15 degrees' of banking here means the plane is tilted over by 75 degrees. (I'm sure that's not standard usage though.)