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Perpendicular forces

  1. Jan 10, 2017 #1
    1.JPG


    2. Relevant equations


    3. The attempt at a solution

    I was of the understanding that the slope can be calculated as mgsin20 whereas the force acting straight down through the lift is mgcos20 but the answer is mg divided by cos 20??
     
  2. jcsd
  3. Jan 10, 2017 #2

    Doc Al

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    Hint: What force components act in the vertical direction? What must they add to?
     
  4. Jan 10, 2017 #3
    in the vertical direction we have weight acting downwards and lift acting upwards? so does it look like this?
    2.JPG
     
  5. Jan 10, 2017 #4

    Doc Al

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    The weight acts downward. Good! But only a component of the lift force acts vertically. What is that component?
     
  6. Jan 10, 2017 #5
    So only the 'top' part of the weight is what we're after? And the angle is also 20 degrees? So it looks like this?

    3.JPG

    in which case lift would then be Lift = mg/cos20

    I see!

    In that case, how does this differ to my notes that say the perpendicular force to the slope would be mgcos20? Is it because we're dealing with a component only?
     
  7. Jan 10, 2017 #6

    Doc Al

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    Not sure what you mean. There are two forces acting on the wing: Weight, which acts down. And the lift force, which acts at the angle shown.

    You need to analyze the vertical components.

    The angle that the lift force makes with the vertical is 20 degrees. So what is the vertical component of that force?
     
  8. Jan 10, 2017 #7
    The vertical component is weight (mg) right? So to get lift, we use trig to end up with Lift = mg/cos20? Is that what you mean by vertical component?

    4.JPG
     
  9. Jan 10, 2017 #8

    Doc Al

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    No. I simply mean: What is the component of the lift force in the vertical direction? You know the angle to the vertical, so how would you find the vertical component? (It will be in terms of L. You'll then use it to solve for L.)
     
  10. Jan 10, 2017 #9
    okay now i'm lost, i thought the mg/cos20 IS the vertical component? if not, can you write it, it'll probably click for me that way.
     
  11. Jan 10, 2017 #10

    Doc Al

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    I hope I'm not adding to your confusion!

    Not quite.

    You had this correct! (Didn't see it earlier.)

    But to be clear, here's how to figure out what L is.
    (1) vertical component of L = L cos20 (This is what I was trying to get you to say!)
    (2) vertical component of weight is just mg (of course, since it's vertical)

    ΣFy = 0
    L cos20 - mg = 0

    Thus:
    L cos20 = mg
    L = mg/cos20

    Does that make sense?
     
  12. Jan 10, 2017 #11

    haruspex

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    You should start by considering accelerations. You are told the direction of flight is (continuing) horizontal. So in which direction can you be sure there is no acceleration? For that direction, you know that the sum of forces is zero. What is the component of lift in that direction?
     
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