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How do i solve for the fnet?

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the net force acting on each object indicated in the following diagrams.
    21noa6b.png
    2. Relevant equations
    Cosine, Sine law

    3. The attempt at a solution
    I'm currently doing an online course, but the content is EXTREMELY confusing. I have only been provided the cosine and sine law in the course and nothing else. I don't want anyone to answer the question for me, but possibly guide me through this. I've only been able to get as far as to attempt the tip-to-tail strategy by drawing a line in the diagram above (the line on top of the 35 degrees).
     

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    Last edited by a moderator: Apr 28, 2017
  2. jcsd
  3. Oct 22, 2016 #2

    kuruman

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    Hi den388 and welcome to PF.

    Your drawing is a bit confusing. I assume that you have only two forces acting on the block, 32 N straight up and 38 N down and to the right. Is that correct? What I find confusing is which angle is 35o? Is it between the "East" axis and the 38 N vector?
    Anyway the tip-to-tail method will not carry you very far. Do you know how to break a vector into its components?
     
  4. Oct 22, 2016 #3
    Yes. There are only two forces acting on the block 32N straight up and 38N down. The diagram was given to me like that, but I'm pretty sure the 38 degrees is located between the dotted line and the 38N line. I'm sorry to say that the course hasn't taught me how to break vectors into it's components. I can do a bit of researching if you'd like. I apologize for the confusing diagram :frown:
     
  5. Oct 22, 2016 #4

    kuruman

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    OK, but you didn't answer my other question, do you know how to break a vector into its components?
     
  6. Oct 22, 2016 #5
    I'm sorry.. i don't.
     
  7. Oct 22, 2016 #6

    kuruman

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    Do you know the basic trig relations in a right triangle?
     
  8. Oct 22, 2016 #7
    Yes i do. I just checked online and breaking vector components was basic trig relations. I apologize.
     
  9. Oct 22, 2016 #8

    kuruman

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    OK, Let's look at the 32 N force first. Call it F1. Let's change names for the directions to more conventional names. We will call "East" +x, "West" -x, "North" +y and "South" -y. To go from the origin to the tip of the 32 N force, you need to move only in the +y direction and not at all in the x-direction, positive or negative. We say that this force has a zero x-component and a +y-component that is 32 N. One way to summarize this is to say F1x=0 N and F1y = 32 N.

    Now for the 38 N force. To go from the origin to its tip you need to move in the positive x-direction by amount F2x and along the negative y-axis by amount F2y. To find what these are, draw a right triangle as follows: Start at the tip of the vector and draw a line segment perpendicular to the positive x-axis. Use trig to find the length of the two right sides and tell me what they are and how you got them.
     
  10. Oct 22, 2016 #9
    2cih5ra.png
    Did i draw it correctly? I'm assuming that the Fnet is the dotted line as there's more force being applied downwards.
     
  11. Oct 22, 2016 #10
    Sorry, I'm just curious why I have to go along the negative y-axis.
     
  12. Oct 22, 2016 #11
    2rzc9b7.png
    Sorry, I read what you explain to me carefully and resulted in the diagram above. Im still curious wehther why I have to go along the negative y-axis.
    1t1s2x.png
     
  13. Oct 22, 2016 #12

    kuruman

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    You can use the law of cosines and your answer is correct. I was going to show you a different way of doing this by resolving the vectors into their components. Maybe that's comes later in your online course so you're good for now. Be sure to come back if you have more questions.
     
  14. Oct 22, 2016 #13
    Thank you so much! I'm actually learning a lot from you! I've actually had a second question that was also quite tricky because it involves multiple forces.
    209n5tw.png
     
  15. Oct 22, 2016 #14

    kuruman

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    This one is easy if you look at it the right way. You can add the forces in any sequence you like and the result would be the same. What would you get if you added the 8.0 N "up" force and the 10.0 N "down" force to get a single force? Once you have that, you will have two forces to add just like before, except the one of the two is a combined force, but that doesn't affect the end result.
     
  16. Oct 22, 2016 #15
    I'm still having issues with creating the triangle. I was wondering if you could possibly teach me how to create the triangle. Cause what still confuses me is why couldn't I just create a triangle much like this?
    167m8h1.png


    And from what I've noticed the diagram looks much like the first diagram if you combined the forces as you've said. Though i hate actually hate how they've labeled the degree as it is extremely confusing.
    2cgd6rd.png
    Could you also tell me if my logic is correct here? (The 25N is the combine force and the 10N is from my understanding from the first example)
     
  17. Oct 22, 2016 #16

    kuruman

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    Forget the 17.0 N force for the time being. It will come in the picture later.
    Please answer this question first
    When you add 8.0 N "up" and 10.0 N "down" what do get?

    Once you get that, we will take it from there.
     
  18. Oct 22, 2016 #17
    bfpfp.png
    You get a total of 2N going down. Is this what you were trying to show me?
     

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  19. Oct 22, 2016 #18

    kuruman

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    That's exactly it. Now combine the 2 N down with the 17.0 N force to get the net force.
     
  20. Oct 22, 2016 #19
    I see!! I'm starting to get this!! Thank you so much!! I've even able to solve the last one on my own! You were a great help! 20b1x52.png
     
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