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Help me understand x and y components, sin cos, etc.

  1. Sep 19, 2006 #1
    I'm really having difficulty understanding how to approach problems in my physics class due to 1. never having taken physics before, and 2. having a professor that I cannot understand (english is not his native language, and he speaks very soft/fast).

    We're now doing tension in class, and I'm having trouble with the first problem :(

    Given this diagram (block is at rest), what is the tension on the left string?

    [​IMG]


    The only thing I can come up with is that since acceleration is zero, then Sum of Forces (Fnet) = 0.

    The book says something like tension in the x direction is Tcos(theta) and tension in the y direction is Tsin(theta).

    What does that MEAN? Does that mean they are the x and y components? please help!!
     
  2. jcsd
  3. Sep 19, 2006 #2
    Yes, they are the componenets of the force on the block due to the string (tension).

    The force actually acts along the string, 50 deg from the horizontal, towards the wall. But since force is a vector, you can split it up into components to make things simpler. Draw a right triangle with base angles 50 and 90. The "force vector" will be the hypotenuse and the two sides will represent the components.

    If that doesn't help, read this page... http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec5
     
  4. Sep 19, 2006 #3

    radou

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    Homework Helper

    I suggest you look at this link first:

    http://www.walter-fendt.de/ph11e/equilibrium.htm

    Try to express the condition of equilibrium [tex]\vec{F}_{net} = \vec{0}[/tex] in a graphical way. Further on, note that every vector has a x and y component. So, everything that applies for your vectors, applies for the components, too.
     
  5. Sep 19, 2006 #4
    neutrino,

    Ok, I now understand that the hypotenuse is the force vector and the right and bottom sides are vectors (adjacent and opposite of theta).

    But how does this help me in determining the tension on the left string which is on the other side of the ring? How does calculating the x and y components help me find T?

    Sorry for the stupid questions, but I'm having trouble putting two and two together.
     
  6. Sep 19, 2006 #5
    The block is not accelarating, which implies that the net force acting on it is zero. As I said earlier, force can be considered a vector, F = Fxi + Fyj. If the net force is zero, so must be it's components.

    It's known that there is no accelaration along the y-axis (also along the x-axis), and then there's gravity acting downwards. Therefore,

    Fy(string) + force due to gravity = 0 (i.e. the y-comp of the net force is zero)
     
    Last edited: Sep 19, 2006
  7. Sep 19, 2006 #6
    I'm assuming that both strings are made of the same material and tension in both are the same. In that case, all you need to do is solve for T by the above method.
     
    Last edited: Sep 19, 2006
  8. Sep 19, 2006 #7

    Fy(string) + force due to gravity = 0 (i.e. the y-comp of the net force is zero)


    Force due to gravity = 147N, I understand this.

    But how do I get Fy(string) ?

    Jesus I hate physics, does every university throw problems like these at students and offer no help after "lecture"? Maybe I should consider transferring. After all, I'm studying computer technology and will never be exposed to this material after this semester.
     
  9. Sep 19, 2006 #8
    I'm sorry for not being clear enough. I meant the y-component of the string force. It acts upwards, while gravity downwards.

    So, Tsin50 - 147 = 0
     
  10. Sep 19, 2006 #9
    so Tsin(50) = 147
    T = 147/sin(50)

    T= 191.89 N for the Y component of the string force. Do I then plug that value into some equation for T(left) to get the actual value of tension on the left string?

    Thanks, things are slowly starting to make sense.
     
  11. Sep 19, 2006 #10
    T is the tension and Tsin50 is the y-component (Tcos50 is the x-component). So you're answer is 191.89 N.
     
  12. Sep 19, 2006 #11
    That is weird, I try entering that value into my online physics page and get this:

    [​IMG]

    hmm, are you sure there are no more steps involved? Thank you for your help!
     
  13. Sep 19, 2006 #12
    Any ideas? I've only have one more attempt before the software denies me any more attempts - is there something I'm still missing? thanks
     
  14. Sep 19, 2006 #13
    Sorry about that. The x-component actually balances the tension from the left string, which should leave you with

    Tension of left string = Tcos50 = 191.89cos50
     
  15. Sep 20, 2006 #14
    neutrino,

    Great, that worked! I have one final question. Upon completion of this problem I was presented with a flowchart involving the steps to solving this problem:

    [​IMG]

    how do you differentiate between your sin/cos formulas and the formulas in the attached picture? I guess I don't understand how they come up with those three formulas; again, thanks for the help.
     
  16. Sep 20, 2006 #15
    The "Trigonometry formula" says that if you take the ratio of the components (in a certain way), you end up with the tangent of the angle. In you're case, Tsin50/Tcos50 = tan50 - this is must be familiar to you from trigonometry. The second part says that the forces acting along the x-direction add to 0, and the third part says the same about the y-direction. Remember, the minus sign indicates that that particular force acts along negative x-direction, or y-direction as the case may be.
     
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