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Understanding Stability and Equilibrium

  1. Oct 1, 2007 #1
    And probably the rest of the lessons I missed.

    We were asked to research stability and equilibrium. So far, I know that equilibrium is a state an object is in when no unbalanced forces are acting upon it.


    I got lost when I read after the scale diagram. I searched some more and I found some problems

    (1. An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4 N and points in the negative y direction. Find the magnitude of the third force acting on the object. http://www.physics247.com/physics-homework-help/simple-equilibrium.php)

    that I really had no idea how to solve because all I knew was that equilibrium was that balance!

    So could someone please help me understand? I know it looks like I haven't tried hard enough but I really did. Don't be mad.
  2. jcsd
  3. Oct 1, 2007 #2
    hey, welcome

    now dont fret
    think about it in the form of a graph. the x axis is horizontal and the y axis is vertical
    now, if you pull and object in a "Y" shape (that is on either side of the two pointy parts of the "Y"), it goes straight forward.

    but in this question, you have two axis, and if both were of equal force, they would go 45 degrees at a gradient of 1.

    however, the question says that you have two different forces! so the third force would not be straight down the middle, rather towards the larger force as it has a larger pull.
  4. Oct 2, 2007 #3
    Oh gosh. I really have no idea what that meant. I REALLY need this. Uhm... if someone could talk to me on AIM or Yahoo something. Volunteer... anyone?


    I am trying to do this problem.
    http://www.physics247.com/physics-homework-help/simple-equilibrium.php (((Number 2)))

    I realized that there wasn't an answer. This is similar to what we're doing in class right now.
    So far, this is what I have. I don't know if it's right or not, if it is, I really have no idea how to move on.

    Since the weight is distributed evenly, then each rope is 3.33kg, right? So I made the little graph but I'm not sure what to put for the weight. 3.33 or 10. I didn't understand the class discussion. BUT. I have a picture of my work.

    Last edited: Oct 2, 2007
  5. Oct 2, 2007 #4


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    For the first problem, you need the net force on the object to be zero. Have you studied i, j vectors?

    You have three forces.
    [tex]\vec{a} = 6.5\vec{i}[/tex].
    [tex]\vec{b} = -4.4\vec{j}[/tex].
    You want to get [tex]\vec{c}[/tex].

    You know that:
    [tex]\vec{a} + \vec{b} + \vec{c} = 0[/tex].

    Solve for [tex]\vec{c}[/tex]. what is its magnitude?
  6. Oct 2, 2007 #5
    We did the first condition of equilibrium. I'm not familiar with i, j vectors. I'm not too familiar with magnitude. Basically, I entered a three month old physics class and I just... fell into this.
  7. Oct 2, 2007 #6


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    Do you know how to get the sum of two vectors... ie the resultant?
  8. Oct 3, 2007 #7
    Oh yes. Just add the two vectors. Like 5N east + 6N east = 11 N east.
  9. Oct 3, 2007 #8


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    cool. How about 5N north + 6N east?
  10. Oct 4, 2007 #9
    Oh yeah. I just learned that.

    5N north + 6N east =

    25 +36 = square root of 61 north east
  11. Oct 4, 2007 #10


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    Cool. the magnitude is correct. and you're right it is north east. but you should also be able to calculated the exact angle. so you have a right triangle with one side 5, the other side 6. and the hypoteneuse sqrt(61).

    You need the angle between the 5 and the sqrt(61). so you can use trig to get this angle. you can use arctan(6/5) = 50.19 degrees. So the direction is north 50.19degrees east.

    Now for your problem, do the same type of thing. Get the magnitude of the sum of the 6.5N force which is along the x-axis and the 4.4N force which is along the -y axis.
    Last edited: Oct 4, 2007
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