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Poloygon of forces to demonstrate that the crate is in equilibrium

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the Polygon of forces to demonstrate that the 200kg crate is in equilibrium on the slope

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I have attached a picture so you understand the problem a bit better.

    I have worked out that the weight acting directly down is:

    F=mg
    F=200*9.81
    F= 1962N

    The Force acting back down the slope is:

    Wx=W(weight) sin theta
    Wx=1962 sin 25
    Wx= 829.18N

    What other forces are acting on the crate?

    Thanks, Joe
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2011 #2

    tiny-tim

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    Hi Joe! :smile:
    What polygon did you get that from? :confused:

    You're missing the point of the question …

    it is asking you to name the three forces on the crate,

    then draw a triangle of them,

    then use that triangle to calculate the forces (from the given W). :wink:
     
  4. Aug 31, 2011 #3
    My book says:

    'Another method of looking at equilibrium involves using the polygon of forces. This
    involves taking all the force vectors that are acting on an object and adding them end to
    end. If the polygon is completely closed then the object is in equilibrium'

    So i thought i had to work out all the forces acting on the crate?

    Thanks, Joe
     
  5. Aug 31, 2011 #4

    tiny-tim

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    Is this your ICS book again? :redface:

    If you were given the magnitude and direction of all the forces, then yes could put them in a polygon (in this case a https://www.physicsforums.com/library.php?do=view_item&itemid=99"), see whether it joins up, and if it does then the body is in equilibrium.

    But this hardly ever happens :rolleyes:

    usually, you're given some of the forces, you're told that it's in equilibrium, and you have to find the remaining forces (so you draw a polygon with an "unknown" side or angle, and then find that side or angle).

    In this case, you're told that it's in equilibrium, and you're expected to draw a triangle (with all angles and one side known), find the lengths of the two "unknown" sides, and so find the normal force and the https://www.physicsforums.com/library.php?do=view_item&itemid=39" force.​

    (But this is a bad example, since there's really only one https://www.physicsforums.com/library.php?do=view_item&itemid=73"

    the question obviously expects you to split it into the normal and friction components, but doesn't actually say so. :frown:)
     
    Last edited by a moderator: Apr 26, 2017
  6. Aug 31, 2011 #5
    Where would the triangle sit on this occasion?
     
  7. Aug 31, 2011 #6

    tiny-tim

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    Sit? :confused:

    You can draw it anywhere, so long as its three sides are parallel to the three directions (vertical, normal, and slope).
     
  8. Aug 31, 2011 #7
    Hi tiny tim :)

    ive attached a picture of an attempt.

    Thanks, Joe
     

    Attached Files:

  9. Aug 31, 2011 #8

    tiny-tim

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    Hi Joe! :smile:

    Yes that's the correct triangle, and the correct results.

    (But I'm not totally convinced that you got the result the right way …

    did you use components of the force, or did you just use geometry?)
     
  10. Aug 31, 2011 #9
    If im being totally honest i dont completely know what im doing haha , it wont be the last time i say that.

    The only thing i can tell you is i went in paint and drew some straight lines roughly to what i thought the triangle would look like and worked out the forces by calculating the lenghts of the triangle as written.

    If theres a proper way, please enlighten me:)
     
  11. Aug 31, 2011 #10

    tiny-tim

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    ok, that's correct then!! :biggrin:
    yes, that's right, you drew the sides of the triangle parallel to the directions of the three forces
    yes, you treated this as a geometry problem, which is correct :smile:
     
  12. Aug 31, 2011 #11
    Since i have to use the polygon of forces to demonstrate its in equilibrium will this mean i should do a scale drawing to prove this? or do you know a better method to answer the question
     
  13. Aug 31, 2011 #12

    tiny-tim

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    You could use a scale drawing (that certainly would work),

    but the normal method is to use trigonometry …

    the important thing is that you have changed the original physical problem to a geometry problem

    (and geometry includes trigonometry)
     
  14. Aug 31, 2011 #13
    Thanks ever so much, i have one question left and thats pretty much all the maths done for the course so i can relax little:

    The crate at the end of the 15m journey is unhooked from the cable. At this point
    it is now at the bottom of a conveyor belt which is used to raise the crate to the
    next level of the factory at a constant speed.
    When the conveyor belt was first installed safety checks were carried out to
    determine the limit of angle for safety. It was found that at an angle of 30° the
    crates started to slip down the slope.
    Calculate the coefficient of friction between the crates and the belt.

    Now the book has the most vague coverage of this topic that i don't even have a clue where to start

    This is all with the same image first uploaded:)

    I greatly appreciate all your help so far, i rather pay money to you than ICS but unfortunately if i ever want out of my current job and to work on better aircraft then i need to get through this:)
     
  15. Aug 31, 2011 #14

    tiny-tim

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    ok, you can ignore everything up till …
    The way you do a problem like this is to assume that it doesn't slip (in other words, it's in equilibrium).

    So at angles of 10° 20° or 30° it's in equilibrium, but at 31° it isn't.

    If you drew diagrams for each of 10° 20° and 30°,

    they would all have the same vertical side, but the other two sides would vary.

    (btw, I should have said earlier that you should always label your triangles, for example ABC, so that you can easily refer to them in the equations)

    The rule for static https://www.physicsforums.com/library.php?do=view_item&itemid=39" is that the friction force divided by the normal force is always ≤ µs (the coefficient of static friction) …

    so how would you translate that physical rule into a geometrical rule for the triangle? :wink:
     
    Last edited by a moderator: Apr 26, 2017
  16. Aug 31, 2011 #15
    Thanks,

    ≤ µs (whats this mean?)

    EDIT. just hunted through my course and:

    Friction=pressing force*constant

    I guess the pressing force is the force of weight and gravity so 1962N
    How can i calculate friction without the constant (coefficient of friction???) and
    vice versa?
     
    Last edited: Aug 31, 2011
  17. Aug 31, 2011 #16

    tiny-tim

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    Haven't you done https://www.physicsforums.com/library.php?do=view_item&itemid=39" yet?

    µs is the symbol for the coefficient of static friction

    µk is the symbol for the coefficient of kinetic (or dynamic or moving) friction :wink:

    (and the crate is stationary relative to the belt, so the static coefficient is the relevant one)
     
    Last edited by a moderator: Apr 26, 2017
  18. Aug 31, 2011 #17
    my course shows neither of those symbols:/ and ive scanned through the friction page but its going in one ear and out the other.

    How can you calculate friction without the coeeficient, makes no sense that a formula cant be applied without the other

    It says that ideally you need to calculate it by trials and without knowing the materials used specifically you cant use the table. its a horrible question haha
     
  19. Aug 31, 2011 #18

    tiny-tim

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    Yes, the trial basically consists of putting it on a surface, and tilting the surface until the cart starts to slip.

    As I said, you calculate the coefficient as equal to the friction force (along the slope) divided by the normal force (perpendicular to the slope).
     
  20. Aug 31, 2011 #19
    Thanks,

    Im not sure why the friction force is is along the slope, is there a direction of the force?

    Anyway:

    829.18N/1778.18N

    Coefficient of friction = 0.466 (thats at 25%)

    At 30%:

    Slope:
    F= w sin theta
    F=1962 sin 30
    F= 981N

    Normal:
    F=w cos theta
    F=1962 cos 30
    F=1699.14N

    Coefficient of friction = 981/1699.14
    Coefficient of friction = 0.577 (30%)
     
  21. Sep 1, 2011 #20

    tiny-tim

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    Hi Joe! :smile:

    (just got up :zzz: …)
    Yes, the https://www.physicsforums.com/library.php?do=view_item&itemid=540" is always the direction opposite to which the object "wants" to move.

    So the https://www.physicsforums.com/library.php?do=view_item&itemid=39" is always parallel to the surface between two bodies (or to the tangent plane if the surfaces are curved). :wink:
    Yes, that's fine, :smile:

    However you could have done it more quickly …

    there was no need to find either of the forces …

    all you needed was the formula µs = Wsin30°/Wcos30° = tan30° = 0.577 :wink:
     
    Last edited by a moderator: Apr 26, 2017
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