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How does a purse hook/hanger works?

  1. Jan 8, 2015 #1
    Hi.

    Could anybody tell me how does a purse hook works?
    guide.gif
    At first, I tried to ask myself why the hook will topple if we apply downward force at the end of the rod labelled "opening" (I know opening means different in the figure). I'd say that the base of rotating head (bottom-right corner) which is in contact with the table may acts as a fulcrum/ pivot point while we are "transferring" torque through the "opening" to the rod that is situated to the right of the rotating head. The same explanation would be given by me if I were to answer why the hook topples if we apply downward force on the rod situated to the right of the rotating head.

    Now let's apply downward force on the hook, again I tempted to say we are "transferring" torque all the way to the rod that is situated to the right of the rotating head. However, I know it is wrong!

    The fourth situation that I can think of is applying downward force on the ball-shaped/spherical end. This time the bottom-left of the rotating head will act as pivot point. But why? I start to think about center of gravity, line of center of gravity etc. , but how?

    I need help on explaining how it works. What a funny tool!

    Thanks.
     
  2. jcsd
  3. Jan 9, 2015 #2

    CWatters

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    It's just a funny shape pendulum. This is more obvious for the ball version.

    For the other version recognise that the pivot point moves if the "pendulum" is deflected. I suggest redrawing it an angle. The direction the pivot point moves acts to increase the restoring force due to gravity. This version should be more stable than the ball version.

    Edit: Your ideas about torque are basically correct. If you deflect a pendulum gravity provides a torque that tends to restore it to the vertical position.
     
    Last edited: Jan 9, 2015
  4. Jan 11, 2015 #3
    Hi, sorry for late reply.

    I still have trouble understanding the physics behind the purse hanger. I tried to search other sites for more information but what I found was that if we discuss it in depth, we are probably the very first person to do so on the Internet LOL.

    Anyway, you mentioned "ball version", and I think you are talking about this:
    images?q=tbn:ANd9GcQstUyclTEXhTCzaSxl9KWpu-nWDmGxZpnszjCmKyPpDPppdW75.jpg
    I am interested in knowing more about what you said:"I suggest redrawing it an angle.". I hope you can tell me how to draw the free-body diagram for the purse hook showing how it works like why when you apply downward force on parts other than the hook (the place where you hang your handbag etc.) (I mentioned four situations, one is hanging at the hook, while the other three is the situation which will cause imbalance ), the whole hanger will be in an imbalance state( or I should say the pivot points move).

    Thanks.
     
  5. Jan 11, 2015 #4
    I think you're somewhat over thinking the problem. Look at it this way:
    Obviously, if the hook was just a straight rod down, and you'd hang something on it, you would not start arguing about torques and stuff, right?
    Now imagine you introduce a tiny sideways U-bend alongside the rod. Does that change the situation? Not really, provided the U-bend can take the pull.
    Now, what if you make the bend bigger? No real difference to the small bend, which was no different from the straight rod.
    If you make the bend big enough, you end up with your hanger there.
     
  6. Jan 12, 2015 #5

    CWatters

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    Actually I meant a hook like yours that had a ball instead of the "rotating head" shown on your diagram. But that's not important.

    See the diagram which shows it knocked at an angle.
    guide.jpg

    Newtons laws say that for an object to be in equilibrium (eg not accelerating) the forces acting on the object must sum to zero. This is also obvious from his equation F=m*a. If a =0 then F=0. This also applies in rotation eg if the object isn't accelerating in rotation then the torques acting on it must sum to zero.

    So looking at the diagram do the torques sum to zero when it's knocked to the angle shown? No. The two forces (due to gravity and the normal force) cause a couple or torque that accelerates the hook in the clockwise direction and this restores it back to the vertical position.

    Something similar happens if it's knocked in any direction, it always returns to the vertical position. In this respect it's identical to a pendulum.
     
    Last edited: Jan 12, 2015
  7. Jan 12, 2015 #6

    CWatters

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    The fact that the pivot point moves is an interesting complication. Note that It always moves such that the restoring torque is increased. This contributes to the hooks stability. Canoes and ships are designed so that the bottom of the hull is shaped to produce a similar effect. On the other hand round objects, like a floating log, do not have this property.
     
  8. Jan 15, 2015 #7
    Thanks for helping.

    I think there is one way we can actually figure out hanging things on the hook without making it topples. It is by using center of gravity.
    http://www.grc.nasa.gov/WWW/k-12/airplane/cg.html

    We can say that torque that topples purse hook is just the motion caused by the downward force that missed the center of gravity/mass, even it is a little miss. Intuitively we know that the cg of the hook is somewhere around the center of the whole body of the purse hook (from side view) and somewhere above the hook where we hang thing. The cg doesn't necessarily to be a part of the body of the hook (like that of the bent body of high jumper).

    This animation may be helpful, though it shows centroid of U-shaped beam.:
    http://web.mst.edu/~mecmovie/appA/m17_03_centroid_ubeam.swf

    Again, thanks for helping.
     
  9. Jan 16, 2015 #8

    CWatters

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    The aircraft isn't a great analogy because an aircraft is free to rotate about any point where as the hook is constrained. Look at how a pendulum works. That's much more relevant.
     
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