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Dynamics: Planar Kinetics of a Rigid Body(Work and Energy)

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    The operation of this garage door is assisted using two springs AB and side members BCD, which are pinned at C. Assuming the springs are unstretched when the door is in the horizontal (open) position and ABCD is vertical, determine each spring stiffness k so that when the door falls to the vertical (closed) position, it will slowly come to a stop. Use approximate numerical values to explain your result.

    media%2Fa83%2Fa83c7300-246d-4782-afe2-f4822f59eca7%2FphpPISPZY.png

    Where do I start? And can anyone please provide me with a guided solution?
     
  2. jcsd
  3. Nov 13, 2013 #2

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    Hello naynaitmus,

    Welcome to Physics Forums!

    Are there any variables missing from the problem statement, such as the mass of the door, the height of the door, length of BC and angle of ABC? Many of these variables might not be necessary if the initial length of BC (as shown in the picture) is very small, which it looks to be (it's difficult to tell from the picture though). But I think one way or another you're going to need variables for the height and mass of the door. (They don't necessarily need to numerical with units, but you'll need variables to express them in your final answer.)

    Anyway, from what I gather from the title of this thread, I'm guessing that you are supposed to ignore friction and any external forces, and assume that conservation of mechanical energy holds.
    • What's the difference in gravitational potential energy of the door between its up and down positions?
    • Given that the door starts from rest (when open) and ends at rest (when closed), what does that tell you about the difference in Kinetic energy between its up and down positions?
    • Therefore, how much energy must the springs absorb?
    • How does the potential energy held by a spring relate to k? (And don't forget that there are two springs involved with this problem.)
     
  4. Nov 13, 2013 #3
    1. The mass of the door, the height of the door, length of BC and angle of ABC should be given abitrary values.

    2. Gravitation potential energy would be larger when it's opened and smaller when it's closed due to the height.

    3. The difference in linear Kinetic energy would be zero and only the kinetic energy existing would be the 1/2Iw^2 ?

    4. The springs must absorb the total of both the kinetic energy and gravitational potential energy of the door when it's closed.

    5. Potential energy due to spring 1/2kx^2

    Did I get it right?
     
  5. Nov 14, 2013 #4

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    Well, if the door "slowly comes to a stop," I think we can ignore the final rotational kinetic energy just as well as the linear kinetic energy.
     
  6. Nov 14, 2013 #5
    One final question, how does the angle ABC come into play though?
     
  7. Nov 14, 2013 #6

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    I don't think it does really, if the initial BC is short (when the door is up). It's just a little difficult to tell for sure from the picture.
     
  8. Nov 14, 2013 #7
    I think ABCD makes the angle, with B being the end point of the spring and C being the end point of the arm connecting one another. But yes, thanks a lot though I'll figure it out from here :)
     
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