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CIE A'level question on toppling of connected solids

  1. Apr 18, 2017 #1
    α IMG_1872.PNG IMG_1862.PNG 1. The problem statement, all variables and given/known data
    The problem my question is about is 4. (iii). Second image First image is markscheme



    2. Relevant equations


    3. The attempt at a solution
    The CIE markscheme has h= 0.944 as the least possible value for h. They find this value for h by setting up an equation for the centre of mass of the complete objects and then stating that the centre of mass of the hole object is above the common circumference ( on the side of the cylinder). This all seems clear to me, except for the second equation where the cos() is involved. If anyone could explain that, I would be very grateful .Yet when I first tried the problem myself I attempted to find the value for h by deducing that the moment of the weight of the cylinder about the centre of the common circumference of the complete object must be greater than the moment of the cone about the same point and I found my h to be equal to 0.4. Here is my moments equation:

    Let W be the weight of the cone.

    3Wx(h/2) = W x ( 2.4x0.25)

    Solving for h gives :

    h=0.4

    If anybody could tell me where I went wrong I would be again very grateful. Finally if anybody wants to check with the markscheme for that exam( the first part labeled with iii above the second image)

    Thank you
     
  2. jcsd
  3. Apr 18, 2017 #2

    BvU

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    Hi mori,

    Well, now that that has been cleared up: what is your question ?
    [mock surprise:smile:]Ah ! it is hidden in your attempt at solution: I do it soandso and don't see what I do wrong

    You assume the cylinder isn't leaning against the cone or pulling at it. In view of the final result 0.944 it is pulling. Of course: it has to lever itself as well as pull up the cone

    But the main point is that the moments with respect to the axis of rotation should balance
     
  4. Apr 18, 2017 #3
    Thank you for your reply! The axis of rotation would be at the point on the ground on the common circumference of the complete object ( the point I labeled A in my sketch) , right ? What would the moments of the respective weights be about the axis of rotation? Thank you for any further effort!
    IMG_1873.JPG
     
  5. Apr 18, 2017 #4

    BvU

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    Do you understand what they calculate with
    upload_2017-4-18_14-34-38.png
    and with
    upload_2017-4-18_14-34-55.png

    ?

    In answer to your question: ##3w\left (\left (2.4+h/2\right )\cos\alpha - \sqrt{0.7^2+2.4^2}\ \right ) \ \ ## and ##\ \ w \left (2.4* 3/4 \cos\alpha - \sqrt{0.7^2+2.4^2}\ \right )\ \ ##, respectively -- as you can easily calculate yourself.
     
  6. Apr 18, 2017 #5
    Thanks again. No I don't understand what is meant... at least I am not 100% certain. Please explain :smile:!
     
  7. Apr 18, 2017 #6

    BvU

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    It all starts with
    if you don't know what VG stands for, you are in the dark.
    But if you see how VG is related to ##2.5/\cos\alpha## you might see the light !
     
  8. Apr 18, 2017 #7
    Is VG the distance of the centre of mass of the complete object from the vertex of the cone along the axis of symmetry? And yes it all makes sense then. And looking back on my original post I think I did mention that suspision. Yet, referring to my sketch above, I do not understand how the equation of moments would be if Imwere to take moments about the point I labled A. More specifically, for the moment of a force I need the perpendicular distance from a point to where the force ismapplied. If you could clarify what the distance is I am looking for I would be very grateful. :smile:
     
  9. Apr 18, 2017 #8

    BvU

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    Yes. And its projection on the plane is ##VG\cos\alpha##...
     
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