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Stress and extension question

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data
    A winding engine lowers a load of 5 tonne at the rate of 1.4m/s2 for a distance of 10m. Assuming that the load is suspended from a steel rope 12mm in diameter calculate the stress and the extension in the rope after travelling this distance. Assume Esteel=200 GN/m2 and g=10m/s2

    2. Relevant equations

    I've used the following equations to base my answer around:-
    Strain=extension/original length

    3. The attempt at a solution
    My attempt to answer this is as follows::surprised
    A=3.142 * (6)^2
    = 113.1mm2
    = 113.1*10^-6 m2

    F=5*1000*g where g=10m/s2
    = 5*10^4

    S=F/A = (5*10^4)/(113.1*10^-6)
    Stress=442 MN/m2

    using:- E=stress/strain
    E= 200 GN/m2
    strain= 2.21*10^-3

    My main query is where on earth does the 1.4m/s2 come into the answer and am I anywhere near the correct solution? This has been bugging me for quite some time now so any help would be much appreciated :smile:
    Last edited: Dec 6, 2006
  2. jcsd
  3. Dec 6, 2006 #2


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    Remember that 1.4m.s-2 is an acceleration.
  4. Dec 6, 2006 #3
    Thank you for the advice what a complete idiot to miss out something so basic :-)

    So I need to use Force=mass x acceleration to determine the actual 'downward pull', for want of a better word, of the load.

    In that case if I have (5*10^4) * 1.4 the force would be 7 *10 ^4.

    Would the rest of my calculations be correct if I plugged this value in?

    Last edited: Dec 6, 2006
  5. Dec 6, 2006 #4


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    Careful! There are two forces acting on the mass, tension acting upwards and weight acting downwards, causing the net acceleration. Therefore;

    [tex]mg - T =ma[/tex]

    Where the tension is the force acting on the steel cable. Does that make sense?

    Also note that one tonne = 1x103kg
  6. Dec 6, 2006 #5
    I do apologise for being such a numpty it's been far too many years since I last attempted anything like this so you'll have to bear with me.

    Right, I presume what you're saying is mass x gravity - T = mass x acceleration
    In that case, 5000x10 - T = 5000x8.6
    T = 7000

    That's presuming an upward acceleration of 8.6 m/s2 opposing gravity at 10m/s2 (net acceleration 1.4m/s2).

    w.r.t conversion of tonnes I was trying to change the weight into Newtons, hence 1x10^4 and I think by doing so have completely confused myself. Sorry!
  7. Dec 6, 2006 #6


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    No problem, that's why we're here :wink:
    Your almost there, but the acceleration in the equation, is the net acceleration i.e 1.4m/s2. Do you understand why?
  8. Dec 6, 2006 #7
    Hmmmm not sure! I may be on completely the wrong track here but these are my thoughts..... if the object was accelerating at 10m/s2 then it would have no force against it so tension in the cable would be zero. However, if the object was stationary it would be working against the full gravitational pull so tension would be m*g. As our object is accelerating at 1.4m/s it has m*1.4 less amount of force needed to hold it (if that makes any sense) so the actual Tension would need to take that into account => mg - ma = Tension ?
  9. Dec 6, 2006 #8


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    So, you are proposing that the acceleration is the difference between the acceleration due to gravity and the actual acceleration? (just making sure I've read your post correctly)
    Last edited: Dec 6, 2006
  10. Dec 7, 2006 #9
    Yes, I think :-) And your explanation is a hell of a lot simpler!
    Last edited: Dec 7, 2006
  11. Dec 8, 2006 #10
    why are you taking the acceleration to be 8.6 m/s^2.
    Why is that so? That is wrong.

    Using newtons second law.
    f=ma. f=net force acting on the body
    the two forces acting on the object are its weight and tension in the rope.
    so f=mg-T

    hence, mg-T=ma
    the net acceration given is 1.4 m?s^2

    The tension force is there in the rope because the object is accelerating otherwise there would be no tension in the rope. i.e if a=0 m/s^2
    Last edited: Dec 8, 2006
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