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Correlation between metal strength and melting point?

  1. Dec 1, 2011 #1
    I am unsure of this matter and so I am curious, do stronger metals more than likely have higher melting/heating points? Does a metal that takes more force/pressure to break or snap have a higher melting/heating point as well? Do metals/alloys like steel or titanium have higher melting points than a softer metal due to their strength? or is it deeper than that?

    Also, although not technically related I thought another matter was not worth a new thread, I was just interested in how to calculate, if its possible the energy used in moving an object based on weight and distance/time it took to travel.

    Thank you for your time.
     
  2. jcsd
  3. Dec 1, 2011 #2

    Simon Bridge

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    1. Its one of those yes and no things - you can look up tensile strengths and melting points for different metals and see for yourself.

    2. you do work whenever you accelerate an object equal to the change in kinetic energy, which is also the applied force multiplied by the distance moved.
     
  4. Dec 1, 2011 #3
    1. So its not really related? strong metals may melt quicker than softer metals and its really just specific to the metal?

    2. So how do I find the change in kinetic energy? Lets say I picked up say a bowling ball, assuming all I got was its weight and the speed I accelerated it to, distance it went etc could I find out how much energy I was dealing with?
     
  5. Dec 1, 2011 #4
    It's a bit more complicated than this since tensile strength is itself a function of temperature as was found at great cost when the 'Liberty Ships' broke up due to cold induced brittle failure.
     
  6. Dec 1, 2011 #5

    Simon Bridge

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    Like the guy said - the strength of materials also depends on temperature - and depend differently for different materials so material a is stronger than material B at temp T1, but it is reversed at T2. Like I said - go look.
    You can do it directly from difference of kinetic energy:
    [itex]W=\frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/itex]

    ... for your example, you'd have to find the initial speed from kinematics.

    [itex]v_1^2=v_2^2-2ad[/itex]

    But I'd reason like this:

    [itex]W=Fd[/itex] but [itex]F=ma[/itex] so [itex]W=mad[/itex] ;) and your example has all these. (Note: the KE method will produce the same formula.)
     
  7. Dec 1, 2011 #6
    Hm these formulae are a bit beyond me. Do you mind breaking it down as basic as possible? without any abreviation?
     
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