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Homework Help: What is tensile strength?

  1. Oct 1, 2008 #1
    Is there a way to turn this tensile strength into maximum tension which is in Newtons?
    Last edited: Oct 2, 2008
  2. jcsd
  3. Oct 1, 2008 #2


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    Welcome to PF.

    Tensile strength is the breaking point of the material.

  4. Oct 1, 2008 #3
    So it's a quantity somewhat like Tension?

    Is there a way to convert this into Newtons?
    I don't know if it's possible because this is lbs/cm2 but Newtons are kg/m * s2
  5. Oct 1, 2008 #4


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    It's maximum tension ... splat.

    What is the diameter of the strands SpiderGuy uses? That's the cross section area you need for a strand?

    Figure if he is going to be scooping up Kirsten Dunst their weight together would be what? Is there enough strength to hold their weight?

    If there is , then what happens when he shoots it our 30 meters or so and swings across traffic going maybe 15 m/s? Does it satisfy the mv2/r centripetal force?
  6. Oct 1, 2008 #5
    I don't have the diameter of the strand. Just what I have in that first post.

    I figure that Spiderman + Mary Jane would be about 300lbs which seems well below the 20,000 lbs/cm2. Am I right to assume this? This would be about 300 pounds = 136 kg. (Multiply this by gravity (9.8 m/s^2) and it's a combined weight of about 1333 Newtons.[I dont know if this is vaild information but it's nice to have it])

    If the rope is 30m or 3000cm, then from the 20,000lbs/cm^2 that rope should be able to hold....

    Using these assumed values, the Centripetal force would be (136)([15m/s]^2)/30

    which is 1020 N.

    If all of this is done correctly, they will get to safety?
    Last edited: Oct 1, 2008
  7. Oct 1, 2008 #6


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    Then for your answer just work out what diameter he would need. Is this reasonable for a thread?
    You might be surprised by how small the answer is - spider silk is very strong.
  8. Oct 1, 2008 #7
    Say the cord is about .5cm in diameter (i'm considering that his spider web-y stuff would be pretty thin)

    how would i go about figuring about the weight this could hold?

    I figure I'd find the area of the cross section which would be (3.14)(.25cm)^2

    and I get .2cm^2

    what does this tell me?

    from this, how do I figure how much weight a 30m cord can hold?
  9. Oct 1, 2008 #8


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    If it can hold 20,000lbs/cm2 then 0.002cm2 can hold
    20,000lbs/cm2 * 0.002cm2 = 50lbs

    ps. your arithmateic is out dia=0.5cm is an area of 0.2cm2,
    Think about a square half a cm on a side must be about about 1/4cm in area - it's always worth doing a rough answer in your head to see if it's reasonable before hitting the calculator.

    The length of the rope doesn't matter - it's only the thickness that sets the strength.
    Well in theory anyway, in rock climbing it does matter!
    Last edited: Oct 1, 2008
  10. Oct 1, 2008 #9
    I know! I'm sorry the calculation was off (i fixed it). I just wanted to convert the answer into meters...or something (it's dumb mistakes like that that really get to me)

    so, the length of the rope doesn't matter. Thanks!
    .2 * 20,000lbs^cm2 would be 4000lbs.

    How does this factor back into centripetal force? Fc= mv^2/r say at a rate of 15m/s and his web is shooting off 30m and he's swinging across traffic.

    The centripetal force would have to be less than what the rope could hold. Correct? Either that, or spiderman would have to swing at a faster rate.

    (I know i'm asking a lot of questions. I really appreciate everyone who's helping!)
  11. Oct 1, 2008 #10


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    The centrifugal force makes things a bit more complicated. And yes the length of the rope would matter since it's the radius of the swing.

    If they were swinging in a vertical circle then at the top there would be weight pulling down and force pushing up - so there might be no tension in the rope (you can try this by swinging a weight on a string at just the right speed) but at the bottom of the circle their weigth and the force would be in the same direction and so they would add.
  12. Oct 1, 2008 #11
    I think I get what you're saying. The total force on Spiderman and Mary Jane would be

    Fnet= Fc + Fg?

    The centripetal force plus their mass times gravityi

    But isn't it that at the bottom of a vertical circle, the Tension (or in this case, the tensile strength) - Fg = Fc?

    The rope would have to be strong enough to withstand this force, correct?
    Last edited: Oct 1, 2008
  13. Oct 1, 2008 #12


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    Slower. Faster is greater force.

    Don't get too married to my numbers. I was just throwing out things off the cuff.

    Say it was as big around as a 6mm bb from an air rifle. That's a little over 1 x 10-4 m2

    Figure too the weight and volume involved in the stuff. How much of the stuff does he have to carry around? How much does a 30m long strand with a cross sectional area of 10-4 weigh or can he carry around in his spider glands?
    Last edited: Oct 1, 2008
  14. Oct 1, 2008 #13
    Wow. Now I'm very confused. Well, that or overwhelmed with the amount of information I need to know. Summary so far.

    The radius of his spider silk is crucial to determining the maximum amount of weight this string can hold. But then, how much will this silk weigh?

    For spiderman and Mary-Jane to get across safely their combined weight has to be less than the the maximum weight the rope can hold. And The tension of the rope has to be more than the centripetal force?

    Is this correct?
  15. Oct 1, 2008 #14


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    Pretty much. I'd say you want to really phrase it as centripetal plus the weight of the two.

    If you are only worried about 1 swing worth, then ignore string weight and volume of cord. I was suggesting that more for getting around town for an evening of delivering pizzas.
  16. Oct 1, 2008 #15

    Fc + mg = T

    this T has to be less than the value I am given.

    But my answer for T will be in Newtons, so how do I compare this to 20,000lb/cm^2?
  17. Oct 1, 2008 #16


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    Determine the strength of the strand. That's Tmax

  18. Oct 1, 2008 #17
    Okay. Doing that Equation above. Assuming Spiderman and MaryJane's combined weight is 300lbs (136 kg). Velocity = 15m/s the cord is 30m long = radius of the circle.

    Fc + mg = T

    m* v2/r + m* g = T

    136* (15^2)/39 + 136 * 9.8

    = 2352.8 N = 529 lbs.

    Now I just have to find the diameter of the string that could support that amount of weight.

    By using A= (pi)*r2

    then that A * 20,000lbs/cm2
    I hope this is right.

    Once again, thank you so much to everyone who has helped by posting :)
  19. Oct 1, 2008 #18
    The last piece of info:

    what do you think is a reasonable diameter for a silk strand. I understand it's going to be pretty small. But then again, spider silk is very strong
  20. Oct 1, 2008 #19


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    If you know how much it needs to hold, then figure out the minimum.
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