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Stress in rope with pulley and weight

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Im trying to work out the stress of a fishing line after connecting a fishing line to a pulleyboard and adding 100g weights one at a time onto the fishing line to create strain and stress.

    I know the formula is Stress = Force divided by area

    And i have already worked out the force acting on the fishing line for each 100 gram weights which are added. If i have done this right it is 0.1 (100 g) multiplied by 9.81 (gravity) to give 0.981.

    Also i have the diameter of the fishing line as 35 microns (3.5 x 10 to the power of -5)

    So the area of the cross section of the circular fishing line is needed and then needs to be multiplied by the length of the fishing line which is 1.2 metres.

    Can anyone help me out with this please?

    Thank you so much

    2. Relevant equations

    Stress = Force / Area

    Area of circular fishing line = pie x radius squared x length of fishing line (1.2 metres)

    3. The attempt at a solution

    I keep on getting a 8 billion pascals answer :S
     
  2. jcsd
  3. Nov 2, 2007 #2

    Astronuc

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    Staff: Mentor

    Yes.

    No. Why multiply by the length. Area * length = volume.

    Take the force 0.981 N and divide by the area pi*d2/4 m2.

    1.02 E9 Pa. or 1.02 GPa

    Compare to 1 atm = 0.101325 MPa. So this is about 10000 atm of pressure/stress or 147 ksi (for a 100 g load), which is a rather large stress.
     
    Last edited: Nov 2, 2007
  4. Nov 2, 2007 #3
    So would that surfice to go on and work out the young modulus of that fishing line seeing as i already have the strains?

    I thought you may have had to work out the cross sectional area of the fishing line and multiply by the length to find the entire area of the fishing line, but i think you may be right.


    And thank you so much :)
     
  5. Nov 2, 2007 #4

    Astronuc

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    Staff: Mentor

    35 microns is thin.

    That is 1.38 mil or 0.00138 inch.

    For tensile stress, one is interested in the load (force) divided by the cross-sectional area.

    Young's modulus has units of stress, since strain is dimensionless.
     
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