Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Tension: Force + Velocity. Breaking Strength.

  1. Oct 23, 2007 #1
    [SOLVED] Tension: Force + Velocity. Breaking Strength.

    1. The problem statement, all variables and given/known data
    The tension at which a fishing line breaks is commonly called the ``breaking strength''. What minimum breaking strength is needed for a fishing line to stop a salmon that weighs 90.0 N in 18.0 cm if the fish is moving horizontally with an initial velocity of 2.5 m/s? Assume the acceleration is constant.

    2. Relevant equations
    F=mg, F=ma, V^2 = Vo^2 + 2a(X-Xo)

    3. The attempt at a solution
    I'm not really sure how to tackle this problem. At first I simply assumed the minimum breaking strength of the string would have to be at least > 90N. However the fact that it's moving at 2.5m/s says to me that the 18cm of distance traveled increases the force created by the fish. But I'm not sure, algebraically, how to relate Force to Velocity. Also, since the fish weighs 90N, that implies its weight in relation to gravity. How would this translate when analyzing only movement in the X direction?
    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 23, 2007 #2

    Ok. This problem is trivial and I apologize for wasting space on your forum, but maybe some day it will help someone else who gets caught up on the details...

    I neglected to notice that final Velocity would be 0. Once I realized this I saw I could use one of the kinematics equations to derive acceleration.

    Here's what I did...

    (Vf^2-Vi^2)/(X-Xo) = 2a
    --> 0m/s - (2.5^2m/s)/(0m - .18m) = 2a
    --> a = 17.361m/s^2

    Now I can derive the mass of the fish by using F in the y-direction (since we know gravity is 9.8m/s^2)...

    --> 90N = m(9.8m/s^2)
    --> m = 9.18kg

    Now, put it all together for the x-direction...

    F = (9.18kg)(17.361m/s^2) = 159.443 N
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook