Solving Compression Problem: Water Skier & Tow Rope

  • Thread starter Thread starter shaka23h
  • Start date Start date
  • Tags Tags
    Compression
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 5K views
shaka23h
Messages
38
Reaction score
0
A 57-kg water skier is being pulled by a nylon (Young's modulus 3.7 x 109 N/m2) tow rope that is attached to a boat. The unstretched length of the rope is 18 m and its cross-section area is 1.9 x 10-5 m2. As the skier moves, a resistive force (due to the water) of magnitude 190 N acts on her; this force is directed opposite to her motion. What is the change in length of the rope when the skier has an acceleration whose magnitude is 0.81 m/s2?


Ok here is what I was able to do so far I know that

F = Y(Delta L/L0)A

When I solve for Delta L I know that Delat L = FL0/ Y(A).

I just don't know how to factor in the acceleration and most importantly how to use the 190 N resistance force?

Please let me know and thanks a lot.

 
Physics news on Phys.org
oh ic ic.

That explains it so much better.

Thank you so much.