1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rope and Table

  1. Oct 27, 2009 #1
    A rope of length L sits at rest on a horizontal table with length L/3 hanging over the edge. The rope is mass M. What is the normal force acting on the rope? (I keep having people tell me that it is the product of the mass of only the rope that is on the table with the acceleration due to gravity and I am having a problem with this.)

    This is not a homework question. It was part of a test question involving the rope that starts sliding of the table with friction involved as well.
     
  2. jcsd
  3. Oct 27, 2009 #2
    The normal force is the total rope mass times g. It is evident from the force balance: the reaction of table compensates the whole mg.
     
  4. Oct 27, 2009 #3
    Thank you. Now, someone please answer to me the following question. If the rope starts to slide off (not at rest anymore), that is, the center of mass begins to accelerate, how does the normal force change during the movement or does it change at all (while the rope remains in contact with the table)? I am just looking for a qualitative answer.
     
  5. Oct 27, 2009 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    What do you think?
     
  6. Oct 27, 2009 #5
    Here is what I think. I know that the center of mass of the rope starts to accelerate. The acceleration has a downward component, call it a_y. Then I know that there is an imbalance between the normal force and the force due to gravity. Let m be the mass of the rope that is hanging over the edge of the table (this changes with t, time or with x, the distance the rope (on top) has moved across the table). So the net force on the center of mass in the y direction is M*a_y or m*x'' where x'' is the acceleration of the rope (on top) across the table. To get the normal force N, N = M*g - m*x'' . So, the normal force is proportional to the product of the distance x, and the acceleration x'' .

    Keep in mind the x and x'' correspond to only the acceleration of the rope on top of the table moving horizontally; these are the same as y' and y'' of the rope hanging over the table if you just consider the rope to only move in one path, that is, it doesn't sling off the table realistically.

    The reason I need the correct normal force is to know how the friction changes so I can find x as a function of t; to solve the correct differential equation (If I can). When I did the problem on the test, I said the normal force was constant, Mg, but after the test, I ask my professor what it should really be and she told me it should just be the mass of the rope on top times g. But, I really don't know what to think so... please help.
     
  7. Oct 28, 2009 #6
    Anyone? Just a tiny little hint? Please?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook