1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Forces and Pulley System

  1. Sep 19, 2006 #1
    Block A has weight w_A and block B has weight w_B. Block A is positioned on the horizontal surface of a table. Block A is connected to a cord passing over an easily turned pulley that has Block B hanging down from the pulley. Once Block B is now set into downward motion, it descends at a constant speed.

    a. Calculate the coefficient of kinetic friction between Block A and the tabletop.

    I drew Block A's force diagram. For this part, I am assuming Block A is not in motion as yet; it is just sitting on the tabletop. (Is this correct?)

    Well, F_fr = mu_K * F_N and the normal force = w_A.

    F_fr = mu_k*w_A

    mu_k = F_fr/w_A

    But I think this is somewhat incorrect. Musn't I express mu_k with the known variables give. What is the Value for F_fr? :confused:

    b. A cat, also of weight w_A, falls asleep on top of block A. If Block B is now set into downward motion, what is its acceleration (magnitude and direction)? Express a in w_A, w_B, and g.

    Well, I am unsure of the F_fr expression. This is what I did for this part.

    For Block A, the expression is now F_T - F_fr = [(2*w_A)/g]*a
    For Block B, ---------------------w_B - F_T = (w_B/g)*a

    Adding these two expressions, I get

    w_B - F_fr = [(2*w_A)/g]*a + (w_B/g)*a
    w_B - F_fr = a[(2*w_A + w_B)/g]

    a = [w_B - F_fr]/[(2*w_A + w_B)/g]
    a = g*[(w_B-F_fr)/(2*w_A + w_B)]

    But I don't know what to put for F_fr as noted in the first part.

    Are my reasonings and math that I did do correct?

  2. jcsd
  3. Sep 19, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    One may assume block A has zero velocity.

    To find F_fr, what is F_fr opposing? What is the significance of Block B descending at constant speed?
  4. Sep 20, 2006 #3
    F_fr = w_B, but when I plug this value into the expression for a that I found, acceleration would be 0. According to Newton's First Law, a body that has constant velocity and, therefore, 0 acceleration is acted on by no net forces. Where do I go from here?
  5. Sep 20, 2006 #4
    Soaring -- didn't your last post just answer your own question??? In your first post, you wanted to know what F_fr was. So, what is it?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook