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: Another Friction Problem

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Block A in the figure View Figure weighs w_1 and block B weighs w_2. The coefficient of kinetic friction between all surfaces is u_k.

    Find the magnitude of the horizontal force \vec{F} necessary to drag block B to the left at constant speed if A is held at rest (figure (b)).


    2. Relevant equations
    F = ma
    a = 0

    3. The attempt at a solution
    I attempted by drawing two FBDs for A and B. I have for

    B: w_2 downward, F_aonb downward, normal force upward, F to the left dragging, and fk_ground
    A: w_1 downward, normal force upward, F_bona, and fk_ab

    Then I solved for the normal force and I got n = (w_1+w_2)/2 and then I plugged it into the fk = ukn, but well wrong answer.

    I have a few questions because I don't really understand the whole picture. There's two frictional forces acting on B right? From A and from the ground? Is this correct? Also, the normal force for both B and A...are they the same or are they different? They're different right?
     

    Attached Files:

  2. jcsd
  3. Feb 11, 2007 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Draw the FBD's again and identify the forces on each block. The normal force on block A is the force upwards of B on A.
     
  4. Feb 11, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You left out the friction force of A on B. Note also: the force you call F_aonb downward is the normal force between A and B. A and B exert two forces on each other: a normal force and a friction force.
    Seems like you are counting the normal force twice: F_bona is the normal force. And you left out the tension in the rope pulling on A. (You don't need to analyze the forces on A to solve this problem, but it's good exercise.)

    I assume you mean the normal force between the ground and B. How did you solve for it? (Consider vertical forces only.)

    Absolutely correct.
    Yes. The normal force between A & B is different from the normal force between B & ground.
     
  5. Feb 11, 2007 #4
    Okay, the friction force from A to B is going in the same direction as the friction force from the ground to B right? Because friction forces are always in the opposite direction from the acceleration and perpendicular to normal forces..? If that's not wrong, then this is what I'm getting:

    n_gb: n_ab + w_2

    F: fk_ab + fk_gb

    Is this correct? Am I going in the right direction, no pun intended?

    Edit: Oh! And n_ab is w_1 right? Because the normal force on A = n_bona = w_1? That's what I'm getting from my FBD of A...
     
    Last edited: Feb 11, 2007
  6. Feb 11, 2007 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, this seems correct....
     
  7. Feb 11, 2007 #6
    I got the answer. Many thanks for both the help. ^^
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook