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!

Equilibrium on slopes.

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    two boards of length 5m and 8m are hinged together and the hinge supported 4m above the ground, thus forming a double inclined plane 13m long measured along the slopes. Two blocks of equal weight are placed on the boards and connected by a cord passing over a frictionless pulley at the apex. If the coefficent of friction between blocks and boards is 0.3

    a) show that the system is in equilibrium.

    http://img525.imageshack.us/img525/3209/q10fs0.jpg diagram

    2. Relevant equations


    3. The attempt at a solution
    Block A: The left one.
    FN=mgcos53.13deg
    Frictionforce=mgcos53.15deg*(0.3)
    =0.18mg
    Force=mgsin53.13deg
    =0.8mg

    Block B: The right one.
    FN=mgcos30deg
    Frictionforce=mgcos30deg*(0.3)
    =0.2598mg
    Force=mgsin30deg
    =0.5mg

    If these calculations are correct then I dont see how they are in equilibrium.
     
    Last edited: Oct 22, 2007
  2. jcsd
  3. Oct 22, 2007 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember that the static friction force on each block is less than or equal to uN.
     
  4. Oct 22, 2007 #3
    it doesnt state that the coefficient is static(implying equilibrium).

    it only gives a coefficient of friction. After calculating the forces and friction forces and finding the net result it doesn't seem they are in equilibrium.

    Can someone show where I am going wrong in my calculations?
     
  5. Oct 22, 2007 #4
    What about mgsin(theta) component on each block? Further on one of the blocks the frictional force will be up the incline and on the other down.
     
  6. Oct 22, 2007 #5
    yup i have accounted for all those factors.

    If I use block A I get:

    0.8(downwards force along slope) - 0.18(friction force) + 0.2598(friction force of block B) - 0.5(force of block B along slope) = 0.48
     
  7. Oct 22, 2007 #6
    There must be some fundamental thing I am leaving out or that the system is not in equilibrium at all.

    According to the professor it is.
     
  8. Oct 23, 2007 #7

    learningphysics

    User Avatar
    Homework Helper

    The way I do these problems is to first assume that the system is in equilibrium (ie a = 0)... then see if the frictional forces come out to < uN... if they do, then they are indeed in equilibrium... if they don't, or can't... then the system isn't in equilibrium...

    don't assume the frictional force is uN... just let the frictional forces be f1 and f2...

    The system tends to try to accelerate to the left... so the left side has the fricion upward... right side has the friction downward...

    for the left block.

    mgsin(theta1) - T - f1 = 0
    mg*4/5 - T - f1 = 0.

    f1 = T - mg*4/5

    what is the equivalent equation for the right block? what is f2?
     
  9. Oct 23, 2007 #8
    T-mg*4/8 - f2=0
    f2=T-mg*4/8

    I think I see what is going on though.

    If we assume the system accelerates to the left: The net force will be
    mg*4/5 - f1(which is 0.18mg) - f2(which is 0.2598mg) - mg*4/8 = Negative answer!

    By F=ma we get a negative acceleration.

    If we assume the system accelerates to the RIGHT the net force will also be negative! This also gives negative acceleration. Therefore the system must be in equilibrium. I was just getting my signs mixed up which you cleared up. I was thinking the right side friction was going upwards.

    Thanks lot learningphysics!
     
  10. Oct 23, 2007 #9

    learningphysics

    User Avatar
    Homework Helper

    Ah... good. I was actually not thinking that way exactly. But that's better than what I was thinking! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Equilibrium on slopes.
  1. The slope (Replies: 2)

  2. Is it in equilibrium? (Replies: 1)

  3. Frictionless slope (Replies: 13)

  4. Friction and slope (Replies: 17)

Loading...