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Moment at a point when the object is sliding

  1. Oct 1, 2015 #1
    1. The problem statement, all variables and given/known data
    i cant understand why when the total moment at B is positive , the object is not sliding ? can someone help to explain ?

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Oct 1, 2015 #2


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    I agree there is something wrong with the note.
  4. Oct 1, 2015 #3
    Dependent on h, a and μ, if you increase the force P, the block will either start to slide or overturn, depending on which of the following cases will be reached first:

    Case 1: P > μ ⋅ W (horizontal forces / sliding)

    Case 2: P ⋅ h > a ⋅ W → P > a/h ⋅ W (momentum around B / overturning)

    Imagine Case 2 is reached first: If the block starts turning what happens with the distances a and h?

    Of course there is also the indifferent case μ = a/h, I didn't take into account, but in my opinion that's not scope of the note.
  5. Oct 1, 2015 #4
    P ⋅ h > a ⋅ W → P > a/h ⋅ W

    why overturning will occur ?
  6. Oct 1, 2015 #5
    ΣMB = P ⋅ h - W ⋅ a > 0
  7. Oct 1, 2015 #6


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    Sure, but goldfish' question is why the book says that if the net moment is positive then sliding will not occur. As you say, it can occur.
  8. Oct 1, 2015 #7
    I tried to clarify the point, I supposed the author of the book was trying to point out, but maybe I missed it too.
  9. Oct 1, 2015 #8
    the moment of P.h and a.W are counterclockwise, right? so , when let say when P.h = -8 , a.W = -5 and -aa respectively .
    case 1, -8-(-5) = -3
    case 2 , -8-(-11) = +3
    in case 1 , since the total moment at B is negative , it will start sliding , right ? P ⋅ h > a ⋅ W is also satisfied here.
    in case 2 , since the total moment at B is positive , it will not start to slide, right ? P ⋅ h <a ⋅ W here.
  10. Oct 2, 2015 #9
    Considering the moments around point B

    P ⋅ h → clockwise
    W ⋅ a → counterclockwise

    Case 1:

    ΣMB = P ⋅ h - W ⋅ a + N ⋅ a = 0 (as the block still touches the ground with the whole bottom, N is located exactley beneath W → the block doesn't turn as the net moment is 0)
    ΣFx = P - Fkmax = P - N ⋅ μ0 = P - W ⋅ μ0 > 0 (the force P exceeds the maximal possible friction force → the block starts to move in horizontal direction (sliding))

    The sum of the moments around B must be zero, else it would turn (and if the net moment would be negative it would turn counterclockwise, so "into" the ground)

    Case 2:

    ΣMB = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also Fk) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
    ΣFx = P - Fk = P - N ⋅ μ = P - W ⋅ μ = 0 (the force Fk < Fkmax, which means as reaction force it will have the same value as P → the block doesn't slide)

    The sum of the moments is larger than 0, this means turning, but as the force P doesn't exceed the maximal possible friction force Fkmax (= N ⋅ μ0 = P - W ⋅ μ0), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.
  11. Oct 2, 2015 #10
    1.)why the net moment is negative , it would go 'into' the ground?

    2.)is it possible that the object start to turn/overturn while it's not sliding ?

    3.)is it wrong that the author doesnt take the N.a into calculation ?
  12. Oct 2, 2015 #11
    According to the picture you posted, clockwise moments are defined as positive. If the moment around point B is negative, the block would turn counterclockwise around point B.

    Yes it can and I think that is the point of the whole text you posted. If μ0 > a/h the net moment around B will be > 0, before the force P > Fkmax → no sliding, but turning

    I don't know what the author wrote before the paragraph you posted, but the picture shows the instant, when the block starts to turn. The only spot the block is touching the ground with is the right bottom edge (B). That means all the reaction forces have to affect this edge, also N. But if the sum of all moments is calculated around this point, the lever of N is zero. So based on the situation in the picture the moment produced by N is zero, so it's not wrong.
  13. Oct 2, 2015 #12
    why when If μ0 > a/h the net moment around B will be > 0???
  14. Oct 2, 2015 #13
    why μ0 is used in case 1? and μ is used in case 2? the value of μ remain constant all the times , right ?

    why when the object is turning , the N will shift to point B ? but not beneath W ?
  15. Oct 2, 2015 #14
    when P.h greater than W.a , it will start to turn , right? in this condition , the total moment at B= positive .... why the book gave when it is positive , the object is not sliding / overturning ? sliding and overturning are different things , right ? why the author mix them together ?
  16. Oct 3, 2015 #15
    μ0 shall be the static friction factor which remains constant all the time, using it it's possible to calculate the maximal friction force Fkmax. If P < Fkmax the sum of all forces in horizintal direction is ΣFx = 0 = P - Fk → P = Fk → P = N ⋅ μ = W ⋅ μ. If instead μ0 would be used, then the block would accelerated to the left (because P < Fkmax). So the μ without the index 0 indicates, that P can still be increased without starting to move the block.

    Because at point B will be the only point the block is touching the ground. But you can try it yourself, take for example a tall bottle and push it close to the bottom - it will slide. If you increase the distance of your finger to the ground/desk, there will be a point, when the bottle doesn't start to slide, but turn (and fall, if you don't stop).
    Last edited: Oct 3, 2015
  17. Oct 3, 2015 #16
    I'd say "not sliding / overturning" means "overturning and not sliding". As already mentioned I think the point of this paragraph should be to show that dependent on the height h (and the breadth a and the friction represented by μ0), the block will either turn OR slide (except at the indifferent case μ0 = a/h). But honestly, reading what you've posted, I'm not sure about the quality of this notice (but I only saw one paragraph, so I don't want to judge it to soon). Maybe of course, I don't get the point of the author either.
  18. Oct 3, 2015 #17
    in order to prevent the block sliding to the left before the P is large enuf to 'counter' the frictional force , so we use μ instead of μ0 ? if we use μ0 , the object will slide to the left , it will looks weird?
  19. Oct 3, 2015 #18
    Of course it wouldn't slide to the left, because the friction force is a reaction force. It's just the formalism for the paper, because the equations are not correct using μ0 (as then the calculated net force would point to the left).
  20. Oct 3, 2015 #19
    how did u come to the equation μ0 = a/h ?
  21. Oct 3, 2015 #20
    ΣFx = P - Fkmax = P - N ⋅μ = P - W ⋅μ → P = W ⋅μ
    ΣMB = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a
    → W ⋅μ ⋅ h = W ⋅ a → μ ⋅ h = a → μ = a / h

    If μ < μ0 → μ0 > a / h ... turning
    If P > Fkmax → μ0 < a / h ... sliding
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