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Homework Help: Friction and acceleration

  1. Aug 9, 2007 #1
    A block rests on a wedge inclined at angle B. The coefficient of friction between the block and plane is mu (but lets just call this M). The wedge is given horizontal acceleration A. Assuming tan B > M (essentially that the block will slide down), find the minimum and maximum acceleration for the block to remain on the wedge without sliding.

    I said the horizontal projection on to the surface would be -mA(cos(B)) and it's normal would be mA(sin(B)), thus friction would be MmA(sin(B)). For the block, friction is -Mmg(cos(B)) and force due to gravity would be -mg(sin(B)).

    I said the minimum would be where the friction of the block due to gravity equals the other forces. So:

    -Mmg(cos(B))= -mA(cos(B)) + MmA(sin(B)) - mg(sin(B)), which simplifies to:

    (divide by cos(B) to obtain tan(B))

    -Mg = -A + MA*tan(B) - g(tan(B))
    (g(tan(B)) - Mg) = A(M*tan(B) - 1) or..

    A = g(tan(B) - M)/(M*tan(B) - 1)

    the book says if I plug in 45 degrees, I should get:

    A= g(1-M)/(1+M)... which is similar, but I am thinking I may have just gotten lucky or put in a wrong negative sign somewhere.
  2. jcsd
  3. Aug 9, 2007 #2
    From FBD, i got similar exp:

    {I don't take care of signs, but instead I just collect all the forces that are on one say, and so on..}
    A is force due to the acceleration, and W is mg.
    and, I also got same answer, when I plugged in 45 deg.

    M[W.cos(B)+A.sin(B)] is the frictional force.
  4. Aug 9, 2007 #3

    Doc Al

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    Staff: Mentor

    I'm having some difficulty following your reasoning. In any case, the normal force between wedge and block is not simply mg(cos(B)).

    Another hint: The difference between minimum and maximum has to do with the direction of friction.
  5. Aug 9, 2007 #4
    -Mmg(cos(B))= -mA(cos(B)) + MmA(sin(B)) - mg(sin(B)), which simplifies to:

    he took another part to the other side
  6. Aug 10, 2007 #5


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    Homework Helper

    Asdifnlq, if you can't figure out an analogous starting expression
    for the MAXimum acceleration, then your insight might've been luck.
    To reduce the need for physical insight, use algebra :
    set up one axis parallel to the acceleration (horizontal),
    and sum the vertical Force components to zero.

    Two of the pieces are proportional to N.

    No, RootX, the Normal Force is not mg(cosB) + ma(sinB) , either.
    rather N(p-q) = mg , so N=mg/(p-q) ... you find the p and q.

    How do you expect to be confident of your signs,
    if you don't take care of them
    and don't distinguish cause (Force) from effect (ma)?
  7. Aug 10, 2007 #6
    I set it up in terms of horizontal and vertical forces now and I think I have the right answer:

    M*N*sin(B) + Ncos(b) - mg = 0 for vertical forces, or essentially the friction forces vertical component + the normal force component - the weight will be zero.

    Nsin(B) - M*Ncos(B) = ma for horizontal forces, or essentially the horizontal of normal - horiz of friction will be equal to the horizontal force applied

    I then moved the mg over in equation 1, divided equation two by equation 1, giving: (I also divided the normal force out of the equations)

    [sin(B)-M*cos(B)]/[M*sin(B)+cos(B)]= a/g

    then I just multiplied both sides by g and brought it into term of tan(B) by multiplying the left side by [1/cos(B)]/[1/cos(B)]. I do it because I like tan(B) better :[ Then for the max I would make the force of friction point down the ramp (reverse directions).

    Thanks for the help!
    Last edited: Aug 10, 2007
  8. Aug 10, 2007 #7


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    Why do you think that N = W cosB ?
  9. Aug 10, 2007 #8
    I wrote down the equations and saw that the Normals would cancel out, so I didn't really bother with what it is actually equal to. But I figure it doesn't actually equal that because there are more accelerations involved that just gravity.
    Last edited: Aug 10, 2007
  10. Aug 10, 2007 #9


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    Isn't the friction Force = mu N ?

    ... good edit.

    So, looking at eq 1,
    N = mg/(cosB + MsinB) .

    best wishes.
    Last edited: Aug 10, 2007
  11. Aug 10, 2007 #10
    I believe so. Did I place it down differently somewhere?
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