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Block on an accelerating inclined plane

  1. Oct 8, 2004 #1
    I have a problem (it was homework, but I already turned in my wrong answer) dealing with a block stationary on an accelerating inclined plane.

    There is a force applied horizontally to the inclined plane, which has an angle of 35 degrees above the horizontal. A box of mass 0.5 kg is on the plane, which has a mass of 2 kg. Coefficient of friction between the surfaces is 0.8. I need to find the maximum and minimum force on the plane to keep the block stationary.

    I've tried for several hours to solve this problem to no avail...

    I assume there are four forces on the block....gravity, normal force, friction force, and F' (which comes from the F on the wedge). I decided F' is 0.2 * F, since the block is 1/5th the total mass of the system.

    I thought the normal force was (m * g * sin(35)) + (F' * sin(angle)) .. but I doubt that's correct and it all goes downhill from there. Any help would be appreciated, thanks!
  2. jcsd
  3. Oct 8, 2004 #2

    Consider two possibilities: the applied force may be small enough to permit the block to slide down the incline or it may be large enough to cause it to rise up the incline. Once you have this figured out and you know that the block always stays on the incline and that friction opposes relative motion, you should be able to draw two freebody diagrams and find the max and min values of the applied force.

  4. Oct 9, 2004 #3
    OK...the entire system is going to be accelerating horizontally due to the applied force, so there will be some horizontal force F' on the block..which is unknown. The force of gravity, mg, is known and is 4.9 N downards. The normal force is F(gravity) * sin(35) ... if the block is not accelerating. Since it's accelerating I guessed that I have to add F' * cos(35) to the normal force. Then force of friction = (coefficient) * Fn.


    Fg = 4.9 (-90 degrees)
    Fn = 4.9(sin(35)) + F'(cos(35)) (55 degrees)
    Ffric = (0.8) * Fn (145 degrees)

    Fg + Fn + Ffric + F' = 0

    This seems to be enough information...but I have tried solving the problem many times with no success. The book gives the answers -1.57 and 80 something N for the final F. ( = F' * 5 I think)

    Have I got something wrong in these equations, or am I doing the algebra incorrectly?
  5. Oct 9, 2004 #4

    Doc Al

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    Staff: Mentor

    There are only three forces on the block: weight, friction, and the normal force. (Any added force on the inclined plane will affect the block by increasing the normal force between the block and plane. Do not add a separate force F'. That force acts on the inclined plane, not the block!)

    Now apply Newton's 2nd law to the block:
    (1) Vertical forces must just balance
    (2) Horizontal forces must produce an acceleration

    Hint #1: If you can find the acceleration the block must have, then you can figure out what the force on the plane must be to create that acceleration.

    Hint #2: Comparing the minimum force situation to the maximum force one, what's different about the friction that the plane exerts on the block?
  6. Oct 10, 2004 #5
    Haha Doc Al's Hint #2 says it all doesn't it :biggrin:
  7. Oct 11, 2004 #6
    Thanks guys, I managed to figure out the answers! :rofl: But I'm still having some conceptual problems.

    I got the vertical components of friction and normal forces, setting them to 0 when added with weight. This allowed me to get the magnitude and direction of all 3 of those forces.

    Then I added the horizontal components of normal and friction, getting a non-zero answer. This answer was 1/5th of what they have in the book...so I'm guessing this is the way to do it.

    However, what exactly is this value I'm getting? Was my original calculation of the normal force not actually the normal force, but only the part without this extra horizontal component?
  8. Oct 11, 2004 #7

    Doc Al

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    Assuming you mean that the sum of the vertical force on the block must be zero: right!

    The sum of the horizontal forces on the block must equal ma for the block, and this is just 1/5 of F = (M + m)a for the block + incline (assuming there is no friction between the incline plane and the surface it slides on).

    I don't know what you mean by "part" of the normal force. The normal force is the force that the block and plane press against each other--this of course depends on how the thing is accelerated. Here's a simple example: Hold a book in your hand. The normal force = mg. Now accelerate your hand upward--the normal force increases as you squash the book against your hand, or drop your hand down quickly--and feel the normal force decrease. Make sense?
  9. Jul 14, 2005 #8
    ok, now i have a similiar problem to this too...

    An inclined plane that makes an angle of 28° to the horizontal is mounted on wheels. A small block of mass m = 1.2 kg rests on the plane, held there by a coefficient of static friction µ = 0.73.

    The plane is accelerating to the right. What is the minimum acceleration in order that the block slides down the plane?
  10. Jul 14, 2005 #9


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    Draw the free body diagram for the block, and apply Newton's 2nd Law, you need to solve for a.
  11. Jul 15, 2005 #10
    According to my problem. and my FBD:

    sum of forces of x comp: -Wsin(theta) - f = ma
    y comp: N = Wcos(theta)

    my question is that if the plane is accelerating to the right, what kind of force will act upon the block? B/c i've been told there are only three forces, gravity, friction and normal force.

    I've, or tried by
    applying F= ma to both the vertical and horizontal components, and solve for a, but its still wrong, help!!!
  12. Jul 15, 2005 #11

    Doc Al

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    It looks like you are taking your components parallel (x) and perpendicular (y) to the plane. That's not a good idea in this problem since the plane is accelerating. These equations are incorrect.

    Instead, use vertical and horizontal components with respect to an inertial frame (the ground).

    Those are the only forces action on the block.

    Show what you did using vertical and horizontal components. Write Newton's 2nd law for the vertical forces (which are in equilibrium) and horizontal forces (which produce an acceleration). Note: Since the mass is just about to slide down the plane, the static friction is at its maximum value.
  13. Jul 15, 2005 #12
    FBD for my block: i've been told to set both horizontal and vertical components to ma:

    x comp: f - Wsing(theta) = ma
    y comp: N - Wcos(theta) = ma

    so i plugged in one in another, didnt work

    i have friction force acting up the incline
  14. Jul 15, 2005 #13

    Doc Al

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    Who told you that? The only acceleration is horizontal.

    Again, you are taking components parallel and perpendicular to the plane. Don't do that!

    Of course it won't work. Read my previous post and do what I suggest.

    That's good. :smile:
  15. Jul 15, 2005 #14
    im just lost, i guess i'm just confused as to how to set the vertical and horizontal components with respect to the ground.

    you have N and mg as the big vertical forces. and friction pretty much the horizontal component.

    but how do i incoporate the angle of the plane into it?
  16. Jul 15, 2005 #15

    Doc Al

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    We'll sort it out. See my comments below.

    While mg is a vertical force (it points straight down), N and the friction are not. N and the friction do have components in the vertical and horizontal directions.

    For example, N points straight out of the surface. So what angle does it make with the vertical and horizontal axes? Draw a diagram; do a little trig.
  17. Jul 15, 2005 #16
    well, now i have:

    and I am not sure at all

    mg sin(theta) - f = ma cos(theta)
    N = ma sin (theta) + mg cos (theta)

    tried that, got another wrong answer, but is this what you mean?
    Last edited: Jul 15, 2005
  18. Jul 15, 2005 #17

    Doc Al

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    Nope. Let's start over. Once again, I'll ask you to find the vertical and horizontal components of the three forces. Do it now:

    1) Weight (mg):
    horizontal component = ?
    vertical component = ?

    2) Normal force (N):
    horizontal component = ?
    vertical component = ?

    3) Friction ([itex]\mu N[/itex]):
    horizontal component = ?
    vertical component = ?

    Don't go any further until you've found these components correctly.
  19. Jul 16, 2005 #18
    This is the same problem I've been having (my numbers are different, mass is 1.8kg, lower left corner is 28 degrees, and my static coefficient is .74. I can't figure out where I've gone wrong. :(

    1) Weight (mg):
    horizontal component = 0
    vertical component = (mass of block)(9.81m/s^2)

    2) Normal force (N):
    horizontal component = mg cos 45
    vertical component = mg sin 45

    (I picked the angle as 45 since I have my coordinate system set with the x-axis level with the ground, and the normal force is perpendicular to the surface, is this correct?)

    3) Friction
    horizontal component = sin 62 = fx*13.0669
    vertical component = sin 28 =fy/13.6609

    (I don't know if this is right, but for the magnitude of the friction (just above, the 13.6609 mumbers)is u * N)

    sigma Fx = fx - Nx = ma
    sigma Fy = -mg + Ny + fy

    What I can't figure out is how to set those equal to each other. I don't set them both to ma, do I? I can't figure how how to solve for the unknown acceleration between the two. Do I really have two unknowns?



    [Edit: Oops, sorry, forgot to say what exactly I couldn't figure out)
    Last edited: Jul 16, 2005
  20. Jul 16, 2005 #19

    Doc Al

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    This is OK.

    No, not correct. Two problems here:
    (a) You just arbitrarily picked 45 degrees!? The angle of the incline is given (nothing to do with 45 degrees). Hint: if the angle of the incline were 1 degree, then the normal force would make an angle of 1 degree with the vertical.
    (b) You don't know what the normal force is! It's not equal to mg. Leave it as a variable and SOLVE for it (once you write your force equations).

    Here's one problem: The magnitude of the friction depends on the unknown normal force. Leave it as [itex]\mu N[/itex] in your equations and SOLVE for the unknowns later.
    Right idea, but you need to get the correct force components. The second equation is incomplete: sigma Fy = 0, since the the mass is at the verge of slipping down, but is still in equilibrium.

    Once you get the equations written correctly, the only unknowns will be N and a. Two equations, two unknowns. Solve!
    Last edited: Jul 16, 2005
  21. Jul 16, 2005 #20
    Thank you so much!! I figured it out! I think my main problem was making the wrong assumption on the angle. I feel so stupid now. :)

    Thank you!

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