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Homework Help: Newton's Forces - Block Down An Incline

  1. Oct 17, 2007 #1
    [a]1. The problem statement, all variables and given/known data[/b]

    The block shown in Fig. 5-27 lies on a smooth plane tilted at an angle = 25.5° to the horizontal, with µk = 0.17.

    5-27.gif

    Figure 5-27

    (a) Determine the acceleration of the block as it slides down the plane.
    ________ m/s2 (down the plane)
    (b) If the block starts from rest 9.50 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?
    _________ m/s

    2. Relevant equations

    F=m(a)
    Ff=Mk(N)

    [c]3. The attempt at a solution[/b]

    I drew a Vector Diagram so that gravity was pointing directly downward from the box, so that means that the Fg vector is at an angle of -64.5 from the Horizontal, (or 25.5 degrees from the Y axis, whichever you prefer!), and then so Fg has horizontal and Vertical components.

    Fy(net)=m(a)=Fa+Fgx-Ff
    Ff=.17(Fnormal)

    ??? I'm very confused..it seems like I need to know the mass to solve the problem, but the other half of me doesn't even know where to begin.
     
    Last edited: Oct 17, 2007
  2. jcsd
  3. Oct 17, 2007 #2
    First, you need to break down the forces into components just as you said.
    forces in the Y direction will be mgcos(theta)-Normal Force=0 since the box is not rising or sinking.

    This tells you that mgcos(theta) is therefore equal to normal force.

    Now forces in the X direction will be mgsin(theta)-Kinetic Friction=ma (it equals ma since it is moving)

    We know that kinetic friction = Normal force*friction coefficient (by definition). We also know from above that normal force is equal to mgcos(theta).

    So continuing to solve forces in the X direction - you will have mgsin(theta)-force of kintetic friction=ma. Here we will use mgcos(theta) to stand in for normal force.

    mgsin(theta)-mgcos(theta)(0.17)=ma

    now notice how the masses will cancel since they are on both sides.

    gsin(theta)-gcos(theta)(0.17)=a

    edit: I think your confusion might come from a weakness in drawing clear free body diagrams. I will try and sketch one for you in a paint program.
     
    Last edited: Oct 17, 2007
  4. Oct 17, 2007 #3
    Hope this isnt too confusing. I am not a graphic artist.

    http://img141.imageshack.us/img141/686/fbdue2.gif [Broken]

    The key point is that when you draw the mgsin(theta) force - make sure it is parallel to the slope. Some beginners draw it as a flat 90 degree line and they can't visualize the forces correctly.
     
    Last edited by a moderator: May 3, 2017
  5. Oct 17, 2007 #4
    Hey!! Wow, you helped me so much!! I got the answer right on the first try!! =) (the acceleration is 2.715 by the way).

    I realized what I did wrong, by the way. I didn't realize that the forces in the Y direction equaled zero, although I should have. And I was discouraged by the fact that we didn't know mass, but it ended up canceling out at the end. Thanks again!! =)

    I even got the 2nd part too: it was 7.18 by the way. ;)

    Thanks again Bob!! You were great!!
     
  6. Oct 17, 2007 #5
    Glad I could help.
     
  7. Oct 22, 2007 #6
    I want to know about the Forces acting on each other; for 3 blocks sitting side by side
    on a frictionless leveled table top. 2 blocks on the sides are of same mass. The one on the middle is bigger than the tow on the sides.
     
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