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Incline question

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A boy drags a 100N sled up a 20degree slope at constant velocity. If the coefficient of friction between sled and hill is 0.2, what force must he exert at an angle of 35 degrees with respect to the hill?


    2. Relevant equations
    What does 'respect to the hill' mean?


    3. The attempt at a solution
    Im not sure where to start? I drew a FBD: perpendicular to slope I have mgcostheta going down and normal force going up perpendicular to slope. F is going to the right and fs is going down to the left. Is this correct?
     
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  3. Jun 30, 2010 #2
    (a) respect to slope surface, I guess

    (b) I'm not sure if I imagine well your image (what is F?). Can you attach it?

    regards
     
  4. Jun 30, 2010 #3
    F is the force exerted. The boy is dragging the sled up the hill.
     
  5. Jun 30, 2010 #4
    Normal force is OK.

    F is going "up to the right" with 35 degrees angle respect to the sled surface (it does mean 55 deg to the horizon).

    ragards
     
  6. Jun 30, 2010 #5

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    Hello Mizzy,

    Welcome to Physics Forums!

    You should reevaluate that force. You will have the gravitational force, of course, but this force should point really straight down (i.e. not perpendicular to the slope). What do you think about its magnitude?
    Okay, that's good. :approve: What's its magnitude, by the way?
    Hmmm. I guess it depends on what you mean by "right"

    The slope itself is 20o from the horizontal, and the boy is pulling at an additional 35o with respect to that, making the pulling angle 55o with respect to the horizontal. (I think this answers your first question.)
    The direction of that last one sounds okay to me, I think. Do you mean that Fs is perpendicular to the surface?
     
  7. Jun 30, 2010 #6
    yes gravitational force is going straight down, but y component is perpendicular to slope.

    Normal force is 100N, right?

    Yes, Fs is perpendicular to the surface.

    I need to find acceleration in this question, right?
     
  8. Jun 30, 2010 #7

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    Sorry, 100 N is the gravitational force that points straight down (really straight down -- not perpendicular to the slope's surface). You'll have to combine that with a trigonometric function to find the part of the normal force's magnitude associated with gravity.

    But be careful! The gravitational force and the angle of the slope are not the only things which affect the normal force's magnitude. There is something else too.
    The problem statement says, "boy drags ... sled ... at constant velocity." So that implies... :wink:
     
  9. Jul 1, 2010 #8
    ok. I've found acceleration and I got 6.9m/s^2. This is using the angle 55 degrees.

    Is that correct?
     
  10. Jul 1, 2010 #9

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    Sorry, no. The problem statement says that the boy and sled are moving at a constant velocity. This directly implies that the acceleration is zero. (If the acceleration was not zero, the velocity would change as a function of time.)

    Given that, here is a hint: First find all the forces involved in the problem (three of these individual forces are dependent on the magnitude of the pulling force F). Then perform the vector sum of all the forces and set that equal to zero (don't forget that you're summing vectors and not scalars). Solve for the magnitude of the pulling force F (you already know its direction).
     
  11. Jul 1, 2010 #10
    ok. tell me if i got all the forces right. Normal force, gravitational force, mg cos theta, mg sin theta, force of friction, the x and y components of the pulling force.

    is that right?
     
  12. Jul 1, 2010 #11

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    Sounds good so far. :approve:
    What forces are these? I don't understand.
    Yes, those are fine. But realize that all the forces have both x and y components. It just so happens that the x component of the gravitational force is 0, but the x component still exists (it just happens to be 0). All the other forces have non-zero x and y components.
    I only see 4 forces. Consider trying it this way (keep in mind that each of these forces has both x and y components):

    o Pulling force F. The magnitude of this force is what you are looking for. The direction has already been determined.
    o Gravitational force mg. We've established that this force points straight down.
    o Normal force Fn. This force points perpendicular to the slope (which is in a partially upward direction). The magnitude of the normal force is influenced by the slope angle, the gravitational force, and the pulling force (the pulling force F is not parallel to the slope's surface, so it has an effect on how hard the sled pushes against the slope -- thus it affects the magnitude of the normal force).
    o Frictional force Ff. Points parallel to the slope's surface, which is in a somewhat downward direction. It is a function of the magnitude of the normal force Fn and the friction coefficient. Since the magnitude of Fn is a function of the F, the magnitude of the frictional force is also (indirectly) a function of F.

    Do you see how I come up with that?
     
    Last edited: Jul 1, 2010
  13. Jul 1, 2010 #12
    ok, i think i'm lost.

    Maybe my FBD is different from yours. I broke the gravitational force into its x and y components.
     
  14. Jul 1, 2010 #13

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    Theoretically you can do that; i.e. rotate everything 20o, or some other angle, including gravity and everything. You'll get the same answer, but it won't necessarily make make the problem any easier. It might even make things more complicated. But again, you could do that if you want. Doing so doesn't change the magnitudes of the vectors or their relative directions (relative to each other), which is why it is okay to do that.

    [Edit: My point is that rotating or not, you still have 4 forces: Pulling, Gravity, Normal, and Friction. Their relative directions are fixed, and essentially given in the problem statement.]

    As a rule of thumb, I try to keep gravity pointing in the negative y direction unless there is a compelling reason to do otherwise. But maybe that's just me.

    So anyway, (rotating or not) what expression did you get for the magnitude of the normal force (as I mentioned before, this should be a function of F)?

    [Edit: is it possible for you to post an image of your FBD?]
     
    Last edited: Jul 1, 2010
  15. Jul 1, 2010 #14
    For normal force, i have N = mg cos theta

    i tried attaching the image but it was too big. I'll try again.
     
  16. Jul 1, 2010 #15
    ok here is my diagram.
     

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  17. Jul 1, 2010 #16

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    There's more to it than that. Remember, there is a component of the pulling force pulling up on the sled, in a direction away from the surface.

    Let me describe a different situation (we can go back to the original problem later). For a moment, imagine a block sitting on a bathroom scale (and the scale might be sitting on a sloped surface). The bathroom scale is measuring the normal force. At this point, the scale is registering mgcosθ. But now grab the block and lift up on it. How does that affect the scale's reading? :wink:

    [Edit: 'Just looked at your FBD, and it looks fine. :approve:]
     
  18. Jul 1, 2010 #17
    The scale's reading will be lighter.
     
  19. Jul 1, 2010 #18

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    Yes, :approve:. Now back to the original problem.

    The normal force is the sum of the parallel components of all the other forces (and in the opposite direction of this sum).

    You got one of them. mgcosθ is the part of the gravitational force that is parallel to the normal force.

    The frictional force has no part that is parallel to the normal force (because the normal force and frictional force are perpendicular to each other).

    But there is a part of the pulling force F, that is parallel to the normal force that needs to be considered too. This needs to be part of the sum that makes up the normal force.

    (By the way, this might be a good time to try out that rotation idea, at least in your mind, for this particular step. Imagine rotating the whole figure such that the normal force points straight up [you can rotate it back later -- this is just to find the normal force]. Now the normal force points straight up, in the positive y direction. It's x component is zero. Find the y components of all the vectors [including the normal force] and set the sum equal to zero.)

    You'll find that the normal force has an mgcosθ which you've already found, and also another term which is a function of the pulling force F.
     
  20. Jul 1, 2010 #19
    ok. I solved for F, and I got 75.7N. Is that correct?

    I really appreciate your help. Thanks so much.
     
  21. Jul 2, 2010 #20

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    Sorry, but I double checked and 75.7 N is not the right answer (magnitude of the pulling force). :frown:

    If you show your work maybe I can help figure out what went wrong. Start with your expression for the normal force, since that's the first step in solving this problem (remember, you should expect that your expression for the normal force is some function of the pulling force F). :smile:
     
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