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Where Can I Get Online Tutorials?

  1. Dec 17, 2003 #1
    My teacher doesnt really teach out of a book, but just out of his head. I have no clue on how to do this stuff. I have to redo a test and one of the questions is...

    A 6-kg mass is placed on a plane inclined at 37degrees. It is attached to a 10-kg suspended mass. If the coefficient of kinetic friction is 0.33, at what rate will the masses accelerate? If the suspended mass is 1.2 meters above the ground, how long will it take to fall? At what speed will it strike the ground?

    Every question is different, and has missing stuff and formulas dont really help me much when you always have to change something or find something if something is different everytime.
     
  2. jcsd
  3. Dec 17, 2003 #2

    chroot

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    Travis,

    Do you know how to draw a free-body diagram and sum the forces applied to an object?

    - Warren
     
  4. Dec 17, 2003 #3
    It has a diagram on the test with the block and a pully with a weight hanging from it.
     
  5. Dec 17, 2003 #4

    chroot

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    That's not a free-body diagram -- that's just a picture of the apparatus.

    If you represent the 6kg mass on the plane as a dot, and draw each force (gravity, friction, and the force due to the 10kg mass pulling on it) as a vector sticking out from the point, you'll have a free-body diagram.

    Then, you can sum (add) the vectors together to arrive a net force applied to the mass. From that, you can use Newton's second law of motion to find its acceleration.

    Does any of this sound familiar to you?

    - Warren
     
  6. Dec 17, 2003 #5
    Sorry, I dont understand it.:frown: I know how to plug everything in to a formula if I have all the information but If its missing information then I dont have a clue on what to do.
     
  7. Dec 17, 2003 #6

    chroot

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    That's okay! Let's see if we can make some sense of it.

    First, let's analyze something simple and familiar: a game of tug-o-war. Let's say two teams attach a rope to a bowling ball, and then they begin pulling on the bowling ball in opposite directions.

    If both teams pull equally hard, what happens? Nothing, of course -- the bowling ball stays perfectly still.

    We can represent the people pulling on the bowling ball with little arrows called "vectors." Here's an example of a diagram, showing the teams pulling on the bowling ball:
    Code (Text):

        <-------o------->
       
    (The bowling ball is the little 'o' in the center.) There's one force in each direction. The length of the arrows represents the strength of the force. If the arrows are equally long, then the two teams are pulling equally hard.

    With me so far?

    - Warren
     
  8. Dec 17, 2003 #7
    Yes, isnt is also true that the ball will never be perfectly horizontal even if you pull the ropes as hard as possible. It will always hang somewhat, I think I remember that from some pervious classes.
     
  9. Dec 17, 2003 #8

    chroot

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    (Don't worry about the ropes not being exactly horizontal -- that's a small detail we're better off neglecting!)

    Okay, so you can represent forces with little arrows. Great! Now what? Well, next you need to learn how to add the vectors together. Why would you want to do that? Well, because the movement of an object (like the bowling ball in the middle of the tug-o-war game) depends upon the total force applied to it. Another word you'll hear is "net," which means "total."

    Think about this: if the two tug-o-war teams pull equally hard, the total force on the bowling ball must be zero. If it had a net force applied to it, it would move. So these two forces, equal in magnitude but opposite in direction, add to zero. If you want, you can think about them as numbers on the number line: one team pulls with force +5 newtons (to the right), while the other pulls with force -5 newtons (to the left). -5 + 5 = 0, of course; there's no net force.

    Check: what if one team pulled with force +3 N and the other with force -5 N? What would be the net force on the bowling ball? -5 + 3 = -2 N. The ball has a net force of 2 N applied to it, pointing to the left.

    With me so far?

    - Warren
     
  10. Dec 17, 2003 #9
    Yes im starting to understand!
     
  11. Dec 17, 2003 #10

    chroot

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    Woohoo! Okay, this next part is a big step, so read it very carefully.

    We need to generalize our little arrows from all pointing along a straight line, to pointing at any direction on a plane.

    Are you familiar with the term "components?" I can describe any vector on a plane with two numbers: how much is goes to the right, and how much it goes up.

    For example, if I have a vector of length 5, pointing to the right, like this:
    Code (Text):

        -------> 5
     
    its components are (5, 0). That means 5 units to the right, and zero units up.

    How about a vector with length 5, angled up at 45 degrees from the positive x-axis?
    Code (Text):

            .  
           /
          /    
         /    
        /
       
     
    Well, it's components are (look at this carefully!)

    [tex](5 \cos 45^o, 5 \sin 45^o)[/tex]

    Do you understand this? You might have to think a bit about right triangles, and remember what the sin and cos mean. If you don't understand this, let me know and I'll show.

    If you do understand, here's an exercise: Can you tell me the components of a vector that is 10 units long, pointing 60 degrees from the positive x-axis?

    - Warren
     
  12. Dec 17, 2003 #11
    [tex](10 \cos 60^o, 10 \sin 60^o)[/tex]
     
  13. Dec 17, 2003 #12

    chroot

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    Great! You got that exactly correct (and you even used our math typesetting too!)

    Okay. Now, once you have some vectors written out in components, you can add them together like this:

    [tex](a, b) + (c, d) = (a+b, c+d)[/tex]

    Just add all the horizontal components together, then add all the vertical components together.

    The first thing you need to do is write out all the forces applied to your 6 kg mass.

    Since it's on an inclined plane, let's consider that plane to be our x axis. In other words, we'll represent "up the plane" by a vector pointing due right, and "down the plane" by a vector pointing due left.

    So, if I draw a diagram of the forces applied to the mass, it'll look like this:
    Code (Text):


       <------o--->  
             /
            /    
           /    
          /
         v
     
    The vector pointing left is that 10 kg mass pulling it down the ramp. The vector pointing right is friction (remember that friction always opposes motion -- if the block is being dragged down the ramp, the friction is trying to pull it up). The slanted vector hanging down is the effect of gravity. Gravity pulls the block straight down, or 37 degrees from our y-axis. Think about this for a moment, and make sure you understand why it's slanted to the left, and not the right. Due left means down the ramp, and gravity is definitely pulling it down, not up.

    If I write these three vectors down in components, they look like this:

    [tex](-10 g, 0)[/tex] (the 10kg mass pulling down)

    [tex](\mu_k N, 0)[/tex] (friction trying to pull it up -- we'll get to that N business in just a bit -- it means "normal force.")

    and finally

    [tex](- 6 g \sin 37^o, - 6 g \cos 37^o)[/tex]

    Think about this very carefully, and draw a triangle like this to convince yourself:
    Code (Text):

         <------o--->  
               /|
              / |  
             /37| -mg cos 37
            /   |
           v----|
     -mg sin 37 |
                |
                |
     
    Okay....... still with me?

    - Warren
     
    Last edited: Dec 17, 2003
  14. Dec 17, 2003 #13
    Ok I understand that m is mue and yes I got most of what you said.
     
  15. Dec 17, 2003 #14

    chroot

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    Careful: I used "m" to represent mass, 6kg. I used [itex]\mu[/itex], spelled "mu," as the conventional symbol for the coefficient of friction.

    And I know this is a lot to learn at once -- in fact, probably too much. I'm a bit worried that the problem your teacher assigned is just not at the right level for you yet, but that's okay. If you can follow along just a bit longer, we're almost to the solution. You may want to read or re-read all this a few times, and ask questions if you get stuck.

    Okay. Now, for that pesky N, which I've been ignoring up to this point. N is the "normal force." The block presses down onto the plane, and the plane pushes back up to keep it from falling through. The normal force is how much force the plane is exerting up on the block.

    Think about the case where the plane is exactly horizontal (its inclination is zero). All of the weight of the block pushes down into the plane. The normal force is just mg.

    Think about the case where the plane is exactly vertical (its inclination is 90 degrees). None of the weight presses against the plane at all. The normal force is zero.

    You can thus conclude that the normal force is

    [tex]N = mg \cos \theta[/tex]

    and [itex]\theta[/itex] is 37 degrees in this problem. (This is a pretty slick way to quickly recall whether a quantity should be cosine or sine -- the cosine function is 1 at 0 degrees, while the sin function is 0 and 0 degrees. If you don't like being so slick, you can still draw little triangles.)

    So there's really one more sneaky little vector to add -- the normal force -- but it points straight up off the plane. That's vertical in our coordinate system -- and thus it does not affect the horizontal movement at all. (That's why I decided to use a coordinate system aligned with the plane in the first place.)

    Now, we're going to assume the plane is sturdy, and only allows the block to slide -- it can't tear a hole in the plane and fall through it. In other words, the block's motion is constrained along the x-axis. We're only going to concern ourselves with the x-component of the total force.

    What's the total horizontal (down-the-ramp) force applied to our 6 kg mass? Add up all the horizontal components of all the vectors:

    [tex]F_x = -10 g + \mu_k N - 6 g \sin 37^o = -10 g + \mu_k 6 g \cos 37^o - 6 g \sin 37^o[/tex]

    That's the total horizontal force acting on the block. Does this make sense? There's a term with a negative sign (down the ramp) from the 10 kg mass, there's a term with a positive sign (up the ramp) due to friction, and there's another term with a negative sign (down the ramp) due to the 6 kg mass's own gravity.

    Can you plug all that into a calculator and tell me how much total horizontal force there is on the block? Hint: it should be negative, meaning down the ramp!

    - Warren
     
  16. Dec 17, 2003 #15

    chroot

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    And once you have the total down-the-ramp force, [itex]F_x[/itex], you can use it to calculate the down-the-ramp acceleration by using Newton's second law:

    [tex]F_x = m a_x[/tex]

    And you'll have your answer to the first part of your problem.

    - Warren
     
  17. Dec 17, 2003 #16
    I got -10
    -10+6sin(37)-6sin(37)

    So
    m=6kg
    Theta=37Degrees
    Mu=0.33
    a=1.2
     
    Last edited: Dec 17, 2003
  18. Dec 17, 2003 #17

    chroot

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    How did you get that?

    - Warren
     
  19. Dec 18, 2003 #18
    With [tex]F_x = -10 g + \mu_k N - 6 g \sin 37^o = -10 g + \mu_k 6 g \cos 37^o - 6 g \sin 37^o[/tex]
     
  20. Dec 18, 2003 #19

    chroot

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    I don't understand how you got from

    [tex]-10 g + \mu_k 6 g \cos 37^o - 6 g \sin 37^o[/tex]

    to

    [tex]-10+6sin(37)-6sin(37)[/tex]

    - Warren
     
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