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Newton's Law Flat Curve

  1. Feb 7, 2008 #1
    [SOLVED] Newton's Law Flat Curve

    1. The problem statement, all variables and given/known data
    You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.470 kg suspended from the ceiling of the bus by a string of length 1.88 m is found to hang at rest relative to the bus when the string makes an angle of 28.0 with the vertical. In this position the lunch box is a distance 50.0 m from the center of curvature of the curve.

    What is the speed v of the bus?
    Take the free fall acceleration to be g = 9.80 m/s^2.
    2. Relevant equations
    x direction

    3. The attempt at a solution
    =(.47)(9.8)/cos(28) = 5.21
    Last edited: Feb 7, 2008
  2. jcsd
  3. Feb 7, 2008 #2
    What is the question?
  4. Feb 7, 2008 #3
    hahah thanks
    What is the speed v of the bus?
    Take the free fall acceleration to be g = 9.80 m/s^2.
  5. Feb 7, 2008 #4
    BTW, you gotta quit driving that fast on side streets!

    So, draw a Free Body Diagram of the lunchbox. Though the problem doesn't really say so, I think you have to assume the CG of the lunchbox is at the 1.88 m point.

    A lunchbox would normally hang straight down. What causes this one to move out to 28 degrees?
  6. Feb 7, 2008 #5
    acceleration around the curve, or frictional force?
  7. Feb 7, 2008 #6
    OK, so let's look at this a little.

    Friction of WHAT on the lunch box? What's it rubbing against?

    When you say acceleration around the curve, what do you mean as far as direction?
  8. Feb 7, 2008 #7
    centripetal acceleration. i gave the formula for that.
    the lunch box doesnt have friction. it has a force tension in the string.
    am i correct in saying that?
    to be honest, i haven't a clue where to start.
  9. Feb 7, 2008 #8
    You're making a good start; you're thinking.

    So, there's no friction on the lunchbox; OK.

    Now, you've named centripetal acceleration and what the lunch box "sees" in the reference frame of the bus is centrifugal acceleration which is a fictitious force that is equal to centripetal force except points the other way. In other words, centripetal force causes the bus to move toward the center of the curve while the lunchbox tries to go straight. But, if you're sitting on the bus, it looks like there is a centrifugal force pushing the lunchbox out.

    Now I see you tried to solve for centripetal force (an = v^2/R) but then you put some pi in the formula? Since you're trying to find v, you only need an and R.

    So, what's R?

    Have you been told that, when you draw a FBD, the geometry of the forces is proportional to the magnitudes of the forces? So draw the FBD of the lunchbox and show the forces that act on it, including the Centrifugal force. Pay close attention to the geometry; that's the key to this.
  10. Feb 7, 2008 #9
    my FBD has Fn opposite W on the Y axis. then at 28 deg from vertical i have Ft from the string-ceiling bus accompanied by Fc in opposite direction.
    is that all so far?
    i have tried getting components to sum the forces, but have had no luck getting anywhere.
  11. Feb 7, 2008 #10
    OK, try it this way.

    BTW, I know mass is gonna cancel out everywhere, so I'm gonna leave it out. If that confuses you, put it in.

    Draw a horizontal line (bus roof). Draw a verical line down 9.8 cm (this will be gravity). Draw a horizontal line to the right of the end of the 9.8 cm line; go 5.2 cm. (This line represents centifugal acceleration approximately) Put a big dot representing the lunchbox at the end of the 5.2 cm line. Draw another line from the starting point on the roof to the lunchbox. Now you have a right triangle.

    If you measure that angle up at the roof, it will be about 28 degrees. The vertical will be the acceleration due to gravity, the horizontal will be the acceleration due to centrifugal force. The last line (hypoteneuse) sums the two forces. It represents the tension in the string which has to cancel both gravity and centrifugal.

    Now, it's up to you to go back and use trig to get better accuracy on the centrifugal acceleration than what I gave you as a simple approximate sketch. OK? Let me know what you get.

    By the way, if you are a good draftsman, you can solve this simply by accurate drawing and measurement.
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