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Nonequilibrium Applications of Newton's 2nd Law Problem

  1. Oct 18, 2005 #1
    I along with a buddy of mine in the same course have been trying to figure out the following question. We spent at least an hour and a half to two hours and nothing makes sense. Any help would be greatly appreciated.

    Here's the question as it's written:

    One way to measure the acceleration of your car is to see how far a pendulum swings from the vertical. That is why so many highly intelligent drivers have objects hanging from their rear-view mirrors. You notice that when you accelerate, the fuzzy dice (0.25 kg) you have hanging swings 15 degrees from the vertical. Find the acceleration that produces this amount of deflection.

    We began by drawing a picture of a 15 degree angle with a mass at the end of the hypothesis. We've never seen a problem like this in our class before.

    If it has acceleration, it is not in equilibrium. So we looked through the whole chapter and this is what we discovered; "If an object is not in equilibrium, then Newton's second law must be used to account for acceleration." And they give us these two equations: (SUM)Fx=m(ax) and (SUM)Fy=m(ay).

    We've done problems similar to this finding forces using the x- and y-components, but we're drawing a blank here.

    Any help is great appreciated. TIA, Bryan
  2. jcsd
  3. Oct 18, 2005 #2
    You might have to think in terms of angular acceleration, because the fluffy dice are not moving in a linear fashion :P
  4. Oct 18, 2005 #3
    The thing is, this problem comes from a chapter that does not discuss angular acceleration. Angular acceleration is discussed 3 or 4 chapters after this one. :(
  5. Oct 18, 2005 #4
    It is in equilibrium at an angle [itex]\theta[/itex] to the vertical, which means the net forces in X and Y are zero. You'll wantto list these forces (tension, acceleration of car, and acceleration of gravity) and solve the system of equations for the tension to find the acceleration of the car.

    Remember Net F = 0.
  6. Oct 18, 2005 #5
    I've solved questions like this when I needed to find tensions in TWO strings when a weight is hanging from them at an equilibrium.

    With two strings; therefore, two tensions, it was easy to find.

    (SUM)Fx = 0 --> F2x + (-F1x) = 0 --> F2x = F1x
    (SUM)Fy = 0 --> F1y + F2y = w

    then solved for F2 and then found F1.

    But I don't know how I can do an equilibrium problem (if this truly is one) with using only one set of x- and y-components.
  7. Oct 19, 2005 #6
    Forces in the X direction,

    Car accelerative force + Tension in the string in the x direction (at 15 deg)

    Y direction,

    Gravity force + Tension in the string in the y direction (15 deg)

    Express them in terms of m,g and theta and then solve the system to find T, the tension.
  8. Oct 19, 2005 #7
    Here is what we came up with. I still think we are far off base:

    After drawing the vector and free-body diagram which included W, Wperp, Wparallel, FN, Angle, and Acceleration, we came up with these equations.

    The problem doesn't say anything about air resistance, so we neglected it.

    Fnet = Wparallel - friction
    Fnet = ma
    FN = Wperpendicular

    sin(15) = Wparallel/W --> Wparallel = Wsin(15)
    cos(15) = Wperpendicular/W --> Wperpendicular = Wcos(15)

    Fnet = Wparallel - friction (friction = 0)
    ma = Wsin(15) - 0
    ma = mgsin(15)
    a = gsin(15)
    a = (9.8m/s^2)sin(15)
    a = 2.54 m/s^2

    That cannot be right. We were given mass, and it eventually cancelled out.

    We're completely stumped

    Thanks for all the help so far.
  9. Oct 19, 2005 #8

    Doc Al

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    Staff: Mentor

    It's not right, but not because the mass cancels.

    Do this:
    (1) Identify the forces acting on the dice. (I see two forces.)
    (2) Find the vertical and horizontal components
    (3) Apply Newton's 2nd law to the horizontal and vertical components. (Hint: The only acceleration is horizontal.)​
    Step 3 will give you two equations. Solve them to find the acceleration.
  10. Oct 19, 2005 #9
  11. Oct 19, 2005 #10
    Theres a bunch of those problems in this forum. I helped with one or two in the past two weeks if you want to search..
  12. Oct 19, 2005 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Sure, it's the same problem. But I suggest you force yourself to try and solve these kind of problems on your own before looking at someone else's solution. (It may not seem like it, but it's a bad habit. It's very easy to deceive yourself into thinking you can solve a problem while reading the solution. And then the test comes, or a problem you haven't seen before, and... oops... you've got nothing.) The hints given in this thread should be more than enough for you to solve this problem.
  13. Oct 19, 2005 #12
    Yea, I completely understand and when the test does come, I'll be prepared. I'm the kind of person who likes to work backwards with problems. If I know the solution then it will register with me on how that solution came about. Hard to understand I suppose.

    But I appreciate all the help
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