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More Plumb Bobs

  1. Oct 22, 2006 #1
    I seem to be specializing in plumb bobs this week:rolleyes:

    A box car has a freely hangling plumb bob (from the ceiling). It is accelerating up an incline. which makes an angle b with the horizontal. The plumb bob makes an angle c with the perpendicular to the ceiling. Find the acceleration of the box car.

    This didn't seem that tough, but my solution is so different from the books that I thought I would ask about it. I hope someone has some time to look this over. I've attached a diagram for my work.

    From the box car ceiling, I have two angles. One is just b, the same as the incline, which is the angle the bob would make with the perpendicular if there is no acceleration. The other is a, which is the angle caused by the acceleration. So a+b = c, the angle of the perpendicular to the ceiling.

    Fy = T cos a - mg = 0
    Fx = T cos a = ma

    Solve these to get a = g tan a.

    a = c - b, so the final solution is: a = g tan (c - b). Simple and nice. I wish it were right, too :frown:

    The books answer is quite complicated:

    a = g ((cos b)(tan c) - (sin b))

    I tried simplifying this with some trig identities to get my expression, but it doesn't seem to be possible.

    Thanks, as always, for any help.
    Dorothy
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2006 #2

    Andrew Mason

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    I suggest you use a for acceleration and call the angle [itex]\alpha = c-b[/itex].

    You are forgetting that there vertical component of acceleration so the vertical force is equal to mg + may:

    [tex]F_y = T\cos\alpha = ma_y + mg[/tex]

    [tex]F_x = T\sin\alpha = ma_x[/tex]

    [tex]\frac{a_x}{g + a_y} = tan\alpha[/tex]

    and [itex]a_x = a cos b; a_y = a sin b[/itex]

    See if you can get that to work out.

    AM
     
  4. Oct 22, 2006 #3

    andrevdh

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    The box car (and the bob) is accelerating up the incline. This means that the bob is accelerating in both the x and y directions. So your equation

    Fy = T cos a - mg = 0

    should be changed to

    [tex]F_y = T \cos(a) - mg = ma_y[/tex]

    you get the other component of the acceleration [itex]a_x[/itex] for considering the force components acting on the bob in the x direction. To get the acceleration along the incline you need to combine these two acceleration components.

    The direction that it is not accelerating in in this case is perpendicular to the incline. So the problem "might" be solved easier if you choose the x-direction along the incline and the y-direction perpendicular to it.
     
    Last edited: Oct 22, 2006
  5. Oct 22, 2006 #4

    andrevdh

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    Slight change to previous post.
     
  6. Oct 22, 2006 #5
    Argh, what a dumb mistake! Thank you both very much.

    Dorothy
     
  7. Oct 22, 2006 #6
    Thanks again... What I ended up doing was this:

    Sum Fy = T cos c - mg cos b = 0
    Sum Fx = T sin c - mg sin b = ma

    And solving these gets the same answer in the book... And it makes sense now, thanks to you guys.

    Thanks so much!
    Dorothy
     
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