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HELP: Force and Inertia

  1. Nov 2, 2004 #1
    1) Two men are carrying a uniform plank of mass 5.6 kg and length 2.3 meters. A box of mass 108.4 kg is placed a distance 0.9 meters from the first man. Find the force applied by the second man.

    Here what I did:
    first man = force 1, second man force 2
    F1 + F2 - mg = 0
    F1 = (1.4)(5.6)(9.81) = 76.9104
    F2 = mg - F1 = ((108.4)(9.81)) - 76.9104 = 986.4936

    But the correct answer is 443.58. What did I do wrong?


    2) A hollow ball of mass 1.9 kg and radius 0.1 meters begins rolling a distance 1.1 meters down a plane inclined at an angle 26 to the horizontal. Find its speed at the end.

    Here is what I did:
    Potential energy = (1.9)(1.1sin26)(9.81) = 8.988
    KE = (1/2) (I) (w)^2
    I=(2/3)(1.9)(0.1)^2 = 0.0127
    PE = KE so, 8.988 = 0.00635w^2 ----> w = 37.622
    w=v/r ----> v = (37.622)(0.1) ----> v = 3.76

    But the correct answer is 2.38. What did I do wrong? Please help. Thanks. :smile:
     
  2. jcsd
  3. Nov 2, 2004 #2

    Pyrrhus

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    1) Looks like a Equilibrium Problem.

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 [/tex]

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    Where's the weight of the box??, you didn't include it.

    2)

    Kinetic Energy must include the kinetic energy for the linear movement

    [tex] K = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mv^{2}_{cm} [/tex]

    remember

    [tex] v_{cm} = r \omega [/tex]

    [tex] K = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mr^2 \omega^2 [/tex]

    Applying Conservation of Mechanical Energy

    [tex] mgdsin\theta = \frac{1}{2}I_{cm} \omega^2 + \frac{1}{2}mr^2 \omega^2 [/tex]

    Solving for [itex] \omega [/itex]

    [tex] mgdsin\theta = \omega^2(\frac{1}{2}I_{cm} + \frac{1}{2}mr^2) [/tex]

    [tex] \sqrt{\frac{mgdsin\theta}{\frac{1}{2}I_{cm} + \frac{1}{2}mr^2}} = \omega [/tex]

    and

    [tex] r \sqrt{\frac{mgdsin\theta}{\frac{1}{2}I_{cm} + \frac{1}{2}mr^2}} = v_{cm} [/tex]

    Note: [itex] I_{cm} = \frac{2}{3}mr^2 [/itex] (Hollow Spherical shell)
     
    Last edited: Nov 2, 2004
  4. Nov 2, 2004 #3
    still getting the wrong answer :frown:

    T1 = F1(0)
    T2 = F2(2.3)
    T3 = -Mg(0.9)
    T4 = -mg(2.3)
    (T is torque, M is mass of box, m is mass of plank)

    sum of T = F1(0) + F2(2.3) - Mg(0.9) - mg(2.3) = 0
    F2 = Mg(0.9) + mg(2.3) - F1(0)
    F2(2.3) = (108.4*9.81*0.9) + (5.6*9.81*2.3) + 0
    so, the Torque of F2 is 471.05

    T = Fr
    so, (471.05) = F2 (1.4) -----> F2 = 336.46 Newtons

    but the right answer is 443.58
     
    Last edited: Nov 3, 2004
  5. Nov 3, 2004 #4

    Pyrrhus

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    buffgilville, for equilibrium problems we assume the weight of the body is at its center of gravity, this is because we consider uniform density and uniform gravitational field over the object, so the center of mass and center of gravity are in the same location. For an uniform rod its center of mass (Center of gravity) is half its length. This means the weight of the planck will act on its center of gravity.
     
  6. Nov 3, 2004 #5
    Cyclovenom, I'm still not getting the right answer. Please help :frown:
     
  7. Nov 3, 2004 #6

    Pyrrhus

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    Understand what you're doing, rather than merely pursuing a number.
    What i said about center of gravity is indeed your error.
     
  8. Nov 3, 2004 #7
    Cyclovenom, you said that the center of mass (center of gravity) for an uniform rod is half its length;so, the weight of the plank will act on its center of gravity. That means that the axis of rotation is at the center of mass and r is the distance from the axis of rotation.
    So,
    T1 = F1(1.15)
    T2 = F2(1.15)
    T3 = -Mg(0.25)
    T4 = -mg(2.3)
    (T is torque, M is mass of box, m is mass of plank)

    sum of T = F1(1.15) + F2(1.15) - Mg(0.25) - mg(2.3) = 0
    F2(1.15) = Mg(0.25) + mg(2.3) - F1(1.15)
    F2 = -722.36

    I'm still getting it wrong, I read the whole chapter on static equilibrium, but I still am not getting it. Can you please help me work out this problem? Maybe that would help me understand better. (I am learning this on my own because my professor hasn't gone over this yet.)
     
  9. Nov 3, 2004 #8

    Pyrrhus

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    Ok, Here's the problem solved.

    I will only need

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    info:
    [tex] d = 0.9 m [/tex]
    [tex] L = 2.3 m [/tex]
    [tex] m_{box}=108.4kg [/tex]
    [tex] m_{planck}=5.6kg [/tex]

    I will apply the torque condition where you applied it. (on the extreme we are not finding) The lever arm for the weight of the planck it's half the length of the planck, because that's where it acts (It's not necessarily the half, it depends on the origin for the torque sum, but the weight of the object acts on its center of gravity which varies its location depending on the body)

    [tex] F_{man}L - m_{box}gd - m_{planck}g \frac{L}{2} = 0 [/tex]

    [tex] F_{man} = \frac{m_{box}gd + m_{planck}g \frac{L}{2}}{L}[/tex]

    [tex] F_{man} \approx 443.58 N[/tex]
     
    Last edited: Nov 3, 2004
  10. Nov 3, 2004 #9
    Okay, I think I get it know. Thanks a BILLION Cyclovenom!!! :smile:
     
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