Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two mechanics problems

  1. May 15, 2005 #1

    I just want to verify that my thought process for these problems is valid, any help would be appreciated. (I have pictures attached for each problem for clarity)

    1) You have mass1 attached to a spring with a certain spring constant k, it rests on a tabletop with a force of friction with a coefficient [tex]\mu [/tex]. Attached to mass1 is an ideal cord that extends over the table over an ideal pulley which holds mass2. a) If the spring is compressed a certain amount, and then released (spring stretches an amount x, smaller than max displacement), how far does mass2 move down? b) How much energy is dissipated by friction?

    a) Ok, so I set my coordinate system as positive for mass1 to the right, and for mass2 positive is down. I'm assuming the accelerations for both blocks is equal?

    For mass1 [tex] = \sum F_x = kx + F_T - \mu m_1g = m_1a [/tex]

    For mass2[tex] = \sum F_y = m_2g - F_T = m_2a [/tex]

    So I solve for a in both equations, set them equal to each other, and I get:

    [tex] F_T = \frac{m_2(m_1g + \mu m_1g - kx)}{m_1 + m_2} [/tex]

    [tex]a = m_2(g - m_1g + \mu m_1g - kx) [/tex]

    Ok, times to stop for [tex] m_1 [/tex] and [tex] m_2 [/tex] are equal, but I don't know where to go from here.

    b) I think I'm supposed to use the work energy theorem to figure out how much energy was dissipated by friction...but are the initial kinetic and potential energies of the system 0?

    2) STATICS problem:
    You have a uniform beam of length (l), mass M, supported by two wedges, Q and P. There are two boxes, W1 and W2 on top of the beam; each box has a different distance from the center of mass of the beam. a) What masses of W1 and W2 is the system stable? b) At which configuration does the normal force at Q goes to zero?

    Ok, the second picture attached should help on the setup of the problem.

    My solution:
    Ok, so I need the [tex]\sum \tau [/tex] around P to = 0, as well as the net force:

    (counter-clockwise positive)
    [tex]\sum \tau = W_1 A + Mg B - W_2 C - F_Q D = 0 [/tex]

    [tex]\sum F = F_Q + F_P - W_1 - W_2 - Mg = 0 [/tex]

    This is where I get stuck, and I'm not even sure these are right. Thanks for the help!

    Attached Files:

    Last edited: May 15, 2005
  2. jcsd
  3. May 15, 2005 #2
    In question 1:
    a) I would rather suggest the use of work-energy relationships. There is work done by a non-conservative force (friction) which is equal to the difference in mechanical energy of the system.

    [tex]W_{nc} = \Delta ME = \Delta (KE + GPE + EPE)[/tex],


    [tex]KE = \frac{1}{2}mv^2[/tex],

    [tex]GPE = mgh[/tex],

    [tex]EPE = \frac{1}{2}kx^2[/tex].

    Denoting by s the displacement, then we setup the equation:

    [tex]-\mu m_1gs = (0 + m_2gs + \frac{1}{2}kx^2) - (0 + 0 + 0)[/tex]

    Simplify this and solve for s.

    b) If you found s you can also determine

    [tex]W_f = -\mu mgs[/tex]
  4. May 15, 2005 #3
    Thanks, that helped, and ideas for problem 2?
  5. May 15, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    [tex]\sum \tau = W_1 A + Mg B - W_2 C - F_Q D = 0 [/tex]

    Two of your distances in this equation are incorrect. All distances must be measured from the same rotation axis. Your other equation is OK. Stability can be achieved for many different values of W1 and W2, but there will be a relatioship between them. The normal force at Q will go to zero when W2 is sufficiently heavier than W1. At that condition, one term in both your equations will drop out
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook