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Dynamic friction and spring force

  1. Feb 22, 2010 #1

    KFC

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    1. The problem statement, all variables and given/known data
    A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 [tex]m/s^2[/tex]. By how much does the spring stretch if it pulls on the sled at 30.0 degrees above the horizontal?


    2. The attempt at a solution
    The following is how I solve the problem by myself.

    First, the object is only moving along the horizontal level and there is friction. I assuming the total force which pulls on the sled at 30 degree is F and the friction is f. The force affecting the spring is [tex]F_k = F_x = F\cos(30) - f[/tex].

    For friction, it should be
    [tex]f = \mu (mg - F\sin(30)) = \mu (mg - F_x\tan(30))[/tex]

    and from Newton second law
    [tex]F_x - f = ma[/tex]

    so
    [tex]F_x = ma + f[/tex]
    gives

    [tex]F_x = \frac{ma + \mu mg}{1+\mu\tan(30)}[/tex]

    that is, according to Hook's Law

    [tex]\Delta x = F_x/k = \cdots [/tex]

    I wonder if my calculation is corrected? I think about it for several times, it seems make sense. But one of my friend come to me few minutes ago. He said "the sled is experiencing a total force (along the horizon), which is [tex]F_x - f = ma[/tex], so the spring will also experience the same (total) force. Hence, x = ma/k instead of [tex]F_x[/tex]/k". This seems also make sense, so which one is correct? WHY?

    Here I attach the figure of the problem (given by my TA instead of from the text)

    http://img697.imageshack.us/img697/4535/figt.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 22, 2010 #2
    This is almost correct. The extension of the spring is not related to the Fx that you wrote, it's related to the the magnitude of Fspring:

    http://img97.imageshack.us/img97/807/sledu.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Feb 22, 2010 #3
    A horizontal spring with stiffness 0.7 N/m has a relaxed length of 18 cm (0.18 m). A mass of 17 grams (0.017 kg) is attached and you stretch the spring to a total length of 21 cm (0.21 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 18 cm (0.18 m)?

    I've attempted this problem several times, to no avail.

    These are the equations I've been using.

    Maximum speed = 0.5(mv^2) = 0.5(k)(x^2)
    where k = spring constant and x is maximum stretch

    V at any point = sqrt((k/m)*(xmax^2 - x^2))
    k is spring constant, m is mass, x is equilibrium position and xmax is it's stretched distance
     
  5. Feb 22, 2010 #4
    Good Evening shnav,

    This is a different problem, so you'll need to start a new thread.
     
  6. Feb 22, 2010 #5
    ok keep an eye open for it, thank you, I didn't know
     
  7. Feb 22, 2010 #6

    KFC

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    [/URL]

    Thank you dr_k. May I ask you one question? What physics it related to such that the extension of the spring is related to Fspring instead of just the x component? My point is there is no stretch along y direction, so why can't we conclude that the y component of Fspring doesn't contribute to the extension? Thanks.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 22, 2010 #7
    http://img96.imageshack.us/img96/2180/springs1.jpg [Broken]

    For example, in the above picture, the block experiences a force, externally exerted by the spring, along the dashed line, its direction depends upon whether the spring is compressed or extended.

    Calculate the magnitude of [itex]\rm F_{spring}[/itex]. This magnitude is proportional to [itex]\rm k |\Delta s|[/itex] , where [itex]\rm |\Delta s|[/itex] is the one-dimensional compression/expansion of the spring along the dashed line.

    For your problem, when you do your FBD, and apply Newton's 2nd postulate, you will, as you've done, break [itex]F_{spring}[/itex] up into components. Then isolate and find the magnitude of [itex]\rm F_{spring}[/itex] , which is then proportional to [itex]\rm k |\Delta s|[/itex] .
     
    Last edited by a moderator: May 4, 2017
  9. Feb 22, 2010 #8

    KFC

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    Thanks. Just because the force should be along the dash line, the force exerted by the spring in my case should be [tex]F_x[/tex]. Take the last figure given by you as example, the spring is compressed or stretched by along the dash line, hence, the force from spring should also along the dash line. And in the problem, the spring is along the x direction (horizontally), so the force of spring should also along this direction, right?

    Or I put in this way, Hook's law is given by
    [tex]\vec{F} = -k\vec{s}[/tex]

    where [tex]\vec{s}[/tex] is along the same direction as the spring force (on which the spring compressed or stretched). Hence, if the spring is along x direction, the spring force should be [tex]\vec{F}=F_x \vec{\hat{x}}[/tex], not [tex]F_{spring}[/tex] as shown in the figure given by you, right?

     
    Last edited by a moderator: May 4, 2017
  10. Feb 22, 2010 #9
    Reading the problem that you posted, I envision a picture that looks like this:

    http://img132.imageshack.us/img132/8369/springs2l.jpg [Broken]

    Does your text show a different picture?

    Perhaps we're talking about two different pictures.
     
    Last edited by a moderator: May 4, 2017
  11. Feb 22, 2010 #10

    KFC

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    Oh, thanks for reminding me. I think my book gives me a wrong picture, the one shown in the book is different from your. And I read the problem carefully again, seems your picture makes more sense. Sorry to misunderstand you.


     
    Last edited by a moderator: May 4, 2017
  12. Feb 22, 2010 #11
    Can you draw the picture in your text, and post it? My comments in this thread are w.r.t. the picture I just posted. If the picture in your book is different than mine, then we were solving different problems.
     
  13. Feb 22, 2010 #12

    KFC

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    Yes. I just add the picture in the first post. Actually, this is a practice given by my TA instead of my the text. I don't know where does this problem is from. I guess my TA gave me a wrong picture. Thanks dr_k, anyway.

     
  14. Feb 23, 2010 #13
    Wow. That is not the picture I envisioned when I read

    This pictures shows x(t) satisfying a non-homogenous 2nd-order ode, but since you're given the value of [itex]\rm \ddot{x}[/itex], the displacement is merely (for a stretched spring)

    [itex] \rm \frac{F(\cos\theta + \mu_k \sin\theta)-m (\mu_k g+a_x)}{k} [/itex]

    if you know the magnitude of this F.

    This is where a black-board would have come in handy, so we could have agreed on what problem we were trying to solve. :)


    After this much work, why not solve the problem for this picture:

    http://img132.imageshack.us/img132/8369/springs2l.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. Feb 23, 2010 #14

    KFC

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    OK. Here is what I did based on my picture. First, I am trying to find the total force on x direction

    [tex]F_{xtot} = F \cos\theta - f = ma_x[/tex]

    here [tex]f[/tex] is the friction, which is given by

    [tex]f = \mu_k (mg - F\sin\theta)[/tex]

    Hence,
    [tex]F_{xtot} = F \cos\theta - \mu_k (mg - F\sin\theta) = ma_x[/tex]

    From Hook's law, the force on the spring satisfies

    [tex]F\cos\theta = - k x[/tex]

    namely,
    [tex]F \cos\theta = \mu_k (mg - F\sin\theta) + ma_x = - kx[/tex]

    Hence,

    [tex]x = \frac{\mu_k F\sin\theta - m(\mu_k g + a_x)}{k}[/tex]

    that close to what you gave except that there is no [tex]F\cos\theta[/tex] in the expression, am I doining anything wrong?
     
  16. Feb 23, 2010 #15
    For the "new picture", we don't have a description of the problem, so we'd have to make up our own. So I'll make a few assumptions...

    Suppose the block is moving to the right with increasing speed, and the spring is stretched [itex] \rm \Delta x [/itex] from its equilibrium position:

    http://img109.imageshack.us/img109/2784/springs3.jpg [Broken]

    Newton's 2nd Postulate implies:

    [itex]\rm F_{{net}_x} = F \cos\theta - k\Delta x -f_k = m a_x[/itex]

    [itex]\rm F_{{net}_y} = F \sin\theta + N - mg = 0[/itex]

    Solve for [itex]\rm \Delta x[/itex]
     
    Last edited by a moderator: May 4, 2017
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