1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force Method for Two body system

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Two bars masses m1 and m2 connected by a non deformed light spring at rest on a horizontal plane. The coefficient of friction b/w the blocks and the table is k. What minimum horizontal CONSTANT force F has to be applied on m1 so as to just shift m2.
    (For reference Problem 1.123 Irodov).

    2. Relevant equations

    a)Fext = m*a(cm)
    b)M2 - M1 = Wext + Wnon-conservative
    where M is total mechanical energy.

    3. The attempt at a solution

    I got the correct answer by applying the second theorem mentioned above:
    Supposing spring constant as a and maximum extension of spring as x and keeping our observations from the time force starts to act upto the time the block m2 first moves.
    ½a*x2= F*x + (-km1gx)
    and x = km2g/a
    However I find it hard to digest my own assumption that force will be minimum if the final velocity of m1 is 0. I thought a method involving forces would be easier to understand but it is too complex to work out.Would really appreciate any insight and help that gets me started on a force method.
     
  2. jcsd
  3. Mar 15, 2010 #2
    Try superposition. Consider the force required on m2 alone, then add the spring knowing it will store energy (the spring will compress a particular amount before m2 moves), then add m1.
     
  4. Mar 15, 2010 #3
    Thanks for the reply. I thought that they were pulling m1 away from m2 instead of compressing the spring but I guess it's just the same.
    I got this:
    Force required to start block m2 = km2g
    Hence maximum extension the spring should have = km2g/a

    Now analyzing for m1-
    Fnet = m1A ( A= acceleration)
    This gives
    F - ax - km1g = m1A where x is a general extension of spring.
    Therefore A = (F - ax - km1g)/ m1
    This I rearranged a bit to have velocity as function of disatnce
    Since A = v*dv/dx
    so (F - ax - km1g)dx =m1*v*dv
    Integrating
    Fx - ax2/2 - km1gx = m1*(v2/2)

    Which is same as the energy equation I wrote earlier. Not surprising but I have still not put in the limits. For x it goes from 0 to max. extension.
    The limits for velocity however raise the same question I put up last time. I have my answer if I put it from 0 to 0. There is only one reason I am thinking of it. For F to be minimum the right hand side of eqn. needs to be minimum. Since we have a square term(v2) the minimum value happens to be zero.
    Need a bit more help with the limits of velocity. But thanks a lot!
     
  5. Mar 18, 2010 #4
    Recall there are two types of friction; static and kinetic. The system of masses and the spring have to satisfy the static frictional force in order to start moving. What is the initial velocity of the mass when the applied force equals the static frictional force? I think this will answer your question regarding velocity.
     
  6. Mar 28, 2010 #5
    Sorry for such a late response. I hope you are still interested.
    I thought when they said coefficient of friction was k it meant that coefficients for static and kinetic friction were equal and equal to k so is it right to talk of static and kinetic friction when their magnitudes are identical ? And though the initial velocity of both blocks is 0, the final velocities too are zero which is not so obvious to me. Were you implying something else?
     
  7. Mar 28, 2010 #6
    The static and kinetic coefficients of friction are typically not equial with static greater than kinetic. Therefore, a larger force is required to initially move the object and less of a force is required to maintain a constant velocity. The force applied to break loose the object will be in equilibrium with the static friction force hence final velocity is zero. If an infinitesimal amount of additional force is applied then the object will accelerate.
     
  8. Mar 28, 2010 #7
    Well the block pulled does accelerate.
     
  9. Mar 28, 2010 #8
    Well, think of the system as having a light (meaning massless) rigid rod replacing the spring. Then consider the rigid rod replaced by a light, flimsy spring. Do you think there's a reason why a spring constant was not given?
     
  10. Mar 29, 2010 #9
    Do you mean that the spring here is not important? That would be strange, I guess.
    I was wondering about the point you made about static and kinetic friction. Here they have given a single coefficient of friction so how can we quantify any increase or decrease in friction? Alternatively I would think of it as a situation where kinetic friction equals static friction. Of course that is false but that's what the problem tells us, isn't it?

    I was wondering if the final answer lay in the mathematical part of the problem rather than a qualitative answer.
     
  11. Mar 29, 2010 #10
    Show how you arrived at your answer explicitly so your mathematics can be reviewed. I raised the question about the spring because I surmise you are thinking in terms of energy rather than force. Think of a very light, very stiff spring attached to a mass and you push on the spring in order to move the mass. If 5 N of force is required to move the mass, the spring will compress a certain amount (store energy) then stop compressing. The applied 5 N of force continues and therefore transfers through the spring to the mass. If the spring constant (stiffness) changes, the compression will change but the applied force required to move the mass is still eventually delivered to the mass. What changes is the time required to move the mass with an applied constant force. As for friction, typically static and kinetic coefficients are different. The problem appears to be simplified by assuming the two coefficients are equal.
     
  12. Mar 30, 2010 #11
    I wrote out the math earlier but it is clearer to me now. Well here it is:

    Acceleration of the pulled mass = (F - kx - µm1g)/m1
    And a = v*dv/dx
    This will give m1v*dv = (F - kx - µm1g)dx
    When I integrate this
    m1(vf2-vi2)/2 = Fx - kx2/2 - µm1gx
    where the limits of x go from 0 to max extension which we keep x for a while.
    Hence force is
    F = m1(vf2-vi2)/2x + kx/2 + µm1g
    vi = 0
    so F = m1(vf2) + kx/2 + µm1g


    Since the velocity term is positive (vf2) hence it's minimum value is 0.The value of x is the max. extension of spring required which is µm2g/k. Now it is clear that the velocity term is the only part which can be altered, the rest are constant no matter what the situation.

    When we put these assumptions in
    F = µm2g/2 + µm1g
    = µg ( m1 + m2/2 )
    Is this it?
     
  13. Mar 30, 2010 #12
    Your analysis has some math errors. However, it's not necessary to explicity solve in terms of energy. The force required to move m2 is

    [tex]f_2=km_2g[/tex]

    The spring is going to stetch (or compress) until the force pulling (or pushing) on the spring equals f2 whereby m2 will begin to move. The force required to move m1 is

    [tex]f_1=km_1g[/tex]

    Therefore, the total force applied is

    [tex]F_{applied}= f_2+f_1[/tex]

    Now, consider the three cases:

    m1 > m2
    m1 < m2
    m1 = m2

    Each case requires a different minimum constant force.
     
  14. Apr 1, 2010 #13
    Sorry I don't see the math errors you pointed out.
    You wrote
    Fapplied = F1 + F2
    It was what I thought earlier but you see if we apply this force on the block m1 it would have some velocity when the block m2 just shifts, something we have debated for quite a while that it shouldn't be. And as for the analysis for the cases you mentioned(m1>,<,=m2), aren't they accounted for if the expression worked out for minimum force contains these terms?
     
  15. Apr 1, 2010 #14
    You wrote:

    so F = m1(vf2) + kx/2 + µm1g


    Check the units of each term; one does not equate to the units of force. Anyway, consider m2 with a light spring attached. You are asked to pull on the spring with sufficient force to just move m2. This force is

    [tex]f_2=km_2g[/tex]

    The spring will stretch a certain distance until m2 begins to move. What happens after m2 begins to move? Well, the applied force is infinitesimally larger the [tex]f_2[/tex] for a infinitesimal duration of time (an impulse) then can be reduced to [tex]f_2[/tex] such that m2 has a constant velocity (net force is zero since static and kinetic friction forces are equal in this problem). This always has to be the condition to move m2.

    Now, attach m1 to the free end of the spring. The two masses are not rigidly coupled so the system does not behave like a single mass when the spring is being stretched. The force required to move m1 is

    [tex]f_1=km_1g[/tex]

    Once m1 begins to move the spring will stretch a distance x from an additional applied force that has to be

    [tex]kx=f_2[/tex]

    whereby the condition is met to move m2. Now, the system behaves as a rigidly coupled system because with a constant force the spring does not stretch once equilibrium is achieved. I was wrong before about the different cases; sorry about that.

    I hope this clarifies my view.
     
    Last edited: Apr 1, 2010
  16. Apr 7, 2010 #15
    Hello! Sorry again about the late reply; been busy for a while.
    Yes I was wrong with that eqn. I should have divided m1vf2/2 by x.
    You wrote :
    Once m1 begins to move the spring will stretch a distance x from an additional applied force that has to be
    kx = f2


    whereby the condition is met to move m2. Now, the system behaves as a rigidly coupled system because with a constant force the spring does not stretch once equilibrium is achieved.

    I agree with you as far as conditions for moving m1 and m2 are concerned. But do you think they are independent enough to be added straight like that? If you apply that force on m1 [kg(m1+m2)] then m1 will still have some velocity when extension of spring is enough to just shift m2.
    Consider this:
    Invoking work-energy theorem on m1(with assumption that a constant force kg(m1 + m2) acts on it)-
    K2 - K1 = Wall
    1/2 mv2 - 0 = kg(m1+m2)x - kgm1x - 1/2 Kx2 (K=spring constant; k=coeff. of friction)
    where the second term on RHS is work of friction and the third term is obtained by integrating the work of non-constant spring force. Now x is max. extension of spring which is km2g/K. If we put this in eqn. above we get 1/2 mv12 = k2m22g2/2K2 .
    This should mean that m1 should have some velocity at that moment. But haven't we discussed this before - m1 should have zero velocity at this point?
     
  17. Apr 7, 2010 #16
    The mass m1 will have zero acceleration when the spring reaches maximum extension (this is the point in time where m2 will begin to move); the velocity will not be zero but the net force will be zero on the system. For m1 to have a zero velocity after a constant force is applied another force must be applied to m1 to slow it down to zero velocity.
     
  18. Apr 8, 2010 #17
    And then that constant force would be subtracted from the original constant force?
     
  19. Apr 8, 2010 #18
    I believe I see your point with m1 having a zero final velocity. Initially, an applied constant force must act on m1 such that m1 begins to move, f1 = km1g. Now, in order to get m2 to move the spring must stretch a distance x such that km2g = bx (where b is the spring constant). So, an additional constant force of bx must be applied to m1 to stretch the spring such that m2 begins to move. When the spring reaches the distance x then m1 will be at rest (spring restoring force equals applied force) hence the final velocity of m1 is zero.
     
  20. Apr 11, 2010 #19
    And I see it too. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force Method for Two body system
  1. Many body system (Replies: 4)

Loading...