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Homework Help: Equation of motion with drag and external forces

  1. Jul 26, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to crudely approximately the distance individual masonry units fly when a blast load impacts a masonry wall. I'm a structures guy so I can calculate when the wall will failure, but I am having trouble with the calculus associated with the equation of motion.

    My structure is subjected to a linearly decreasing blast loading,
    F(t)=A*Po*(1-t/td) when 0<t<td
    F(t)=0 when t>td
    where td is the duration of the load

    The drag force exerted on the units is given by

    2. Relevant equations
    The equation of motion, introducing constants A and B to simplify the equation, is:

    u'' + A*(u')^2 = B*F(t)

    The acceleration, u'', and velocity, u', as well as the load F(t), are all a function of time

    3. The attempt at a solution
    The equation has derivatives of u, so I introduced v=u', v'=u'', so the equation of motion becomes:
    v'+A*(v^2)=B*F(t) which is now first order and should be easier to solve....
    v'=B*F(t)-A*(v^2) but I am not sure how to proceed.

    Any ideas on how to continue with this analysis?

  2. jcsd
  3. Jul 26, 2010 #2


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    Homework Helper

    The only thing I can suggest is a numerical solution. You have a nonlinear equation, these things are nasty.
  4. Jul 27, 2010 #3
    Hi hunt_mat, do you have any suggestions on which numerical method method to use?
  5. Jul 27, 2010 #4
    Ok, so I've been working on this problem such that I can solve it with a numerical method. I am stuck because I am trying to isolate a variable, v, but I am left with a quadratic equation with two roots. Here's what I have:

    u'' + A*(u')^2 = B*(1-t/td)
    Introduce v=u', v'=u''


    Rewrite in terms of v'=dv/dt
    dv=[ B*(1-t/td) - A*(v^2) ]dt

    Integrate both sides with initial conditions v=0 @ t=0:
    v= [ B*{t-(t^2)/(2*td)} - A*t*(v^2) ]

    Introduce C=B*{t-(t^2)/(2*td)} and D=A*t to simplify, and rewrite (again!):

    D*v^2 + v - C = 0

    I am not sure what to do at this step as I am left with the solution to a quadratic equation....which root do I pick?

    Any ideas?
    Thanks! :)
  6. Jul 27, 2010 #5
    I can help you if you explain your equation:

    u'' + A*(u')^2 = B*F(t)

    u'' is the acceleration? (x''(t))
    u' is the velocity? (x'(t))

    and if the force is a function of time as well dont you have 2 variables in one equation (u and F) or do you know F(t) and it acts on the same object, we can rewrite it as F(t)=m*a(t)=m*u'' ?

    also where di you got that original equation u'' + A*(u')^2 = B*F(t) ? :)
  7. Jul 27, 2010 #6
    gomunkul51, I'll start from the beginning:

    u is defined as displacement
    u' is defined as velocity
    u'' is defined as acceleration

    I have an object that is excited by a time varying force. In this particular case, the force, F(t), is a linearly decreasing force with initial magnitude Fo and acts for a duration of td. For a time of 0<t<td, the force F(t)=Fo*(1-t/td). However, after time t>td there force no longer acts and the object travels under its own inertia (F(t)=0)

    Inertia forces of the object resists its motion: Fi=mu''
    Drag forces also resist motion: Fd=0.5*Cd*A*rho*(u')^2 - note that Cd, A, and rho are all constants

    Writing the equation of motion for 0<t<td yields:
    Fi + Fd = F(t)
    mu'' + 0.5*Cd*A*rho*(u')^2 = Fo*(1-t/td)

    Divide by the mass, m
    u'' + 0.5/m*Cd*A*rho*(u')^2 = Fo/m*(1-t/td)

    A and B were defined based on the above equation because most of the value are constants....

    I am most interested in the solution procedure for time 0<t<td as I figure the solution becomes simply when the force, F(t)=0 at time t>td.

    Let me know if you need any more info
  8. Jul 28, 2010 #7
    OK, so you can basically write it like this:

    C = (1/m)*0.5*Cd*A*rho

    u''(t) + C*(u'(t))^2 = (Fo/m)*t

    and you are interested in the solutions from t=0 to t=td.

    *it could more completely written as:

    u''(t) + C*(u'(t))^2 = (Fo/m)*t - H(t-td)*((Fo/m)*t)

    *H(t-td) is the Heaviside (step) function that starts at t=td.

    *but writing the equation with a step function is redundant if you are interested only in the solutions t=0 to t=td.

    This equation: u''(t) + C*(u'(t))^2 = (Fo/m)*t could be solved it terms of Airy/Bessel functions and calculated to any desired accuracy with a computer program.

    The answer from Maple is:

    u(t) = (1/3)*ln(C^2*m*(-_C1*AiryAi(-(-C*F[0]/m)^(1/3)*t)+_C2*AiryBi(-(-C*F[0]/m)^(1/3)*t))^3/(F[0]*(AiryAi(1, -(-C*F[0]/m)^(1/3)*t)*AiryBi(-(-C*F[0]/m)^(1/3)*t)-AiryAi(-(-C*F[0]/m)^(1/3)*t)*AiryBi(1, -(-C*F[0]/m)^(1/3)*t))^3))/C

    *pretty large, but could be calculated with the same program :)

    P.S: if the force Fo is constant, the answer is analytically simple :)
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