1. Limited time only! Sign up for a free 30min personal 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!

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


    User Avatar
    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook