# Equations of motion in a free fall with friction

torito_verdejo
Homework Statement:
A point mass ##m## is dropped from a hight ##z##. Its motion is subject to gravity force and a friction force ##F_f=-\lambda \dot{z}##.
Write the equations of motion for this system.
Relevant Equations:
$$F_f=-\lambda \dot{z}$$
I'm stack at the very beginning. If I use Newton's second law to find acceleration and integrate until I find the position, I must face

$$v(t) = \int_0^t g-\frac{\lambda v}{m} dt'=gt-\frac{\lambda }{m} \int_0^t\frac{\partial z}{\partial t}dt$$

But this last term feels pretty weird. I don't know how to interpret it, even though

$$\frac{\partial z}{\partial t}dt = dz$$

If I extract velocity from friction, we get

$$v(t)=\frac{m}{\lambda}(g-a(t))$$

which once again defines a function by its derivative. Should I be solving a differential equation?

In any case, I face two possible ways and I wonder which one, if any, is the correct one. Thank you in advance for the help.

Last edited:

Homework Helper
Gold Member
2022 Award
The simple answer is that yes, this is about solving differential equations. Have you studied these?

Your last equation could better be written:

##\ddot{z} = g - \frac{\lambda}{m}\dot{z}##

Note: this is, by definition, the "equation of motion" for the system.

In the first part of your post you effectively integrated this, using the initial conditions, to get:

##\dot{z} = gt - \frac \lambda m z##

Note that your "weird" integral was just the integral of a derivative, which is given by the fundamental theorem of calculus. Note that the derivative should be an ordinary time derivative, not a partial derivative.

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• torito_verdejo and Delta2
Homework Helper
Gold Member
Yes you are called to solve a first order linear differential equation here. The first step is to gather the unknown function and all its derivatives to the same side of the equation and write the equation as $$\dot v+\frac{\lambda}{m}v=g$$.
Now we notice that if we multiply the equation by the factor ##e^{\frac{\lambda}{m}t}## then the LHS becomes the derivative of the product ##e^{\frac{\lambda}{m}t}v(t)## so we ll have
$$(e^{\frac{\lambda}{m}t}v(t))'=ge^{\frac{\lambda}{m}t}$$
and from this last equation you can integrate both sides and after some algebraic manipulation you can solve for the unknown function ##v(t)##.

• torito_verdejo and PeroK
Staff Emeritus
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