3
$\begingroup$

I was told forces can depend on time, location and velocity, but never on acceleration. At first I thought this must be wrong, thinking of $F=m\cdot a$. But I think/hope I now got what was meant, what my misunderstanding was:

Firstly: $F=m\cdot a$ has nothing really to do with it directly.

The forces that apply to an object are always the same, no matter what the current acceleration of that object is.

But these forces are (or can be) dependent on the objects velocity (e.g. some kind of friction) and location/time (which is obvious)?

Is this right or am I wrong again?

$\endgroup$
  • $\begingroup$ Which force are you talking about? for example gravity force is not depended on the velocity of the object. it totally depends on the type of the force. $\endgroup$ – Mobin Apr 11 '15 at 19:02
  • $\begingroup$ Any force actually, that's why I added the "can be" dependent. $\endgroup$ – MarocJ Apr 11 '15 at 19:03
  • $\begingroup$ Yes it's like this, forces create acceleration in the object. and each force can be dependable on so many things. like gravity it depended on the mass and the distance of two objects while electro-statical force depends on the charge of two objects. $\endgroup$ – Mobin Apr 11 '15 at 19:06
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/52024/2451 and links therein. $\endgroup$ – Qmechanic Apr 11 '15 at 19:27
  • $\begingroup$ They might be talking of the broader $F=\frac{dp}{dt}$ definition. $\endgroup$ – Tetradic Apr 11 '15 at 19:40
7
$\begingroup$

An example of a force that depends on position (of the particle) is the force due to a spring:

$$F_x = -kx $$

An example of a force that depends on velocity (of the particle) is the force due to a dashpot

$$F_v = -c\dot x$$

Now, consider a hypothetical force that depended only on the acceleration of a particle:

$$F_a = -d \ddot x$$

The differential equation of motion would then be

$$m\ddot x = -d \ddot x \Rightarrow (m + d)\ddot x= 0$$

Then, unless $m = - d$, the particle's acceleration must be zero.

Next, consider the case that there are position, velocity, and acceleration dependent forces on the particle. The differential equation then becomes

$$(m + d)\ddot x + c\dot x + kx = 0 = m'\ddot x + c\dot x + kx$$

That is, the acceleration dependent force would have the effect of changing the inertial mass of the particle from $m$ to $m' = m + d$. Such a force could be realized by, e.g., an electronic control system.

Typically, a force that does not depend on the particle's position, velocity, etc. but may be time dependent is a driving force and would appear on the right hand side of the equation of motion:

$$m' \ddot x + c\dot x + kx = F_{ex}(t)$$

Finally, the state dependent forces might have time dependence (the system would be time variant).

$$m'(t) \ddot x + c(t)\dot x + k(t)x = F_{ex}(t)$$

$\endgroup$
3
$\begingroup$

$F = ma$ when mass is constant: I think that's a common misconception. It doesn't really represent the law correctly (unless mass is constant).

Newton's second law of motion states, $F = d/dt(mv)$ the external forces acting on an object in an inertial frame is equal to the change in linear momentum (the measure of motion)

I can see why he made that statement: from the equation, the input parameters can be time, mass, velocity (change in position with respect to time).

$\endgroup$
2
$\begingroup$

Because in Newton's second law, you only have the differentials $dx, dv$ and $dt$ (hence why force can only be a function of $x , v$ and $t$. You never have $da$ where $a$ is acceleration (unless Newton's second law was a third order differential equation then you would have $da$). So mathematically you can't really solve Newton's second order differential equation if force was a function of acceleration.

$\endgroup$
0
$\begingroup$

To clarify the law should be stated

$$\sum F(t,x,v) = m\,a $$

which is used to solve problems by re-arranging into $$a(t,x,v)=\frac{1}{m} \sum F(t,x,v)$$

The point is that various forces depending on time, position and velocity are used to define accelerations. So yes, you are correct.

A less obvious case is in problems of nverse dynamics. Here all motions are prescribed the forces are found by $$F_{unknown}(t) = m a(t) - \sum F_{known}(t)$$. You can interpret this as a force dependent on acceleration (motion) but in reality what you have is motion dependent on time only, and hence forces are functions of time.

$\endgroup$
0
$\begingroup$

Unless you are talking only about classical mechanics, I think that it is not true. For example, the Abraham-Lorentz force depends on jerk (time derivative of acceleration).

$\endgroup$
0
$\begingroup$

I was told forces can depend on time, location and velocity, but never on acceleration.

That depends on what one means by "force". The concepts of "force" as used in Newton's second law and in Newton's third law are distinct but related concepts.

That force and acceleration are intimately connected with one another entire the point of Newton's second law, $\boldsymbol F = m\kern{1.5mu}\boldsymbol a$. In words, force is the product of mass and acceleration).

Newton's third law, $\boldsymbol F_{a\rightarrow b} = -\boldsymbol F_{b\rightarrow a}$ (in words, the force exerted by body $a$ on body $b$ is equal but opposite to the force exerted by body $b$ on body $a$), addresses force in a different way. The forces in Newton's third law are distinct from the forces in Newton's second law.

The concept of "force" in the context of Newton's third law may well be what your instructor was talking about when he said "forces can depend on time, location and velocity, but never on acceleration." Before electrodynamics, there was no classical force that was a function of acceleration. Electrodynamics changed this. Accelerating charges radiate, even classically.

Finally, that forces are vectors is what provides the connection between the concept of force in Newton's second law and the concept of force in Newton's third law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.