0
$\begingroup$

If no net force acts on a massless object can there be any acceleration of the object?

My attempt:

$$F=ma\\ \implies a=F/m\\ \implies a=0/0 $$

$\implies a$ can be anything.

$\endgroup$
  • 3
    $\begingroup$ You cannot reduce the physics problem to a mathematical operation. The relation F=ma expresses a conclusion from observation of the world and inference of what would happen in the absence of no force, plus some definition of what is understood by force and mass and acceleration. So the answer is there. You ask, what is the acceleration of an inexistent body from the action of an inexistent force? I guess you can figure the answer out. $\endgroup$ – rmhleo Jun 8 '16 at 9:27
  • 2
    $\begingroup$ I think that a simple answer is that no, you can't do that. From a mathematical standpoint (which it seems to me is from where you have approached the question), just because division by zero is undefined does not mean that the answer can be anything you want. $\endgroup$ – Matt Jun 8 '16 at 16:07
  • 2
    $\begingroup$ You can't divide by $m$ if $m=0$ either in math or physics, so your second equation is just a bunch of letters that don't mean anything. $\endgroup$ – garyp Jun 8 '16 at 16:31
  • $\begingroup$ It is already known that all massless particles move at the constant speed of c. $\endgroup$ – David White Apr 8 at 21:02
0
$\begingroup$

No Bro. *even though the object is mass-less and the net force is zero but also then it can't accelerate. net force can be zero only if any force(s) is balancing the applied forces i.e. the resultant force will have no direction so there is no reason and direction left for the object to accelerate for.In the figure the ground can be ground or any kind of support.that's just an example. *enter image description here

**hope you decipher what i am trying to explain to you.Looking for your further suggestions. Even if there is any flaw in my explanation inform me please so that i can make my concepts clear.

:-)

$\endgroup$
  • $\begingroup$ Doesn't a photon experience an acceleration in a curved path? $\endgroup$ – Gareth Meredith Apr 8 at 17:35
6
$\begingroup$

In a theory view point, it is possible. For example, consider figure below. The string is massless and inelastic (rigid). (There is no friction)

enter image description here

The string moves with the mass $m$, so its acceleration is $a=\large{\frac Fm}$. But, the net force acting on the string is zero.

enter image description here

$$\Sigma F_{\text{string}}=F-T=0$$

$\endgroup$
  • 2
    $\begingroup$ Very good example! Probably not what the OP had in mind, but entirely correct! $\endgroup$ – garyp Jun 8 '16 at 16:29
1
$\begingroup$

Simple answer is No.

First of all you cannot apply Newton's Equation (Classical Physics) to any mass-less objects, because, all mass-less objects (like photons) travel at the speed of light, and at that speed we use equations from modern physics (Quantum and Relativity).

Without any force, they will keep traveling at constant velocity. To produce acceleration (to change velocity) some force has to be applied. (Remember that you can only decrease the speed)

Mass-less objects are created and destroyed (most common example is photon; their energy can be converted into another form such as electrical energy). Mass-less objects don't exist just at any speed. Since the moment of creation, they start moving at the speed of light.

Your question somewhat resembles to this one. Do photons have acceleration?

$\endgroup$
0
$\begingroup$

If the acceleration of a massless object is less than infinite, the net force on the object must be zero. An example is a massless frictionless piston manually being held in place within a cylinder featuring an initial high pressure inside the cylinder and atmospheric pressure on the outside. If the piston is suddenly released so that the gas experiences an irreversible expansion, the acceleration of the piston will not be infinite, and the forces on either side of the piston (which will include viscous forces and inertial forces of the gases) will immediately adjust such that the net force on the piston will be zero.

$\endgroup$
  • $\begingroup$ I think your example is similar to that given by @Lucas. $\endgroup$ – sammy gerbil Jun 10 '16 at 18:33
  • $\begingroup$ @sammy gerbil I don't feel that it is (at least I am not able to see a close connection). In the example I presented, there are forces acting on both sides of the object, and these are non-conservative forces exerted partially as a result of viscous stresses and dissipation. Also, in this example, the forces on both sides of the object automatically adjust so that the net force is zero at all times, even though the acceleration is finite. $\endgroup$ – Chet Miller Jun 10 '16 at 21:22
0
$\begingroup$

On a massless object (like an ideal rod) you can specify the acceleration of one end and derive the acceleration of the other end. But you cannot apply forces and derive the acceleration because the inertial properties are zero.

$$ \vec{a}_A = \vec{a}_B + \vec{\alpha} \times \vec{r}_{BA} + \vec{\omega} \times \vec{\omega} \times \vec{r}_{BA} $$

$\endgroup$

protected by ACuriousMind Apr 8 at 20:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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