Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I understand a problem like this has already been asked, but I have not found an answer that makes it clear. A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. In this case, acceleration is equal to zero. The net force does not. How can we explain this?

$$ \sum F = F_{app} = m a $$

The acceleration $a = 0$, but $F_{app} > 0$.

Can someone clarify this for me?

share|cite|improve this question
Yeah, the question is wrong. Either there is a resisting force such as friction, or the box accelerates. What's your source for this statement/question? – Carl Witthoft Apr 16 '14 at 0:28
Thought problem. Imagine pushing a box at a constant velocity. Even with friction, $F_{app} > F_{f}$ and $F_{net} > 0$, but the box would still not be accelerating. Would it be safe to assume that IF there is a force acting, there MUST be acceleration ALWAYS? – Shinobii Apr 16 '14 at 0:30
The answer is simple, the table is on an angle and you are pushing upwards against gravity. – Mew Apr 16 '14 at 13:04
To expand on "Mew's simple solution" (in the above comment) "the table is on an angle and you are pushing upwards against gravity": $\mathbf F_{app} := m \, (\mathbf g\cdot\mathbf v)\mathbf v/\text v^2$; $F_{app} := m \, g \, \text{Sin}[ \phi ]$. If it seems somehow counter-intuitive that the applied force doesn't depend on the velocity $\mathbf v$ remember that the power does: $P = (\mathbf F_{app}\cdot\mathbf v) = F_{app} \, \text v$. (But of course it remains to be seen whether the OP had this sort of "solution" in mind at all.) – user12262 Apr 16 '14 at 18:20
up vote 3 down vote accepted

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity.

It's impossible. Or, don't ignore friction.

When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it.

Another thing I can think of: This argument is missing data. If constant velocity is recorded with respect to table, then there's inertial force to balance your force on box. Meaning, table reference frame is non-inertial.

share|cite|improve this answer
Yeppers. Rookie mistake. The question was worded incorrectly (cannot rely on the Authors these days). Thanks for your response! – Shinobii Apr 17 '14 at 16:30

By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing $F_{app}$ otherwise there will be an acceleration.

share|cite|improve this answer

Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$

$$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$

Second possibility : If your box is spherical,

By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity.

Hence, your ball attains terminal velocity.

$$F=6\pi\eta rv$$

$$v=\frac{F}{6\pi\eta r}$$

share|cite|improve this answer
Seems slightly off topic, but interesting reply! – Shinobii Apr 17 '14 at 16:31
@Shinobii Note that $F=ma$ is only a special case of $F=dp/dt$ – evil999man Apr 17 '14 at 17:19

If you apply a force on the box, and see no acceleration, then the force you apply is equal to the friction force.

Friction is velocity dependent, you cannot say "the friction force is so much" independently of the force you are applying.

share|cite|improve this answer

It takes zero force to push an object along a frictionless plane with a constant velocity. In fact a force can be defined as the rate of change of linear momentum, and zero force applied to a body yields no change on momentum.

When friction is added, only when the applied force exactly matches the friction force the resulting motion is uniform. In situations like this, knowing the motion $x(t)$ and its second derivative $a(t)$ yields the required applied force $F(t)$ to enforce this constraint (the motion) as $$ F(t) + \sum F_{\rm{other\ froces}} = m \, a(t) $$

You are fighting your own intuition here whereas to push a table with what appears as constant speed you need to apply a force, but in reality you are a) fighting friction and b) the speed is not exactly constant but changing (slightly).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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