# Acceleration is zero, for non-zero net force

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?

• 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. – Kenshin 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
• @Shinobii Do you mean to say a force is applied briefly and then removed? In that case during the period of time that the force is applied the object is accelerated but after the force is removed the object moves at a constant velocity. The key is that the acceleration at any given moment is proportional to the force at that same given moment. – jgerber Nov 7 '16 at 5:22

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.

• 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.

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.

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

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

• 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.

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).

First of all, acceleration= change of velocity. Now you are applying force which means you are changing its initial velocity to a final velocity. Initially the box is at rest or in a certain velocity. After applying force when it moves in another constant velocity it does mean that in the mean time velocity has been changed i.e acceleration is actually there. In short, force is applied and a constant velocity appears after that means that a change of velocity from rest to that constant velocity has occurred. So, actually acceleration is there.

## protected by ACuriousMind♦Nov 7 '16 at 1:19

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