# Can an accelerating body be at rest?

A block of mass $$1~\rm kg$$ is lying on a table. It is stationary because the gravitational force is balanced by the force from the table. But the body should have an acceleration $$g$$ and consequently its change in momentum should be non-zero. But how come the body is at rest? Are there other examples where a body is at rest but is accelerating?

• You may enjoy reading a general relativity approach to this question: physics.stackexchange.com/a/226564/123208 ... or you may find that it just adds to your confusion. ;) May 7, 2020 at 12:10
• Just close your eyes and imagine that you are sitting in a chair of an accelerating rocket with acceleration $a = g$. This is famous Einstein equivalence principle. In this respect we ALL on Earth are accelerating. May 7, 2020 at 12:39
• Why do you think the body should have acceleration $g$? Newtonian gravity is NOT an acceleration. Masses interact with forces on each other, not accelerations. May 7, 2020 at 12:58
• @PM2Ring yes that helps. I wanted to know whether you can SAY that the stationary body is accelerating. As it turns out from both answers it is acceptable. thanx. Some people are highly confused on this page ;). May 7, 2020 at 14:03
• This is essentially a duplicate of my questionhttps://physics.stackexchange.com/q/264128/113699....have a look here for a few answers May 16, 2020 at 16:21

The body on the table is not accelerating because, as you said, the force of the table counteracts the force of gravity. The confusion here is perhaps from the term "acceleration due to gravity" but that doesn't mean that it has to cause an acceleration (at least once a steady state is reached).

An accelerating object can be momentarily at rest. For example, if you throw a ball into the air straight up, at the top of its trajectory it will be at rest.

• "...ball into the air straight up, at the top of its trajectory it will be at rest." but its change in momentum will give you the acceleration. but seemingly not the body on the table problem. May 6, 2020 at 9:08
• It's a question of the net force: $a=F/m$. But you can't just consider gravity. You have to consider all of the forces on the object as $F$. The change in momentum is a consequence of $a$ not of $g$ (directly). May 6, 2020 at 9:12

To elaborate on @rghome answer let’s comparie the situation where a mass is “at rest” on a table versus another mass being “at rest” at the same height at the top of a purely vertical trajectory,

In physics a body is “at rest” to an observer when it is stationary (motionless) in the reference frame of the observer, that is when it’s velocity is zero. The mass on top of the table is at rest in the reference frame of an observer standing still on the floor next to the table. In the same frame let another object of the same mass be launched vertically upward from the floor next to the table so that the top of its trajectory happens to be at the exact same height as the mass on the table. At that instant both masses are at rest (have zero velocity) to the observer.

But at that instant the net force acting on the mass on the table is $$F_{net}=-mg+N=0$$ where $$N$$ is the upward normal force of the table. From Newton’s second law

$$a=\frac{F_{net}}{m}=0$$

The only force acting on the launched mass is the downward force of gravity

$$F_{net}=-mg$$

And it’s acceleration is

$$a=\frac{-mg}{m}=-g$$

Both masses are momentarily at rest but only one is accelerating.

So a body may or may not be accelerating when it is at rest. It depends on whether or not it experiences a net force.

Hope this helps

• no it does not help. you just restated my question in mathematical terms. I think the block on the table IS accelerating otherwise it wont't have a weight equal to $mg$. "Both masses are momentarily at rest but only one is accelerating. So a body may or may not be accelerating when it is at rest. It depends on whether or not it experiences a net force." Both sentences contradict each other. Is the block on the table accelerating or not. May 7, 2020 at 8:30

Consider the case of a mass attached to a simple Hooke's Law spring, resting in an equilibrium situation (horizontally, vertically, on an incline--it doesn't matter). Then, displace the mass from the equilibrium situation collinear with the spring, defining the $$x$$ direction, by the amount $$A$$ and release at t=0. The displacement from equilibrium will be described by $$x=A\cos\Omega t$$ where $$\Omega$$ will be the angular frequency, a constant dependent on the spring constant and the mass.

If you take time derivatives to find the velocity and the acceleration, you will find that when the velocity is zero, the acceleration will be a maximum.

The body on the table is at rest and not accelerating in the reference frame of the table. A person doing a bungee jump next to it will at some point be able to say the body is at rest and accelerating with g. Many more reference frames are possible, inertial and noninertial, each giving a different answer.