How can an object have an acceleration of 9.81m/s^2 when there is ...

Whovian

Well, it has zero velocity. I don't quite understand. (The formulas were saying that "the sum of the forces on an object is mass * acceleration, or the sum of the accelerations depending on if you define acceleration to be the total change in velocity or the result of the force, in which case the total acceleration, or the change in velocity, is just the sum of the accelerations." A little wonky.)

DrGreg

Yes. If a rock falls, there's 2 ways to look at it.

1. The rock is accelerating downwards relative to the ground.
2. The ground is accelerating upwards relative to the rock.

The revolutionary aspect of General Relativity was to base it on (2) instead of (1).

student34

Wow, thanks!

NewtonsHead

It's all about the reference frames.

ehild

At the instant when you release a stone it has zero velocity, but a very short time Δt later the velocity is gΔt. Acceleration is defined as change of velocity over change of time: a =Δv/Δt. From zero to Δt the velocity changed by gΔt, divided by Δt it is g.

Newton's second law says that the acceleration is equal to the resultant force divided by the mass of the object.
The Earth pulls all objects with the force of gravity G=mg where g is about 9.8 m/s^2. If the object is released it accelerates downward: ma=G, a=G/m=g, with acceleration g. If the object is supported, sits on a table for example, the table exerts a force N (normal force) on it that prevents the object from falling: the object does not move, no change of velocity, the acceleration is zero, as the sum of the downward force of gravity and the upward normal force cancel: N-G=ma=0.

ehild

A.T.

Here the 2 ways visualized:

mikemeg

I could be wrong, but I think we're going wayyyyy beyond what the original poster was talking about. A lot of questions are not particularly deftly phrased in textbooks.

"g", commonly referred to as "the acceleration due to gravity" can be used for calculations without the object actually accelerating. You can think of it as "a falling object would accelerate at 9.81 m/s²," and that gives you a sense of the gravitational field. Dimensionally, m/s² is the same as N/kg. So you could equally think of "g=9.81m/s² near the earth's surface" as "there's a gravitational force of 9.81N for every kg of mass near the earth's surface".

In other words: "there's a 5kg rock. The acceleration due to gravity is 9.81" tells you that the weight of the object, mg, is 5kg*9.81m/s² = 49.05 N.

A.T.

I don't think so. In post #5 he clarified that he actually asks about Einstein's General Relativity where objects at rest on the Earth’s surface have a proper acceleration upwards of 1g.

soothsayer

He also clarified that he is in first year physics, so I don't think GR is going to help him understand much.

I'm not sure where the confusion comes from, but I don't think previous posts are really helping a lot.

Student34, can you tell us who or what told you that an object has an acceleration of 9.8 m/s² even if there is no change in velocity? It's just not true. Acceleration is change in velocity.

If the rock is falling, it will be accelerating at 9.8 m/s², but if the rock is resting out the ground, the force from gravity, which normally accelerates things at g, is being balanced out by the force from the ground. You know the gravitational force near Earth's surface for an object is mg, where m is the mass of the object, and g is 9.8 m/s², but the rock is at rest, so acceleration is zero, which means the normal force from the surface of the Earth has to equal exactly mg as well, so there are two forces on the rock that we can calculate using g, but they are in exact opposite directions so there is no net force, no acceleration, no change in velocity, the rock just sits there. Just because we're still evoking g = 9.8 m/s² in a free body diagram of a rock sitting at the Earth's surface, doesn't mean it is ACTUALLY accelerating, we just use the value for g to determine the magnitude of the normal and gravitational forces.

A.T.

He was specificaly asking about Einstein's view on acceleration in post #5
That is coordinate acceleration. But the OP is confused about statements regarding proper acceleration.

arydberg

see my webpage

http://www.maverickexperiments.com/fma/Force.html

soothsayer

I think you're right now. I was thinking back to the thread after I posted this and I think I see now where the confusion arose.

Right, I was only approaching it from a Newtonian perspective.

It seems like the OP had a moment of clarity regarding Einsteinian gravity, so I'll assume the question has been dealt with.

Paramadman

It can't. As phrased the query is meaningless.

Paramadman