Clarifications regarding frame of reference.

• sankalpmittal
In summary: This 0.0339 Newton net force is small compared to the force due to gravity, which for a 1 kg book is about 9.81 Newtons. That it is small why we can ignore the rotation of the Earth -- at least in some problems.In summary, the conversation discusses the concept of inertial frames of reference and how the acceleration of a book resting on a table with respect to the Earth is not zero due to the Earth's rotation. It also mentions other forces that act on the book, such as the buoyant force and the gravitational force from other objects in the universe. The overall summary is that the book is undergoing uniform circular motion and the net force on it
sankalpmittal
I have just completed my class 10th. I was presently reading the class 11th H.C Verma physics textbook and here is what I confronted which seems to me as if I am unable to fathom :

From textbook : The book is at rest with respect to Earth - when the book is kept on the stationary table. The a of book with respect to Earth is 0. The forces acting on book are :
(i.) The gravitational force W exerted by Earth on book
(ii.) the contact force N by table

Is sum of W and N 0 ? A very accurate measurement will give the answer "NO " ! The sum of two forces ain't 0 although book is at rest w.r.t. earth. Earth is not strictly an inertial frame. However sum of W and N is not too different from zero and we can say that Earth can be assumed as an inertial frame of reference to a good approximation. Thus for routine affairs , "a = 0 if and only if F=0" is true in Earth frame of reference. This fact was identified and formulated by Newton i.e. Newton's first law of motion. If we restrict that all measurement will be made from the Earth frame , indeed it becomes a law. If we try to universalize this to different frames , it becomes a definition. We shall assume that unless stated otherwise we are working from an inertial frame of reference.

Now I can't understand that how come acceleration of book with respect to Earth isn't 0 ?! Please see bold part of quote above. Also please provide an intuitive understanding of frame of reference. Also , please explain the quote a bit more comprehensively.

Thanks in advace.

:)

as you look at the book on the table (dr Suess book?) does it move? if no then the sum total of forces on the book are zero by virtue of f=ma when a=0 then f=0

sankalpmittal said:
Now I can't understand that how come acceleration of book with respect to Earth isn't 0 ?! Please see bold part of quote above.
The bolded parts of that quote do not say that the acceleration of the book with respect to the Earth is zero. Here's the pertinent part of that quote from the text:
The forces acting on book are :
(i.) The gravitational force W exerted by Earth on book
(ii.) the contact force N by table

Is sum of W and N 0 ? A very accurate measurement will give the answer "NO " ! The sum of two forces ain't 0 although book is at rest w.r.t. earth. Earth is not strictly an inertial frame.

(Aside: I truly doubt the book says "The sum of two forces ain't 0")There are other forces acting on the book. For one, there's the buoyant force exerted by the atmosphere on the book. There's also the gravitational force exerted by the Sun (and by the Moon, and Jupiter, and every other object in the universe) on the book.

To keep it simple, I'll ignore these other forces. Assume the only objects of interest are the Earth, the table which is resting solidly on the Earth, and the book which is resting solidly on the table.

Key question: What is the book doing from the perspective of an inertial observer?The answer is that the book is undergoing uniform circular motion. The Earth is rotating on it's axis, one revolution per day. Suppose the book has a mass of 1 kg and suppose the table is at the equator at sea level, 6378 kilometers from the center of the Earth. The book is undergoing uniform circular motion with an angular velocity of one revolution per day (one revolution per sidereal day to be pedantic). Calling this angular velocity ω, $\omega = \frac{2\pi}{\text{sidereal day}} \approx 72.92\,\text{microradians}/\text{second}$, the acceleration of the book as observed by our inertial observer is
$$a = \frac{v^2} r = r\omega^2 \approx 0.0339\,\text{meters}/\text{second}^2$$
This means the net force on this 1 kg book is this about 0.0339 Newtons -- not zero. It can't be zero. If it was zero the book wouldn't be undergoing uniform circular motion. This 0.0339 Newton net force is small compared to the force due to gravity, which for a 1 kg book is about 9.81 Newtons. That it is small why we can ignore the rotation of the Earth -- at least in some problems.

D H said:
The bolded parts of that quote do not say that the acceleration of the book with respect to the Earth is zero. Here's the pertinent part of that quote from the text:(Aside: I truly doubt the book says "The sum of two forces ain't 0")There are other forces acting on the book. For one, there's the buoyant force exerted by the atmosphere on the book. There's also the gravitational force exerted by the Sun (and by the Moon, and Jupiter, and every other object in the universe) on the book.

To keep it simple, I'll ignore these other forces. Assume the only objects of interest are the Earth, the table which is resting solidly on the Earth, and the book which is resting solidly on the table.

Key question: What is the book doing from the perspective of an inertial observer?The answer is that the book is undergoing uniform circular motion. The Earth is rotating on it's axis, one revolution per day. Suppose the book has a mass of 1 kg and suppose the table is at the equator at sea level, 6378 kilometers from the center of the Earth. The book is undergoing uniform circular motion with an angular velocity of one revolution per day (one revolution per sidereal day to be pedantic). Calling this angular velocity ω, $\omega = \frac{2\pi}{\text{sidereal day}} \approx 72.92\,\text{microradians}/\text{second}$, the acceleration of the book as observed by our inertial observer is
$$a = \frac{v^2} r = r\omega^2 \approx 0.0339\,\text{meters}/\text{second}^2$$
This means the net force on this 1 kg book is this about 0.0339 Newtons -- not zero. It can't be zero. If it was zero the book wouldn't be undergoing uniform circular motion. This 0.0339 Newton net force is small compared to the force due to gravity, which for a 1 kg book is about 9.81 Newtons. That it is small why we can ignore the rotation of the Earth -- at least in some problems.

Very well explained ! Thanks.

That's really mind-bashing of the book's author. Well he mustn't say that acceleration is not zero. I am not saying that he is wrong , of course he is cent percent correct. But still he must have given some intuitive idea of the word "why ?" regarding this discussion.

1. What is a frame of reference?

A frame of reference is the set of coordinates used to describe the position, motion, and orientation of an object. It is the perspective from which observations and measurements are made.

2. Why is it important to specify a frame of reference?

Specifying a frame of reference is important because it allows for consistent and accurate measurements and observations. Without a defined frame of reference, it is difficult to compare data or make predictions about the behavior of objects.

3. How do you choose a frame of reference?

The choice of a frame of reference depends on the situation and the purpose of the observation or measurement. It should be a stationary point or object that is easily identifiable and does not move with respect to the objects being observed.

4. Can a frame of reference change?

Yes, a frame of reference can change depending on the observer's perspective or the movement of the objects being observed. For example, if an object is moving relative to one frame of reference, it may be at rest in another frame of reference.

5. What are some common frames of reference used in science?

Some common frames of reference used in science include the Earth's surface, the center of mass of an object, and the origin of a coordinate system. Other frames of reference may be used depending on the specific field of study and the type of motion being observed or measured.

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