Clarifications regarding frame of reference.

AI Thread Summary
The discussion centers on the concept of frame of reference in physics, particularly regarding a book at rest on a table. While the book appears stationary relative to Earth, the forces acting on it, including gravitational and contact forces, do not sum to zero due to Earth's rotation, which means the book experiences a small net force and acceleration. This challenges the notion that the book's acceleration with respect to Earth is zero, as it is actually undergoing uniform circular motion. The conversation highlights the importance of understanding inertial frames and the implications of Earth's non-inertial nature. Overall, the discussion emphasizes the need for clarity in explaining these concepts in physics education.
sankalpmittal
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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.

:)
 
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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.
:smile:

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.
 
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