What is actually weight?

When a book is kept on the table than gravity of earth is attracting the book downwards and gravity of book attract earth this is action reaction pair

Now here are two more force acting normal reaction force which is pushing the book upwards and an action force pushing the table downwards

So what basically weight is the force that is pushing the table or the force by which earth attract the book (force by which book attract the earth) although all of them will have same magnitude but still?

• The answer to this question depends on who you ask. A lot of people would say the weight of the book is just another way of saying the force of gravity that the Earth exerts on the the book but not everyone - see this lecture by Prof. Walter Lewin. Commented Feb 22, 2022 at 2:21
• I think it depends on whether we mean the actual weight or apparent weight of the object.
– Dodo
Commented Feb 22, 2022 at 10:21

There seem to be at least two schools of thought.

In terms of attraction in a gravitational field
One definition of the weight of a body is the gravitational attraction on a body and so weight equal to $$mg$$ where $$m$$ is the mass of the body and $$g$$ is the gravitational field strength.
With this definition weight varies only in so far as the gravitational field strength, on Earth $$g = 9.8 \,\rm m/s^2$$ approximately, varies.
When you and the scales are accelerating, the scales are reading your apparent weight.
When you are orbiting the Earth you appear to be weightless although you must still have a weight otherwise there would not be a force present causing the centripetal acceleration to keep you in orbit.

In terms of the reading on a spring balance
After watching 8.01x - Lecture 7 - Weight, Weightlessness in Free Fall, Weight in Orbit you are left in no doubt as to Professor Lewin's definition; the weight of an object is the force exerted by an object on some “scales” whether it be a push or a pull.
The object “feels” its weight because of the Newton's third law reaction on it, the force exerted by the scales on the object.
So when you are in a lift and the cable breaks you are weightless (the scales register a zero reading) and you feel weightless (the lift floor is not pushing up on you).
If by some means the lift accelerated downwards with an acceleration greater than $$g$$, the scales, now on the ceiling of the lift, would register a reading and so you have a weight.
The quantity $$mg$$ Professor Lewin never calls weight; he calls it gravity or gravitational attraction or something similar.

Many textbooks on physics say weight is the force acting on the object due to gravity. I strongly disagree.

Weight is what you feel when a contact force counteracts the effects of gravity, that is, the force that pushes the book upwards, so weight is the quantity that is measured by, for example, a spring scale, more generally an accelerometer.

You can't feel gravity, when you are in free fall, you feel weightless, just as if you were in orbit, or standing still in space. This is the equivalence principle. You can only feel weight if there's a normal force holding you, because it's a contact force, and thus can be measured by an accelerometer.

On the other hand, you can feel weight on a thrust accelerated rocket, or on a rotating space station, far from any significant gravity field.

• So why do we feel weighted by hanging on a string as their is no no normal force only tension force on the string not on object Commented Feb 22, 2022 at 19:51
• @VanshikaSejwani, the tension on the string - assuming its fixed on some solid structure, which is in turn fixed on Earth's surface - is a contact force, and thus it conveys proper acceleration against the acceleration of gravity, and thus you feel weight.
– Arc
Commented Feb 23, 2022 at 0:31

You need to distinguish between cause and effect.

The weight of the book is the name we give to the value of the gravitational attractive force between the book and the Earth.

The weight is the cause of the effects you describe in your question.

In the absence of a table, the weight will cause the book to be accelerated towards the centre of the Earth. However, the Earth is solid, so the book lies on the surface of the Earth, compressing the material of the book and the Earth. That compression causes a repulsive force between the book and the Earth which exactly counters the weight so the book lies in equilibrium on the ground.

If you introduce a table, the principles remain the same. The Earth attracts the book and that attraction causes a compression of the material of the table, the book and the Earth. That sets up forces that counter the compression, leaving all three objects in equilibrium.

What is actually weight?

The weight usually means the gravitational pull exerted by a massive body (like a planet) on an object. But you might find different definitions for the weight! More important than formal definition of weight is to understand principle of forces and how to apply Newton's laws motion.

In your example, there are two forces acting on the book: (i) gravitational force exerted by the Earth ($$\vec{w}$$), which pulls the book towards the center of the Earth, and (ii) normal force exerted by the table ($$\vec{n}$$), which acts in the direction perpendicular to the contact surface.

Since the book is at rest (i.e. in equilibrium), the first Newton's law of motion applies, which states that the vector sum of all forces acting on the body equals zero

$$\vec{w} + \vec{n} = 0 \qquad \rightarrow \qquad \boxed{\vec{n} = -\vec{w}}$$

It must be noted that the normal force does not always equal weight! A ramp (inclined plane) is one example in which normal force does not equal weight.

The reaction forces act on different objects and should not be considered when analyzing the book. According to the third Newton's law of motion, each force has its reaction pair, which in your example are: (i) gravitational force exerted by the book ($$\vec{w}^\star$$) on the Earth, which pulls the Earth towards the center of book, and (ii) normal force exerted by the book ($$\vec{n}^\star$$) on the table, which acts in the direction perpendicular to the contact surface. Since action-reaction pairs are equal in magnitude and opposite in direction, it follows

$$\vec{w} = -\vec{w}^\star \qquad \text{and} \qquad \vec{n} = -\vec{n}^\star$$