External work in raising a stone

Question

Suppose a person is raising a stone upwards against the force of gravity. The common methodology that is followed is this: Since gravity pulls by $mg$ force,the person has to put $mg$ force in opposite direction to make the stone move with constant velocity making the external work of agent $mgh$. But here is a doubt of mine. Since a person is holding the stone,there must be a normal reaction between the palm of the hand and stone. Now, the forces acting on stone are gravity $mg$,normal reaction $N$ and force of agent $F$,so the required equation should be $N+F=mg$. I don't know why normal reaction is taken into account in normal methodology. Or i am being wrong somewhere?

Answer 1

The normal force between the hand and the stone is the same thing as the "force of agent".

"Normal force" is just a blanket term that applies to the component of the force of interaction between two surfaces that is perpendicular to the surfaces at the point of contact. So the "force if agent" in your case is a normal force, but putting both $N$ and $F$ is double counting the single force the person exerts on the rock.

Answer 2

@Biophysicist is right. I am adding some more points to clarify your doubt. The basic concept is that the Normal Reaction force is electromagnetic in nature, i.e., it arises due to interaction at the atomic/subatomic level. It can not simply be calculated by any predefined equation like $\frac{kq_1q_2}{r^2}$ unlike gravitational force. Gravitational force can be calculated using a predefined equation like $\frac{Gm_1m_2}{r^2}$. The $mg$ is actually $m(\frac{GM_{earth}}{R_{earth}^2})$. I want to clarify one misconception that $N = mg$ because of action reaction pair. It is not true. The hidden piece of information in Newton's 3rd law is that the action and reaction forces must be of same nature. Electromagnetic force and gravitational force are not of same nature. Hence $N$ and weight are not action reaction pairs. Rather, the force exerted by earth on stone and the force exerted by stone on earth are action reaction pairs. $N$ is named normal reaction because it is due to contact force between two surfaces. So, contact force exerted by stone on ground and normal reaction exerted by ground on stone are action reaction pairs. Now, coming back to calculating $N$. $N$ can be calculated by observing the state of the object. When the stone is kept on the ground, we can observe that it is in rest. So, we can tell "no matter what be the complex molecular/atomic interactions, as I can see the object at rest, the net external force must be $0$ ". The external forces are $N$ (normal reaction on stone from ground) and $mg$ (force exerted on stone by earth). So $N - mg = m(0) = 0$ or $N = mg$. When we lift the stone, we can similarly tell "I don't care about the complicated interactions at atomic level, as I can see the object at motion, the net force must be according to Newton's 2nd law". So, $N - mg = ma$ or $N = m(g+a)$. The term used for this kind of behaviour of $N$ is "self-adjusting" nature. To understand this imagine you and your friend sitting on a bed. Suppose your weight is $50 kgwt$ and that of your friend is $70 kgwt$. Both of you are at rest. If $N$ would be $70$ for you then you would bounce but that does not happen. That's self adjusting nature.

Answer 3

1. What you are saying seems correct to me that there is already the Normal force with the same magnitude as the gravitational force mg on the particle/mass you are lifting. BUT it is only right till the time you hold it. It is justified as long as you hold it(just hold it still in one place, and the mass is not moving). Now, since the acceleration is zero, the only forces acting on the mass you are holding are normal and gravitational.

Newton's second and third law satisfied (action-reaction pair and the net force is 0 since net acceleration while only holding is 0)

2. When you are trying to lift the mass, you need to accelerate it from its rest position in your hand, it has some inertia(inherent property of mass). SO, we need to apply a force on it, which you termed "extra" apart from the normal reaction(which you denoted as F)

In all, the action-reaction pair cancel out each other whether you are holding your mass or accelerating it.

But, there is the force you apply to it to change its momentum. Let that force be F(which you said extra-to lift the mass-to accelerate it from rest resulting from action-reaction)

NOTE: You are applying a net force of N+Force(to accelerate from rest)>mg if F>0. So, there is a net force.

The equation will be: N+F-mg=ma(net)

N+F=m(g+a)

Since N=mg

F=ma(net)

So, your N component is resolved by the mg.

Also, in your equation,(which is wrong), N+F=mg, you are stating that if F is not equal to 0, then N is less than mg, but N and mg are action-reaction pairs.

It wont sink in at once-read it again, Good Day!

EDIT 1: See to make an object move at a constant velocity, it first needs to accelerate to that velocity in some earlier time, then the net force =0 can be done via the intervention of you.

EDIT 2: As S Das pointed out, I would first like to say that I know what I had done. I did it just to make this thing simpler for the questioner. I have put it in the comments section that N and mg are not action-reaction pair.