# [Physics] Lifting a weight and reaction force from ground

forcesfree-body-diagramnewtonian-mechanics

If I am standing on my own without holding anything, the reaction force provided from the ground is equal to my weight.

This makes sense to me as this would be the reason why I am not falling through the ground at an acceleration of $9.81 ms^{-2}$.

Now, if I lift a mass as shown below:

according to this website, magnitude of the vertical component of the reaction force is the sum of the masses and my own weight.

I'm confused here. I thought the reaction force from my hands is upwards but the weight of the masses is acting on my hands (Newton's Third law) so how is it that:

$$F_R = F_G + F_{GČ}$$

1. For the Weights:

There is one system in which you would analyze the dynamics of only weights in your hand, and then consider all the external forces acting upon it. In this scenario, there is gravitational force and the reaction force from your hands. That's it.

1. For yourself plus the weights:

In this system, you will now consider your own dynamics and look at only the external forces acting upon yourself. As again, there is gravity! But, note that gravity does not see you or the weight in isolation. Hence, $F_G = g (m_{body} + m_{weights})$, is the gravitational force. And then you are still standing on ground, which means there is a reaction force.

So you see, that equation is for system 2, and your doubt is in system 1.

I think the key-point is that in rigid-body dynamics, you ignore internal forces (since the body is rigid) and look at the dynamics governed by all external forces.