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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:

enter image description here

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

See link here.

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:

$$F_R = F_G + F_{GČ}$$

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Your confusion stems from not clearly defining your (two) systems.

  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.

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Think of your body like a super strong iron crate, let's call it crate 1. Now the weights are a very heavy crate, called crate 2.

Now let us say you're standing in a hot air balloon and it has enough hot air to carry you, crate 1 into the air. However not enough to carry both crate 1, and crate 2. Crate 1 is on the floor of the hot air balloon and crate 2 is next to it. So the hot air balloon doesn't take off. Now if you stack crate 2 on top of crate 1 would the hot air balloon take off?

If the total force acting on the floor decreases then it would take, if it stays the same it wouldn't.
So if you would say that the force that crate 1 exerts on the floor is only the weight of crate 1 than it would take off, thus we must conclude that:
Crate 1 exerts both the weight of crate 1 AND crate 2 on the floor of the balloon. According to newton every force has an equal and opposite force when the system is in equilibrium. Thus the floor (of the balloon) must exert the force of crate 1 + crate 2.

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