Question about something i just read

[Johnny5]

Lets say you're standing still and your weight is 200 lb.

 

The floor is pushing you upwards with a force of 200 lb.

 

Now lets say you push back with a force of 200 lb.

 

Now is the force 400 lb or 200 lb?

 

Vector analysis will help.

 

The earth pulls me down' date=' and creates a reading on a scale which we will call W. If I slowly varied the mass of the earth, I could change the reading of the scale. For example, if I take my scale, and me to the moon, and stand upon it there, the reading on the scale will be less than what it read on the earth. So that W is affected by the mass of what is underneath the scale.

 

The earth pulls things towards its center of mass, via gravitational force.

 

Let the x axis be horizontal, and let the y axis be vertical. So j^ points into the sky, and -j^ points right at the center of mass of the earth.

 

The planet earth pulls my body down, and the harder it pulls, the greater the reading on the scale. This force is being denoted as W. Specifically we have:

 

[math'] \vec W = w_1 \hat i + w_2 \hat j + w_3 \hat k [/math]

 

 

The earth isn't pulling me horizontally in any direction in the tangent plane. Therefore w1=0 and w3 = 0. The only component of W points in the negative j hat direction. Hence we have:

 

[math] \vec W = -w_y \hat j [/math]

 

where [math] w_y [/math] is a positive number.

 

Ok so that covers my weight, which has been handled as a force. Now, the tangent plane is fixed to the earth. There is no acceleration in the -j hat direction, because the surface of the earth is there, and it is pushing me up in just the right amount, to cancel out my weight. Let us call this upwards force the normal force, and denote it by N. We have this now:

 

[math] \vec N = n_1 \hat i + n_2 \hat j + n_3 \hat k [/math]

 

And there are no components of N in the tangent plane. Therefore we have:

 

[math] \vec N = n_2 \hat j [/math]

 

And [math] n_2 [/math] is a positive number.

 

Assume these are the only forces on me. Therefore, the sum of all the forces upon me is W+N:

 

[math] \sum_{i=1}^{i=2} \vec F_i = \vec W + \vec N [/math]

 

And this vector sum is equal to ma, where m is my inertial mass, and a is my acceleration in this reference frame, which is attached to the earth.

 

I am not accelerating, I am at rest in this frame, therefore:

 

 

[math] \vec W + \vec N = 0 [/math]

 

That is:

 

[math] \vec W = -w_y \hat j [/math]

 

+

 

[math] \vec N = n_2 \hat j [/math]

 

is equal to zero, from which it follows that

 

[math] w_y = n_2 [/math]

 

You say that the floor is pushing me upwards with a force of 200lbs. Yes, that is the normal force, the floor is reacting to my weight.

 

My weight is 200 lbs.

 

The net force upon me is zero.

 

You ask if the force is 400 or 200, that isn't a perfectly clear question, but the force is not 400 lbs simply because I am not accelerating in the frame. The net force upon me is zero.

 

My weight is 200 lbs, but that isn't the only force upon me. The other force has the same magnitude, but opposite direction.

 

As for which of the two my sensory perception detects, that is a good question.

[J.C.MacSwell]

Vector analysis will help.

 

The earth pulls me down' date=' and creates a reading on a scale which we will call W. If I slowly varied the mass of the earth, I could change the reading of the scale. For example, if I take my scale, and me to the moon, and stand upon it there, the reading on the scale will be less than what it read on the earth. So that W is affected by the mass of what is underneath the scale.

 

The earth pulls things towards its center of mass, via gravitational force.

 

Let the x axis be horizontal, and let the y axis be vertical. So j^ points into the sky, and -j^ points right at the center of mass of the earth.

 

The planet earth pulls my body down, and the harder it pulls, the greater the reading on the scale. This force is being denoted as W. Specifically we have:

 

[math'] \vec W = w_1 \hat i + w_2 \hat j + w_3 \hat k [/math]

 

 

The earth isn't pulling me horizontally in any direction in the tangent plane. Therefore w1=0 and w3 = 0. The only component of W points in the negative j hat direction. Hence we have:

 

[math] \vec W = -w_y \hat j [/math]

 

where [math] w_y [/math] is a positive number.

 

Ok so that covers my weight, which has been handled as a force. Now, the tangent plane is fixed to the earth. There is no acceleration in the -j hat direction, because the surface of the earth is there, and it is pushing me up in just the right amount, to cancel out my weight. Let us call this upwards force the normal force, and denote it by N. We have this now:

 

[math] \vec N = n_1 \hat i + n_2 \hat j + n_3 \hat k [/math]

 

And there are no components of N in the tangent plane. Therefore we have:

 

[math] \vec N = n_2 \hat j [/math]

 

And [math] n_2 [/math] is a positive number.

 

Assume these are the only forces on me. Therefore, the sum of all the forces upon me is W+N:

 

[math] \sum_{i=1}^{i=2} \vec F_i = \vec W + \vec N [/math]

 

And this vector sum is equal to ma, where m is my inertial mass, and a is my acceleration in this reference frame, which is attached to the earth.

 

I am not accelerating, I am at rest in this frame, therefore:

 

 

[math] \vec W + \vec N = 0 [/math]

 

That is:

 

[math] \vec W = -w_y \hat j [/math]

 

+

 

[math] \vec N = n_2 \hat j [/math]

 

is equal to zero, from which it follows that

 

[math] w_y = n_2 [/math]

 

You say that the floor is pushing me upwards with a force of 200lbs. Yes, that is the normal force, the floor is reacting to my weight.

 

My weight is 200 lbs.

 

The net force upon me is zero.

 

You ask if the force is 400 or 200, that isn't a perfectly clear question, but the force is not 400 lbs simply because I am not accelerating in the frame. The net force upon me is zero.

 

My weight is 200 lbs, but that isn't the only force upon me. The other force has the same magnitude, but opposite direction.

 

As for which of the two my sensory perception detects, that is a good question.

 

OK, let's get back to the point. Out in space if you push against me with F then I will push back with F and there will be no 2F resultant force on either one of us. If a bar is in between us and the bar does not move it works the same way. You push with F, I push back in the opposite direction with F and cancels it out for the bar . You are accelerated one way at F=ma and I am accelerated the opposite at F=ma.

[Johnny5]

OK, let's get back to the point. Out in space if you push against me with F then I will push back with F and there will be no 2F resultant force on either one of us. If a bar is in between us and the bar does not move it works the same way. You push with F, I push back in the opposite direction with F and cancels it out for the bar [/b']. You are accelerated one way at F=ma and I am accelerated the opposite at F=ma.

 

Ok, I think I know what the source of confusion is.

 

Let us suppose that I am Ivan Drago, and you are Rocky Balboa.

There was this machine in Rocky III that could measure the strength of Ivan Drago's punch. We could discuss energy, force, power, let's deal with impulse. Instantaneous force. Let this be F1 for me, and F3 for you.

 

Suppose there is an inert brick wall somewhere. I can punch it, and it will barely accelerate, for all intent and purpose, the wall can be treated as infinitely massive.

 

Let us suppose I punch this wall. The moment my hand is fully extended, contact is made. There is a force F1 upon that wall.

 

However, I will also hurt my hand.

The force of the wall upon my hand is F2.

 

These forces have equal magnitude, opposite direction, so F1=F2 but:

 

[math] \vec F1 = - \vec F2 [/math]

 

 

Now suppose this takes place in space, replace the wall with you.

 

In the case where you do not simultaneously punch my fist with your fist, I will exert F1 upon you, and you will exert F2 upon me, and F2=F1.

 

But, if you also simultaneously punch, so that both our fists meet at a common point in space at the same time, when both our arms are fully extended, and the maximum force you can punch with is F3, then the magnitude of the net force upon me at the moment our fists contact each other is given by:

 

F3+F2

 

or F3+ F1

 

Since F1=F2, as was stated earlier.

 

Since you can punch just as hard as me... F3=F2=F1

 

The net force on me therefore is

 

F3+F2 = F2 + F2 = 2 F2 = 2F1

 

That's where the factor of two comes in. You said that we both push each other. Is this clear?

[swansont]

No. Newton's third law dictates that the force you exert must equal the force exerted on you. You can't exert F and feel 2F back.

[Johnny5]

No. Newton's third law dictates that the force you exert must equal the force exerted on you. You can't exert F and feel 2F back.

 

That's not what I am saying. I tried to make it clear. The net force on me is 2F, and the net force on him is also 2F.

 

My maximum punch can deliver F.

His maximum punch can deliver F.

 

All I am saying is that if we punch each other simultaneously, the net force upon either of us is given by F+F=2F.

 

If only one of us punches the other, then we each have an external force of magnitude F applied on us.

[J.C.MacSwell]

That's not what I am saying. I tried to make it clear. The net force on me is 2F' date=' and the net force on him is also 2F.

 

My maximum punch can deliver F.

His maximum punch can deliver F.

 

All I am saying is that if we punch each other simultaneously, the net force upon either of us is given by F+F=2F.

 

If only one of us punches the other, then we each have an external force of magnitude F applied on us.[/quote']

 

OK, here is the problem. If you punch me then in laymans terms you are the one using force. This is not the correct use of the term in physics. (or engineering)

 

So let's be very specific. In physics you cannot exert a force on me without me exerting an equal but opposite force on you. If you want to add "additional" force to make it 2F thats fine but it also has to come from both you and I.

 

So that's 2F from you and an equal but opposite 2F from me. Not F from me and F from you.

[Johnny5]

Ok well we have drifted off the topic. Originally, my question was, is the verbal formulation of Newton's second law (at the site I quoted) correct. Then eventually we progressed to the analysis of a rigid bar, where there was a force applied to one end of the bar, and I think I wanted to analyze the motion of the whole bar.

 

Then you said all of the force goes into translating the center of mass, and all of the force goes into rotating the object. I said that violates conservation of energy, and that some of the force goes into translating the center of mass, and some of the force goes into rotating the bar. I do not insist that I am right, I'm not sure.

 

So which is right?