How Does Newton's Third Law Apply to Friction and Air Resistance?

[Lim Jan-Wei]

Homework Statement

Homework Equations

principle of moments
Newton's 3rd law

The Attempt at a Solution

Okay. I am not asking how to solve this question. I just have a little misunderstanding of the concept of the Newton's Third Law. I will get straight to the point. For example, in this question, should the reaction from the peg C which is perpendicular to the rod be equals to the sum of forces acting on the rod ( which includes friction ) resolved in the direction of the reaction or is it just equal to the force acting at the point of contact at C ( aka R=W sin θ ) Equilibrium of forces tells me that the former is true but intuitively, I do not see how the frictional force acting on the rod near the ground can affect or can be transferred to the point of contact between the peg and the rod. Can someone clearly enlighten me and help me clear this confusion?

And...this may be a little stupid to ask please bear with me. I have been wondering why I do not experience a large pain on my palm when I push the air around my surroundings. Well due to Newton's third Law, a force that I exert on the air molecules should conjure an equal reaction force from the air molecules on my palm right? Or is it not? And if the former is true, since the force acting on my palm is equal to the force I conjure and since there are millions of air molecules I act upon, shouldn't I feel an extremely large force acting on my hand? Uhmmmm...well I am confused, so can someone enlighten me on this topic? Thanks!

 

[William White]

Your body has evolved to not feel pain when acted on by forces and pressures felt in everyday life; so waving your hand around is not going to cause enough stimulus in your nerves to be sensed (waving your hand is not generating a large force) Your senses are simply not attuned to the forces felt in everyday life (though there are some very unfortunate people with terrible disorders that cause normal stimulus to generate unbearable pain)

If you stick your hand out of the window of a fast moving car and wave it around, you will feel the air. If the air is full of moisture, you will definitely feel that! Walk up a mountain in Scotland in the dreech weather, and the water molecules and ice particles hitting your face at over 100 mph can feel like razor blades and very uncomfortable - and make it very difficult to stand up! I've been on skis in Scotland and been blown uphill (happens at about 70mph winds)

 

[fireflies]

Resultant of parallel force on a rigid body may help

 

[fireflies]

The mg at C will divide into mg cos theta against the normal of the plane, and the other component mg sin theta will be acting along the plane. That is along CB

This force will be causing the rod to slip down until it crosses the static limiting value of frictional force. We know, frictional force acts against the force applied. So, here frictional force minimizes the mg cos theta (until it is less than the maximum value of frictional force. If it crossed the maximum value then it will start to move and slide down)

 

[Chestermiller]

Have you drawn a free body diagram of the rod showing the forces acting on it, including the normal force perpendicular to the ground and the tangential frictional force acting parallel to the ground? Please write down the force balance equation that this diagram leads to for the component in the vertical direction.

Chet

 

[fireflies]

mg sin theta against normal and mg cos theta along the plane actually, since the theta is with vertical line. I mistook it

 

[fireflies]

I was trying to solve out the question. I solved it out, but a question came to my mind. Can I ask the question here?

 

[Chestermiller]

Hi fireflies,

Actually, I was responding to the OP's original question, rather than to your response. But, if you have a question that is closely related to the topic of this thread, please feel free to ask. On the other hand, if your question may be somewhat off-topic, I suggest starting a new thread.

As things stand now, the OP has yet to respond to anyone. Hopefully he will.

Chet

 

[fireflies]

Thanks. Well, it is quite close.

While I divided the forces into components, I divided the weight at C. Then one of the components went through B. I divided it again at B, (into parallel and against the normal).
Well, I knew that weight acts along centre of gravity. But I used to think that it can be divided into components at any particle of the object. But while solving this, now I think that it can't be. The full weight can be divided into conponents only at the centre of gravity. Is it right?

 

[fireflies]

And if it is the case, then what difference is it going to make if the peg
was not at C (centre of gravity here), rather a bit upward or downward the rod?

 

[William White]

The rod would fall over.

To stop the rod falling over you have to change the angle of the rodTry it with a ruler resting against your finger. move your finger up and down - the ruler falls. But if you change the angle of the ruler, you can again balance it with your finger at the position.

 

[fireflies]

The rod would fall over.

To stop the rod falling over you have to change the angle of the rodTry it with a ruler resting against your finger. move your finger up and down - the ruler falls. But if you change the angle of the ruler, you can again balance it with your finger at the position.

Is it an answer to my question?

That means, by anyhow I need to make the peg's position at the centre of the gravity?

 

[fireflies]

Well, I tried with my finger and a book. Keeping the sameangle if I take the finger upwards then it does not fall, but if I take my finger downwards, at a time it falls. Like a ruler, we can keep it with a finger at top, right? Certainly the centre of gravity is not at the top.

 

[William White]

is the centre of gravity between your finger and the ruler, or 'outside' the peg and the ruler?if the centre of gravity is 'outside' where the rod is balanced, it will move, like a see-saw.

http://images.rapgenius.com/7c47bafebdf1d95c9269fc7a4b98c769.329x335x1.jpg

the skinny man could balance the fat man by moving the 'peg'

 

[fireflies]

Ruler means the upper point or lower point?

When the centre of gravity is between the upper point and finger then it alls like see-saw. I understand the reason. There force is downwards. And when the centre of gravity is downwards the finger (near to the ground) it stands

 

[fireflies]

Well, more questions are buzzing in my head. I better open up a new thread

 

[fireflies]

http://images.rapgenius.com/7c47bafebdf1d95c9269fc7a4b98c769.329x335x1.jpg

the skinny man could balance the fat man by moving the 'peg'

It acts like a lever then?

 

[Chestermiller]

Thanks. Well, it is quite close.

While I divided the forces into components, I divided the weight at C. Then one of the components went through B. I divided it again at B, (into parallel and against the normal).
Well, I knew that weight acts along centre of gravity. But I used to think that it can be divided into components at any particle of the object. But while solving this, now I think that it can't be. The full weight can be divided into conponents only at the centre of gravity. Is it right?

If you are taking moments, which is required in this problem to establish the reaction force at C, then the effective location where the gravitational force on the rod is acting matters. If the problem could be solved without taking moments, then this would not be necessary.

Chet

 

[Chestermiller]

So far on this problem, I haven't seen one single free body diagram for the rod from anyone. On such a free body diagram, I haven't seen one indication of the forces drawn. No wonder so much difficulty is being encountered.

Chet