i have seen newton's third law as every action has equal and opposite reactions and they act on different bodies,  my concern is is it only when both the bodies are at rest as while standing on water i m giving a force mg to water and the water should give me back mg keeping me in equilibrium,where m i going wrong.  i understand that its the normal reaction of water that should balance my weight to keep myself in equilibrium but my concern is why the normal reaction is not equal to mg,i understand it very well that as my foot presses water,the water yields,but my query is why all these things happen.
123916
I try to stand on water surface but I just drown. A book resting on the table because of the W(gravitational Force)  and normal force of the table acting on the book oppositely in equal magnitude being canceled out.  When it comes to a fluid (air, water, oil etc)  If I put my two feet on the surface I will simply drown. Why that?  See,  the weight of my body is acting on the surface and according to the third law at that instant there will be an equal and opposite force acting on me from the surface. So should not I be able to stand?  If you are about to say that fluid molecules have surface tension and they are not steady then I would like to say the little portion of weight of any point of my body that pushes the tiny fluid molecule, that tiny part should cancel that little amount of weight. By this each and every point of my body (along with my Centre of gravity) part should have a net 0 force which should make stand on the fluid surface theoretically. But in practical case it doesn't happen. Why that?  I can stand on frozen Lake but can't stand when it is not frozen.
68857
I was confused with this equation like why force is equal to $ma$ and not $mv$? Then I read the same question here on stack exchange, it's answers almost made sense but I still can't visualise the situation where a thing is moving at a constant velocity, would it then have no force? But if I think this scenario in my head, it would definitely have force when it hit something. So I wonder if I am mis-associating the terms in the equation?
46954
If an elephant can apply 250N force and a rabbit can apply 25N force , and if they are pulling each other , then elephant will pull with 250N force and according to newtons third law the rabbit should also pull the elephant with 250N force.then from where do the rabbit got the extra 225N i.e.(250-25=225) force??? can anyone please explain.
84533
I have been researching for a while looking for what to include more than just deriving the equation of Newton’s law of cooling and a website gave me inspiration by stating what if the temperature of the environment is changing (± C). In conclusion, I wasn’t sure if that person carried out the mathematical process in the right way and the only place I could ask for accurate information is here so please let me know what you think about the mathematical process he went for and if it is right or wrong. Thanks a lot. Website:
112045
When a body lies on the surface of the Earth it is under the influence of gravity. The force on the body due to gravity causes it to exert a force on the ground and the normal reaction acts in the opposite direction causing the resultant force on the body to be zero.  However, how can the body exert a force on the ground when it does not have any acceleration? Since force equals mass times acceleration how does a body without acceleration experience a force?
45177
I'm having trouble understanding how a  (not the  that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I don't see how this alignment gives rise to "attraction" or force.
42196
When applying a force outside of the center of mass of the body, the body will get both linear and angular momentum. Right?  Does the linear velocity from this force equal to the linear velocity from the same force, which is applied on the center of mass?  $$v_{\text{center}} = v_{\text{non-center}}\quad\text{? (no vectors)}$$  If it does: How is it that the energy applied to the system is equal? In one we have both rotation and linear movement, and in the other just linear movement.
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Newton's first law states that in an inertial frame, a body at rest continues to be at rest, and a body in constant rectilinear motion continues its motion, unless an external force is applied upon the bodies. My question is how far have these assertions been tested experimentally? More precisely, is there any need to test this statement, or is it taken as a "definition" of sorts?  I assume the body at rest is a trivial case. On the other hand the case of the body in constant motion would be considerably harder to test, since no external force on a body is an idealization. But I am still curious whether there have been any attempts to test the case of the moving body.
98721
Just a clarification question based on an example I read about.      Can normal force do work on an object?   The answer is yes, with an example being a person jumping. The normal force causes work to be done.  However, I'm wondering, is that actually normal force, or is that the reaction force from applying force to the ground? There is a difference between the reaction force and normal force. I'm not sure if technically that example is correct. If it isn't, can someone provide a different one where normal force does do work?
98799
If a force is applied to a body which does not act through its centre of mass, it rotates about its centre of mass and not any other point. Why?
46753
Since Newton's laws are defined for point particles, I'd like to derive some laws of motions for rigid bodies only by considering a rigid body as a system of particles such that the distances from every particle to every other particle doesn't change with time. I think I derived that the force applied on one particle of a rigid body must be the same for every other particle of the rigid body in one dimension by the following:  Consider two particles on a line $P_1$ and $P_2$ both with masses $dm$ and positions $x_1$ and $x_2$. Let's say that a force $F_1$ acts on the particle $P_1$. By Newton's second law we get: $$F_1 = dm\frac{d^2x_1}{dt^2}$$ By the definition of a rigid body, the distance between $P_1$ and $P_2$ doesn't change with time. Define $r$ as this distance ie. $r = x_1 - x_2$. Therefore: $$\frac{dr}{dt} = 0$$ Taking the derivative of both sides we further get that $$\frac{d^2r}{dt^2} = 0$$ $$\frac{d^2(x_1 - x_2)}{dt^2} = 0$$ $$\frac{d^2x_1}{dt^2} = \frac{d^2x_2}{dt^2}$$ By Newton's second law this is the same as: $$\frac{F_1}{dm} = \frac{F_2}{dm}$$ (where $F_2$ is the force acting on $P_2$), and since $dm \ne 0$ finally: $$F_1 = F_2$$  These steps can be done for arbitrary amount of particles, and so we get that in one dimension, if a force is applied on one of the particle of a rigid body, every other particle of the rigid body experiences the same force.  The problem is that I cannot do a similar proof for two dimensions by defining the distance $r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, but I'm sure that it can be done, and that if done torque, moment of inertia and center of mass would arise. Can someone do a similar proof for two dimensions, if it can be done like this at all?
22550
While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
31966
I hear people say that , $\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?  I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.  Derivation:  Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$ Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$ Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$ Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step): $$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$  Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.
124138
Yesterday my friends and I were solving this easy-looking question:  A metal cube is placed in an empty vessel. When water is filled in the vessel  so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube: (A) Will Increase (B) Will Decrease (C) Will remain the same (D) Will become zero  At first we thought the answer to be (B) considering the buoyant force acting opposite to the weight of the cube. But then, a question popped up in my mind: &quot;What causes this buoyant force even if no fluid was actually displaced?&quot; So I referred to the Archimedes' Statement:  Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. — Archimedes of Syracuse  But according to the question, no such fluid was displaced by the block ! So, here are my questions:  What is the basic reason behind buoyancy? Is it the pressure difference in fluids or the 'liquid-displaced' theory?  What must be the answer to the question (That I was solving) ?  Considering the case that you may say- &quot;The reason behind buoyancy is the pressure difference in the fluids&quot;; Will there be no buoyant force applied to the iron cube as there is no fluid layer present below the cube i.e. no pressure difference?   (Considering the iron cube is VERY dense and thus not allowing any fluid layer to push its way below it)  Or you might add another statement to the above question that-&quot;Yes, buoyant force is applied as there is always a fluid layer present below it...&quot; .Please do explain yourself.  If your answer is according to the Principle given by Archimedes of Syracuse , will there be no buoyant force on the cube as the cube didn't actually displace any fluid? Just like the case where you may consider it to be a part of the vessel itself.
737
I'm going to provide my argument for why I think the tension in a rope should be twice the force exerted on either side of it.  First, let's consider a different example. Say, there is a person named A and a block in space. A pushes on the block with a force of 100 N. Then, the block will also push A with a force of 100 N by Newton's third law. Now, consider the case where instead of the block, there is a person B who is also pushing on A with a force of 100 N while A is pushing on him. A will experience a force of 100 N because he pushed on B, AND another 100 N because he is being pushed by B. Hence he will experience a force of 200 N. Similarly, B also experiences 200 N of force.  Now, back to the original problem. There are two people A and B in space with a taut rope (no tension currently) in between them. If only A is pulling and B is not, then I agree that the tension is equal to the force A exerts. This situation (in my opinion) becomes analogous to the above if B is also pulling. So, say both of them pull from either side with a force of 100 N. Then the rope at the end of B will pull B with a force of 100 N (this pull is caused by A). By Newton's third law, the rope will experience a pull of 100 N. But B is also pulling his end of the rope with 100 N. Therefore, the tension should be 200 N. Similarly, the end of the rope at A must pull A with 100 N of force (because B is pulling from the other side) and hence experience a force of 100 N itself by Newton's third law plus another 100 N because A is pulling on the rope.  Apparently, the answer is not this (according to my searching on the web). So, could anyone tell me why this reasoning is wrong? Thanks.  EDIT : So apparently people don't agree with my first example, leave alone the second. This is to the downvoters and the upvoters of the highest-rated answer: You all agree that if only A pushes B with a force of 100 N, then A and B both will get pushed by a force of 100 N in opposite directions, right? Then, in the case where B is also pushing with a force of 100 N, it doesn't make sense that the answer would be exactly the same. It doesn't seem right that no matter what B does, B and A will always experience the same force as they would have if B hadn't applied any force.  EDIT 2 : I'm going to provide here a link to a question that I posted:  According to the answer and the comments there, the reason as to why my first example is incorrect is different to the one provided here. So maybe you should all read the answer and the comments provided by the person and reconsider what you think.
110918
When we drop a ball, it bounces back to the spot where we dropped it, due to the reaction forces exerted on it by the ground. However, if a person falls down (say, if we push them), why don't they come back to their initial position where they started their fall?  According to Newton's 3rd law of motion, to every action there is always an equal but opposite reaction. If we take the example of ball then it comes back with the same force as it falls down. But in the case of a human body, this law seems not to be applicable. Why?
113968
I have a bit of misconception about weight which I want to clarify.   The air pressure is explained as the weight of the air column above our head acting per unit area. But since air is not continuous how can the weight of all the air molecules (above our head) be acting on our head? I mean we would only feel the weight of the molecules near the surface of our head (if not then why not?) but how do we feel the weight of molecules so far away?  I have edited this question and asked  follow up question (so that it remains specific).