For example, in a hanging massless rope, or in a massless rope hanging   over a frictionless pulley, the tension must be the same at all  points, because otherwise there would be a net force on at least one tiny piece, and then $F = ma$ would yield an infinite acceleration for    this tiny piece.   What happen if the tension is different at some point?How it yields an infinite acceleration?
122276
Suppose I have an Atwood machine, that is, two different masses connected with an inextensible, massless rope over a pulley. Assuming no friction between the rope and the pulley, the heavier mass will accelerate towards the ground, the lighter mass will accelerate towards the pulley, and the rope will accelerate towards the heavier mass. These three accelerations will be equal in magnitude. But this makes no sense to me. Force causes acceleration. But there is no force acting on the rope. And even if there was, the acceleration of the rope would be infinite because its mass is 0. So why does the rope accelerate? And how can the magnitude of this acceleration be finite?
59540
There are several conditions that must be met in order for a force to be . One of them is that the curl of that force must be equal to zero? What is the physical intuition behind this? If you can, please explain it to me via the magnetic force fields because i have read that time-varying magnetic fields are not conservative because they do not meet that condition.I do not understand this.
97581
Let us consider a homogeneous rope hanging from the ceiling. I will call the vertical direction $x$ and the horizontal displacement $y$. When we apply the second Newton's Law to a portion of mass $\Delta m$ and proceed in the same way we do for a  we get $$\mu\frac{\partial^2 y(x,t)}{\partial t^2}=\frac{\partial}{\partial x}\left[T(x)\frac{\partial y(x,t)}{\partial x}\right].$$ The difference now is that the tension is not constant. Defining $x=0$ at the free end of the rope and orienting it upwards we have the tension $T=\mu gx$. Therefore, $$\frac{\partial^2 y}{\partial t^2}=gx\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}.$$  So, my issue is that I have seen in a couple of physics books () that the wave speed of this rope is simply  $$v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}.$$ In my opinion these books are either:  i) Cheating the students: they know it is wrong but assume it is right just to show some nice features (the speed would increase as the wave goes upwards).  ii) Had misconception assuming that the usual wave equation are still valid.  iii) Is doing some obscure approximation, which is another cheat if you do not reveal it.  My questions are:  1) Is there an expression for the wave speed for the "wave equation" above?  2) Is there an approximation leading to the wave speed given above?
104554
If $r=0$ in the well know equation $F= G\dfrac{m_1\cdot m_2}{r^2}$, it will not follow that the force will be infinite?  May someone please clarify it to me?
114011
Sorry for posting such a simple problem but I'm having trouble understanding this concept.  In the picture below, we have a force, $ \vec F_{app}$ acting on two objects, A and B.  Mass of block A: $ M_a = 4kg $  Mass of block B: $ M_b = 6kg$  The acceleration of the two blocks can be found using F=ma  $$ 20N = 10kg * a $$ $$ a = 2m/s^2 $$  If we find the force Fba on block B from block A we get:  $$ \vec F_{ba} = 6kg * 2m/s^2 $$ $$ \vec F_{ba} = 12N $$  What I am confused about here is why don't we say that the total force $ \vec F_{app} $ applied onto block A is also applied to block B (20N). Is the force somehow distributed into the blocks?
18610
In my textbook, the definition of tension was given that  is the reactive force which exists when string is being stretched at its both end. After it there was a case given that to calculate acceleration and tension produced in string while it's attached with different masses at both its ends and passed over a frictionless pulley. First body is being hung and second is placed over horizontal frictionless surface with relatively smaller mass. It is given in my book that there are three forces acting on the body placed on horizontal smooth surface viz., Weight of body, the reaction force to weight and tension in string pulling body towards pulley's wheel. How can there be tension in string according to the definition given my book however there is no acting force (pulling force) on the body that pulls it away from wheel.  There are different definitions for tension. The one in our book is creating problem however second one, that is, " tension describes the pulling force exerted by each end of a string, cable, chain, or similar one-dimensional continuous object " is al right.  Since acceleration is lowered due to Tension then acceleration produced in the body must be given by, $a=\frac{Net-Force}{m}$ $=&gt;$ $a=\frac{m_1g-T}{m_1}$ since $W_1$ $&gt;$ $T$.  If tension in string is the reactive force to the weights pulling the string apart then according to Sir Isaac Newton, $W_1=W_2=T$.
43138
In the arrangement shown in the figure, the ends P and Q of an inextensible string move downwards with uniform speed $u$, pulleys A and B are fixed. With what speed does the mass M move upwards?    My attempt: I resolved the component of $u$ along the $y-$axis. That is $u\cos\theta+u\cos\theta=2u\cos\theta$. But the answer is $\frac{u}{\cos\theta}$.  My questions  why can’t I simply resolve vector into components and solve ? I’m unable to determine my mistake. What am I missing here?
20274
Not a homework question instead its a doubt in concepts about vectors related to tension force I was learning about vectors and i encountered a problem in breaking vector of tension force t, as I have given all details already please refer to it, I am sorry to write question in image, but it my best possible handwriting , please dont discard it🙂 To explain my issue a little: In lower two diagrams I have drawn vector t in string t Which is pulling from both sides. But when I breaks vector t in separate cases, I gets vectors also in opposite direction(img 2 and img 3) Since single force cant have vectors in all four direction[two in 1 case and 2 in other] I concluded one of my fig 2 or 3 is wrong. Which is wrong as per you? Why do you the option you which you did not chose is wrong? Also if you can theoritically or by intution explain how tension is pulling at both ends I would be grateful tp you.
82301
Intro:  In completing , he presents an experiment (go to 42:44) which leaves me baffled.    Experiment:  (I recommend watching the video; see link above.)   There is a $2$ kg block with 2 identical strings attached to it: one at the top, the other at the bottom. The top string is attached to a "ceiling", and the bottom to a "floor". Professor Lewin "stretches" the system (by pulling on the bottom string) with the block not accelerating. One string snaps.     Prediction:   Initially, the top string has a tension of approximately $20$ N, to counter the force of gravity. The bottom string has no tension at all. Then, when Lewin pulls the bottom string, it gains some tension $n$ N. To counter act the force exerted by the bottom string, the top string exerts now $20 + n$ N. I assume that the string with more force will give out sooner, leading me to conclude that the top string will break.     Results:  (This was conducted by Lewin, not me; see link above.)   Trial 1: Bottom string breaks. Trial 2: Top string breaks. Trial 3: Bottom string breaks.     Additional Notes:  The results don't seem consistent. If I was right, I'd expect all 3 experiments to be right; conversely, if I was wrong, I'd expect all 3 experiments wrong, with one exception: the results are more-less random and one result isn't preferred over the other.    Question:        Why was my prediction incorrect?   Was there a flaw in my logic?   Why were the results inconsistent?
33025
A $ 2.0 kg $ box of cucumber extract is being pulled across a frictionless table by a rope at an angle $ \theta=60° $ (from positive direction of $ x $ axis, we have taken horizontal surface of table as $ x $ axis) The tension in the rope is $ 12N $ and causes the box to slide across the table to the right with an acceleration of $ 3.0 m/s² $ But the direction of net force is along the rope. The direction of acceleration isn't. Why is that?
17307
Suppose there is one rod and a force F is applied in two cases: Case 1: exactly at the middle on center of mass Case 2: at one end of rod   In which case would the center of mass of rod be accelerated more? Or would acceleration of its center of mass be the same in each case? I have worked towards this and found case 2 would impart a greater acceleration. I just want to check my answer and see if I am making a mistake here.(The question just came into my head and I don't have any source to check the answer from) $$My$$ $$solution:$$ In case 1 the rod would undergo a translational motion and C.O.M. would get an acceleration of $F/M$. In case 2 the rod would rotate about its other end and a torque would act which would be equal to $Fl$. Equating this to $I\alpha$ we get $\alpha$ = $3F/Ml$ and since C.O.M. is at $l/2$(rod is of uniform density) acceleration would be equal to $3F/2M$. Is my solution and thinking correct?
59889
I am something of a dilettante in physics, so please forgive me if the answer to this question is painfully obvious. The question is simple, can something that theoretically has no mass exert a force. I have been tossing around this and other similar questions in my head for a while now and have not really found any concrete answers to my inquiry. I am thinking about how light seems to be able to push objects but yet has no mass, however I expanded the question to be more encompassing in hopes of further learning.
88522
Tension, for me, is a tricky thing.  After finishing a related chapter of my book and watching a video, I still can't get a hang of it.  Here is a situation:   My knowledge is that tension, just like normal force, happens in  just the same way, but with the difference of a string attached.  It is understandable that in the picture, the string attached to the hanging mass has a tension directed upward, because it's weight is directed downward. (Action and reaction)  But what intrigues me is that applying the same logic, the horizontal mass is being pulled to the right, it would have been understandable if the tension is directed to the left.   also Can I think of tension as the reaction force when you pull, and normal force when you push?  update If I was to think tension as a reaction force in the first picture for the horizontal mass, then can the frictional force be the action force? (Ehh, I don't think that makes sentence. Friction is always reaction)
7488
As a 8th grader I am still a newbie at science and I have problems understanding the forces that I can not imagine and one of these "forces" is the moment.  We are told that,  $$\text{moment}= Nm$$ but how can a certain force (i.e. $N$) be applied when there is $m$? How does it depend on the length of the considered object?  And how can 1 moment be defined on the basis of that representation?  And what would be the effect if the moment value is increased by the newton part as opposed to the effect when the value is increased by the length part.  Or is the result the same in both cases?
63905
It is true that under equilibrium condition, no matter what pivot point one choose, the resulting net torque will always be zero. I wonder if this principle apply to non-equilibrium scenarios as well, namely, the net torque of the system being the same regardless of choice of pivot point? For example, if we put a sphere on a inclined smooth ramp, we would expect the sphere to slip down the ramp without rolling. However, if we choose the point of contact between the sphere and the surface of the ramp as the pivot point, we will have a torque due to gravity. Apparently this contradicts our assumption. What's the mistake here?
117026
Lets imagine a binary system of two astronauts in space connected to one another via light rope.  The rope is taut and they're spinning round and round with their axis of rotation being the the axis perpendicular to the their centre of mass.  Now, my question is this. Lets say they each let go of the rope; they move off at tangents.  Is angular momentum conserved ? And what is their subsequent motion ?  My hypothesis : They will move off at constant velocity at a tangent to their circular motion. I believe that because they now under the effects of no external force by the rotational analogue of newton's second law (dL/dt = T) their angular momentum will be of equal magnitude but opposite sign.  Does this make sense ?
47567
I don't understand the fact that let's say hypothetically I was pushing a block on a frictionless surface, and I kept applying the same force on the object how it would keep accelerating! Wouldn't there have to be a certain velocity where it would stop getting faster? If not then what is actually acting on the object to cause it to accelerate infinitely?