A ball on a stationary frictionless wedge would naturally roll down because of the component of its weight along the plane. But I read if the wedge is accelerated fast enough it would prevent the ball from doing so. For this to be true there must be a new force that acts on the ball and whose magnitude will depend on the acceleration of the wedge (since a certain acceleration must be reached). But I'm having trouble comprehending this new force. Is it the 'reaction' force to that of causing the acceleration or is it simply inertia and the ball is just trying to be the way it was? Or is it something else entirely?  It also said that with enough acceleration the ball would slide up the ramp and fall over. I can visualize this but can't seem to explain it.
120593
If I increase the force $F$, only the normal force $N1$ acting on $m1$ would increase which has no component along the plane, ie. along $m1gsin\theta$, so how would applying this force prevent m1 from sliding?  When I view it from the accelerated frame of the wedge,   It begins making sense. How is this possible? I'm really confused.
32545
I know this may seem a weird question, but it always bothers me. My physics book (Resnick,Halliday,Walker), and also various sites never say anything beyond acceleration.   But when a moving body is being acted by a variable force , its acceleration will definitely change: it will either increase or decrease. Then there will be rate of change of acceleration with respect to time. So, why don't books mention this? What is the cause for not measuring $\frac{d\vec{a}}{dt}$ ? If it exists, what is the use of it?
104570
In most of the mechanics textbooks , the  is stated without any justification or an experiment to confirm this law. Also after introducing this law forces are also decomposed without any justification. Please suggest any links or books related to this. Any help will be appreciated. Thanks in advance.
7099
Does the force of kinetic friction increase with the relative speed of the objects involved?    I have heard and read that the answer is no.  This is counter intuitive, and is a big part of why the "Airplane on a Treadmill" question is so interesting.  What phenomena are at work with kinetic friction, and why does it not increase with relative speed?
68035
Assume we have a ball with diameter 1 meter and mass 1 kg rotating at 99,5% speed of light (Suppose I'm an observer placed above the ball, I see under me a disk wich external  speed is 99,5% the speed of light)  Then we accelerate such ball to 99.5% of speed of light, to make clear from wich position we look at the ball when it's moving so fast I did a picture (there's a black ring around the ball, as you se the rotational axis is vertical to our observer)    I'm interested in a full explanation of what we observe (measured rotation speed, color etc.) and why that happens, as far as I know we should not measure something moving faster than light but I'm not sure how relativity constraints would alter the "measured state" of the ball as long as we accelerate it.  What we see change while the ball accelerates?  EDIT:  A precisation, I assume the ball is spherical when I start rotating it before even moving, this is because commenters made me to note that shape may change.
57331
Consider there is a box of mass $m$ at rest on the floor. Most books give an example that we need to do a work of $mgh$ to lift the box $h$ upward.  If we analyze this work done, the external force acting on the box by us should be equal to the weight of the box. Therefore the net force is zero which in turn there is no acceleration. If there is no acceleration and the initial velocity of the box is also zero, how can the box move upward?
25817
I know that this machine does not work, via thermodynamics. I am asking for an analysis in terms of mechanics and magnetism.  Anyway, so here is the machine:   (source: )    The magnet (the red ball) pull the ball up the ramp, and then it drops the ball through the hole, which then rolls down, and go ups the ramp again. Thermodynamics shows that this can not work.  From a mechanics and magnetism perspective, what happens when you do this, and why can't it happen?  I have a source saying that it would work if it were frictionless and we didn't try to extract energy from it (), so in a certain sense, it is very close to possible (they didn't provide a full analysis though.)  Another Image:
67721
The coefficient of kinetic friction can be calculated using $\mu_k = F_k/F_n$. A change in angle does not affect the normal force, but doesn't a change in angle affect the friction? As in if the angle is steeper, wouldn't the object have a greater propensity to move down the slope, thus there will be less friction?
70064
I'm having trouble understand the concepts for this problem. Here's the problem:     A block is placed on a plane inclined at angle $\theta$. The coefficient of friction between the block and the plane is $\mu = \tan \theta$. The block is given a kick so that it initially moves with speed $V $horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?   We can see that $F_x = mg\sin \theta - mg\cos (\theta )\mu = 0$, if we gave the block an initial velocity along the plane, it should be subjected to a friction force opposite its velocity, where $f = mg \cos \theta \tan \theta = mg\sin \theta$   This however, is the answer my book gave:     The normal force from the plane is $N = mg \cos \mu$. Therefore, the friction force on   the block is $N = (\tan \mu)N = mg \sin \mu$. This force acts in the direction opposite to   the motion. The block also feels the gravitational force of $mg \sin \theta$ pointing down the   plane.   Because the magnitudes of the friction force and the gravitational force along the   plane are equal, the acceleration along the direction of motion equals the negative   of the acceleration in the direction down the plane. Therefore, in a small increment   of time, the speed that the block loses along its direction of motion exactly equals   the speed that it gains in the direction down the plane. Letting v be the speed of   the block, and letting vy be the component of the velocity in the direction down the   plane, we therefore have $$v + vy = C$$   where $C$ is a constant. $C$ is given by its initial value, which is $V + 0 = V$ . The final value of $C$ is $V_f + V_f = 2V_f$ (where $V_f$ is the final speed of the block), because the   block is essentially moving straight down the plane after a very long time. Therefore,   $2V_f = V \rightarrow V_f = \frac{V}{2}$   I don't really understand what this is trying to say. The bolded part is what's giving me problems. I think it just may be poorly written, I think I need a clearer explanation.
109368
So, in my physics textbook it's written that the kinetic energy possessed by a body is equal to the amount of work done by an opposing force to stop the body, now that got me thinking that when a ball hits a wall, the wall which is the opposing force does no work  because there is no displacement, so to say isn't the kinetic energy possessed by the ball zero?
93513
I can't seem to find a specific answer to this anywhere.  I understand that in a rocket there is a chemical reaction that causes gas particles to leave the rocket at high velocity. By Newton's third law, and the conservation of momentum, this caused the rocket to be propelled.   What is missing is a physical explaination of what exactly causes this force on the rocket, as most answers annoyingly miss this final bit!  People often use a skateboard-bowling ball analogy. However, in this analogy the person throws the ball and the ball provides an equal and opposite force on the person as it is thrown. The problem is, rocket doesn't 'throw' the exhaust out as it directly doesn't accelerate the particles - this is a result of a reaction. So what exactly causes the force on the rocket itself? Is it the case that in the explosion some gas particles collide with the rocket base, and the nossle is designed as to maximise collisions that will provide an upward force?   I have seen people say this is wrong, or suggest it is right. Wikipedia says:   'About half of the rocket engine's thrust comes from the unbalanced pressures inside the combustion chamber, and the rest comes from the pressures acting against the inside of the nozzle'  I want to be able to understand this in terms of particle collisions. I know there is a force due to the physical laws but people don't seem interested in the mechanics of the force itself.   Back to the analogy: if someone where to throw a bowling ball over your skateboard, you wouldn't move, just as particles leaving a rocket, without collision of any kind, wouldn't cause the rocket to move (I know that's not possible, but hopefully makes my point). Hope this question makes sense.
79613
In school we solve a lot of physics problems involving rolling cylinders/hoops/wheels. Often it is specified in the question statement that "rolling can be assumed to occur without slipping". Then we know we can use the relationship between translational acceleration and rotational acceleration: $a = \alpha r$.  But can we know from a particular set up when the object can be assumed not to slip when rolling? Assume we know the usual macroscopic details of the set up such as the mass and radius of the rolling object, the angle of the plane, friction coefficients etc.. without deep insight into specifics of the materials of the wheel and the plane.
27801
Let's say that we have a car travelling at a constant velocity (V) on a flat plane , let's assume that it's applied force or the force applied from its engine ( Fapp ) balances the frictional force ( Ffrict ) , hence the power due to Fapp is equal to Fapp×V , now let's assume that the same car now moves up an incline with the same engine power as previously stated and at a constant velocity ...but this time it will travel at a much lower constant velocity since the force which the engine exerts is much higher now in order to balance Fparalell and Ffrict , but what I don't understand is what caused this reduction in velocity , if the engine applies an equal force to that of the counteracting forces shouldn't that mean that there is no net force hence no change in velocity ,also if p=FV and if we now apply a greater force for the same velocity V why does the power increase , I know the answer can be simply explained using mathematics but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time ?
501
Suppose a wheel is rolling smoothly on a horizontal plane i.e., it is rolling without slipping.  Now let's take the two cases of the horizontal plane:   It has friction It is frictionless   In the first case, as of my knowledge friction doesn't come into play when there is no slipping. So, will the wheel roll forever? If it comes to rest is the answer friction? If not what brings it to rest?  In the second case, there is no friction. So, here also will the wheel roll forever? And if it comes to rest what brings it to rest?
65676
Consider a cylinder that rolls without sliding on an inclined plane. If it's placed at the top of the plane, with its center of mass at a height $h$ from the bottom, it will have a potential energy $mgh$ (considering the bottom of the plane as the zero point). Then, the cylinder will start rolling down the plane, as its weight does work on it, causing its potential energy to be transformed into rotational and translational kinetic energy. I'm confused about the origin of the rotational kinetic energy of the cylinder. Since the static friction and the normal force are applied at the point of contact with the plane, neither of them can do work on the cylinder (which is why its mechanical energy is conserved), since the cylinder is, by constraint, not sliding. This seems to imply that the cylinder's rotational kinetic energy comes from the work done by its weight. However, I don't understand how this happens, since a rigid body's weight can be seen as being applied on its center of mass, which means it can exert no torque on it. Instead, the only torque exerted with respect to its center of mass comes from static friction. Moreover, if the inclined plane was frictionless, there would be no torque exerted on the cylinder, so it would not rotate. Yet, I don't see how this situation could be distinguished from the previous one by only looking at the cylinder's mechanical energy (since the normal force would still be doing no work). So, my question is: how is gravitational potential energy converted into rotational kinetic energy, and what is friction's role on this?
81544
Please could someone help with some physics about resistive forces that would slow a ball down which is rolling across a floor. I have found conflicting bits of information about the effect friction has. If the ball grips the floor am I correct in thinking that there is no kinetic friction since the surfaces do not move relative to each other? There would be static friction of course but in terms of slowing the ball down would there be no frictional loses between the floor and the ball. My reasoning is that work done against the balls motion would be the frictional force multiplied by the distance it acts across but there is no kinetic frictional and static friction cannot do any work by definition as it is static? Is that right? Some bits of information I have found said friction does oppose the motion but I don't see how it can if the ball is not slipping as its rolling?
69291
Suppose i have a hollow ring and i take it in some place where there is no gravity and there is no air. Then i just start to rotate it and then because of it's inertia it will continue rotating and won't ever stop because there is no friction.   Question - Suppose i take a point on it of mass $dm$ and i see that it is also rotating around the COM of course. Which force is really providing it the centripetal acceleration? It can't be tension because the tension by the adjoining masses on will cancel out. Then what force is it?