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A central force is just one that acts in the direction of a particular central point. I think you need to distinguish in your mind the difference between a field or potential and a force. A massive body will exert a central force of gravity on another massive body. If you have several bodies together they will each continue to exert central forces ...


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A central force is simply a force that is always directed towards a fixed point in space. Gravity can be treated as a central force in certain circumstances. The gravitational force acting on object $A$ orbiting around object $B$ can be approximated by a central force acting towards the centre of $B$ if $B$ is much more massive than $A$. By Newton's third ...


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There are many notable examples in electromagnetism. To name just three of them: The flow of energy in electromagnetic radiation (usually called the Poynting vector) is proportional to the cross product of the electric and magnetic fields. The torque on a current loop in a constant magnetic field is proportional to the cross product of the (oriented) area ...


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I'm not sure why the cross product in the Loren force would be particularly "famous", but another (almost trivial) example, that you probably encountered before the Lorenz force, is of course angular momentum ;) PS I would have commented this simple answer, but I do not yet have enough reputation to do so!


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Actually, I'm not sure about magnitude of forces acting on pins. But, what I AM sure of, is that in case (b) walkways has more degrees of freedom, because of several contact points involved with gap, and thus (b) configuration is highly unstable. In theory walkways moves independently of each other in case (b) inducing random stresses to either of pins and ...


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Look at it from the side, and make some free body diagrams: In the first case (left), the load paths for each walkway meet at the rod after the pins. Each pin only supports one walkway (blue and red) and the rod supports both (purple). In the second case (right), the entire weight of the bottom walkway goes through the top walkway support resulting in the ...


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I think that part of the misunderstanding is due to the use of the term centripetal force in a context where more that force is involved. Take the vertical circle motion as an example. The circulating mass is under the action of two forces, the gravitational attraction due to the Earth and the tension in a string. These two forces produce a net force on ...


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Is this a question about rock climbing? If so, then the answer is never to use twin ropes, because they're a total pain, and their theoretical advantages never show up in actual climbing. (And note that (1) twin ropes have different physical specs, and (2) the twin ropes will usually not share the load at all equally.) But anyway, as a pure physics question,...


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"Centripetal force" is only really a thing when something is traveling along a circular arc. For an object that is moving in a circle, the net force required in a direction perpendicular to the instantaneous direction of travel is given this name. If the only force acting on the object is the pull (tension) of a string, then the component of this force that ...


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"Centripetal" is Latin for "towards the center." A centripetal force is not a particular type of force like a frictional force or a magnetic force. It's just a force that makes an object go in a circle. The word "centripetal" describes the direction of the force, not the type of force. When a car drives around in circles on level ground, the centripetal ...


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The work of a force that is constant (which is the case for all 4 forces acting on this block), is calculated like this: Work = Magnitude of the force x Displacement in the direction of the force. Work done by applied force: The magnitude is 200N and the displacement in it's direction is 12m. Hence W=200*12=2400J Work done by the normal: the displacement ...


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Allowing for a tidier knot than you have drawn, so the knot itself has no effect on load (i.e. the knot doesn't have any 'give' in it), then things will be as follows: While the weights are falling freely the load cells in both systems will record zero load (for simplicity I am assuming a weightless rope). When the masses reach the bottom of the rope/ropes, ...


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pushing requires energy In fact, pushing does not require energy. You can lean a ladder against a wall and without consuming energy it will tirelessly push against the wall until the wall crumbles or the ladder is moved away. If a human being wants to push on the wall with their arm then they will find that human arms are inefficient machines and in this ...


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You are correct the deceleration is independent of the mass. But your final equation is incorrect for $p$. Once the applied force is removed, the only force acting on the mass is the friction force, which acts in the direction opposite to motion. Then from Newton's second law $$a=\frac{F_{f}}{m}=\frac{-μ_{k}mg}{m}=-μ_{k}g=\frac{dv}{dt}$$ $$dv=-μ_{k}gdt$$ ...


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You have to look at the net forces on the ball and the paper individually, and not just the action-reaction pair. Gravity exerts a downward force of $mg$ on the ball and the paper momentarily exerts an equal and opposite force of $mg$ on the ball per Newton's third law. The net force on the ball is momentarily zero and the ball is momentarily stable. But ...


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Downward gravitational force applied on the sheet.Reaction force on the Ball in upward direction.But thin sheet can not sustain such heavy force and breakdown and losses its continuity and comedown by gravity .At that instant Metallic ball only experiences downward gravitational force so it's fall down.


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Answer of both case is different, in first case while the man is in air, he is in free fall with respect to earth, he apply force when he just touches he earth, there will be the impulse experience by the earth due to change in momentum of your body from $mv$ to $0$, While in the second the force is constant that is the weight of the body.


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When a man falling it must carriyng some velocity before touching ground after touching ground its velocity change so change in momentum creating force on the ground.But the man who lying in ground not changing velocity so he is not applying any force.


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When the falling man hits the ground his impact force will be greater than his weight. On way to determine the impact force is to apply the work energy theorem, which states that the net work done on an object equals its change in kinetic energy, or $$Fd=\Delta KE$$ Where $F$ is the average impact force, and $d$ is the stopping distance. Example: a 75 ...


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It averages out. All of the time that he's falling, he puts no force on the earth's surface. He is in free fall from the earth's force, and the earth is in free fall from his force. Once he touches the earth, most of the kinetic energy he built up while he was falling converts to pressure on the earth's surface. Edit: Pressure is force per unit area. If ...


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No the force applied is not the same. In the first case the force that is exerted on the earth is because of the exchange of momentum of body with the earth over a short instant of time.It means that when you fall and hit the ground the body will have zero velocity. It means you decelerate over a very short instant of time. This negative acceleration times ...


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No, the force applied in the two cases won't be the same. The falling person will exert a force on contact to the earth which will be much higher than his weight. The force will be impulsive and is given by $$\vec F=\lim_{\Delta t\to 0}\frac{\Delta \vec p}{\Delta t}$$ That's because in the process of falling the person has gained a lot of momentum and on ...


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If you take the question at face value- so A and B are everyday objects (cellphone, brick, bust of Heisenberg in bronze, etc), and they are thrown by a human rather than some device that takes them to near light speed, then the answer is that B experiences the electromagnetic force during the collision. Although you state that the objects carry no net charge,...


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I am assuming the ball to be not properly infiltrated.Now because of this when you get the ball to the floor of pool,because of the pressure from the water above the ball the ball gets pressed and assumes the shape of a flat disk.Buoyancy force will affect a body only if the body has some fluid underneath it to push it upward,in all the other cases it will ...


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For the first question, (e) is also a correct response. For the second one, you are meant to intuit that the pressure inside the beach ball is no greater than atmospheric pressure—otherwise the “thin” plastic would presumably “stretch”. I’d say that premise is a bit thin, and the reasoning a bit of a stretch, but otherwise the given answer is correct.


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The key distinction here is that the plastic is not stretched—this means that the pressure inside of the unstretched ball equilibrates with the external pressure. Assuming the air behaves like an ideal gas, Boyle’s Law indicates that $p_1 V_1 = p_2 V_2$, so the volume the air occupies decreases as the ball goes down thanks to the pressure increase from the ...


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So, if it is not an action reaction pair, then how is any force being exerted by the table on the ball? Also, is the ball applying any force on the table? This is the crux of the issue. There are two completely separate things happening here. One is gravity on the ball, the other is the ball on the table. The first is Newton's Universal Gravitation. ...


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If a ball is kept on a table then there is gravity acting on it as well as a normal reaction force by the table on the ball. This is correct. But as both the forces are being exerted on the same object i.e. the ball, then it cannot be called a- action reaction pair under Newton's third law of motion because it requires the forces acting on two different ...


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For something like a tennis ball, one or two assumptions are in order: 1) Assume some amount of "flatness" of the ball when it is most compressed against the wall. 2) Estimate the distance from the center of the ball to the wall at this point. Call this distance "d". Note that before the collision, the distance from the center of the ball to the surface ...


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It is not in fact infinitely small. If you look at slow motion video of the collision, you can see that the ball and the wall are in contact for reasonable amount of time. During that time, energy of motion is stored as elastic energy od the ball (some is naturally lost) and then converted back to the kinetic energy of the ball which is now moving in the ...


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You have to be more precise with the calculations. First consider the time interval $t < t_0$. There we have $\frac{dv}{dt} = 0$ and therefore $v(t)=v_0 = const.$ and $x(t) =x_0+v_0t$. A very similar result you obtain for the time interval $t > t_0 + \delta t$ (but different initial conditions!). In the time interval in-between you have a constant ...


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There is a difference between $\int Fdx$ and $\int \vec{F}.d\vec{x}$.Basically a difference of dot product which is capable of calculating work by a variable angle.


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Indeed, it is the same formula. $$W=\int_C \vec{F}\cdot d\vec{x} $$ Check that this is the same as $$\int_C |\vec{F}|\cdot \cos(\alpha)\cdot dx $$ This integral is a path integral, that means that you're summing all the contributions along all the tiny portions of the path of your trajectory. Both $F$ or the angle can vary along the path. Anythin inside ...


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You cannot break a stick by purely pushing or pulling it. You need multiple forces to create enough stress in the stick that it breaks. From the picture, the stick was most likely broken by applying two opposite torques about the center (similar to how you would break it using your hands).


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You are dealing with two different scenarios here; When I punch a wall When I punch a balloon Newton's III Law of motion doesn't say that the force applied on the wall should be equal to the reaction of the balloon or viceversa. I think you are mixing both scenarios. Indeed the force is different in both situations but action and reaction in each of those ...


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Where does the other part of the force go for the funnel? It's supported by the horizontal component of the sloping walls of the funnel, not the bottom of the cylindrical portion of the funnel. See figures below. The bottom of the cylinder to the left supports the entire weight of the fluid above it. The weight of the fluid outside the cylindrical part of ...


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If you're looking for intuition about the relationship between forces and connections, I think the best you can do is to think long and hard about the following elementary example (adapted from Moriyasu's book 'Elementary Primer for Gauge Theory'): Consider a vector at position $x$. Call its length $f(x)$. We want to know how this length changes as we go ...


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It presses on the funnel. Imagine that the lower cylinder of the funnel extended all the way to the top. So you had water in the central cylinder pressing on the floor, and you had a separate conical area with the center cut out, full of water. You could drain the center cylinder and the rest of the water would still be there, pressing down on the funnel. ...


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The acceleration a = 0 is for the motion of the car when the wall doesn't exist, that it , there is no external force acting on the car which leads to it having a constant speed. When the car collides with the wall, it experiences a force opposite to the direction of it's velocity which causes it's velocity to slow down over a very small period of time. ...


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All of the effects of Special Relativity (SR) that apply to time and distance carry over to other quantities that are expressed in terms of time and distance. For example, velocity, acceleration, force, and so on, all need to be transformed when moving from one reference frame to another in an appropriate way that reflects the transformation of time and ...


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You are overlooking the fact that when h is zero it means that the brick is just touching the uncompressed spring. IE the spring is not yet taking the weight of the brick. If you release the brick it doesn't settle down to the equilibrium point at which the spring is taking its weight- it goes the same distance beyond that point and oscillates up and down ...


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If h is zero though, the problem is just find the equilibrium of a spring of spring contact k supporting a brick of mass m No it is not. The place where the mass comes to rest will not be the equilibrium position. At equilibrium the mass will still have kinetic energy, so it will overshoot equilibrium and move farther to the $2mg/k$ that you have found to ...


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Let a charge $+q$ be at the point $(a, 0)$ and a charge $-q$ be at the point $(-a, 0)$. Then the electric field at a point $(x, y)$ is \begin{equation}\tag{e1}\label{e1} \vec{E} = q\vec{r}\left(\frac{1}{r_1^3} - \frac{1}{r_2^3}\right) - qa\hat{e}_x\left(\frac{1}{r_1^3} + \frac{1}{r_2^3}\right), \end{equation} where $\vec{r} = x\hat{e}_x + y\hat{e}_y$, $\vec{...


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To move an object you must apply a force where $F = m\vec a\ and\ \vec a = d\vec v/dt$. To move it instantaneously implies that dt --> 0 and the force would go to infinity. Hence instantaneous acceleration or deceleration is impossible. A non-accelerating object can only move with constant velocity in a perfectly straight line. Any curve in the line implies ...


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No, not unless the wall moves. If the force is not applied through a distance, no work is done.


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For a fixed mass (and non-relativistic speeds), acceleration is proportional to force (Newton's Second Law). This is important to the example I would like to give that helped me to explain this. Personally, I'm not a big fan of the "mass on a string" example, because the tension in a string is not that easy to visualise without getting confused with the ...


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Intuitive explanation to why force perpendicular to velocity results in circular path This is not always true, sometimes it may result in a circular path given the boundary conditions and the forces involved. In short, my question is exactly how and why a stone attached to a string follows a circular path when its velocity is perpendicular to force? It ...


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F=dp/dt and we get F=mdv/dt, which is equal to Ma, whenever there is change in momentum the body will simultaneously applied and experience the force, that's what Newton third law describe, now coming to your question it is OK acceleration is zero before collision with wall, but during the Collison momentum changes like this Final momentum - initial ...


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Many are pointing out the importance of friction of the car against the road and the need to apply a constant force in the real situation to mantain the car moving uniformly. I think this is not the point of the question, an object (let's say a micrometeoroid) moving in the vacuum of space and hitting a "wall" (let's say the solar panel of an artificial ...


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Friction and air resistance are irrelevant. If a body is coasting through space (no friction and air resistance) at a constant speed relative to the wall, then hits the wall, the wall speeds up and the car slows down, so they both accelerate.


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