# Tag Info

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I cannot comment on the full design because the question is lacking on details, but I can explain more about the situation where you hit a mass with a ball and a spring reacts to it (like the sketch shown) Consider the friction force as $f=\mu m g$ and the equations of motion $$m \frac{{\rm d}^2x}{{\rm d}t^2} + k x \pm f = 0$$ The sign of $f$ depends on ...

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You are quite right. A fully filled iron block cannot move because the force cannot be transmitted by the atoms due to their fixed positions. But the atoms in the layer you push will apply a force on you. | | | | <------ ------> | force exerted on atoms force exerted on your hand | It is similar ...

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The iron cube will push you back with the same force. It's Newton's third law, if you push something it will apply the same force on you but in the opposite direction to the direction you pushed it in. When you push a car which has, lets say a ball, the ball move towards you not because it is feeling a reaction force but because of its inertia. The reaction ...

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As far as I understand, what reactions refers to is the total force applied back due to electromagnetic repulsion between the atoms and your hands. In simple terms, you iron cube, should have electrons. Then those, electrons and your hands electrons should repel each other as $-$ $-$ $=$ repulsion.(charges) I don't understand what do you mean by ...

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The problems solved by the method of images are, at their heart, boundary value problem involving the divergence of a scalar field, and those problem have uniqueness theorems that says there is only one configuration of fields in the volume that generates a particular set of boundary conditions. So if you find any method of generating a field that meets the ...

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Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system. I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on ...

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When the light hangs off the ceiling by means of the electrical cable, the cable actually acts as a very stiff spring. The light bulb provides a downward force due to its weight $mg$. The cable acts like a spring, slightly idealised here as a Hookean spring which provides the counter force (what you called the force of tension) of $k\Delta y$ with $k$ the ...

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The simple word is "damping". Initially when you hang an object from a string (spring, etc), it will move - side to side, and up and down. While it is moving, there will be a changing force on the object - after all it is accelerating / decelerating. This shows up as a force in the string that changes with time. In all "real" systems, there is also a ...

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Let's simplify this problem a bit more. Suppose there is no air resistance and the ground is flat and smooth. If we throw this bouncy ball (which I'm assuming will undergo elastic collisions) at some angle to the ground, $\theta$, then it will have two components of velocity; that perpendicular to the ground, $v_y$, and that horizontal to the ground, $v_x$. ...

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The point is that the ball gets a tangential hit by the ground. This changes the angular momentum of the ball. Consider a ball thrown with a horizontal speed v. It should also not rotate. Right before hitting the ground, the ball has an angular momentum of $$L=mvr$$ This is a result of $\vec{L}=\vec{v}\times\vec{p}$, which is also valid for linear ...

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You are correct in calculating the force needed to accelerate the ball so it will fly the right distance when it leaves the hand. But you forgot the fact that the ball has weight - so even if you are not accelerating, there is a force of 2 N. You need to add this force to the one you calculated. Another way to look at this is with conservation of energy. ...

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When the hand exerts a constant force $F$ on the ball over $s=0.2m$, work is done on the ball: $W=Fs$. This work imparts on the ball a kinetic energy, which during flight is completely converted to potential energy (because at the height $h=2m, v=0$), $E_{pot}=mg(h+s)$. So that $Fs=mg(h+s)$, which works out at: $F=\frac{mg(h+s)}{s}=22N$.

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It's a general consequence of Newton's Laws of Motion that an object is stationary when no net forces or momenta act on it. Avoiding net forces or momenta is precisely what makes perfectly balancing a rectangular block on a triangular one so difficult. Above is a rectangular block with mass $m$, perfectly balanced on a triangular block. I've added the ...

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As @rmhleo pointed out in his answer, the frictional force doesn't depend on the surface area, because no matter which part of the object is in contact with the other surface, the total normal force (and thus the total frictional force) is unchanged. However, that assumes a couple of simplifying conditions: namely, that the two surfaces are consistent in ...

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Yes, frictional force does not depend on the area. This is clear from a simple mental experiment, think of a block resting on a surface. Assume it is a prism whose faces have different areas. Whichever face it is resting on, the friction force will be the same: when on the smaller surface, the contact is reduced compare with the case where the resting face ...

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You say that you "can actually feel a substantial difference when I lift her, until her feet are not in contact with the ground anymore". That makes perfect sense i.e. she is also participating in the lifting process by pushing off the ground except when she's being difficult. Once she's off the ground, as long as you're holding her in the same way and she ...

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From Making yourself heavier Yes, it is possible to make yourself harder to lift. By shifting your center of gravity , which is usually referred to as your hips, down you can make it much harder to be lifted. It depends on which kind of lift, but this usually does the trick. Shifting your center of gravity does not make you heavier. It shifts the point ...

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All of this is all right, but the problem is that I'm taking a course on electrodynamics and the teacher said that the work $W_{\mathrm{ext}}$ done by one force external to the system is $$W_{\mathrm{ext}} = \Delta K + \Delta U,$$ that is the change in the total energy of the system. I don't know where this comes from It follows from the work-energy ...

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So why if applied force is increased the normal force can't withstand the increased force? The other answers answer this well. Also is there any possibility that in a situation the frictional force could always cancel the applied force? In that case you would need a material (surface) which is infinitely strong. If you put a heavy stone on a ...

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To add a little to Asher and Gert's answers: The atoms on surfaces are constantly in motion. If you're used to thinking of solids as being really rigid and difficult to deform, try this viewpoint instead: think of them like a bunch of grains of sand held together by tiny Slinkies. The atoms can move out of the way and slide back into place, or even be ...

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Why can't friction withstand any force? Because the amount of friction that inter-surface interactions can provide is limited, not unlimited. In the idealised force diagram below, only the object's weight $mg$ provides a vertically downward (aka 'Normal') force, causing a friction force between the object and the plane. This friction force is ...

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Your image shows what is going on at the microscopic level between two surfaces. To understand why friction works, you have to look smaller, at the atomic level: and when you get to that point you're no longer taking about "friction" as we know it, but about physiochemical interactions between atoms and molecules. Those interactions are mediated by ...

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Definitions and assumption: Mass of the hoop $m$, radius of the hoop $R$, force exerted by the string $F$, angle of rotation $\theta$ ("theta"), pivotal point P. It is assumed the string fits around the hoop without friction. Inertial Moment of the hoop: Around its centre axis the moment of inertia is $I_c=mR^2$ and with the Parallel Axis Theorem the ...

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Did you see section 4 of the calculator "warnings"? "Your index is too large" (index = ratio of diameter of spring to outer diameter. You have a ratio of 40 - a very thin wire wound on a very large diameter. And since length inside hook needs to be decreased by the diameter in order to get the length of the "actual spring", you basically have "almost no ...

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If you look up composite pendulum you will find the equation $$\ddot{\theta} = - \frac{m g \ell}{I+m \ell^2} \sin\theta$$ $$\Omega^2 \approx \frac{m g \ell}{I+m \ell^2}$$ where $m$ is the mass, $I$ is the mass moment of inertia about the center of mass in the direction of rotation and $\ell$ is the center of mass distance to the pivot axis. Combined ...

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The magnitude of static friction adjust its value to the applied force hence static friction is called self adjustable....

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Assuming the image you gave is a side view, and that the front view looks like what I have just added below, here is my answer. This will behave as if all the mass was concentrated on the center of mass, which I assume is the one you point out, and I depict here in the picture as a red star. So when you have a horizontal acceleration as represented, you ...

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$A$ is correct, assuming the forces are the force of gravity $F_g$, the normal force $N$, and the friction force $f$. $D$ is not ideal because you should draw $N$ in this case. You might be confusing yourself about $E$ because you are drawing $F_g$ as well as the component of $F_g$ in the direction parallel to the plane ($mg\sin\theta$). One thing you can ...

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Briefly, yes. The box would speed up momentarily, until the propellant from the rocket hits the back end of the box and slows it back down again. Once the propellant is exhausted, the box will again be motionless (neglecting the minuscule jiggling caused by thermal motion of the propellant gas inside). The controlling principle here is that the total ...

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When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...

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In (d), the net force on the object is into the plane, and so it should be accelerating through the surface of the ramp. Obviously this does not normally occur. There is a unbalanced force going into the plane. But that vector won't change the fact that the object is at rest. If there's an unbalanced force on an object, then the object will ...

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You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...

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Recall that force is equivalent to, $$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$ Where $p$ is the momentum, and $t$ is time. Momentum is given by, $$p=m \cdot v$$ Where $m$ is mass, and $v$ is velocity. When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small. When you hit the ground ...

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The simple answer to both questions is, yes, you exert a force on Pluto and on the universe. By simple, I mean that the Pluto question is treated as a "two body" problem (and nothing else). This answer is obtained using the formula given by dotancohen. For the universe question, it should be easy to see that the same process can be applied to you and any ...

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To find the minimum velocity at the bottom-most point, we find the minimum velocity at the uppermost point. This minimum velocity is the one such that the centripetal force is equal to the force of gravity. Any lower and gravity will pull the object down and out of the loop, any higher and a faster velocity would be required to generate it (thus, not a ...

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A tension less than zero isn't really physical; this is the point where the string stops being taught and the object doesn't make a complete circle. Therefore, when finding the minimum velocity for an object to make it around a loop, we solve for when the tension at the top is zero as it is the minimum possible tension for the object to keep going in a ...

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I'll just answer the part of your question about terminal velocity (better to call it terminal speed...). The terminal speed of a falling object is not caused by anything special about gravity. It is caused by the fact that for an object falling straight down there are two main forces acting on it: gravity (which points down as is constant to a good ...

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I write this because it is too long for a comment Situation You are putting $k$ watts (=joules/second) of kinetic energy into a body. We will show the acceleration decreases with time. $$E=\frac{1}{2}mv^2$$ so $$v=\sqrt{\frac{2E}{m}}$$ so $$\frac{dv}{dt}=\frac{m}{\sqrt{\frac{2E}m}}\cdot\frac{de}{dt}$$ (I can't be bothered to tidy this up - remember I'm ...

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Weirdly I just wrote about this: Magnetic force storage or amplification question Anyway. For the first one! YES the equations for force due to magnetism and force due to gravity are both very similar, they take the form: $$F=\frac{k}{x^3}$$ where $x$ is the distance from the source of gravity/magnetism. I am a mathematician, so I am completely happy ...

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Does a body always rotate purely about its center of mass? Well, that depends. The first assumption you need is that the body is rigid. Violate this assumption and all bets are off the table because you can't even necessarily classify all motions as "rotations": for example if a long thin board starts twisting sinusoidally into/out-of a helix shape, ...

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Antonio -- sure: your mass affects Pluto. Here's a line of thought that might get your head around this amazing concept: You probably agree that the Earth as a whole affects Pluto .. right? Simply, consider the earth chopped up in to little pieces each about the size and mass of yourself .. say 100kg. (Funnily enough .. there's actually NOT THAT MANY of ...

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A body in free motion does not necessarily rotate about the center of mass. The center of mass might have straight linear motion in addition any rotation. The general motion is a screw motion with a rotation about some instantaneous axis and parallel translation at the same time. Consider an arbitrary body rotating by $\vec{\omega}$ and at some instant the ...

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I think this formula would work better F=MxA Since you are hovering you need to consider the acceleration of gravity in your calculations. So you would get F= M x 9.832 meters per second squared. Use kilograms for the mass in this case

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There is a confusion of what that $v$ means. You are thinking about the velocity of the drone, which is stationary this $v_{drone}=0$. But in your equation, to calculate the power needed by the wings, you have to consider the velocity of the motor providing the thrust (propelling air downwards at a certain rate) to keep the drone floating which is making ...

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Choosing your system carefully and drawing free-body diagrams is crucial! It is often easiest to start with defining your system as one big block. You can do this because the engine and wagons are all connected by rigid steel locks that cannot compress or stretch, meaning that if one car moves, the others have to move along at the exact same rate. If we ...

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Power equals work/time. Work is force times distance, so you can simplify to force times speed, when the force is constant and the force is causing the speed. In your situation there is a force due to gravity that would do work on your drone, and what you need is for the drone to do work to counter balance that. In other words, power is always work over ...

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The problem is that finite wings inherently induce drag (called lift induced drag). Even if you could assume the wing infinite you still have parasitic drag. So you will always need to supply energy to keep the plane flying at a constant speed (and as a consequence to keep generating lift). Without drag it could, in principle, keep flying without power. It's ...

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In principle, you are right that it is possible to keep an object in the air without expending any energy. Think about a blimp, you can fill it with helium, and then just let it float. It doesn't have to do anything to float, it just floats because it is lighter than air at ground level (the air gets less dense as you go up). You can even make heavier than ...

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At any given moment the thrust is equal to the force experienced by the rocket at that moment. That's the definition of thrust. To get the equation of motion right, you need to consider the instantaneous force (variable) and mass (also variable). The acceleration is then given by $$a(t)=\frac{F(t)}{m(t)}$$

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