# Tag Info

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I know of two reasons for why we should consider gravity to be a force. The first is purely classical and Newtonian: tidal forces. Gravity is solely responsible for producing tidal forces, and they cannot be considered a fictitious force, whereas the usual acceleration due to gravity in some sense can always be thought of as fictitious. The way you know ...

1

To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0.$$ But if you insert the actual values from your ...

1

I might be erring something basic here, so downvotes are welcomed, but I would love if they include comments to correct this answer, or just erase it. I do not believe the Coulomb law has been tested beyond the order of a few meters. Arguing that light remains unchanged across the universe should be irrelevant. The reason is that the electrostatic and ...

4

There have been lots of experimental attempts to test the validity of Coulomb's $r^{-2}$ law. Many of these are reviewed by Tu & Luo (2004), and is where I am getting the numbers quoted below. Somewhat equivalently, experiments have looked at trying to set an upper limit to the photon mass, which is testing the hypothesis that rather than a $r^{-1}$ ...

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If gravity isn't a force, then why do we learn in school that it is? Because it is a force. It's just not a force in the Newtonian sense, wherein work = force x distance. When you drop a brick the "force" of gravity doesn't add any energy to the brick. Instead it converts potential energy into kinetic energy. This is different to what you do if you ...

0

For an object to move in a constant velocity the zero acceleration then the Forces should be zero. The $$\mu=\frac{65*g*\sin 30}{65*g*\cos30}=0.57735$$ Your answer is perfectly true. $$\mu<1$$ if your question is why does the object still move even when force is zero. Example is Terminal Velocity.in air. The reason the object comes down is that there is ...

0

If you want to research the question more deeply, I would suggest you take a look at the Solar Wind. This is composed of charged particles (mostly protons) emitted by the Sun. The flow and behaviour of the Solar Wind has been studied quite deeply, not least because it affects greatly satellite operations, spaceflight, radio transmission and other important ...

2

Coulombs law as well as Amperes law and similar mathematical formulations of two centuries ago, were incorporated within the strict mathematical format of Maxwell's equations . The apparently disparate laws and phenomena of electricity and magnetism were integrated by James Clerk Maxwell, who published an early form of the equations, which modify ...

0

Let $\theta$ is the slope angle. The normal force is $F_{N} = mg\cos{\theta}$, which you have calculated. Now there are three important points to consider. The object is in constant speed, so $F_{net} = 0$. There are three forces acting on the object $F_{friction}, ~F_{gravity} ~and ~ -F_{normal }$ whose net sum is zero to justify (1). $F_{friction}$ by ...

0

Answer 1:The drag force in the car during the forward acceleration is mainly inertia and friction. The air resistance is taken into consideration only at high speeds. The air resistance will increase due to fluid dynamics but the frictional force will remain constant. Rolling friction is proportional to Weight of the car not the speed and ...

0

In simplest terms and using Newton's mathematics: F = m * a. or Force F = mass m * acceleration a. Example #1 - A body on Earth. Now on the planet Earth, the gravitational acceleration "g" is about equal to 9.8 meters/second^2. So let's substitute a=g in the above equation. Then the force required to keep an object of mass m AT REST near the surface of ...

1

For the forces between elementary particles we have Feynman diagrams, where there exists a mediating particle for the interaction. In the simplest diagrams: for the strong it is the gluon, for the weak it is Zs and Ws and for the electromagnetic it is the photon. Here is Bhabha scattering, where the electron and the positron ( attractive force) are first ...

1

The rider feels "forced down" because the object to which they are attached is accelerating upwards. Because the acceleration is opposite to gravity, the normal force, $\mathbf{F}_{N}$, being exerted on the rider must increase in magnitude (relative to the "at rest" magnitude on a horizontal track) in order to produce a net upwards acceleration. Thus, it ...

1

The key is this: how is the force applied to the rider? Gravity pulls directly down on the passenger's mass, but what keeps the passenger from heading towards the center of the Earth? When you're in the passenger seat of a sports car, and the driver floors it, you feel pushed back into the seat. But what's actually happening is that the seat is pushing you ...

2

Actually A.Zee's book on "QFT in a nutshell" has a very nice explanation on this on chapter I.5, I will breifly sketch it (this is a very rough skeptch), $$Z=\int DA e^{iS(A)} =e^{iW(J)}$$ where W(J) is given by, $$W(J)=-1/2 \int \int d^4xd^yJ(x)D(x-y)J(y)$$ where D(x-y) is the photon propogator and J(x) and J(y) refer to two lumps of matter Plugging in ...

0

It's important to consider exactly where the forces of friction are being applied in a typical rising elevator. In most cases (I would hope), the elevator car isn't just scraping along the walls of the elevator shaft on its way up. That would make for a very noisy and expensive ride to the top, I would think. Elevators are more complex from a kinematics ...

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Yes, friction force is $F_r=\mu N$, where $N$ is the normal force exerted by the floor on the object, Here $N=m(g + a)$ So yes friction force also increases

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Answer 1: The Air drag increases as the car accelerates, the Air drag is usually a complicated expression which can be approximated to be $$F_d=bv$$ where $b$ is a parameter that depends on the the structure or the area of the car, and $v$ is the velocity, The friction or rolling friction in particular doesn't depend on speed, It'll remain constant Answer ...

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You should not worry about the force exerted during the collision, because it depends on how stiff the object and the surface are, and because it does not affect the outcome. Remember $F=ma$? OK, you know the mass $m$, so what you are asking is what is the acceleration $a$. So what is acceleration? It is the change in velocity $\delta v$ divided by the time ...

0

The equation you gave is just the final speed after a collision. To assume the surface does not move is to say that its initial speed is zero, and its mass very large. If it were not, after the collision it would move (like a car would start moving if hit by another one moving). The equation assumes both objects are free to move after the collision. It also ...

4

I would like to take a slightly different angle on this question and point out that most physicists believe that gravity is in fact a force. The great triumph of modern particle physics, the standard model, contains the strong, weak, and electromagnetic forces. These forces are represented in the standard model by the presence of force carriers (spin 1 gauge ...

1

Some people may think this is a simple question but much of physics research starts with a simple question such as "The school clock shows 3:00pm. Is it really 3:00pm? No. Because light has to travel to my eyes." A person never knows what benefits exist till the research is carried out. Here are some reasons the saucepan is tilted. Quicker hand grasp. ...

2

Well that depends. As always, you should be very careful with such reasoning as Due to the fact that gravity is related to the square of the distance should not the gravitational sum of every particle exceed the force when calculated by the center of mass. because this is a problematic statement. In general your (Newtonian) gravitational potential is ...

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HINT: We can use Kinematics Equation in Circular Motion to find Centripetal Acceleration, Then the acceleration multiplied by mass will give us force required. Hope this helps!

1

Goutham is quite correct in some ways but overlooks something. Look at the diagram below: We known the centre of gravity (COG) of the empty cylinder is $z_1=10\:\mathrm{cm}$ and the mass of the empty cylinder is $100\:\mathrm{g}$. If we fill the cylinder up with water to height $2z_2$ then the COG of the water is $z_2$ and the mass of water is (assuming ...

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Tim B. takes the position that this an example of lying to children. I completely disagree; in my view, what this is an example of is idealization, which is something that every model must do, in every branch of science. As George E.P.Box once wrote: Essentially, all models are wrong, but some are useful. It isn't lying, it's called doing science. ...

1

The flask become most stable when its centre of gravity is at the smallest height. If you start pouring water, you will notice that the effective centre of gravity gets down to a lower postion. As you keep on filling, it would be at the lowest height for some level of water and rises again, afterwards. You will have to find that point of minimum height. Just ...

0

The only thing that makes both the cart and the mass accelerate in the same magnitude is the existance of the geometrical constraint, that is the massless string has no change in length. If such a constraint doesnot exist, this may not be the case. Now, to help you understand what i mean, lets think of a different situation, involving a non rigid string. ...

1

Two North's don't want to align next to each other - it is a lower energy state for one North end to be next to South's on the other compasses. Each compass needle is a little magnet, that aligns with the Earth's field to reduce its energy. Close together, those little magnets are stronger than the Earth's field, so they respond to each other

1

I believe there is no definite answer to this question and it has puzzled physicists for a long time. In fact, it has puzzled them so much that after the grand work of Newton's, many people still sought different formulations. The Cartesian school for instance thought that "only pre-existing motion can trigger a change of motion" akin to what people witness ...

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Newton's first law states that there exist very special reference frames (that we are going to call inertial henceforth) where any point particle not subject to external forces (interactions) moves in straight lines, i. e. the equation of motion is $\dot{\textbf{p}}(t)=0$. Newton's second law states that, in the above mentioned reference frames (and only in ...

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Suppose a force $F$, acting on object $A$, produces acceleration $a_A$, and the same force $F$, acting on object $B$, produces acceleration $a_B$. Suppose a different force $G$ produces accelerations $a_A'$ and $a_B'$. Then Newton's law implies $a_A/a_B=a_A'/a_B'$, which is not tautological.

1

Let's look at this problem from the point of view of equations of motion, see diagram below: Firstly let's make a few assumptions. Ball and cube are of same weight ($mg$) and same size. Simple friction model $F_f=\mu F_n$ holds and $\mu$ is independent of speed. Both objects are completely stationary (no sliding, rolling or tumbling) at $t=0$, at which ...

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Even if we restrict ourselves to a Newtonian conception of the world, forces do not exist. An essential thing that is not emphasized enough when teaching physics, is that physics (in all its wonder) is nothing but a mathematical model of the reality we perceive. Whether you are considering Newtonian, mechanics, relativity, or quantum mechanics. There are ...

6

If we observe our school syllabus, almost all the physics that we learn is Newtonian physics. Everything from force to the laws of motion are all based on Newtonian ideas. And the general theory of relativity is a modern concept which in fact is more true. But you know the GTR is a difficult concept to understand for a child. So to make the course simple ...

0

This is a very common question asked by students like me in mechanical engineering. The friction is irrespective of area means the friction generated, the vector F (The letter with which we denote) will form whether area in contact is less or more But the ability to stop is determined by the Number of friction vectors developed per area of contact. On the ...

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It's an example of "lie to children". https://en.wikipedia.org/wiki/Lie-to-children Because some topics can be extremely difficult to understand without experience, introducing a full level of complexity to a student or child all at once can be overwhelming. Hence elementary explanations are simplified in a way that makes the lesson more understandable, ...

0

We know that is not regular energy nor is regular matter. Dark energy must clump as per space is distorted from its flat symetry (space is dragged like a blanked would be so more cubic space dragged by a massive object than by one less massive) in places where you have concentrated mass like neutron stars and blackholes. There seems to be the posibility ...

0

Whenever one applies a sideways force trough the center of gravity of an object, that force has two components: 1) a direct force that tries to overcome friction and slide the object, and 2) a torque that uses friction to produce a rotation of the object by lifting its center of gravity over the leading edge. A short, flat object will tend to slide ...

106

Because Newtonian gravity, where it indeed is considered a force, is a good enough approximation to the situations you consider in middle school (and beyond). General relativistic effects are very weak at the ordinary scales we humans look at, and it would be overkill to introduce the full-blown machinery of general relativity (which demands a considerably ...

0

Your situation is dynamic in that the ball accelerates under the force of gravity, initially at rest but then reaching a steady state, terminal velocity, similar to what a skydiver experiences when jumping from an aircraft. The terminal velocity is reached when the force of gravity is in balance (equilibrated) with the viscous drag force imposed by the flow ...

0

There has been an apprehension about whether the cyclist can take a turn without steering. In my opinion he can. There are two principles he can use. These are By shifting the line of normal reaction sideways to the line of force acting upon the center of gravity. By trying to rotate the cycle sideways. For simplicity consider the case of a standing man ...

0

Rather than starting with the object-surface interaction and adding a lift force, start with the object-lift interaction and add the surface later. If you have an object that weighs 10 Newtons and pull up on it with a force of 8 Newtons, then there's an imbalance: there's a net downward 2 Newton force and thus a net downward acceleration. Your object ...

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The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1) $$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$ rather than just \tag{2} {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. ...

0

What would happen here is that the object would not be lifted off of the surface. Instead, you would be relieving some of the normal force that is exerted by the table. Instead of $\vec{R} = -M\vec{g}$, you'd have $\vec{R} = \vec{F}-M\vec{g}$, where $|\vec{F}| < Mg$. The net force would still be zero. Imagine the following: If you put a block on ...

0

Well, since the resisting force is the same as the force applied in pulling the rope, then then the body will be in equilibrium. I.e acceleration =0... In the case of increasing force from 0 to X, there will be acceleration until the resisting force reaches the magnitude XN used in pulling the rope, which halts the motion of the body

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from the figure: liquid that cannot be compressed... so forget compression of springs. Molecularly, particles in a liquid have much less space to move around (short mean free paths) compared to particles in a gas. The result is that any applied pressure to these particles is instantaneously transferred through the fluid. The mechanism by which this happens ...

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