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## Hot answers tagged forces

24

The problem is in your assumption that the force is $F = 2GMm/r^2$. This is true for the force on a point mass from a sphere or another point mass, but not otherwise. What you need to do is sum each the force on each particle from every other particle. For a continuum object, $$\vec F = \int \rho \vec g\, dV$$ where $\rho$ is the density and $\vec g$ the ...

6

Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$ If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ ...

6

Yes, it is possible to magnetically levitate molten metal. This is not due to ferromagnetism however. As seen in the below references, the metal sample is placed within a tapered conducting coil, which carries alternating electric current in the ~400kilohertz range. This sets up a magnetic field gradient inside the coil and causes eddy currents in the ...

5

In your picture 1, the gravitational force is calculated incorrectly. The formula you have used only applies between pointlike masses. You have to divide the object into elements, calculate the contributions of each element and sum up. The picture 2 is only the first step in the whole process, so actually not even your picture 2 is generally correct.

5

Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower. You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or ...

4

Well, the reason it doesn't matter is that work is defined as $$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$ so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity. Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a ...

4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...

4

These are two different effects. Satellites don't fall down because they are moving on a circular orbit. Actually, they are falling down all the time, since circular motion is accelerated (though the velocity doesn't change absolute value, it changes direction!), so it is kind of "falling around the earth". The second question is, why doesn't an astronaut ...

3

As altitude increases, density of air decreases. As air density decreases, the lift of the helicopter blades decreases (for a given angle of attack and RPM). As long as the "real aircraft" is open to the atmosphere (doesn't have a pressurized cabin), performance of the toy helicopter would be the same as if it were outside the "real aircraft".

3

Your thought experiment of dropping an iron ball and a feather need not be in water; in fact, it is more commonly considered in air, but the pertinent facts are the same. All objects, regardless of their mass or composition, are accelerated identically by gravity. But within a particular medium, the acceleration of particular objects might be impeded by ...

3

Let us consider there is no external force and the tyres are rolling smoothly without slipping. Here friction doesn't come into play let me explain you how. Even if friction is present, this friction is not the answer. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer. The correct answer is rolling ...

3

could the Coriolis effect on snowing be so dramatic...? No. The Coriolis effect is only noticeable for objects traveling long distances with respect to Earth's surface for significant periods of time. For example, a ballistic missile fired hundreds of miles or a hurricaine that is hundreds of miles in diameter and lasts for days. Across the street is ...

3

1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $4v$ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. ...

3

A very tempting mental model of an atom, reinforced by many illustrations in books, has protons and neutrons as "large" spheres in the nucleus and electrons as "small" spheres somewhere near the nucleus. If you assume that all of these particles are made of some "stuff" that has roughly the same density (which is the case for everyday solid and liquid ...

3

You're absolutely correct - objects do not need to ever reach earth's escape velocity of 11.2 km/s, and many spacecraft that leave orbit, don't. That being said, note that escape velocity depends on where you are: the velocity that a cannonball 1000 km above the earth's surface would need to escape is substantially lower than that needed by a cannonball on ...

2

The state from $1\text{ atm}$ to $2\text{ atm}$ is normally called decompression or contraction. An equation you can use going from one state to the next is: $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$ Where $P$ is pressure, $V$ is volume and $T$ is temperature. Now if you want to calculate the force you have to know the surface area of what you are ...

2

(NOTE: This is not a definitive answer and is largely guesswork, but it might be useful for further answerers.) At first I suspected it was some sort of weak spring or magnetic device which applied an opposing torque to the arm, so as to make the center reading unique. Searching Google, I found this explanation of the theory of a triple-beam balance offered ...

2

I depends on the pressure ratio between the chamber and ambient . As long as the ratio is critical, the ambient air will flow through the orifice at the speed of sound (with respect to the state in the chamber). For undercritical pressure ratio the flow can basically be found by the Bernoulli Equation.

2

As you said, if feather has a stronger decelaration, is due to air friction (not because iron is more dense than feather or something). The same should hold in absence of gravity: if you give a boost to these balls, then the feather one will stop first. From this last consideration, I'd say that this medium does not affect gravitational force, but just it ...

2

The quote is taken from just above eq. (1.32) in Ref. 1: [...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance ...

2

There is no law that says the sum of forces on a given object must be $0$, that is simply the condition for mechanical equilibrium. If an object has constant $0$ velocity (or, more generally, any constant velocity), then its acceleration ($\frac{dv}{dt}$) is $0$ and, by Newton's second law as you have it, the net force acting on it is $0$. However, if all ...

2

The rope on the left pulls the weight and the pulley towards each other. The rope on the right pulls the floor and the pulley towards each other. Each of these pull downwards on the pulley with force $T$. The total downward force on the pulley is $2T$. The pulley doesn't move because the ceiling pulls upward on it hard enough to keep it motionless. The ...

2

As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe. When work is applied by an external ...

2

As you describe, the definition of work is just: $W=F d$. What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then: $$W=Pt=Fvt=Fd$$ Independent of velocity.

2

I'm going to say the same thing as the author but explain more with words. I am also going to take the external force to be zero and ignore it, since the part you're confused about isn't independent of the external force. The force on particle $\alpha$, due to all the other particles in the system is $$f_{\alpha} = \sum_{\textrm{all other particles ... 2 The formula F=G \frac{m_1 \cdot m_2}{r^2} is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ... 2 By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing F_{app} otherwise there will be an acceleration. 2 Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}\text{As, } v=0 ms^{-1}F=v\frac{dm}{dt}$$Second possibility : If your box is spherical, By Stokes' Law$$F_{viscous}=6\pi\eta rv$$where \eta is coefficient of viscosity. Hence, your ball attains terminal velocity.$$F=6\pi\eta rvv=\frac{F}{6\pi\eta ...

2

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ...

2

The deformation wave will travel in both directions - there's no way for it to "know" the shortest path. And the resulting set of vibrations will interfere with each other in interesting ways, causing complicated resonances. So, let's look at a simpler example: just a thin, large torus. We'll look at two points $90°$ apart from one another; one at \$\theta = ...

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