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It is due to tensile internal forces (and the center of mass doesnt change position). And you are correct, the friction doesnt change. Look at the graph for the forces acting on the small element of paper in contact with the floor. We have two equations to describe vertical and horizontal forces at equilibrium: $N-mg-T\cos{\alpha}=0$ and ...


0

I believe a major factor in it is failure of the central pivot's friction to resist the larger torque generated by the longer lever arms. So, yes, this is a matter of friction, but not only on the feet. What resists "first" is the friction of the central pivot. If that fails, then feet friction comes into play.


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Frictional force is horizontal. Use angle to get horizontal force component due to gravity on paper. That's plenty.


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Consider the following image, we have some fluid volume, $V$, having density $\rho$ and traveling at a velocity $v$ along a pipe with some cross-sectional area $A$. The rate at which the water flows through the pipe is called the volumetric flow rate. This is given by, $$ \frac{dV}{dt}\equiv Q=\mathbf v\cdot\mathbf A $$ where $V$ is the volume of the ...


4

The electromagnetic and weak forces have been unified into the theory of the electroweak force. The recent discovery of the Higgs boson put the icing on this particular cake. The strong force is described by the same type of quantum Yang-Mills theory as the electroweak force, however it is not unified with it. There have been several attempts at unifying ...


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Short answer: not yet. Long answer: The strong, EM, and weak forces have been unified via quantum field theories (QFT), while gravity is understood separately via general relativity. The two theories are currently incompatible with each other in the quantum regime. There are a variety of theoretical efforts underway to unify gravity with the other three ...


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It is important to be clear what is doing work on what. In the diagram below, consider a charge starting at rest at A. We release it, and it accelerates down the potential gradient. Here, the field is doing work on the charge - it is accelerating it and therefore giving it kinetic energy. As a result, the charge will travel through the potential minimum at ...


0

I would see the normal forces as a requirement to fulfil a constraint: staying on the surface of the earth. By defenition, no work is performed by constraint (reaction, normal) forces due to action-reaction. It could be easier to look at a spring 'jumping' off a rigid table. First you compress it with your finger, which stores energy in the spring. The ...


1

if we want to move a charge in an electric field then we need a work of an external force to move it from a low potential energy to a high one Not true. The electric field itself exerts a force to move the charge. Moving the charge from high to low does require work and this work is done by the electric field itself. Any force that causes movement is ...


1

Your mass or your acceleration doesn't tell you anything about the forces acting on your body. F=ma simply states that if we add (vector addition) all the force acting on your body then the effect of the net force will be given by your mass times acceleration. There may be a lot of forces acting on your body but if the net sum of these is zero, the body ...


2

The Fundamental Theorem of Calculus is of course correct, and you are applying it correctly. The statements the rate of change of work with respect to displacement is force and the instantaneous rate of change of work with respect to displacement is force are correct. The thing to keep in mind is that it's not work that is instantaneous, but its ...


1

No matter how ridiculous you may find it, it is true. The most general definition of work is indeed that the infinitesimal work done along an infinitesimal path is just force times the length of the path, i.e. $$ \mathrm{d}W = F\mathrm{d}s$$ Therefore, the amount of work done along a path in space $\gamma : [a,b] \to \mathbb{R}^3$ is the line integral $$ ...


0

Here there are two objects, One is the Person pushing, other is the Table. Both have applied force which is equal to (say) 5N, F=5N. Mass of the Person (say) m=20kg Mass of the Table (say) m'=10kg Acceleration of the Person= a m/s^2** Acceleration of the Table = a' m/s^2 Force of the Person and Table= 5N ---- {According to Newton's Third law} Applying ...


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This question has to be a duplicate, but just in case it isn't ... You are confusing the force Newton's third law with the force in Newton's second law. Newton's third law addresses the individual forces on two different objects subject to a common interaction. Newton's second law addresses the net force on one object. The net force on the book are a ...


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Suppose you and the table are floating in space. If you push the table it will go in one direction and you will go in the other direction. Your and the table's acceleration will be different so you will end up travelling at different speeds. This is obvious from conservation of momentum. The momentum in the centre of mass frame is initially zero, so after ...


3

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as: rolling drag between the tires and the road surface, fluid drag from the air that the car moves through, and ...


1

You have ignored the car friction with air! If we assume the car as an aerodynamic body (air flows on the car surface smoothly and does not separate) then the air friction on the car is proportional to the square of car speed.


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to solve the problem you need at 4 linear equations (or the two linear ones you have plus another nonlinear one). Otherwise, the solution is undetermined, you have more variables that linear equations and the number of solutions is infinite. Here are the two extra equations that you need. 1) the total force equald the applied force (otherwise it will move ...


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All the laws of Physics are derived from the law of coservation of energy,also the law of conservation of momentum and Newton`third law of motion...


3

You are wrong at assuming constant friction. Rolling Friction increases when you increase speed of the car (See the formulae at the bottom). Also, aerodynamic drag increases with the square of speed (See the formula at the bottom). So, at higher speed, the car engine needs to counter higher rolling friction and air drag to maintain that speed. While the ...


0

You are right By Newtons Second law $$\ F = m (v-u)/t$$ Once you have attained the speed of $$\ 100 m/s $$ You only need to supply the force to counteract the friction. But attaining to this speed requires more force and thus more energy. We can conclude that riding on the same road at a constant speed of any magnitude does't require more energy. Let ...


3

Theoretical Answer Consider that you travelled in your car at $10~{km}/{h}$ for one hour, then at $100~{km}/{h}$ for the next hour. First hour of the Journey: You travelled a distance of 10km, so the work done is $$W=F\times s=10F$$ Second hour of the journey: You travelled a distance of 100km, so the work done is $$W=F\times s=100F$$ The work done is ...


0

Take the example of gravity. Gravitational potential energy is defined as $$U_g=-\frac{GM}{r}$$ This means that $$U_g=-\int_{r_1}^{r_2}F_g \cos \theta dr$$ Because $F_g=mg$, $$U_g=-\int_{r_1}^{r_2}mg \cos \theta dr$$ Integrate and you find $$U=-mg\cos\theta \left (\left.r\right|_{r_1}^{r_2}\right)$$ and $$U=-mg\cos\theta (r_2-r_1)$$ and if the object is ...


1

The other answers are great. I decided to plot it, however, because it's nice visualizing these things. Since your biggest doubt is about kinetic energy, be sure to pay attention to the last graph. SYSTEM. Motorcycle going to the left, truck going to the right, bound by an elastic rope ten meters long ($k=100\frac{\mathrm N}{\mathrm m}$). Masses and speeds ...


3

This effect is called Capillary Action. Yes we do in fact observe it in nature in a large scale: How do you think plants are able to "suck up"1 water through its roots and send it to the leaves? One of the major forces responsible for it is capillary action. Here, have a quote from the article mentioned above: Wicking is the absorption of a liquid by a ...


0

Pressure from combustion of gunpowder will far exceed pressure of air trapped when muzzle is touching the wall. The bullet will still be driven into the wall, but not quite as far, since it will have had to expend a small amount of energy against the trapped air ahead of it. Not sure whether you were asking about my credentials or fundamental concepts. At ...


1

Hetrzian calculations assume infinite width for the parts and in real life tires have a finite width. What that means is the if the contact is line contact (like a cylinder on a plane) as opposed to a point contact (like a football on a plane) the pressure distribution is going to be abruptly interrupted at the ends, compared to an infinitely long line ...


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is it possible to consider also the other fundamental forces [...] to be fictitious forces like gravity in the framework of general relativity? No, because the equivalence principle only holds for gravity. If we want a final unification of all fundamental forces, hasn't this feature of gravity to become a feature of the other forces as well? The ...


8

The classical theory of electrodynamics can indeed be written as a geometrical theory in a similar way to general relativity. As it happens there is a question and answer addressing just this, but it's in the Maths SE: Electrodynamics in general spacetime. Classical electrodynamics is an example of a class of theories called classical Yang-Mills gauge ...


8

On the quantum level, force is not acceleration. The concept of "fictitious force" makes no sense on a QFT level, because forces are interactions between quantum states, not the classical forces you might imagine. Quantum forces are not vector fields in space. The notion of "fictitious force" would mean that, e.g., the strong force is something influencing ...


2

Why is torque or moment of couple independent of choice of point of rotation? Torque is given by $\tau =\vec r\times\vec F$. **It depends on r **, which is its distance from axis of rotation to the point where force is applied. A couple is defined as: "Two parallel forces with the same magnitude but opposite in direction separated by a ...


0

Technically, little vacuume can carry huge wight even with little air flow. For example typical vacuum cleaner can lift column of water 2 m high other dimensions will go together with the vacuum surface. This means that it can carry some 80 cm column high of concrete or 25 cm column high of steel, assuming it is perfectly sealed. If not perfictly sealed such ...


2

Your assertion that "torque or moment of couple independent of choice of point of rotation" is only true either (1) of a system of forces $\vec{F}_i$ at positions $\vec{R}_i$ that sum to nought or (2) you shift your torque computing centre along a vector parallel to any nonzero nett force. Otherwise your assertion is not true. The fundamental reason for ...


1

This follows from definition of 'torque' and 'couple' and is a simple matter of geometry. In the diagram above, a couple is applied to a disk of diameter D. That is, a force F is applied to opposite sides of the disk. The torque do to the couple is: $\tau = F\times d_1 + F\times d_2$ where $d_1$ and $d_2$ are the distance to some (arbitrary) point, $O$. ...


1

I would say an axis of rotation, at center mass, would produce rotation maximization and symmetricality.


1

The question (even after the edit) is not very clear. So I will make some general statements about forces, objects and rotation. In order to cause a change in the angular momentum of an object (which is one interpretation of "rotate", although it can mean "stop rotating" too), you need to apply a torque. A torque is a force applied along a line of response ...


5

A net force on an object likes to act through the centre of mass of that object; in this case, the middle of the door. Apply a force to the centre of mass and it will accelerate uniformly because there is an even amount of matter on all opposing sides that can resist the motion with an even amount of inertia. If a force is applied off the centre of mass, it ...


0

If by "slow down" you mean "decelerate the rocket as it moves through space" the answer is no. Once the ship as begun traveling at a constant linear speed $v$ and begun rotation at a constant angular speed $\omega$ it will continue to move with a constant velocity (both linear and angular) in the absence of any net forces or torques. The object will have a ...


0

It can be proven that if you just want to know the motion the center of mass of a system of particles (or a continuous extended body), you can just calculate the vector sum of the force vectors on each particle (or the integral of the infinitesimal force on each infinitesimal bit of mass dm in a continuous extended body), and then treat this total force ...


0

I think there is no influence of rotation, regardless of rotation plane and direction vector. The center of mass will move with the same speed. Newton's First Law of motion should be enough to prove it: When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an ...


1

There are many variables that come into play in an armwrestling match, but one answer as to why a person with a larger hand and wrist has an advantage in armwrestling has less to do with strength, and more to do with leverage. In armwrestling, you and your opponent lock hands and attempt to pin each other to a pad on either side of the table. To pin your ...


2

If there is not enough friction to keep the vehicle in its circular path, it will skid. The force needed for the circular path is the centripetal force: friction (the force keeping the car on the road) must be greater that that. Now the no-slip condition (centripetal force < friction) implies = $\frac{mv^2}{r} < mg\mu$ . Your equation follows by simple ...


0

The centrifugal force (said pseudo force) pushes the liquid towards the walls of the cylinder on its entire length. Since the walls restrict further motion of the liquid from the centre of rotation, and the liquid is incompresible, it cannot squish any further acting on side walls only, and thus the bulk of it will start pushing on the top and bottom of the ...


0

$d$ can be infinite only in math books. Your statement about Coulomb's Law is true. if $F=0$ then $q_1q_2=0$. As $d$ gets arbitrarily large $F$ will get arbitrarily small. You can find a $d$ that makes $F$ as small as any number you choose ... except zero. I'm interpreting your question as being about Coulomb's Law itself, not any practical application ...


2

What Jim is talking about in the comments is a limit: for $d$ becoming increasingly large, $F$ becomes increasingly small. For example, if $q_1 = q_2 = 1\,\text{C}$ and the charges are a distance $d = 1\,\text{m}$ apart, we find that the force is $$F = k \sim 10^{10}\,\text{N}.$$ If the distance is made 1000 times bigger, the force is $$F = ...


1

One way to think of this conceptually is that the inertial mass of the system is in fact different from the gravitational mass. That is, while the net force on the systems are the same, the two systems have a different amount of mass, and so they resist changes in velocity to different degrees. Except that in this case, we are considering a compound ...


1

Yes you are right. To understand exactly why you are right, keep firmly in mind exactly what the "bending moment" and "shear" mean. When we say that the moment and shear are $M(x_0)$ and $\tau(x_0)$ at position $x_0$ we mean that: We imagine the beam cut at the position $x_0$ and we draw free body diagrams for the two sundered sections; In particular, we ...


2

Assuming that you want to replicate David Scott's Apollo 15 version of Galileo's Tower of Pisa thought experiment, I think that something like your experiment could indeed work. As Martin says, the boxes would need to be very low drag shapes: presumably you could make them out of a hollow very heavy metal - natural or depleted uranium would probably make ...


1

In the context of the homopolar motor you can find lots of resources on the Internet: e.g. http://blog.first4magnets.com/what-is-a-homopolar-motor-and-how-does-one-work/ In electromagnetism an EMF can be produced by changing the magnetic flux through a conducting circuit. However you can also produce a "motional EMF" by moving charged particles in a ...



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