# Tag Info

5

To augment what tpg2114 wrote, work is usually defined as: $$W = fd$$ Where $W$ is the total work done, $f$ is the force, and $d$ is the distance (actually it's the $\Delta d$ caused by the force). As long as $d = 0$ no work is being done. For example, if you're standing on the surface of the Earth, energy is not being consumed to keep you from falling ...

2

Since $f = -\frac{dE}{dx}$ you don't need "energy" per se, you just need a gradient of energy between the two states. On the other hand, force is independent of work because it's possible to have forces that do no work.

1

If you know about Lagrangian and Hamiltonian formalisms yo might try to find first the equations of motion. This is done in the paper Relativistic harmonic oscillator. In a nutshell, what is done is the following, a "relativistic" hamiltonian (for slow particles) is set up (we set c=1): $$H = \sqrt{p^2+m_0^2} + \frac{1}{2}k x^2$$ Then the evolution of a ...

3

If you try jumping on a trampoline, you will notice that when you jump up, the trampoline bends and stretches underneath you. It stretches some even if you stand still, but it stretches extra when you jump. The trampoline is elastic. When it's stretched, you can feel it pulling back towards its normal shape. Thus, just before you jumped, the trampoline was ...

2

The string contacts the point on two infinitesimally close points with different slopes. Imagine a small pulley end the two points are the entry and exit point of the string. If the string is between points A on the left and point B on the right (with B lower) then we call the angles of the string from horizontal $\theta_A$ and $\theta_B$. If the mass is ...

0

Not sure i understand fully your question but in general friction can be seen as a force (vector) pointing in the opposite direction of motion with (if there is a motion). Moreover, the force is tangent to the surface of contact. Thus for an object with spherical symmetries (like a pulley, cylinder, sphere...), the force is perpendicular to the radius.

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See the diagram for guidance : http://picpaste.com/p012-Lrn6pmdn.jpg Draw the gravity vector and the centripetal vector, then the resultant of the two, the banking should be normal to this resultant for no side forces. m = car mass in kg g = local gravity rate in (m/s)/s. v = velocity in m/s. r = radius to centre of gravity in metres. A = banking angle in ...

1

You are right in thinking that the car's acceleration is what keeps it in place, but it is important to remember that an object moving at a constant speed in a circle is accelerating (despite not speeding up). The reason for this is that acceleration is defined as a "change in velocity," and velocity is a vector quantity (i.e. it has magnitude and ...

2

Is there an equivalent formulation of classical electrodynamics in terms of action at a distance that is completely equivalent to the formulation in terms of fields (Maxwell's equations)? Yes, but only if special boundary conditions on the fields are assumed. For example, if the fields are purely retarded (wiki: Retarded and Advanced Potentials), one ...

0

If you have a center of mass C located at $\vec{r}_C$ and a force $\vec{F}$ passing through an arbitrary point A (at $\vec{r}_A$) then the net moment of the force about C is $$\vec{M}_C = (\vec{r}_A-\vec{r}_C)\times \vec{F}$$ where × is the vector cross product. Your equations of motion sum up all the forces and moments at the center of mass $$\sum_i ... 0 You more or less know the answer (but perhaps you need to rephrase your question). If the question is "for a given force vectors \vec{F}, when should I consider its effects on linear velocity \vec{v} /linear acceleration \vec{a}? and when should I consider angular velocity \vec{\omega} / angular acceleration \vec{\alpha}?" (1) If a force vector ... 0 To invoke Feynman in the interview, it can be rather unsatisfying to invoke virtual particles to explain forces in terms of something else more wonted to you. Virtual particles are wholly internal to the analysed system and so individually do not heed normal conservation laws - they are "off-shell" (i.e. off the light cone defined by E^2-p^2 c^2 = m^2 c^4) ... 0 The moment of inertia is the rotational mass of the object and it is solely a function of mass distribution - the shape. Usually it does not matter if it is rotating, but if the rate of rotation is high enough the body could deform. This is a concern in the connecting rods of high speed engines. The MI does depend on shape, so it will change if the ... 0 To state it simply, friction is the resistance to motion of an object within a system, in this case a ruler on a desk. As you suggest in your question the normal force to the surface is important to friction, the equation is: Coefficient of friction = force required to maintain constant velocity / normal force however turning the ruler on its side does ... -2 restoring force is refer to the system which bring force back to it normal position with out the effect of distance which the force existed to it Constance force. F1+F2= total force. F= kx 1 You didn't specify in what direction the force of hand is applied, so for simplicity I assume that you are applying the force perpendicular to the desk. Now there are four forces on the book: 1) Gravity (mg) is trying to take the book down; it has a component mg\cos\theta that is perpendicular to the desk and a component mg\sin\theta that is parallel ... 1 When you hit the obstacle (if you don't destroy it) your speed goes down to zero in a quite small time. That gives you the acceleration. It starts when you start hitting and it ends when you come to a complete stop. This acceleration is due to the force that the obstacle generates on the car. Think about: \displaystyle a=\frac{\Delta v}{\Delta ... 2 You are right that in this theoretical problem the breaking force to achieve constant velocity is the same for all velocities. You either stipulate that the bike has the desired speed as a initial condition, or that it coasts with no breaking applied until the desired velocity is reached. In the real world, there will always be some friction or drag forces ... 1 Draw a force diagram and determine the braking force required to counteract the (portion of) gravitational force pulling the bike down the slope. If the net force is zero, delta-v will be zero. I think the problem you were given assumes the bike starts out with velocity v . If it doesn't then you'll need to derive a braking force whose magnitude varies ... 0 If you mean what I think you mean, then your idea won't work as well as you're hoping. Let me illustrate with a rhetorical question: Would you sky dive with a helicopter rotor as your parachute? I certainly hope you wouldn't. Upon brief reflection, you will see that your "propeller-chute" would not provide very much drag as you hurtled groundward. As a ... 1 However, the problem says that the floor pushes the man upwards with a stronger force than his legs. This can not be true. The floor pushes the man upwards with as much force as his legs exert downwards. However, this happens to be more than his weight, which is why he accelerates upwards. 0 You are completely correct. Any non-ideal, real-world force is really a pressure. Any force in the real world must be applied over some area (even if it's very, very small) However, in many problems, it is sometimes practical to model a pressure as a single force applied in a single direction at a single point — in many systems, pressures can be (exactly) ... 0 There are certain so-called "No-go" theorems in physics: http://en.wikipedia.org/wiki/No-go_theorem - one classic one is Earnshaw's Theorem, showing that a collection of point charges interacting with Coulomb forces is always unstable (http://en.wikipedia.org/wiki/Earnshaw%27s_theorem). (Of course there are caveats.) However, it is important to be ... 2 The window and the force have to have the same force but on each other. This is dictated by Newton's third Law of Motion. The rock exerted a force on the window which was not enough to break it. The window, in turn, exerted an equal and opposite force on the rock which was quite enough to make it reverse its direction of motion. This is what is happening in ... 3 The question is asking what is the force exerted by the water on either of the two faces of the plate. The net force will be zero as force on either side sides cancel, so your intuition made sense. The force on a side comes from water pressure across the triangular surface. The pressure at any point on the triangle depends on the depth of that point. ... 0 Carl's first paragraph answers your question (though I disagree with his second paragraph) so this is just an addendum to Carl's answer. It sounds to me as if you are describing an ideal conical pendulum. You're correct that no work is done because the two forces, the string and gravity, act at right angles to the direction of motion so \vec{F}.\vec{dr} ... 0 There is force exerted (along the string) thru the point at which the mass's string is attached to whatever is holding it up. If there weren't, yes the ball would fly off. Gravity is a separate force, and will cause the mass to oscillate about the vertical axis. So in fact you won't be able to achive uniform circular motion. Take a look at the toys you ... 1 If a net force is towards the positive x direction, then the x component of velocity of the object is increasing; period. Now, knowing this, are any of the statements (a) through (d) true? a) It can be moving in the negative x direction Sure, it could be moving in the negative x direction. All that is required is that the velocity is ... 1 The only thing that a net force in the positive x direction means is that the net acceleration is in that direction. It tells you nothing about its position or velocity. a) Imagine placing the x-axis vertically and letting the positive x direction be downward. Now imagine that the constant force is gravity and the particle is a ball. If you throw the ... 1 a), b), c), d) All yes. Imagine a ball that is moving at 10 meters per second to the left (negative x direction). Now, a force in the positive x-direction will slow it down initially, and finally make it go to the right, constantly accelerating to the right. This is all independent of the y-direction, in which it might also be moving. A force in the ... 0 Short answer: If you're talking about the spatial components of force, then yes, through a Lorentz transformation on the force four-vector. By "real force" I assume you mean a non-inertial force, so that you are computing force from an inertial frame, or, more generally, in a freefall frame - i.e. a locally flat Minkowskian tangent space to the spacetime ... 0 Well, take the electromagnetic force...it has been shown that the induced magnetic field around moving charges is a relativistic reference frame effect. See this post How Special Relativity causes magnetism Therefore a static electric force in one frame becomes a magnetic force in another. However, from a Newtownian view, we either allow pseudoforces and ... 0 The magnitude does not change with frequency, because the formula for an even wave is$$ \Psi(x) = A\cdot e^{i(kx - \omega t)} Where A is the amplitude and \omega is the frequency. So, if you change the frequency, the amplitude does not change. 0 If the symmetry axis of the cone lies along the direction of travel, and if you are using the drag equation \begin{align} F_\mathrm{drag} = \frac{1}{2}\rho v^2A \end{align} to compute the drag force, then you should take A to be the area of the base, namely the full area that would be obstructing your vision if you were looking at the object coming ... 1 When I read about this question, I was thinking, well, "a" can't be zero because it should represent your deceleration from 900 miles/hour. The resulting F would be the force you experience when you decelerate. (The deceleration depends on how you hit the obstacle, how long did it take you to stop/slow, across what distance, and the degree of deformity of ... 4 Another way to think about Newton's second law (and the way he originally defined it) is F=\dfrac{d\rho}{dt}, where \rho=mv is momentum and \dfrac{d\rho}{dt} is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of ... 6 You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a F = ma calculation where a is the acceleration of your gas pedal. Instead in the collision you have a force F resulting from the collision and you ... 0 The translational accelleration will be the force divided by the mass. The cross product of the force vector with the vector from the touch point to the center of mass is the torque applied to the object. 1 Scalar fields do transfer momentum in classical physics. Just take a look at acoustic signals in a gas. A strong sound can cause your windows to rattle. A well known example of energy transference by means of sound (pressure waves) is demonstrated with tuning forks. Quantum theory speaks of sound as particles (phonons), the discrete quanta of quantized ... 1 The contact forces with two blocks are N_1 = m_1 g + m_2 g for the bottom block (to the floor) and N_2 = m_2 g for the top block (to the 1st block). The available traction is F^\star_1 = \mu_1 (m_1+m_2)\,g and F^\star_2 = \mu_2 m_2\, g or \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} ...

0

If you remove a bit of moon and leave the rest with the same velocity, it will continue to follow the same orbit. Since you are carrying away some of the mass, as you carry it away it will exert a gravitational force on the moon, which could change the velocity. You are correct that cutting the mass in half will cut the gravitational force in half, but so ...

0

At very low energy one can consider effective forces such as Vanderwaals forces and induced dipole magnetic forces ("split off" from the electromagnetic force), or the nuclear force ("split off" from the strong interaction). "Very low energy" is basically the world you see around you through your own eyes, so I'm sure you can make up more examples yourself. ...

0

I believe your confusion comes from a misunderstanding of the designation of a force as "centripetal". Any calculation of centripetal force is telling you how much force is needed to make a circular motion take place. This doesn't create the force. There is no guarantee that a force of the calculated size and direction actually exists! You need to go ...

1

Since the arrangement is symmetrical when you interchange N and S, I would say that the maximum strength will come when D2 and D3 are equal. Also, the smaller D1 becomes, the more the flux lines will become compressed. Also, a sheet of metal where you indicate would have almost the same effect as making the magnets wider, without an increase in mmf. So I ...

1

In the cases where you have static friction, the forces will always be defined by the looking at the system and applying the constraints(in other words $F_s\le \mu N$ will only give an upper bound). On the other hand when you are dealing with kinetic friction, it can be easily derived from the famous $F_k=\mu N$. As an example, let's solve this problem(As ...

2

All forces act in pairs, so let me start by matching them up: Force on $M_1 = F = - M_1$ on Some force providing device Surface on $M_1 = F_1 = -M_1$ on Surface $M_2$ on $M_1 = F_2 = - M_1$ on $M_2$ The values for the forces horizontal components are found using... $F =$ Given (1 Newton) $$F_{sf} \le \mu_{sf} \cdot F_n$$ $$F_1 \le \mu_1(M_1+M_2)g$$ ...

0

Thickness for a window made of a brittle material (like most glasses) is Thickness = sqrt( pressure * radius^2 / Modulus of Rupture ) Where Modulus of Rupture is roughly the tensile strength, it's listed for most materials in data books or http://www.sgpinc.com/materials.htm Be careful that the pressure and modulus are in the same units . In real life ...

0

1b) fibontic correctly pointed out that your expression for Newton's 2nd law is not correct. It should be $$ma=F_\text{net}.$$ You have an $x$ instead of an $a,$ which is causing one of your problems in part b. By writing $ma=\rho Vg-mg,$ you should be aware that you've already implicitly imposed a coordinate system where up is positive. This is probably ...

0

No. If you take all your masses and and squeeze them into a tiny volume, although the density increases, the normal force is the same. Loosely speaking consider the following: Density + Volume = Mass, $m = \rho V$ Mass + Gravity = Weight, $W= m g$ Weight + No Motion = Equal and Opposite Normal Force, $N=W$ Normal Force + Area of Contact = Contact ...

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This is my attempt to illustrate what happens when the car wheel is turned: Focus on the bit of the car tyre marked with a red spot, and the bit of the road marked with a green spot. If we could look at the contact patch between the tyre and the road we'd see something like the rectangle I've drawn on the left. When the wheel is straight the red spot on ...

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