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For a 2-dimensional phase space, they are the same. More generally, for a $2n$-dimensional symplectic manifold $(M;\omega)$ with symplectic two-form $$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI},$$ the Poisson bracket is given by $$\tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} ... 2 The stiffness k of a coil spring can be expresses as:$$k=\frac{E\,d^4}{16\,(1+\nu)\,(D-d)^3\,n}$$Where E is the modulus of elasticity of the material, d is the diameter of the wire used in the coil, \nu is the poisons ratio of the material, D is the outer diameter of the coil, and n is the number of wraps in the coil. Now \frac{D-d}2 is ... 1 Check out the solution in the image. It will be dependent on the inclination of the plane. Here, a is the horizontal acceleration of the inclined plane. 1 Along with Landau and Lifshitz, there are books which although not explicitly about classical field theory, have good treatments. Chapters 11 and 12 of Jackson's Classical Electrodynamics are about special relativity and field theory, and I would recommend Goldstein's Classical Mechanics as an introduction, where field theory is introduced in some of the ... 1 I don't want to give away the answer directly. So I will provide some hints. A central force in polar coordinates has to be of the form:$$\vec{F} = m\vec{a} = m(\ddot r - r \dot \theta^2)\hat r$$Now try to mess around with your r(\theta) = a\theta^2  I believe you need to specify \dot \theta in order to solve the full equation of motion. So pick for ... 1 If you draw the free body diagram, you will get$$T = mg \sin \theta - \mu mg \cos \theta = mg (\sin \theta - \mu \cos \theta)  So, $T>0$ since $\theta$ is non-zero, $\mu$ is non-zero and $mg$ is also non-zero. It is due to the equilibrium that we get to write the above equation. The 2 forces, namely tension and friction are in same direction as the ...

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As stated, mathematically, any fraction of tension and friction is possible. Maybe this will help you to wrap your mind around this perhaps seemingly paradoxical situation: This math, and the mechanics it describes, is and can only ever be an approximation to the real world. Imagine you had actually balanced things so perfectly that the mass is supported by ...

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In the first case when the box is stationary your statement is correct and you asked no question about that case. In the second case, the box is moving and only the kinetic or dynamic friction is relevant. Assuming the crate you add on top of the box weighs the same as the box, the normal force doubles, and therefore the dynamic friction force doubles. This ...

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