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## 物理代写|理论力学代写theoretical mechanics代考|Force and Counter-Force

The best known example of a force is the gravitational force. If we let go of our book, it falls downwards. The Earth attracts it. Only with a counter-force can we prevent it from falling, as we clearly sense when we are holding it. Instead of our hand, we can use something to fix it in place. We can even measure the counter-force with a spring balance, e.g., in the unit of force called the newton, denoted $\mathrm{N}=\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}$.
Each force has a strength and a direction and can be represented by a vector-if several forces act on the same point mass, then the total force is found using the addition law for vectors. As long as our book is at rest, the gravitational and counterforce cancel each other and the total force vanishes. Therefore, the book remains in equilibrium.

Forces act between bodies. In the simplest case, we consider only two bodies. It is this case to which Newton’s third law refers: Two bodies act on each other with forces of equal strength, but with opposite direction. This law is often phrased also as the equation “force = counter-force” or “action = reaction”, even though they refer only to their moduli. If body $j$ acts on body $i$ with the force $\mathbf{F}{i j}$, then $$\mathbf{F}{i j}=-\mathbf{F}{j i}$$ According to this, no body is preferred over any another. They are all on an equal footing. We often have to deal with central forces. Then, $$\mathbf{F}{i j}=\mp F\left(r_{i j}\right) \mathbf{e}{i j}, \quad \text { with } \quad \mathbf{e}{i j} \equiv \frac{\mathbf{r}{i j}}{r{i j}} \quad \text { and } \quad \mathbf{r}{i j} \equiv \mathbf{r}{i}-\mathbf{r}{j}=-\mathbf{r}{j i},$$ where we have a minus sign for an attractive force and a plus sign for a repulsive force (see Fig. 1.1). Clearly, they have the required symmetry.

## 物理代写|理论力学代写theoretical mechanics代考|Work and Potential Energy

It may be easier to work with a scalar field than with a vector field. Therefore, we derive the force field $\mathbf{F}(\mathbf{r})$ from a scalar field, viz., the potential energy $V(\mathbf{r})$ :
$$\mathbf{F}–\nabla V .$$
But for this to work, since $\nabla \times \nabla V=\mathbf{0}$, $\mathbf{F}$ has to be curl-free, i.e., the integral $\oint \mathbf{F} \cdot \mathrm{d} \mathbf{r}$ has to vanish along each closed path. We conclude that a potential energy can only be introduced if the work
$$A \equiv \int_{\mathbf{r}{0}}^{\mathbf{r}{1}} \mathbf{F} \cdot \mathrm{d} \mathbf{r}$$

depends solely upon the initial and final points $\mathbf{r}{0}$ and $\mathbf{r}{1}$ of the path, but not on the actual path taken in-between (see Fig. 2.1). (Instead of the abbreviation $A$, the symbol $W$ is often used, but we shall use $W$ in Sect. $2.4 .7$ for the action function.) Generally, on a very short path $\mathrm{d} \mathbf{r}$, an amount of work $\delta A=\mathbf{F} \cdot \mathrm{d} \mathbf{r}$ is done. Here we write $\delta A$ instead of $\mathrm{d} A$, because $\delta A$ is a very small (infinitesimally small) quantity, but not necessarily a differential one. It is only a differential quantity if there is a potential energy, hence if $\mathbf{F}$ is curl-free and can be obtained by differentiation:
$$\mathrm{d} V \equiv \nabla V \cdot \mathrm{d} \mathbf{r}=-\mathbf{F} \cdot \mathrm{d} \mathbf{r} \equiv-\delta A$$
For the example of the central and tensor forces mentioned in the last section, a potential energy can be given, but it cannot for velocity-dependent forces, i.e., neither for the frictional noro for the Lorentz forcè (acting on a moving chảrge in a magnetic field). We shall investigate these counter-examples in Sect. 2.3.4.

If there is a potential energy, then according to the equations above it is determined only up to an additive constant. The zero of $V$ can still be chosen at will and the constant “adjusted” in some suitable way. The zero of $V$ is set at the point of vanishing force. If it vanishes for $r \rightarrow \infty$, then it follows that
$$V(\mathbf{r})=-\int_{\infty}^{\mathbf{r}} \mathbf{F}\left(\mathbf{r}^{\prime}\right) \cdot \mathrm{d} \mathbf{r}^{\prime}$$

## 物理代写|理论力学代写theoretical mechanics代考|Constraints: Forces of Constraint, Virtual

We can often replace forces by geometric constraints. If the test body has to remain on a plane, we should decompose the force acting on it into its tangential and normal components-because it is only the tangential component that is decisive for the equilibrium (as long as there is no static friction, since this depends upon the normal component). The normal component describes only how strongly the body presses on the support, e.g., a sphere on a tabletop.

Geometrically conditioned forces are called forces of constraint $\mathbf{Z}$. In equilibrium, the external forces cancel, whence $\sum_{i} \mathbf{F}{i}=\sum{i} \mathbf{Z}{i}$. We now consider virtual changes in the configuration of an experimental setup. In our minds, we alter the positions slightly, while respecting the constraints, in order to find out how much of it is rigid and how much is flexible. These alterations (variations) will be denoted by $\delta \mathbf{r}$. If there is no perturbation due to static friction, then the forces of constraint are perpendicular to the permitted alterations of position, and therefore the displacement $\delta \mathbf{r}$ does not contribute to the work. Since $\mathbf{Z}{i} \cdot \delta \mathbf{r}{i}=0$, we find the extremely useful principle of virtual work: $$\sum{i} \mathbf{F}{i} \cdot \delta \mathbf{r}{i}=0$$
In equilibrium, the virtual work of the externally applied forces vanishes. We do not need to calculate the forces of constraint here-instead, only the geometrical constraints must be obeyed. If only curl-free forces are involved, then the associated potential energy of the total system also suffices. Equilibrium prevails if it does not change under a virtual displacement: $\nabla V \cdot \delta \mathbf{r} \equiv \delta V=0$.

## 物理代写|理论力学代写theoretical mechanics代考|Force and Counter-Force

$$\mathbf{F} i j=-\mathbf{F} j i$$

$$\mathbf{F} i j=\mp F\left(r_{i j}\right) \mathbf{e} i j, \quad \text { with } \quad \mathbf{e} i j \equiv \frac{\mathbf{r} i j}{r i j} \quad \text { and } \quad \mathbf{r} i j \equiv \mathbf{r} i-\mathbf{r} j=-\mathbf{r} j i$$

## 物理代写|理论力学代写theoretical mechanics代考|Work and Potential Energy

$$\mathbf{F}-\nabla V \text {. }$$

$$A \equiv \int_{\mathbf{r} 0}^{\mathbf{r} 1} \mathbf{F} \cdot \mathrm{d} \mathbf{r}$$

$$\mathrm{d} V \equiv \nabla V \cdot \mathrm{d} \mathbf{r}=-\mathbf{F} \cdot \mathrm{d} \mathbf{r} \equiv-\delta A$$

$$V(\mathbf{r})=-\int_{\infty}^{\mathbf{r}} \mathbf{F}\left(\mathbf{r}^{\prime}\right) \cdot \mathrm{d} \mathbf{r}^{\prime}$$

## 物理代写|理论力学代写theoretical mechanics代考|Constraints: Forces of Constraint, Virtual

$$\sum i \mathbf{F} i \cdot \delta \mathbf{r} i=0$$

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