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7

$$\vec{F}_g=\frac{Gm_1m_2}{|\vec{r}_{ij}|^3} \vec{r}_{ij}=\frac{Gm_1m_2}{|\vec{r}_{ij}|^3} |r_{ij}|\hat{r}_{ij}=\frac{Gm_1m_2}{|\vec{r}_{ij}|^2} \hat{r}_{ij}.$$ It's just one way textbooks write it, and is exactly equivalent to the right-most expression, which is probably the most obvious way to write the gravitational force in vector notation.

6

For an open orientable surface there are two possible, equivalent normals: $\vec n$ and $-\vec n$. The usual convention is that you choose a direction in which the perimeter of the surface is traversed and define the positive direction as the direction given by the right hand rule, as shown in the following picture. This can also be done for non-simply ...

6

I'm not sure if it helps you with your students, but maybe gives you some background: I guess the underlying reason for orthogonal basis vectors is that you are implicitly using a euclidean metric that will just have diagonal values. These would e.g. be $$g_\mathrm{\mu\nu, ~euclidean}=I=\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \... 5 In vector notation Newtonian force of gravity is$$ \vec F = \frac{GMm\vec r}{r^3}, $$where r = \sqrt{(x_1 - x'_1)^2 + (x_2 - x'_2)^2 + (x_3 - x'_3)^2} and the radial vector \vec r = \vec x - \vec x'. we can consider the unit vector \hat r = \frac{\vec r}{r} We can the write the vector notation as$$ \vec F = \frac{GMm}{r^2}\hat r = F_g\hat r.  ...

4

The converse is not a "theorem" in that you can't prove that it is true, even assuming Maxwell's equations as axioms. It is simply a "hunch" that $|\vec{S}|$ represents the power intensity and $U=\frac{1}{2}\,\epsilon\,|\vec{E}|^2+\frac{1}{2}\,\mu\,|\vec{H}|^2$ the energy density. What you can prove (and what you already understand) from Maxwell's equations ...

3

A unit vector $v$ is a vector, whose norm is unity: $||v||$. That's all. Any non-zero vector $w$ can define a unit vector $w/||w||$. A basis vector is one vector of a basis, and a basis has a clear definition (it's a linearly independent family of vectors which spans a given vector space). So both have nothing to do. Your confusion may come from the fact ...

3

We don't always use orthogonal coordinate frames. For example working with three phase motors it's sometimes convenient to work with a three axis coordinate system in a plane. Convenience, simplicity set aside, the main reason we most often work with orthogonal reference frames is the concept of dimension. We can express an n-dimensional linear system as a ...

2

All vectors, except $\:\mathbf{r}\:$, are infinitesimals. I wonder if the author (Irodov) makes use of this result anywhere in his textbook. EDIT The infinitesimal rotation of a vector $\:\mathbf{r}\:$ around the direction of a unit vector $\:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right)\:$ by an infinitesimal angle $\:\mathrm{d}\theta\:$ may be represented ...

2

It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...

1

The torque supplied by $F_m$ results in the torque due to $F_v$, so these torques are equal : $\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either $\vec A \times \vec B = (AB \sin\theta) \hat k$ where $A$, $B$ are magnitudes and $\theta$ is the angle between (b) or $(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ... 1 Difference between real and absolute value in general: Look at count_to_10 's answer. For acoustics and preasure measurement: Absolute pressure - pressure against perfect vacuum. Real pressure: Usually defined as the pressure against a reference-environment. Also called differential pressure. For example the pressure of the air inside a football against the ... 1 By convention, for a flat lamina or a plane surface, the area vector is a vector whose magnitude is the area of the surface and whose direction points in a direction perpendicular to the surface. If you have a curved surface, then you have to consider elemental areas, i.e: small patches of area denoted by$dA\$ whose direction is perpendicular to the small ...

1

They are linked by the "law of the right hand": The preferred direction of dℓ⃗ dℓ→ along the loop is that from the palm to fingertips of your right hand when it surrounds the loop. Then, the associated preferred direction of dA⃗ dA→ is indicated by the thumb.

1

The head to tail does work here and the answer is the difference in magnitude. Try imagining a triangle and then collapsing it so that the vertex lies on the opposite side. You'll understand why it is difference in magnitude of the vectors.

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