# Tag Info

0

I'm not sure whether you should ask textbook problems or homework problems here. Just keep in mind that like-charges repel each other. So charge_1 tries to push charge_0 away from it (as both are positive charges). So the force on charge_0 should be in negative Y direction. That was more than a hint, now you tell me the vector form.

3

Torque is a mathematical object called a bivector, produced by taking the wedge product of $\mathbf{r}$ and $\mathbf{F}$. Bivectors can be thought of as area elements of planes; the magnitude is equal to $rF \sin \theta$ and the plane itself contains $\mathbf{r}$ and $\mathbf{F}$. By a great coincidence, in three dimensions, there are three distinct planes ...

2

Torque is a vector whose direction is always out of plane. The same with angular velocity $$\vec{\tau} = \vec{r} \times \vec{F}$$ $$(0,0,\tau) = (x,y,0) \times (F_x,F_y,0) = (0,0,x F_y - y F_x)$$ $$\vec{v} = \vec{\omega} \times \vec{r}$$ $$(v_x,v_y,0) = (0,0,\omega) \times (x,y,0) = (-y\, \omega,x \,\omega,0)$$ I always think of planar quantities as a ...

0

Update: The answer is 120 degrees using the angles between all the axis x = cos^-1 (+/- sqrt(1-cos^2(y)-cos^2(z)) x = cos^-1(+/- 0.493) x = 60 or 120

0

The angles given give you 2 out of 3 of the direction cosines, namely $\gamma_y$ and $\gamma_z$. What relationship do the three direction cosines $\gamma_x$, $\gamma_y$ and $\gamma_z$ fulfill? Recall that they are the components of a unit vector, since they are the lengths of the projection onto the three axes of a unit vector pointing along the boom.

0

The acceleration, $\frac{d}{dx} v$, is always tangent to the velocity vector. You could imagine a circular orbit where the velocity vector always is tangent to the ground itself and where the acceleration is pointing downwards (since the force is pointing down and $a = \frac{F}{m} = \frac{1}{m} F$). Since the constant $\frac{1}{m}$ is just a scalar, the ...

0

As HDE 226868 says in his comment, the relationship between velocity and displacement is important. This problem is about an infinitesimal time period. The average, the total, and the instantaneous velocity are identical during this infinitesimal. Average velocity over any time period equals the total change in displacement divided by the time. For an ...

2

Tension is an internal force in a body, such as a rope, that resists any attempt to pull the rope apart. Simply, tension arises due to intermolecular interactions, and if it did not exist, ropes would fall apart the moment you pull on them. Now, it is necessary to distinguish between internal and external forces for a body. External forces are forces that ...

1

We treat the string/rope like another object. This object exerts forces on other objects such as the hanging mass (in your picture). However, a string, by its very nature, can never "push" another object, it can only "pull" another object. That "pull" is a force which we give the name tension. Thus, tension will point away from the mass in the direction of ...

1

The tension on a string between two objects (note: your top diagram shows t̲h̲r̲e̲e̲ objects) is analogous to the force between two objects elastically colliding. The force exerted by the one end of the string is opposite and equal to the force exerted by the other end of the string; both forces must be parallel to the string and pointed towards its center. ...

0

In your expression for $E_x$, $q_2$ should be minus. Plus is for positive charge & minus is for negative charge. The direction of electric field from a point charge is taken to be radially away from that charge i.e. positive-ness. The direction of electric field is taken to be radially towards the charge i.e. negative-ness. $cos(45)$ is coming because, ...

4

It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form $$\vec E \propto \frac{\vec r}{r^3}$$ which, when taking the absolute values, yields the form you are probably more familiar with $$E \propto \frac{1}{r^2}$$ since $\lvert \vec r \rvert = r$.

1

Your strategy is fine (though I haven't carefully checked if your rotations are correct). Since rotations don't commute, you need to be careful to apply them in the correct order. You say that your rotation about the x-axis, let's call it $R$, followed by your rotation around the z-axis (be careful whether it's around the old or the new z-axis!), let's call ...

4

To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$ where ...

1

You can write $$\text dV = - m v \text d v$$ where $v$ is the velocity.

0

With given vectors for acceleration and velocity, is there a way to determine if a body accelerates or decelerates at a certain time-interval? Given the velocity vectors of a (specific) body (with respect to specific members of a suitable reference system) throughout a trial, in particular the velocity vectors $\mathbf v_{\text{initial}}$ at the ...

1

$v$ isn't referring to either $v_1$ or $v_2$, necessarily; $v_1$ is representing the vector before it moves, and $v_2$ is the vector after this movement. If we are working in polar coordinates (the reason he is using $v_\perp$ and $v_\parallel$), then let's suppose this small movement isn't changing the magnitude of the vector, it is just changing the ...

1

The introductory quote defines $\tau$ and $\nu$ to be unit vectors. That means that their magnitude is 1. $\mathbb{i}$ and $\mathbb{j}$ are unit vectors along $x$ and $y$ axes. That's a standard notation. Some of the comments are stressing the fact that the basic understanding of vectors is missing. I have a feeling that you've jumped into a 'complicated' ...

1

In ordinary vector spaces, the dot product $\cdot$ is a binary operator which takes a pair of vectors $(A,B)$ in the space to the field over which the space is defined. Formally, for a vector space $V$ over a field $K$, the dot product $(\ \ , \ )$ is a bilinear map $$(\ \ , \ ): V \times V \to K.$$ The inner product only has assumes the standard meaning ...

1

The equilateral triangle only tells you the directions of the vectors. The vectors must be floated'' so that they all act on the point. Each vector has its own length. Here's the diagram:

0

Conventionally, the length of a vector is proportional to the magnitude of whatever it represents. The vectors here are representing the direction only. With a representation of this sort, the resultant of 3 vectors forming an equilateral triangle is zero, only if their magnitude is the same. Because their magnitudes differ, they can't cancel each other out! ...

1

You know from the question that the forces are in the same direction as the sides of the equilateral triangle, so they are at 60 degrees to each other. They all have different magnitudes so cannot possibly form an equilateral triangle if placed "end-to-end". To work out the answer, you can calculate the horizontal and vertical components of each of the ...

2

Here is a bare bones easy way to see that coordinate tuples are not 4-vectors. Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation: $x' = x + A$ $y' = y$ $z' = z$ $t' = t$ Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors ...

4

Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place. Co-ordinates are absolutely not vectors, they are labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as ...

17

You are correct. Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to ...

1

A representation of the Lorentz group implies a set of matrices $(S_{\mu\nu})_a{}^b$ (one for each $\mu,\nu$) that satisfy the Lorentz algebra, i.e. satisfies a relation of the form (I am not keeping track of signs) $$[S_{\mu\nu} , S_{\rho\sigma} ] = i ( \eta_{\mu\rho} S_{\nu\sigma} + \cdots )$$ What this means is that there is a vector space, with vectors ...

3

Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as  U(\delta + \omega, \epsilon) = 1 ...

Top 50 recent answers are included