Tag Info

New answers tagged

1

The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - ...


0

You are asking for Galilean formula for adding velocities(since boats speeds are not relativistic ;) ) You always add the speeds but in vector form. This means that you will get a minus if they are opposed and a plus if they are in the same direction. But if they act on different lines, you need to find the third side of the triangle formed having the two ...


2

Your conclusion that I think a pseudovector is actually a function of vectors equipped with preallocated arguments and every transformation on a vector has a counterpart on a pseudovector like $\tilde P $. is, I think, essentially correct; the reason parity is confusing is that we tend to drop the tilde. More precisely, whenever parity considerations ...


2

This all links back to the fact that any time you speak about a transformation, you require all the relevant physics to transform along. E.g. velocity would not transform the way it transforms without requiring the trajectory $x^i(t)$ it is tangent to to transform also. In the sense of the notation you use we have $v \equiv <x^i(t),\frac{d}{dt}>$ and ...


0

Since you are saying an object i am considering it to be a rigid particle.Now,since the particle strikes the surface as in your figure.it gives a downward force on the surface and hence the reaction is obviously upward. As you must be knowing normal reaction is perpendicular to the surface.So break the black arrow in componentsi.e you can imagine the ball to ...


3

The key here lies in two observations: If you can jump hard enough to escape the gravitational pull of the asteroid, this will permanently change the momentum of the asteroid; if you just jump and fall back down, nothing of lasting significance happens (conservation of momentum in the system) You want to change the angular momentum of the asteroid's orbit ...


2

The size of the asteroid does not matter: when you jump, the asteroid will move in the opposite direction. Even if you jump up on earth, the earth moves backwards - but only infinitesimally so. Remember that $$F=ma$$ The mass of the earth is $5.97219 × 10^{24} kg$. If your mass is $100 kg$, then the earth's acceleration will be $6 × 10^{22}$ (almost ...


3

Can we add any two vectors? My book says it is true For example, adding acceleration to velocity (seems impossible). Quite simply, your book meant two vectors of the same type. It's just that simple. As you thought, you can not add two vectors that are "different things"! (If you're just getting started with vectors. Note that indeed vectors have a ...


7

From the commentary to the question, the textbook in question appears to be a mathematics textbook rather than a physics textbook. In mathematics, any two elements of a vector space can be added to one another to yield another member of that space. This is one of the requisites of what it means for something to be a "vector" in mathematics. Specifically, a ...


0

You can ague that you can add any vector, since you can look at a adding vectors with different units as other dimensions. So you example of adding velocity and acceleration, both in three spacial dimensions, will give you a six dimensional vector. An example of this would be phase space. However usually those vectors with higher dimensions do not have any ...


9

I doubt if your textbook makes it explicit, but the only sacred tenet in here is to respect dimensional homogeneity. One can make no sense of the sum of quantities with different dimensions.


0

It sounds like the Student may be jokingly quoting the movie Dispicable Me, where a Character names himself Vector "because he has Magnitude and Direction". That said, the difference is how much infomation you get. If I tell you "the car is going west" that tells you a direction without telling you anything about how fast or how far. If I tell you "the ...


3

This is the diagram you need to draw: The red arrow is the net velocity of the boat - the sum of 25 km/h going due North, and a current of 10 m/s at 60 degrees East of South. (not drawn to scale) Just do simple math to determine the size of the two green segments - then compute the angle (heading) from the arctan of their ratio.


0

Resolve the vector components. Then solve the individual tensions. T3 = mg = 5*10 = 50 T2 sin 45 = T3 T2 sin 45 = 50 Therefore, T2 = 50√2 T1 = T2 cos 45 T1 = 50√2 cos 45 Therefore, T1 = 50 Hence, T1 = 50 T2 = 50√2 T3 = 50 Cheers.


1

You have a few basic formulas for solving this kind of stuff: $$\vec{x}_f=\vec{x}_i+\vec{v}_it+\frac{1}{2}\vec{a}t^2,$$ $$\vec{v}_f=\vec{v}_i+at.$$ These are vector formulas, but all you're doing with the vectors is adding/subtracting them. When you add vectors you add the individual components, i.e. $$\vec{a}=(a_1,a_2,a_3)$$ $$\vec{b}=(b_1,b_2,b_3)$$ ...


1

Recall that when classifying representations of the Lorentz group, we consider \begin{equation} \textbf{N}_{\pm} = \frac{\textbf{J} \pm i\textbf{K}}{2}, \tag{1} \end{equation} where $\textbf{J}$ is the angular momentum (rotation generator) and $\textbf{K}$ is the boost generator. The generators $\textbf{J}$ and $\textbf{K}$ satisfy \begin{equation} ...


3

For simplicity let's work in 2D, and take as our axes two unit vectors $\hat{e}_i$ and $\hat{e}_j$. We'll consider some vector $\hat{F}$: In our coordinates we can write the vector as $(F_i, F_j)$, where $F_i$ and $F_j$ are just numbers. The vector $\hat{F}$ is then expressed as the vector sum: $$ \hat{F} = F_i \hat{e}_i + F_j \hat{e}_j $$ I've drawn ...


1

The prime on ∇′ means that you're differentiating with respect to the primed coodinates (x',y',z') treating (x,y,z) as constants. Presumably, r is the magnitude of the vector that points from P' to P-- this is just a simple difference of vectors. Good luck!


3

Let's consider this sketch of a man with legs asymmetrically astride: weight : ca.80kg (800N, g = ca. 10) height : 180 length of leg: 1m fore angle 45° rear angle ca. 15° Distribution of weight The rear leg is 73.2 cm and is 19 cm behind the center of mass. Let's assume that the legs are in compression only, (if the person were standing on ice or ...


0

Resolution of force, or any vector in general, into its components along a particular choice of co-ordinate vectors $\{e_i\}$ is not specific to any particular assumption about the medium/space. It is convenient to employ an ortho-normal set of co-ordinate vectors (e.g. ${\hat i} \cdot {\hat j} = 0$ in the Cartesian case), but even if the vectors aren't ...


1

A square has its diagonals at right angles. So, find the forces along each diag. i.e., m1 and m3 on m5 for one direction and m2 and m4 on m5 for other. You could actually simpy find difference between, "m1 and m3" and "m2 and m4" and the corresponding charges and use them as resultant mass and charge.


1

Use the fact the masses m1 and m3 apply forces along the diagonal. Similarly for the other diagonal. Then convert to the cartesian solution using the two resultants.


5

There is a definition that $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation is equal to spinor tensor $$ \psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}}, $$ where $\psi_{\dot{b}}$ transforms as complex conjugation of $\psi_{b}$. Why do we assume that $\left( \frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$ represent spinors? You can think about ...


3

First consider the spatial component of the equation above, you get $$ \frac{d}{d\tau} \left( \gamma(v)\overrightarrow{v} \right) = 0 $$ But $\gamma(v) = \frac{dt}{d\tau}$ so you have $$ \frac{d^2 t}{d\tau^2} v + \gamma(v) \frac{d\, \overrightarrow{v}}{d\tau} = 0 $$ Now since $$ \frac{d \, u}{d\tau} = 0 \implies \frac{du^0}{d\tau} = \frac{d^2 t}{d\tau^2} = 0 ...


2

Whenever you're confused about how to calculate some quantity, try going back to the definition. Think carefully about what the definition of average acceleration is: $$\vec a_\textrm{avg}\equiv\frac{\vec v_\textrm{final}-\vec v_\textrm{initial}}{\Delta t_\textrm{elapsed}}.$$ Which velocities does this equation depend on? Which velocities does it not ...


0

I found the answer to be .756N by multiplying .63N and sin(36.87).


0

Here is a big hint... I have to believe you can finish it from here. The key is "similar triangles".


0

If you are just interested in the rate of precession, then $r$ just has to be the distance from the support (tip of the top) to the center of mass - this is the distance that gives rise to the torque through $mgr\sin\phi$ However, if you want to do this as a vector equation, then you have the instantaneous angular momentum (vector) of the top $I\vec\omega$ ...


2

This convention is extremely convenient when doing things that physicist often like or need to do, such as computing a flux through a surface. When the area vector is chosen normal to the surface, one can simply take use an dot product to get what you're looking for. In the context of differential forms, this also turns out to be the natural definition, ...


1

In this context, this is just a natural definition that turns out to be useful. For example, the flow rate of a fluid through a plane is the dot product of the fluid velocity with the area vector. [There is a more mathematically sophisticated way of understanding this, which is that the vector area is really a bivector or two-form. In three-dimensional ...


3

Note that: \begin{equation} \begin{aligned} \langle j | A \rangle & = \langle j |\left( \sum\limits_{i} a_i |i \rangle \right) \\& = \langle j| \left(a_1 |1 \rangle + a_2 |2 \rangle + a_3 |3 \rangle + \cdots + a_j |j \rangle + \cdots \right) \\& = a_1 \underbrace{\langle j | 1 \rangle}_{=\delta_{j1} = 0} + a_2 \underbrace{\langle j | 2 ...


2

If you think in terms of vectors and matrices with indices, $a_i$ is a vector and $\delta_{ij}$ is the unit matrix. The summation on the right hand side of your equation represents nothing but a multiplication of the vector $a_i$ with the latter. Since free indices are preserved, the remaining object carries the index $j$. You can write down an arbitrary ...


1

Is there any physical significance to this matrix The physical (geometric) relevance to the matrix $$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{matrix}\right|$$ with regard to the cross product $\vec{a} \times \vec{b}$ is 1: that the three vectors $\vec{i}$, $\vec{j}$, and ...


1

To make it easy to see the logic behind, imagine that the first vector is directed to $\vec{i}$ and the second to $\vec{j}$. You get $$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & 0 & 0 \\ 0 & b_j & 0 \end{matrix}\right|$$ and according to Wikipedia the product in this case is $ab\vec{k}$. There are some physical ...


2

The cross product is defined to be the vector which is perpendicular to both vectors, so for instance the force exerted on a rod moving in a magnetic field is perpendicular to both its velocity and the field, hence is given by their cross product. Now if you work out which vector is perpendicular to both vectors you get the determinant of the two vectors ...


1

In general the base vectors ($\vec{e}_{i}$) are not constant, e.g., in polar coordinates the radial vector do not point in the same direction (unlike the Cartesian base vectors $\hat{i},\hat{j},\hat{k}$. If one takes the j-th base vector $\vec{e}_{j}$ and consider its change if one moves on the direction defined by the i-th vector, mathematically this is ...


0

Forces are not vectors. Forces are 3D lines, with a magnitude and a pitch. A force acts through a line in space. From geometry you need 4 quantities to fully define a 3D line. This is why you can slide a force vector along its direction line. A magnitude described how much force along this line and you need 1 quantity to describe this (duh!) for a total ...


0

tl;dr: you can't superimpose forces when they're acting on different things. An object is not the same as its center of mass. Under certain conditions (rigid body, force directly in line with the center of mass, etc.) you can assume the object is a point particle at the location of its center of mass and thus you can add up the force vectors because they're ...


1

When you translate-and-add, you do get the correct vector. The point of application is something else, not part of the definition of the vector. Additional information that must be supplied. The addition you refer to in your last paragraph is not "something else". It is vector addition. I'm surprised that more people don't have the confusion that you ...


3

In 3 dimensions, sliding the forces on the line of application until the forces meet does not always work because the two lines of application may be skew. To capture the line of application it is good to work with pairs of forces and torques. (You can transform the line of application of a force by adding a torque acting on the rigid body. This way you can ...


0

I think what you are arriving at, will be rotational mechanics. As long as objects just translate, you can just sum up the forced, and calculate the results for on the object, as you understand. When an object is free to rotate, the story becomes a lot more complex and the point of exertion of a force becomes a lot more important. You can easily test that ...


1

Is the cross product of two vectors always perpendicular to both? No. The cross product is the zero vector if one or both of the vectors is the zero vector, or if the two vectors are parallel or anti-parallel to one another. The angle between the zero vector and some other vector is indeterminate. What you can say is that $(\vec A \times \vec B) \cdot ...


1

The cross product of two vectors, be there any angle between them, is perpendicular to the two vectors. It is, more precisely, perpendicular to the plane containing the two vectors. The direction is determined by the right hand thumb rule/corkscrew rule. The magnitude of the vector is given by $|{\bf A} \times {\bf B}| = |{\bf A}| |{\bf B}| \sin\theta$.


1

Have a look at this article on vector algebra. In this context sense is a technical term. The relevent extracts from the article are: Vectors are quantities that require the specification of magnitude, orientation, and sense. The characteristics of a vector are the magnitude, the orientation, and the sense. The orientation of a vector is ...



Top 50 recent answers are included