Tag Info

New answers tagged

0

There is a principle which states that for a rigid body: The net loading can be represented as a 3D line, a magnitude and a pitch. This is the loading screw or wrench. The not motion can be represented as a 3D line, a magnitude and a pitch. This is the motion screw or twist. For both cases the parallel separation of the point of interest from the screw ...


0

Peak torque is a pretty useless number. You can gear torque up or down anyway you like. What sets the limit is the ability to provide torque at speed which is power. In fact, it is the power to weight ratio that is primary indicator of performance. Imagine an electric motor that makes 5000 lbs*ft when its not spinning, but the torque drops down to zero at ...


0

You are correct. The total angular momentum remains constant. As the objects are not approaching head-on, there must be angular momentum. That momentum remains in the joined object as it spins around its COM. The speed of the spin for the joined object will be proportional to the perpendicular distance and speed of approach, and inversely proportional to ...


1

The kinetic energy E of a mass M moving with velocity V is $$E=\frac12 MV^2.$$ Taking the derivative of both sides with respect to time T gives $$P = \frac{dE}{dT} = MV\frac{dV}{dT}.$$ $P = \frac{dE}{dT}$ is power, and $\frac {dV}{dT}$ is acceleration A. Rearranging, we get $${P\over MV} = A.$$ I.e. at a given velocity and mass, the acceleration A is ...


0

You are right. Torques are not free vectors, the same way that linear velocity is not a free vector and it is associated with a specific point. When the point moves, the components of torque and velocity change according to the same transformation law $$ \begin{align} \vec{v}_B & = \vec{v}_A + (\vec{r}_{A}-\vec{r}_B) \times \vec{\omega} \\ \vec{\tau}_B ...


1

Torque is defined as the cross product of two vectors: r and F. r points from the pivot point to the point where the force F is applied. Both vectors are not bound and stay the same vectors no matter where you put them. However, if you combine them via mathematical operations, you have to stick to their rules (here: T1=r2*F3-r3*F2 etc...). There are no ...


0

First, we define the direction of $\vec{\theta}$ between $\vec{\mu}$ and $\vec{B}$ to be positive if $\vec{\theta}$ is coming out of the screen in the sense that $\theta$ is increasing with the right-hand-rule in the counter-clockwise direction. Then, note that the torque $\vec{\tau}=\vec{\mu}\times\vec{B}$ makes the angle between $\vec{\mu}$ and $\vec{B}$ ...


1

Torque is calculated with $$\sum \vec{\tau} = \sum_i( \vec{r}_i \times \vec{F}_i)$$ where $\times$ is the vector cross product, $\vec{r}_i$ the location of each force $\vec{F}_i$. So based on your diagram you have $$\vec{\tau} = \begin{pmatrix} x \\ \sqrt{2 x +1} \\ 0 \end{pmatrix} \times \begin{pmatrix} F_{1}x \\ F_{1y} \\ 0 \end{pmatrix} + ...


1

You can translate a force F along its vector (line of action) without changing the torque. For example, you can slide $F_1$ all the way to the vertical member - the one that force $F_2$ acts along. Now if you look at the diagram I drew: you can see two different ways to compute the torque. Method 1: Extend $F_1$ along its direction, draw the ...


7

Since torque is defined as the rate of change of angular momentum, the more fundamental question would be whether angular momentum is a vector in SR. The answer is no, because there is no vector cross product in four dimensions. Angular momentum is a rank-2 tensor.


1

I think the lecture does not do a good job of explaining the fundamentals of mechanics. The (torque,force) combo works together the exact same way a (velocity,rotation) work together. These are sometimes called dual vectors, or screws. So if a body is rotating, its velocity on the location you are measuring with $\vec{v} = \vec{r} \times \vec{\omega}$. The ...


0

The force that a current carrying wire experiences in a magnetic field is known as the Lorentz force, in differential form: $$ d\mathbf{F} = \int_{wire} I d\mathbf{l}\times\textbf{B}$$. $d\mathbf{l}$ is in the direction of the current. Since $I$ is uniform along the wire, the integration is just replaced by $\cdot L$, where $L$ is the length of the wire. ...


0

A few items I would do differently. First - make sure everything is in SI units (you are mostly doing that by converting inches to meters etc). And don't try to get 6 digits of precision. 4400 pounds is 2000 kg, not 1995.806. Really. Next, don't assume that a "unit of time" (in terms of your calculations of where the car is next) needs to be a second. ...


0

If you want to take the sum of moments, you have to take **all the external forces ** into account. The summation of moments has no preference for a certain force, allowing it to be in the equation. All forces are in the summation, only their resulting moment can be zero. If we look at the following example, assuming that all the spacings, e.g. ...



Top 50 recent answers are included