New answers tagged

1

does it also leads to ... No, it doesn't. A simple Counterexample: Consider the figure below (the bar $\textrm {AB}$ is on a plane parallel to $\textrm {xy}$ plane) We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$ is the ...


1

Put a coordinate system on the center of mass and place each leg i at $$\vec{r}_i = \pmatrix{x_i, & y_i, & z_i}$$ where $z_i = z_{c}+\theta_x y_i - \theta_y x_i$ describes the vertical deflection of the point, given the center of mass vertical position $z_c$ and the two tilt angles $\theta_x$ and $\theta_y$. Add vertical loads the each point ...


0

The vector sum of the forces will be zero: $$\Sigma \vec{F} = 0$$ If you can assume that the bottom of the block is planer and the weight acts perpendicular to that plane, (i.e., the lengths of the legs are identical) things are much simpler. The weight acts downward and the forces from the legs act upward. The sum of the torques (moments) about any point ...


0

From the equations of static equilibrium, N - mg = 0 in the vertical direction. The tipping is caused by the imbalance in the moments applied around point P. As long as mg * (L / 2) > F * H, then the book case won't tip over. H and L are the dimensions of the book case, so you can't do much about changing them. You can either add more books, which ...


1

You almost solved it. Actually when you increase your force steadily the normal reaction shifts it point of action from directly under the centre of mass of the object to the point P. In other words in the limiting condition the normal force acts at point P. And hence it's torque is 0 about P. Well done.


1

The state of equilibrium is characterised by a minimum in free energy $F=U-TS$ (Helmholtz for simplicity), not a minimum in potential energy. What this means is that while the system is indeed attempting to minimise the potential energy $U$, it is simultaneously trying to maximise the entropy $S$. The balance, i.e. which term dominates, is determined by the ...


1

Energy is a secondary concept, at least in Newtownian mechanics, so let's start with the fundamentals. There exists a force field. A gravitational one, say. An object in this field feels a force. This force "wants" to make the object accelerate. From forces, define work as $force \times distance$ and now we can give precise mathematical meaning to the ...


11

This is really a statistical effect, as pretty much all of thermodynamics. You have two free hydrogen atoms. They tend to move around the space they have, and when conditions are favourable (there's enough energy, the atoms come "close enough" together), they might interact - chemically or otherwise. Now, "enough energy" is the important bit here. When a ...


9

I'm going to take a slightly different approach and say it's because we defined energy to make it so. In other words, systems "try" to find the lowest energy state because energy is a concept humans invented in order to describe what we observe. This is the reason that for any given set of constraints, you might need a different "energy" to describe the ...


17

This is a consequence of the second law of thermodynamics, which states that In a closed system with fixed internal energy (i.e. an isolated system), entropy is maximized at equilibrium. It can be shown that this statement is equivalent to the following: In a closed system with fixed entropy, the energy is minimized at equilibrium. Callen in his ...


-6

It is caused by the fundamental forces, and Tendency of energy to escape Number 2 itself is caused by fundamental forces, so it is the forces. Most questions end up at why/how there are fundamental forces.


74

The anthropomorphic formulation "tries to" is misleading. Under the effect of ambient noise, matter explores the possible configurations around its current state: e.g., two single hydrogen atoms wiggle around and meet. If they happen to bind, this releases energy which goes away, and we say that the energetic state of this new $H_2$ molecule is lower than ...


0

The answer from "introduction to electrodynamics" Have a look at that problem. The answer-from solution manual- :A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV . But we know that Laplace’s equation allows no local minima for V . What looks like a minimum, in the figure, must in fact be a saddle ...


1

I think your confusion is coming from the fact that you are actually using theta for two different things here. Let's use phi for the angle between the velocity of the rod and the magnetic field, and use theta as it is depicted in the diagram. Then your expression should be written as $$qvBsin\phi$$ This comes from the fact that $qvBsin\phi$ is derived from ...


0

There are two primary things that draw the ball into the waterfall (or push the ball upstream), and both of these are much stronger than the two effects mentioned in the question. As a physicist and a white water kayaker, I have a lot of experience with the relevant forces, and they can both be very strong and sometimes life threatening. The first force: ...


0

I think, the comment of lucas is the best answer; I'll elaborate a bit If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that? By temperature you mean the mean energy per particle. We observe experimentally, that if you mix two heaps of particles with different mean energies, after some time ...


0

If you mix two gases, their atoms/molecules exchange with energy, momentum, etc. It is not surprising that a hot gaz heats a cold one. It (heating) can only stop when the mutual energy exchange becomes equal to both gases. So the energy distributions of "subsystems" are equal in the end, but the momentum distributions are still not. P.S. The subsystem ...


1

In a system of many particles, we essentially observe the most probable configuration, and relative fluctuations around it are negligible. Here I will prove that the most probable state of a 2-particle system is this with equal energies. The probability of a state is proportional to the volume of the corresponding part of phase space. If a particle has ...



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