# Tag Info

20

Great photo! Edit: My language is "sloppy" (I like talking physics in "lay person" terms so anybody can understand) but @dcmkee made really nice comment clarifying my answer for the more advanced people. Thanks @dcmkee! Since the plane is in a loop there is significant g's due to centripetal acceleration. The water was being accelerated upward$^{1}$ with ...

19

You can have multiple forces exerted on an object that add to zero. Then there will be no momentum change. Think of the two of us leaning against opposite sides of a door with the same force. The door does not change momentum, nor does either of us. I am exerting a force on my chair as I sit here.

14

Actually Newton's second law is better stated as $$F=\frac{dp}{dt}$$ and this is even valid in relativity, both SR and GR, expressed in the right way $$f^\mu = \frac{dp^\mu}{d\tau}=m\frac{du^\mu}{d\tau} = m u^\nu\nabla_\nu u^\mu$$ (for massive particles) so classically forces are always imply a change in momentum. In QFT the concept of a force is no more ...

14

It doesn't actually have anything to do with the plane being upside down, or even changing from a vertical direction to a horizontal one. It's purely the vertical velocity that's at play here. Imagine water being thrown upward. You know what, imagine a fountain, a really big fountain. As soon as the water leaves the underground pump, it starts falling back ...

14

It has to do with drag of the air. This is related to the surface to mass ratio. The surface of a sphere increases with the square of the radius while mass increases with the cube. So the surface to mass ratio is proportional to $r^2/r^3 = 1/$r. This means that overweight lugers would have a big advantage. They don't want that, so lighter lugers are ...

7

No, all forces involve a change in momentum. In classical mechanics force is defined as a change in momentum. In quantum field theory particles interact via exchanging one or more bosons (see feyman diagrams). These bosons always have momentum and therefore the momentum of the interacting particles changes as well.

7

In this equation $F_N$ is the magnitude of the normal force, and $W$ is the magnitude of the weight. The forces are in opposite directions, yes, but their magnitudes are equal. It would be correct to write this: $$\vec{F}_N = -\vec{W}$$ because $\vec{F}_N$ refers to the full force vector, including its direction, not just the magnitude. (And similarly for ...

5

If the center of mass is not moving, then because there are no external forces it must remain in place. But if the bodies meet somewhere else, then the place where they meet would then be the center of mass, which is a contradiction since the center of mass must remain where it was at first! Edit: A (hopefully) clearer explanation. Suppose that at $t=0$ ...

5

At the moment the picture was taken the plane, with mass $M$ was performing an inside loop, and was almost exactly inverted. It was moving at a speed $V$ in a vertical circle with radius $R$; both of these are chosen by the pilot as he execute the loop. The physics of circular motion requires that the plane experience a force towards the centre of the ...

3

Actual aircraft attitude (inverted with respect to the ground, in this case) is irrelevant. All that matters is that for the few moments long enough to pour the water and snap the picture, the aircraft is experiencing some positive g-load (pilot feels that he is pushed into his seat). The aircraft could be in a barrel roll or a loop. Either way, it is in ...

3

This is a fundamentally pointless question because negative mass doesn't exist (or so we think!), but I'll answer anyway because the answer is so unexpected. Suppose we take our two massive bodies: Then the gravitational force between them is repulsive because: $$F = \frac{G m_1 m_2}{r^2}$$ and $m_1$ and $m_2$ have different signs. But let's ...

3

If I remember correctly they only do this in the turns and they use both arms in the straights. It is the outer arm that is active. This helps them turn in two ways. It helps accelerate the outer side more than the inner which is what is what turning really is. The reaction force at the shoulder also helps them lean into the turn which helps them stay stable ...

3

The buoyant block does exert a force on the water, it's force is equal to the mass of the displaced water, so the pressure of the water immediately beneath the block is exactly the same as the pressure of the water at that height in the rest of the container. Indeed, the mass of the system is just the mass of container with water + mass of block

2

So here, really, lies my question: Is there even a point to arguing about this? Perhaps there is a point in discussing this. In the Newtonian point of view, impulse and change of momentum are different concepts. Why? Force $F(t)$ is a basic quantity describing instantaneous influence of one body on another, in general having a magnitude and ...

2

The natural tendency of a rope is to move to the lowest possible tension. When an object hangs onto a rope, it slides down to the lowest possible point, at which the tension is uniform. Now, consider the case when something hangs onto the rope but does not slide. This is equivalent of having two ropes: one from the ceiling to the body, and one from the ...

2

You have the right idea, but the question asks for situations where there is a force and no work. Centripetal force does do no work but there is a force, so I is true. In III you are exactly right, but it says there is a force and no work, which falls under the question. I think you have misunderstood the question. II is false because a force in the ...

2

There is two situations here: 1)The $m_{s}=0.5kg$ mass is in front of the $m_{b}=1 kg$ mass (applied force is applied to 1kg block directly). 2)The $m_{b}=1 kg$ mass is in front of the $m_{s}=0.5 kg$ mass (applied force is applied to 0.5 kg block directly). Case 1: Since both blocks apply 6N of force on each other. We know that 6N o force is applied on ...

2

The answer can be found in the nature of gravity. It is a force that arises due to curvature of spacetime which underlies everything. A black hole is a specific configuration of spacetime where nothing can leave by definition, not even light. Since there is no concept comparable to curved spacetime underlying the other forces, we observe no such phenomena.

2

The force due to gravity balances the buoyant force exerted on the block The buoyant force is there because of gravity (There is a difference in pressure as we go deeper in to the ocean). There's an easy way to think it through. Imagine a beaker with some water, and it is standing still. There is no internal motion. Consider a small portion of this ...

1

Feynman explains it best in this classic video, but here are some of the essentials. Magnets attract and repel at a distance, and there is really no way of rephrasing that fact which will explain this force in terms of "winds of force" or any similar construct and which will not incur inaccuracies and inconsistencies that will render it completely useless. ...

1

First of all, mathematical definitions of force and momentum aren't really very intuitive or common-sensical. Just ask Aristotle for his common sense laws of forces! The fact that momentum is conserved in closed systems is a highly non-trivial fact, as is the Third Law. The reason that these laws exist at all is because you can't really 'see' or' feel' ...

1

A well executed barrel roll maintains the force balance you experience at rest with "gravity" oriented in the direction you experience as "down" (that is the direction from your head to your feet) due to centripetal acceleration. If you weren't looking outside, you might not realize the roll even took place (if the pilot is good). For those not convinced ...

1

When you solve a problem like this, you are using a system of reference (actually you use one in all problems, but here it is very explicit). In this case, the easiest one is y in the vertical and x in the horizontal. Almost all the forces are already in one of these 2 directions. Namely, you have all the weights pointing downwards, so in the -y direction ...

1

It will in general depend on the shape of the object. If it has a large concentration of mass at the edge you are lifting then the force will be close to its weight; if its mass is concentrated near the other edge then it will be very small. The general case is solved using the law of levers: If $d$ is the distance from the fulcrum to the centre of mass ...

1

If you are lifting on one edge and it is resting on the other edge, and the edges are an equal distance from the center of mass, then the answer is $$\boxed{F = \frac{1}{2} W}$$ If you are lifting with a distance of $\ell_1$ from the center, and the pivot is $\ell_2$ from the center then $$\boxed{ F = \frac{\ell_2}{\ell_1+\ell_2} W }$$ This is commonly ...

1

When the person in the air pulls upward on the person on the ground he is essentially applying an upward force on said person on the ground. This causes a downward force upon the person in the air. With no normal force to counteract this, the person in the air will just pull himself back to the ground.

1

If you know the total force as a function of time, then you know the acceleration as a function of time also $$a(t) = \frac{\sum F(t)}{m}$$ Now you find the velocity and acceleration using direct integration $$v(t) = v_0 + \int a(t)\,{\rm d} t \\ x(t) = x_0 + \int v(t)\,{\rm d} t$$ If the forces are constant then you can convert the integral into ...

1

First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out. This problem is easier if you invoke conservation of energy. Simply equate: At t = 0. -Potential gravitational energy. At the end. -Kinetic energy of the toolbox. -Dissipated energy due to friction. You will find that ...

1

He is saying that if $A$ is the set of configurations of a binary star system, then the sum $\oplus: A \times A \to A$ can be not defined, because if $a_1$ is the configuration at some time and $a_2$ is the configuration some time later, then $a_1 \oplus a_2$ will not be a configuration of a binary star system, since it will have four stars. In other words, ...

1

I am not sure wether I unstood your question correctly. From what I understood, you asked wheter particles are only connected/interacting by forces. Probably this is a matter of taste question but the picture of forces gets very unconvenient when one is talking about paulis exclusion principle. Although it can not be put in terms of a simple force it has a ...

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