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33

The "trick" is that the cane he is apparently holding is actually firmly attached to the platform. A rigid piece goes up his sleave, then to a harness that holds his whole body up. For more about this type of magic trick device, google "broom suspension" or "aerial suspension harness". No electric or magnetic fields were abused here. Image Credit: ...


21

If you look closely at the crocodiles' tails you'll see that they wave their tails from side to side to provide propulsion for the jump. Compare this to a fish swimming: The side to side motion of the fish's tail propels it forward, and the crocodiles are using exactly the same sort of side to side motion to propel themselves upwards.


17

There are some quantities that only come with one sign, for example mass/energy. So in any process if you start with some mass/energy $m$ then whatever happens you can't end up with less than $m$. But other quantities come in both positive and negative magnitudes, and these will cancel. If you add a velocity of 10 m/s North to a velocity of 10 m/sec South ...


17

Olin Lanthrop clearly gave the most plausible explanation. But just for fun, let's just assume this was an electromagnetic trick. Would that be possible? First - let's do this using electrical charge: how much charge would you need to allow levitation, and what would the potential have to be? Some assumptions: 70 kg guy 40 cm levitation (based on apparent ...


14

Alright I'll throw my hat into the ring with an answer. The idea that it's an unsolved problem is totally bogus. When you start to fall to one side or another if you turn the wheel slightly in the direction you're falling the bicycle starts to follow a curved path. There is a force due to friction that deflects the rider's path into a curve: The ...


12

It is always best to draw a diagram to convince yourself of things in a case like this. This is intended to represent a steady state situation: nobody is moving / winning. As you can see, there are two horizontal forces on A: the floor (pushing with 100N) and the rope (pulling with 100N). There will be two vertical forces (gravity pulling down on center ...


12

I've recently started trying to swim the butterfly. Unlike other swimming strokes, there doesn't seem to be any way to "go easy" and do a relaxing length of the pool: if I want to get my face out of the water to breathe, I essentially have to use the water to do a push-up. Now if a scrawny guy like me can lift his chin a few inches out of the water using ...


10

I do not agree with the angular momentum theory: if you were to hop off your bike at speed and let it go by itself, it would not go very far before falling on its side, even less if you put a ~150 lbs sandbag on your saddle. There is indeed an effect caused by momentum, but this is negligible when compared to the actual contribution of the rider. I agree ...


8

AlanZ2223 has given a nice summary of what's going on. I'll just make a couple of points that are orthogonal to his and that wouldn't fit in comments. The electrical force is a non-contact force; it falls off with distance like $1/r^2$. But most of the objects we deal with in everyday life are electrically neutral, i.e., they contain equal amounts of ...


7

The diagram is misleading. Look at this: At any moment in time, you have the following forces on the particle: Gravity Tension in the string When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string ...


6

Consider this for the upper ball: The angle of the inclined plane is simply derived from geometrical considerations. Looks pretty easy to solve, right? Once you know the normal force of the inclined slope, reflect and apply it to the bottom ball to complete the problem.


6

The force on rope is equal for both of them at any time. For winning the game the force on ground is responsible.


6

These "non-contact" forces that are ubiquitous in everyday life are mainly attributed to electromagnetism. Basically the four fundamental forces which are the strong force,weak, electromagnetic, and gravitational all have a sort of realm within which they influence most. The strong/nuclear reigns within the subatomic domain, the weak as well but this kind ...


6

The upwards force comes from the rather violent tail movement. When the rest of the body is out of the water, the tail still acts sort of like a hydrofoil pushing the crocodile upwards, only not with a linear but oscillating motion, and obviously it's rather instable but enough to get the whole animal up in the air for a short while.


6

If I walk 3 kilometers East and 4 kilometers North, how far have I gone from my original position? Only 5 kilometers. Where have the other 2 kilometers gone? If I walk 7 kilometers East and then 7 kilometers West, how far have I gone from my original position? Nowhere. Where have all 14 kilometers gone? This situation avoid vectors entirely. But it goes to ...


6

You should think of the formula the other way around, i.e. $$ \mathrm{d}W = F\mathrm{d}x$$ which means that the infinitesimal work done along an infinitesimal path is just the force $F$ times the length $\mathrm{d}x$ of the path along which the force was exerted. If we are now given a real path $\gamma : [a,b] \to \mathbb{R}^3$, the total work done along ...


6

I assume a steady-state universe and that the bodies have no velocity relative to each other. Yes, they will eventually collide. Gravity has an effect over any distance, including the ~46 billion light-year radius that constitutes the spherical observable universe (the actual size of the universe may be much larger). Of course, the force will not be very ...


5

I think it's best to just quote Feynman here: Friction: the force of friction against a dry surface is $-\mu N$, and again you have to know what the symbols mean: when an object is pushed against another surface with a force whose component perpendicular to the surface is $N$, then in order to keep it sliding along the surface, the force required ...


5

Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically. These two forces combine to give the ...


4

According to Newton's 3rd law of motion, the force exerted is equally distributed between the rope and the ground. Force on the ground can never be greater than force on the rope and vice versa. Therefore each players individually exerts the same force on the rope and ground, but that doesn't means that the same amount of work is done on the rope and on the ...


4

I understand force to be a 1-form, through the following reasoning. Given a time-independent, conservative lagrangian $L$, its differential (a 1-form in the purest sense) is $$ \mathrm{d}L = p_a ~\mathrm{d}\dot{x}^a + f_a~\mathrm{d} x^a $$ where $$ p_a = \frac{\partial L}{\partial \dot{x}^a},~f_a = \frac{\partial L}{\partial x^a}. $$ So the components of ...


4

Fear not! You are not alone in your confusion. While it is common these days to teach that forces are vectors and follow vector addition (as the other answers beautifully present it), historically, you have hit upon a quandary that lay at the center of a dispute throughout much of the 19th century, involving such characters as Newton, Lagrange, Heaviside, ...


3

I am not an expert in such fields, but I'll give you an overview of how I've learnt it. The main point to realize is that, on a microscopic scale, the surfaces we initially thought of as "smooth" contain actually a great many irregular protuberances. Coming back to the surface area between the two objects, one must carefully distinguish between the ...


3

The work done by a force $\mathbf{F}_1$ when a particle travels a path $\gamma : [a,b]\subset \mathbb{R} \to \mathbb{R}^3$ is defined by $$W(\mathbf{F}_1, \gamma) = \int_\gamma \mathbf{F}_1 = \int_a^b \mathbf{F}_1(\gamma(t)) \cdot \gamma'(t) dt$$ I wrote that way just as a way to make clear that the work depends both on the path and on the force. If ...


3

A few thoughts to help you on your way. When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work times velocity). If you are changing the velocity of an object, you are changing its kinetic energy: if it's ...


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 ...


3

So far, it is still an open question as to why bicycles are stable at all. There have been a few ideas put forward, but they have been disproved by construction of non-standard bicycles. The most common explanation is that the wheels on a bike act as a gyroscope, preventing the bike from falling over. A bike was constructed with counter-rotating wheels to ...


3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...


3

Yes, they will collide given the initial conditions. The speed at collision can be calculated. We can presume them to begin with virtually zero gravitational potential energy. We need an assumption of size when they collide. Let's assume a size of 50cm. That way when they collide, the centers will be 1m apart. $$U = -\frac{GMm}{d}$$ When they collide, ...


2

The answer is that, despite our best efforts, we still can't quite put a finger on it. The gyroscopic forces mentioned by bobie in his answer have been proven not to be sufficient to fully explain why a bike stays upright. It's indeed very surprising that physicists are not able to explain the mechanism behind such an (apparently) simple and ubiquitous ...



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