# Tag Info

1

This is a question everyone asks at first because it intuitively seems like a contradiction. However, it is not. Conceptual Examples I think you are not far off but perhaps the third law is the one tripping you up, not the 1st... But anyway, here are some conceptual examples, which might help... Example 1. Consider the particle in the frame for a ...

16

In introductory problems about work you're normally taught that it's force times distance: $$W = F \times x$$ and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is: $$W = ... 3 Newton's first law states that: "An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force." It is often called the law of inertia. So if You want to move an object with zero velocity, at first moment You have ... 1 external force acting on the box by us should be a little more than the weight, otherwise indeed: no acceleration! So the mgh is really a lower limit. We need to accelerate. And, by the time we reach h, decelerate -- so we get this extra little bit of work back. But only in a physics sense. 2 A vortex is defined as a region within a fluid where the flow is mostly a spinning motion about an imaginary axis, straight or curved. Mathematically, this is defined as the curl of the velocity field, \mathbf{v}:$$ \boldsymbol{\omega}=\nabla\times\mathbf{v} $$If we consider the strength of the vortex tube as$$ \Gamma=\int\boldsymbol\omega\cdot ...

-1

Since the work done must be $W=F\cdot d$ And we need to calculate just the distance, because we already have the force. $x(t)=x_0+v_ot+\frac{1}{2}at^2=\frac{1}{2}at^2$ Finally we reemplace that in the fist equation to find the work done. $W=F\cdot\frac{1}{2}at^2$

2

You posed two questions: First, Earth exerts a gravitational force on the car. The car, in turn, exerts a gravitational force on Earth. These are equal and opposite. This can be seen from the fact that the formula for magnitude of the (Newtonian) gravitational force: $$F_g=\frac{GM_1M_2}{r^2}$$ remains of exactly the same form if the masses are switched. The ...

-3

They have to meet at the same time to keep the overall pressure behind the wing equal. The velocity of the air on the top run relatively faster because it has a greater distance to travel (due to the curvature of the wing) and all in the same time as the air of the bottom. Why the same time, because when the air divides it still has to keep pressure ...

0

The forces other than gravitation include strong, electroweak, and electromagnetism. The first two could not produce a black hole because they are both very short range forces. Beyond dimensions much larger than an atomic nucleus, these effects (strong, electroweak) are negligible. Chiefly this is because the bosons that carry those forces soon decay into ...

0

I have a formula able to calculate the binding energy of the deuteron: it needs only to apply the electric and magnetic Coulomb's laws to the deuteron. The formula is the same as above, but with the potential instead of the force, needing the empirical radius: A better theory, e.g. electromagnetic, gives the correct result by using only the fundamental ...

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 ...

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

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. ...

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.

0

This is a wild guess. You say the pain is in the temples, which is the kind of sensation you get if you whip your head around too fast. This therefore has nothing directly to do with the hair. When you have long hair it "swings" around more, and you unconsciously synchronize the head flipping to this swinging. It makes sense to optimize this sortof ...

1

You can no longer consider the ice block as a system once it starts melting, so it no longer remains a rigid body. Consequently, you cannot apply Newton's law on it. Peace ;)

0

When the force acts on a body, it displaces the body through a certain distance in the direction of force or against it. $W = \vec{F} \cdot \vec{s} = |F| |s| \cos \alpha$ Where, $W$ is the work done by a force $\vec{F}$ causing a displacement $\vec{s}$ and $\alpha$ is the angle between the force and the displacement vectors.

0

Not exactly what you were asking for, but any force exerted perpendicular to the direction of motion does not change the magnitude of momentum -- though it does change the direction. Two examples are the force exerted by a uniform magnetic field on a moving charged particle, and the force of gravity on a satellite in a perfectly circular orbit. But ...

20

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.

2

A universal relation is that the force exerted on an object equals the time derivative of momentum. No force, no momentum change, vice versa.

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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.

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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 ...

0

I haven't checked the arithmetic, but the method to calculate the acceleration is right. The easiest way (to my mind) to find the tension (T) in the string is to apply F=ma to the smaller mass block. Hope fully if you do that, you will also find that (T-6.4) will also accelerate the larger mass at the required rate as well.

0

The tension in the string has nothing to do with gravity. Just consider the forces acting in the horizontal direction for each mass separatedly.

0

As the blocks are moving horizontally, the gravity and normal forces cancel each other. So your logic about finding the tension is faulty. You should use Newtons second law on each separate block. To find the tension you could suffice with the last one. If there is friction you could add the friction force to the equation, and think about what this would ...

3

It is the elimination of friction on the ground. The friction on the air is very small, as is the resistance of the rope to twisting. No matter how smooth the floor, the friction will be much higher than the resistance of the hanging weight. This is why air bearings were invented.

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

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 ...

0

Note: Solution attempts to problems like these are susceptible to sign-flipping errors, so we must ensure clarity about what positive and negative values mean. Definitions. Let $N_1$, $N_2$, $f_1$ denote magnitudes only, as requested in the problem statement. As a consequence, the fact that the forces quantified by $f_1$ and $N_2$ are supposed to counteract ...

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 ...

0

Assume mass is distributed evenly on the wire, then mass of differential element is: $$\text dm=\frac{\text dm}{\text d\theta}\text d\theta=\rho\text d\theta$$ for some angular density $\rho$ Since the placement of the wire is symmetric with respect to the particle, only the forces in the direction normal to the diameter needs to be considered. We define ...

0

The forces act on BOTH the bodies involved, not on the same one! That's why the statement of Newton's third law is: The third law states that all forces exist in pairs: if one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal and opposite: FA = −FB Source:Wikipedia. So it isn't ...

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' ...

0

a) The reason its $F=mg\sin\theta$ is because that is how much of the weight is pointing down the slope. This makes since since when $\theta=0$ we would have to apply $0N$ so that it doesn't slide down (or remain at constant speed), and at $\theta=90$ we would have to apply $mg$ to keep it from sliding down (essentially hold the entire mass). b) Same idea ...

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 ...

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 ...

0

If gravity is a fundamental force, then the Higgs mechanism is also. This is is true whether they are related or not. The Higgs mechanism is certainly the source of the inertial mass that inspired Newton to quantify what a force is, and how it behaves. Gravity, like the Higgs mechanism, can add mass/energy to matter in bulk (like constant acceleration), ...

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 ...

5

the physics is the same as to why the pilot and passenger are not suspended on their seat belts: they're pressed to their seats by centrifugal force, the same force that makes water fall upwards

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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 ...

-1

The centripetal force we feel on the surface would immediately disappear, causing us to feel lighter. The centripetal force is equal to $E=mv^2/r$ and the acceleration is given by $a=v^2/r$. This results in a reduction of the acceleration towards the earth’s centre, also known as gravity. Therefore we would feel lighter

0

Can? Gravitational forces be transformed to a Restoring force ;the answer is ''certainly''; but this require a special Technic. Retro-causality Phenomenon: The gravitational forces use as the restoring force. If a body IN A SYSTEM is acted on by the G.f, than the body must be raised with the tension speed of the spring at the speed of -64 feet (S) , which ...

0

This doesn't exactly answer the question, but it does suggest that the key explanatory link is the step frequency. I would like to propose the following "theory". The use of one or both arms allows for an increase in "step" frequency. (Why? Don't know. Perhaps for the same reason that speedwalkers flap their arms. Biomechanics. The Dutch Wikipedia refers ...

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 ...

0

Visualise the scene, suppose the force $F$ is acting on the bigger block($1kg$) towards right. The smaller block($0.5 kg$) is to the right of the bigger block. Both are moving towards right with an acceleration $a$(suppose). Consider the smaller block, The only horizontal force acting on it is $Normal force = 6N$ Due to this force it is moving towards ...

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 ...

0

Ok you have one degree of freedom here. Lets call it the angular position from vertical of the ball $\theta$. From that the center position of the ball is $\vec{r} = ((R-r) \sin \theta, R - (R-r) \cos \theta)$. The angular orientation of the ball is $\varphi = \frac{R}{r} \theta$ due to "gearing" between the two surfaces. The center of the ball tracks an arc ...

0

Fictitious forces arise when you are working with non-inertial reference frames (e.g., rotating reference frames). Your problem statement includes the line What is the frequency of small oscillations? which means you are linearizing the system and no rotations are to be considered in the problem.

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