Tag Info

Hot answers tagged

38

Using the brakes on the front of the bike causes your weight to shift forward. Additional weight allows more force before the tire will slip (skid). If you brake hard enough the back tire of your bike will lift up and at that point all of the mass is distributed on the front tire. Remember the maximum force is $F_{max} = \mu F_{normal}$ and $F_{normal}$ ...


8

On the quantum level, force is not acceleration. The concept of "fictitious force" makes no sense on a QFT level, because forces are interactions between quantum states, not the classical forces you might imagine. Quantum forces are not vector fields in space. The notion of "fictitious force" would mean that, e.g., the strong force is something influencing ...


8

The classical theory of electrodynamics can indeed be written as a geometrical theory in a similar way to general relativity. As it happens there is a question and answer addressing just this, but it's in the Maths SE: Electrodynamics in general spacetime. Classical electrodynamics is an example of a class of theories called classical Yang-Mills gauge ...


8

Suppose you and the table are floating in space. If you push the table it will go in one direction and you will go in the other direction. Your and the table's acceleration will be different so you will end up travelling at different speeds. This is obvious from conservation of momentum. The momentum in the centre of mass frame is initially zero, so after ...


5

A net force on an object likes to act through the centre of mass of that object; in this case, the middle of the door. Apply a force to the centre of mass and it will accelerate uniformly because there is an even amount of matter on all opposing sides that can resist the motion with an even amount of inertia. If a force is applied off the centre of mass, it ...


5

The braking force acts between the tyre and the road. The centre of mass is above this point so there is a rotational effect which increases the force going down through the front tyre and decreases the force going down through the rear tyre. Because the amount of braking force the tyre is able to produce is limited by the amount of force going down through ...


4

In an inertial frame the only force that causes a particle to move in a circular motion is the centripetal force. The reason that a particle does not "fall" into the center is because it has some tangential velocity, so it moves away from the center tangentially as it is falling towards it. The relationship between the centripetal acceleration and tangential ...


4

is it possible to consider also the other fundamental forces [...] to be fictitious forces like gravity in the framework of general relativity? No, because the equivalence principle only holds for gravity. If we want a final unification of all fundamental forces, hasn't this feature of gravity to become a feature of the other forces as well? The ...


4

The electromagnetic and weak forces have been unified into the theory of the electroweak force. The recent discovery of the Higgs boson put the icing on this particular cake. The strong force is described by the same type of quantum Yang-Mills theory as the electroweak force, however it is not unified with it. There have been several attempts at unifying ...


3

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as: rolling drag between the tires and the road surface, fluid drag from the air that the car moves through, and ...


3

This effect is called Capillary Action. Yes we do in fact observe it in nature in a large scale: How do you think plants are able to "suck up"1 water through its roots and send it to the leaves? One of the major forces responsible for it is capillary action. Here, have a quote from the article mentioned above: Wicking is the absorption of a liquid by a ...


3

Theoretical Answer Consider that you travelled in your car at $10~{km}/{h}$ for one hour, then at $100~{km}/{h}$ for the next hour. First hour of the Journey: You travelled a distance of 10km, so the work done is $$W=F\times s=10F$$ Second hour of the journey: You travelled a distance of 100km, so the work done is $$W=F\times s=100F$$ The work done is ...


3

You are wrong at assuming constant friction. Rolling Friction increases when you increase speed of the car (See the formulae at the bottom). Also, aerodynamic drag increases with the square of speed (See the formula at the bottom). So, at higher speed, the car engine needs to counter higher rolling friction and air drag to maintain that speed. While the ...


3

As you seem (correctly) to understand, your free body diagram for the car should be as in the left of the drawing below. The force summation does not close to a polygon. You know neither $F_N$ nor $F_F$, the components of the force on the car from the road. But you know their directions ($\theta$ is the banking angle) and you know that all the forces must ...


3

Lots of excellent answers here, but for fun, lets think about this backwards. Imagine you have the worlds first and only FRONT wheel drive motorcycle, and your rev it up and pop the clutch. What kind of launch do you think you would get with very little weight on the front tire? The reverse is true during braking when the deceleration shifts the weight of ...


3

Braking acts to stop the front tire. Friction acts at the contact patch under the front wheel to introduce a vector force directed towards the back of the bike. Since the force is not directed through the center of mass of the motorcycle/rider system, it introduces a moment or torque that acts to rotate the motorcycle and rider such that the back tire begins ...


3

If you think about it, this is what a plane (or helicopter) does all the time. It "jumps" from one air molecule to the next, pushing them down and using them to stay up. The key is to push enough rubble (air) down with enough speed to offset the pull of gravity. This means that the total rate of change in velocity of the material $m\Delta v$ must be greater ...


3

$\boldsymbol g$ is a vector. You define it as a scalar only when you have mentioned a clear reference frame in which the sign of $\boldsymbol g$ makes sense. You could define a vector field from the law of universal gravitation for two bodies $A$ and $B$: $$\boldsymbol F_{AB}(\boldsymbol r) = -G\frac{m_Am_B}{|\boldsymbol r_{AB}|^2}\boldsymbol ...


3

If you do not define the direction of the force, you need to do the math explicitly. Usually, a letter like $g$ identifies a scalar quantity. If you want to show a vector, there are a number of typographical conventions. I have seen $\mathbf{g}$, $\vec{g}$, $\overline{g}$, $\underline{g}$ ... So when I have tension up and gravity down, I know that the two ...


3

Summarizing some of the comments (since they have a tendency to disappear over time): The force lasts as long as you apply it. Force is not a property of material - momentum is. As you know, momentum of a particle initially at rest will be $F\Delta t$ after you apply a force $F$ for a period of time $\Delta t$. More generally, the change in momentum ...


3

Using a big vacuum chamber air resistance is removed. There is a video here.


2

Q1 does not allow for a conclusive answer. Without knowing whether the force pushes $m_1$ or pulls $m_2$, no statement regarding the contact force between the two masses can be made. The bodies may even be in free fall, then there would be no contact force. Assuming that the force acts on $m_1$, the contact force between $m_1$ and $m_2$ is compressive. You ...


2

Conservation of linear momentum, for a physical system whose particles are initially at rest in a given inertial reference frame is equivalent to the fact that the center of mass of the system remains fixed at its initial position. You see that this constraint is quite week for a system made of a large number of particles as a human body. Starting form a ...


2

Well, for starters, centrifugal force is just a pseudo force. Get it? Let me make it a bit simpler.. See, Centripetal force, which is mv^2/r is acting towards the center and centrifugal force away from the center. I thinking you're thinking why did we use the centripetal force in case of banking of road.. right? Well, we did that because, This is nothing ...


2

I see in the comments section this question became something else entirely, but I'll try to sum up all points in my answer -- forgive me if I repeat what others said. First, why should the proton not accelerate forever? It will surely accelerate as long as the force acts upon it. I think what you mean is akin to the Zeno's Paradoxes, in which basically it ...


2

If the (entire) system does not change if you displace it in space, then the total momentum (of all particles) will be conserved. Momentum is the generator of translation, which is a specific case of Noether's theorem. To relate it to your example of living creatures: imagine cat on very slippery ice and neglect air resistance and friction. If you slide ...


2

Yes, It is possible. But only and only if you are capable of applying a force of more than your weight on the rubble. Though the combined center of mass of rubble and you will still be falling with same acceleration.


2

Not quite. The key is that both the bodies move with the same acceleration. This is proven by Newton's first law and that both the bodies are perfectly rigid. If they move with the same acceleration we can consider both the bodies as one body of mass $M_1 + M_2$. Newtons second law is $\Sigma$$F=ma$, so the acceleration of both the bodies is $F/(M_1+M_2)$. ...


2

I suspect this question is rooted in the widespread misconception that some external force is needed to keep the Earth rotation. That is not the case. Angular momentum is a conserved quantity. A rotating object that is not subject to any external torques will rotate with a constant angular momentum. This is the rotational analog of Newton's first law. An ...


2

I got the answer. There are two separate cases. CASE 1 If $m_1$ is exerting a force on $m_2$, which is in turn causing constant acceleration in both the bodies, then $$a_{net}=\frac{F}{m_{1}+m_{2}} ;~~~~~~\therefore F_{12}=\frac{Fm_2}{m_{1}+m_{2}}$$ CASE 2 If the acceleration on both the bodies is caused by an external force separately, as in the case of ...



Only top voted, non community-wiki answers of a minimum length are eligible