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2

Even classically, forces arise from field being propagated at the speed of light. A physically relevant object is the energy-stress tensor, whose components represent energy density and momentum current density, so indeed momentum can be interpreted as a current that is conserved over time (as a consequence of symmetries). This point of view is also ...


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$F = \frac{dp}{dt}$ means that force is the rate of momentum transfer per unit time. Lets say we have mass $m_1$ moving to the right, and mass $m_2$ is on the left side of $m_1$ with zero velocity. If $m_1$ put a force to pull $m_2$, that force will create the acceleration on $m_2$ and increase its velocity, this also means the change in momentum. At the ...


0

The $d$ in front of momentum and in front of time means infinitesimial change of time $$dt = t_{final} - t_{initial}$$. Therefore the change in momentum over the change in time equals the force! Also momentum is equal to $m\cdot u$ ($u = \text{velocity}$) . So the change in momentum is equal to$$ dp = m\cdot u_{final} - m\cdot u_{initial}$$ . We also know ...


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To you, lets apply the second law. As it states, actually $\sum F=m a$ NOTE THE SUMMATION, which means the resultant force applied. If you're just sitting or not moving, both normal force and gravity exerted in opposite direction, for its magnitude, they equals. You're still having problem with how will the Earth move ha? The third law tells, You give the ...


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I know that these forces will not cancel each other out since they are applied on different bodies and not a single ... That's correct. ... but I ask if net force is not zero then according to the equation, a=Fm, I must have some acceleration produced in my body. First off, it's $F=ma$ (or equivalently, $a=F/m$), not $a=Fm$. One thing you are ...


1

When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...


3

Why am I not accelerated by the reaction force applied by earth on me? Because the net force on your centre of mass is zero. The upward force on your feet is of the same magnitude as the downward force of gravity. Your major leg bones and spine are in compression because of the opposing forces. I know that these forces will not cancel each other ...


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I think you just need to treat two forces separately relative to each body as $\vec{F}_1=-\vec{F}_2$ and $\lvert|\vec{F}_1\rvert|=\lvert|\vec{F}_2\rvert|$. Having in mind that $\vec{F}_1(t),\vec{F}_2(t)$ both are functions of time because the motion goes with constant acceleration.


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I think it helps if you look at the problem from more of an intuitive perspective. When you're standing still on Earth, you're clearly not accelerating, right? (In this example we are ignoring the energy and force exerted by the molecules in our body, which are all moving to some extent regardless of where we are.) If you were accelerating, you'd see some ...


2

In the framework of General Relativity, where the inertial frames are the ones in free fall, you can think that the Earth is accelerating upward, so it is not you who is pushing on Earth but it is Earth that is "running you over" because of its accelerated motion. Luckily enough, if we are standing on ground, we can avoid impulsive forces and our bodies are ...


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In the equation $F_f=\mu F_N$, where $F_f$ is the frictional force, $F_N$ is the normal force, and $\mu$ is the coefficient of friction, $\mu$ is sort of a way of expressing the quantity that you're looking for. You ask why this equation does not include an expression of the "roughness" of a surface, but it's not obvious to me how you would concisely ...


1

By definition, work is $$W[\gamma] = \int_\gamma\mathbf F\cdot\mathbf v\text dt$$ i.e. force times displacement. If an object is not moving, even if subjected to a force, then there is no work done by said force. Observe that, since the object is not moving, the sum of all the forces applied to a body must be zero. For example, a body on rest on a table is ...


1

Work= force x displacement force is weight and displacement is 0 so there is no work (in physics)


-1

Here is my answer. I think it's self explanatory enough: In the last sentence there must be a minor correction: If someone keeps omega1 constant, then for the same speed they have to cycle faster so they can maintain the same power. If someone keeps omega2 constant, then bigger alpha mean less forward speed but less power too.


1

Even I didn't get you but I may help you how much I can by describing your case. Your case have two bodies which are being rubbed against each other in opposite direction with constant acceleration. The definition of friction is, "The resistance which either one of the bodies offers to this motion is called the force of friction and is said to be due to ...


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The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed. ...


0

How the acceleration change may depends on the relation between the two bodies' mass and the coefficient of friction of the surfaces. I will assume the initial friction is exerted by ground. In the following I use U means upper body (your second body), L means lower (your first body), G means ground, 2 stands for 'to'(or any other prepositions). 1.If U is ...


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The force of friction will increase with the addition of the second mass. However, the mass of the system will increase as well, and the net effect is that the acceleration does not change. To see this, note that: $$a={F \over m}={\mu _kmg \over m}=\mu _kg$$ Where $\mu _k$ is the coefficient of kinetic friction between the moving body and the surface it's ...


2

When you push something and it remains at rest your muscles transfer energy through isostatic muscle contraction/respiration. This means that even though the muscles don't move they convert the glucose into respiratory energy for muscle contraction that will be dissipated eventually by heating the surroundings. The only work done is that in contracting the ...


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Fn is the normal reaction force and Fay is the y-component of the force Fa, since the block is not moving in the y-direction (+ve or -ve) and since we know that mg is acting downwards, there has to be some force(s) balancing out mg, and apart from the normal reaction, Fay is also acting upwards, so yea it's Fn + Fay = mg because both Fn and Fay are acting in ...


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That equation expresses the fact that the forces on the block in the vertical direction add up to zero. $F_n$ and $F_{ay}$ are both forces pulling the block upwards, while the force of gravity on the block is $mg$ downwards. We don't want the block to accelerate up or down, and so the net forces in the vertical direction must cancel out.


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Let us denote $\ P$ the point corresponding to the end of the rod with no force applied, then we apply a force $\ \overrightarrow{F} $ on the other end of the rod, such that is parallel to the plane and perpendicular to the rod. Now, with respect to the center of mass, the only force whose torque is different from zero is $\ \overrightarrow{F}$: ...


0

The magnitude of the force between $i$ and $j$ is indeed $$ F_{ij} = \frac{Gm_im_j}{\lvert\vec{r}_{ij}\rvert^2}. $$ The direction of this vector is directed along the line connecting the two points, the same as $\vec{r}_{ij}$ (my notation for the vector difference between the positions of $i$ and $j$). In principle you can compute the magnitude for each pair ...


3

Assuming the interpretation I suggested in the comments is correct. Consider the normal support force. It is an expression of the solidity of the surface that won't allow interpenetration. In order for penetration to not happen, there must be a force to prevent the supported object from accelerating toward the surface. Ultimately the origin of this ...


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The equation governing the motion of body at right is, $m_1g-T=m_1a$ [since $F_N=ma$] ($Equation_1$) The equation governing the motion of body at left is, $T-m_2g=m_2a$ ($Equation_2$) Adding both equations we get, $a=\frac{m_1-m_2}{m_1+m_2}\times{g}$ The net acceleration produced in string at right is due to net force acting on the body at right. This ...


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The tensions are not "reactions" in the sense that they are not the 3rd law partners of the weights of the two masses, so they don't have to be equal to the weights. For the tension to be the same throughout the string, it must be 'light', i.e. massless. Because this equal tension can't possibly balance BOTH the two different weights, each mass will ...


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As the weight hanging off the table falls, it has some acceleration downwards. The force on the body is entirely due to gravity and is known. The string acts as a linkage between the hanging mass and the mass on the table. Really in this case, we have weights $m_1$ and $m_2$ being pulled by a force $m_1g$, which leads to the acceleration of the system being ...


0

The acceleration of any object due to gravity is $g = 9.8m/s$ and this constant does not depend on the mass of the object (or the speed of the object or anything else for that matter, as long as you're on/near earth). Pushing an object away from the earth is another story. While the acceleration of objects towards earth does not depend on their masses, their ...


4

When you are lifting an object, you are exerting a force that balances the force of gravity on the object. By $$ F = m g$$ where g is the acceleration due to gravity, you see that a greater mass causes a greater gravitational force that has to be balanced by the force you apply to the object by holding it or lifting it at a constant velocity. Using the more ...


-1

Are you referring to formulas like $$p=w h$$, where $p$ is the pressure, $h$ the height and $w$ the weight-density? In this case the weight-density is given in eg. $\mathrm{kg}\,\mathrm{m}^{-3}$ and the pressure comes out as $\mathrm{kg}\,\mathrm{m}^{-2}$, so you are of course considering the entire weight (heigth times density).


1

Model the ground as massless critically-damped vertical spring that the particle contacts at zero height. When the particle reaches zero height, it has some KE which is dissipated by the damping mechanism. When in contact with the spring, there are three forces acting on the particle, gravity downward and the damper and spring force upward. The net force ...


1

When an object falls and hits the ground - which forces are involved to change its momentum? Vectorial sum of all the forces acting on the object will cause the change in momentum of the object. When the object was in free-fall, its momentum was already changing due to gravity(assuming negligible amount of air resistance) and then it hit the ground. ...


1

In Newtonian Mechanics, if a body of mass $\mathtt{m}$ is in free-fall, then gravitational force is responsible for acceleration & hence changing its momentum. Simple, right? The equation of motion is $$\mathtt{m}\cdot a = \mathbf{F_g} = \mathtt{m} \cdot g$$ where $a$ is the net acceleration of the body. Things become intricate when you consider a ...


1

The center of mass is defined as it is because the center of mass obeys Newton's second law. Consider a collection of particles, labeled by an index $i$. Then each particle obeys $$\mathbf{F}_i=\dot{\mathbf{p}}_i$$ We decompose the force into the interaction between particles and an external force $$\mathbf{F}_i=\sum_{j\ne ...


2

Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall. Let's see why. Let's decompose the weight of the stick into a component along the stick, and ...


1

I'm assuming the circular disk is a friction-less pulley that's in equilibrium/at rest. If so N1 = N2 and the forces of tension are balanced. Once again assuming the strings are not stretchable the objects will not fall because of the tension in the string.


0

By definition, work is the energy required by a force to displace something. So, you're not doing any work, you're just cancelling out the force being applied to you. If you wouldn't push back then the other force would be doing work by displacing you. So, your work done basically cancelled out the work done by the other force.


2

In the physics definition of "work done" energy is transferred from one object (the one doing the work) to another object or system. When you push against the stationary stone you apply effort but the energy transfer is all internal to you own body - glucose being metabolized, etc. You get tired but you do no work according to the definition above.


1

In your case, no work is done. Intuitively: If you want to move a wall, you could push on it and you might use a lot of force. The wall isn't moving, and you are simply "wasting" your energy in your muscles. You are not doing any work of any value. If you push a balloon you can push it far without any real effort. You might move it a long way but that ...


0

Three parameters are needed for 2D force (as opposed to 6 for 3D, see http://math.stackexchange.com/a/1157906/3301). Composition A force with magnitude $F$ along a direction $\vec{e}=(e_x,e_y)$ going through a point $\vec{r} = (r_x,r_y)$ is described by the three parameters $$ f =(a,b,c)= ( F e_x , F e_y , F (e_y r_x - e_x r_y) ) $$ Decomposition Given ...


0

By the one who's exerting the force. It depends. When a system exerts some force on some other system, it does work on it and loses its energy. The lost energy is transferred to the second object, which in turn does work against the frictional force and loses its energy which is transferred into the molecules or atoms of the surface as heat.. There are a lot ...


0

For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it . Work cannot be done by a body who does not have it's own energy source like an hammer can never do work.


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For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it .


0

Strictly speaking, if a body is moved from a point to another by some force, it is said that the force does some work on the body.


0

Rotation is absolute, only uniform paths are inertial. So the answer is yes, the rotating object (which is the sphere in your example) would experience force. The cube would experience no force at all since it is the sphere which is rotating. 1


0

Probably not the correct way to think about it but this is my reasoning if you applied force on the body you had to spend the energy to do that work therefore you did the work the body gained the energy in some form say you lifted it of the ground the body has gained potential energy while you lost some bio-mechanical energy.


0

If you exert a force on a body and it becomes displaced, then it is said that work is done by you or, by the force on that body.


0

(Classical Physics only) Any massive body has a property known as inertia, thus even a body floating in outer space would require some kind of force to be accelerated. Using Newtons second law, you would find $$\tag{NII} \sum \vec{F} = \frac{\mathrm{d}}{\mathrm{d}t}\vec{p},$$ which for constant mass and one-dimensional motion simplifies to $$\tag{NII'} ...


1

High RPM/low torque is what gearboxes were invented for. You reduce the speed and increse the torque. Yes, if the peak torque of the motor is too low, the car won't move. A nice quick tutorial on DC motors is here, showing the peak torque is at 0 RPM, but that may still not be enough. It sounds like you are thinking that torque is a constant, but that is ...


1

Both cases are true. Basically the force the object will feel is due to the net differences in pressure and viscous forces acting on its upstream and downstream faces. How the pressure field and viscous stress are distributed around the object is as function of a number of factors: the shape of the object, its size relative to the pipe, the wall roughness, ...



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