Tag Info

2

1) You can indeed treat them as a single system, but it is important to see why. The pulley, absence of friction, and the taut string imply that there are no unbalanced internal forces between them(if there were, it would be like 2 smaller interacting subsystems). 2) The problem is exactly one dimensional now-the one dimension being dictated by the taut ...

-1

when two objects are in contact with each other and are at rest, then the friction force is zero, because the direction of friction force is always parallel to the surface of contact, and when the two bodies are in contact and also at rest than only two forces act on the body i:e, weight of the body which is downward and the normal force which is upward, ...

0

The only reason is that we are dealing with vector equations and treating a single system in the second equation violates vector laws. True, it works in this simple case, but to describe in a general case whether it works would not be easy.

2

The $\vec\omega\times \vec r_b$ term appear in the rotating (non-inertial) frame. It is there because the motion of a particle in a rotating frame expressed in the lab contains a "pure" motion on the rotating (non-inertial) frame, plus a term to account for the rotation of the non-inertial frame. In equations, $\vec r'(t)=U(t) \vec r$ where $\vec r'$ is ...

0

You're not being entirely unreasonable for wanting to calculate whether the motor will be strong enough to do what you're trying to do. But, having quite a bit of experience in this area you're talking about, there are a couple of stumbling points I think you need to be aware of. Motor ratings by the manufacturer are merely there to allow you to compare ...

1

Here we completely ignore the comfort of the rider. The rider is rising not to spare his body from the impact, but to spare the rims. Actually, it's not quite right to ignore the comfort of the rider. Think of this from the point of view of Newton's third law: if the rider's body doesn't get high impact, then it doesn't give as high an impact on the bike. ...

0

Of course, Newton's laws are valid on all forces provided that an inertial frame is considered (can apply in non inertial frames too, but let's not discuss that now as it is irrelevant). So okay, if you want to apply 3rd law on friction, there will be an equal and opposite force on the surface too. The best way to see this is a two-block problem where you ...

0

Sorry to say, but I see no connection between those two definitions. All $N$ collinear points in a rigid body by definition gives only 1 same line. However, three orthogonal vectors with origin are semantically equal to 4 non-collinear points forming vertexes of Tetrahedron :

1

1) The Lagrangian is simply a function of generalised coordinates and velocities(see also @sanaris above), which when put into an action and extremised gives you the time evolution of the coordinates(and thus, the velocities via differentiation). You do not know the trajectory, and thus the velocity, and thus the kinetic energy, before you have actually ...

0

Lagrangian is defined as the kinetic energy minus the potential energy You are wrong. Lagrangian has nothing to do with kinetic and potential energy. For example Einstein-Hilbert action is $S=\frac{1}{2k}\int R\sqrt{-g}\,d^4x$. There is no fast-hand way to see in there something like "potential" and "kinetic" energy. There is only one true definition of ...

3

The difference in energy between the two static equilibrium positions may only be some potential energy difference. You may assume the friction force is $F=\mu N$ during sliding, where $\mu$ is the kinetic friction coefficient (taken equal to the static friction coefficient) but since this force is non conservative, the work done this force will not account ...

1

We know that : $$W=\vec{F} .\vec{r}=F.v.\text{cos}(\widehat{\vec{F}\mathbf \ {;}\ \vec{r}})$$ The work of the force is the scalar product of itself and the path : You have from your definition : $$P=\frac{dW}{dt} =\frac{\vec{F}.\vec{r}}{dt}=\vec{F}.\frac{d\vec{r}}{dt}=\vec{F}.\vec{v}$$ If the work is independent of the path, then the work is given by : $$W=\... 12 First, I am assuming that there is no kinetic friction acting on the insect as it moves up the bowl. If kinetic friction were involved, you would have energy dissipation, but I will not consider that here. Your mistake is in assuming that the static friction force is equal to its maximum value during the entire process. \mu N only determines the maximum ... 0 In Lagrangian mechanics, we take the kinetic & potential terms as axiomatic, i.e. we don't use Newton's second law to justify K=\frac12m\dot{x}^2, we just claim K=\frac12m\dot{x}^2. Newtonian, Lagrangian and Hamiltonian mechanics (and a few other options) are equivalent, but they assume different things. 0 You are right. The Newtonian and Lagrangian formulations are equivalent. Either can be used as a starting point for mechanics. Which you think more fundamental is largely a matter of personal choice. My own preference is to formulate Newton's laws from conservation of momentum (the third law contains the physical content, the second just a definition of ... 1 The entries at time t are \psi(x,y,z,t). The arrangement of as a column vector, or as a three dimensional array, is essentially arbitrary. You may find it easier to think of it as a three dimensional array. It is more usual to use vector notation \mathbf x = (x,y,z) = (x^1,x^2,x^3), then the correspondence is \psi(\mathbf x) = \langle \mathbf x |\... 0 You need to know that a motion can be accelerated in two independent ways or any combination of these; either when there is a change in velocity, or when there is a change in direction of motion. The above problem has a constant velocity but the direction of car at every point of the loop changes, which mean the motion is accelerated. Newton's laws tell ... 0 It depends on what exactly you mean by kinetic friction, since there are several forces that may be referred by this term. If in this context it means sliding frictiob, i.e. what sratic friction force becomes when object begins to slide, then by its very definition it is directed against the direction of motion. This does not exclude the possibility that ... 0 Let's consider one molecule that hits the piston. The kinetic energy of that molecule is supplied by breaking a chemical bond during combustion. Just before the molecule hits the piston the molecule's momentum is p = mv, and momentum of piston is p = 0. After the collision the law of conservation of momentum says that sum of momentum must be the same. Hence, ... 2 If you have 3 different points on rigid body you can create orthonormal coordinate system with this equations$$\vec Z(t)=\dfrac{\overrightarrow{R}_{13}\times \overrightarrow{R}_{12}}{\left| \overrightarrow{R}_{13}\times \overrightarrow{R}_{12}\right| }\vec Y(t)=\dfrac{\overrightarrow{R}12}{\left| \overrightarrow{R}_{12}\right| } \overrightarrow{...

0

Your derivation ends up with a correct result, but I couldn't quite follow your derivation. A few remarks that may help you: The forces acting on a small element of a continuuous 1D medium is $-P(x+dx)S$ on one side and $P(x)S$ on the other sides so the net force is $\frac{\partial P}{dx}S$. Of course you may drop the $S$ term since everyting will be ...

1

On each hand, you really have two forces: one from the other hand (10N, pushing out) and one from the arm (10N, pushing in). These are the equal and opposite forces, hence your hands do not move.

2

I think you are confusing two ideas (commonly confused). Equilibrium forces (which cancel) and reactive forces, which may or may not cancel depending on the situation. Think of a book on the table. You push on the book. The book pushes back on you with an equal and opposite reactive force. If you push hard enough, the book will move. According to Newton's ...

2

When you push on your right hand with your left you feel a force back on your right hand. This is the 'equal and opposite' force that comes from Newton's third law. However, at the same time you are pushing with your right hand you are also pushing with your left and feel a force back on your left hand, again due to Newton's third law. While it is true that ...

3

Fix a reference frame $R$ with axes $x,y,z$ and origin $O$ and suppose that the solid body $B$ is moving with a planar motion, let's say, parallel to the plane $x,y$. If $O'$ is a fixed point of $B$, but generally moving in $R$, the velocity of a point $P \in B$ in $R$ satisfies $$\vec{v}_P(t) = \vec{v}_{O'}(t) + \vec{\omega}(t)\times \vec{O'P}(t)\:.$$ ...

0

There is a force at work here: gravity. The argument is only that no force needs to start the sphere moving in one direction rather than another. But this only works because a perfect sphere (like a perfect hemisphere) has no flat surfaces. If you think of two dodecahedrons (or rather one dodecahedron and a hemi-dodecahedron) rather than two spheres, you'...

0

This does indeed work out. However, we have to be careful with how we define "from an object's perspective." You're defining different frames, and conversions between frames can change the values of velocities. The key question is what happens to the frames at the collision. If the frame keeps going along on the path m1 or m2 would have taken, then you ...

0

For statically indeterminate systems you need to consider both the material behavior and the kinematics of deformation in addition to the equations of equilibrium in order to determine the forces. So, when you ask about a "rigid body simulation tool", such a tool would have to use some deformable material model, e.g. linear elasticity, and then take a limit ...

2

I'll let the animation speak for itself. The blue arrow shows the force on the object. For this scenario to happen it is important that the collision is elastic (all energy is conserved). I used a force that is proportional to penetration depth. This way the balls feel a force that is the same during the deceleration and recoil. In inelastic collisions it ...

1

Try considering momentum conservation. The wall won’t be moving before or after the collision with the ball, and presuming the collision is elastic (no energy lost to heat, sound etc.) then the sum of the momenta will be conserved before and after the collision and you should find that the ball must have the same momentum before and after the collision. So ...

0

We'll first write out the equation in its "full" form. $$m\vec{a} =m\vec{g} -cv^{2}\hat{v}$$ We will be using the convention that down is negative. We will also be assume we are dealing with two dimensions here. Therefore, we can expand this out as ma_{x}\hat{x} +ma_{y}\hat{y} =-mg\hat{y} -c||\vec{v} ||^{2}\hat{v} ...

1

Because the reaction force is larger. The force needed to make an object reach a certain speed is the same as the force needed to slow it down from that speed to zero over the same time. This is Newton's 2nd law. So, when two equal and movable objects collide, the (action) force that is required to make one object speed up to a certain speed is exactly the ...

0

The centrifugal force mw^2r will act radially outwards. And this force will be balanced by a component of the Normal force. This is because of the strong grip made by our Jimmy. Now equating these two forces , you will get the value of Normal. Insert this N in the Frictional force formula and balance it by equating it with component of mg and bam.... Is ...

3

Would the centripetal force still provided by the weight of the mass? Why or why not? When you spin it vertically, you would most probably spin it with a constant angular velocity (try it out), which means that that the centripetal acceleration required at any point will be the same. However, in this case, the radial component of force due to gravity keeps ...

0

For an idealised system like whole air in the room is at the back of the person who stands at the door is apparently too high and this link 1. Link has already shows the necessary calculations so I didn't put them here and for more broad calculations you may refer this link 2. Link.

-1

The problem comes from the sign of the drag term. The first form of the equation you have is not exactly correct. When you say $$m \dot v = g - c v^2$$ there is an assumption that $v$ is positive, so the negative sign in front of $c v^2$ ensures that the drag for opposes the motion. Often, sign conventions can get confusing when replacing vectors with ...

-2

I was thinking of this also when I was reading the book "Reality is not what it seems" and I realized that if the speed of light were infinite, the universe would expand to infinity instantly after the big bang so that the universe would just "evaporate" right away instead of diffusing slowly. Realistic mass increase formula also supports this. Because the ...

1

Now this could be naive, but I'm inclined to think that any of the shifting around as a result of changing the angle of declination for a pushup shouldn't matter in terms of the net external force the hands have to produce. I would wager any change of measurement you find on a scale has more to do with fluids shifting in the body, and thus the center of mass ...

1

While other answers are not wrong, I'd like to add that the main reason why someone's feet can't stay on the ground in free fall is that any time there is a slight pressure or bump between feet and floor due to body movement, the contact force will push the feet upwards. Since no other forces are involved, the feet will keep floating away from the floor. ...

0

If all three pulleys are fixed to a wall or something you get some friction depending how easily your pulley turns, and if it is heavy you need some force to turn it , so if the pulleys are light and work well it should not ad much, but it always adds a little.

0

I don't think it particularly important that the classical potential is not exactly equal to the expectation of the quantum potential in all cases. Classical potential is defined in the first instance by integrating force. To derive the force from the potential is circular. Really we have two slightly different potentials, because the classical notion of a ...

0

It is simple: F = T*2*Pi*eff/l Where: F: Force, newtons; T: Torque, newton-meters; l: lead, meters; eff: ballscrew efficiency In most cases, efficiency can be set safely to 90% with enough margin. So, for example, a ball screw with 5mm lead, driven by 2Nm motor will give: 2*2*Pi*0.9/0.005 = 2262N = 231kg

0

$P = \rho g h$ is true in static situations. Your wave example is dynamic and pressures at a given height may not be uniform in such cases. If the water just below the surface did actually have a very high pressure, that pressure would rapidly accelerate the water until the pressure equalized. It wouldn't form a barrier to entry.

0

I have raised a similar question here and finally I ended up answering it. My answer is on the derivation of the $2^{nd}$ law which is the lengthiest of 3. The bonus is that I have done the complete proof using cartesian co-ordinates, so even a high school student with calculus knowledge can understand it.

1

Ensuring that $n$ and $k$ are mutually prime ensures the ball gets out of the well at its first opportunity, and not after. Originally I thought that if $n$ and $k$ are solutions, then $2n$ and $2k$ should be solutions as well. But we need to recognize that by that point, the ball is already out of the well! It got out after $n$ and $k$ horizontal/vertical ...

2

Calculation the Car Wheels Load $F_G=m\,g$ car weight $F_V$ wheel front load $F_H$ wheel rear load $a_A$ car acceleration $a_D$ car deceleration take the sum of the torques about point V you get: $$\sum \tau_V=F_H\,l-F_G\,l_V-m\,a_A\,h_s+m\,a_D\,h_s=0\tag 1$$ or about point H $$\sum \tau_H=F_V\,l-F_G\,l_H+m\,a_A\,h_s-m\,a_D\,h_s=0\tag 2$$ from ...

1

The axiomatic system you are looking for needs the mathematical axioms and theorems used for establishing differential equations and calculus, to be complete, Newton's law's are not enough. Euclidean geometry is simple, as it needs only algebraic relations as a mathematical framework. Classical mechanics needs a more sophisticated mathematical framework, ...

2

The point of the first paper is that $F = ma$ doesn't actually indicate how matter moves, at least not without specifying what $F$ is, which is correct. You can't really get much of use out of Newtonian mechanics, outside of general theorems, if you don't specify the forces. I don't think this is quite the takedown it is, because an axiom system can get by ...

0

For each object for energy conservation you will have $\Delta\text{KE}=-\Delta\text{PE}$. Since both of these terms are directly proportional to the mass, the mass variables on each side of the equation will cancel out. Therefore, you do not need to know the mass of each object to solve this problem. Also, in this case speed means the translational speed of ...

5

First, note that you can't "misdefine" something. Angular acceleration is defined to be the time derivative of angular velocity; that is it. Instead of questioning the definition then, you should be questioning your understanding of the relationship between this definition and other physically relevant definitions. The error is in assuming that the moment ...

Top 50 recent answers are included