# Tag Info

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The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$a=\frac{m_3-m_2}{m_2+m_3}g\tag{1}$$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...

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Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...

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A very simple motivation for writing $\ddot{\bf x}(t) = {\bf F}({\bf x}, \dot{\bf x}, t)$, which might shed some light, is the following. We are given ${\bf x}(t)$ and $\dot{\bf x}(t)$ and we desire to calculate ${\bf x}(t + \delta t)$ and $\dot{\bf x}(t + \delta t)$. Now, ${\bf x}(t + \delta t) = {\bf x}(t) + \dot{\bf x}(t) \delta t$, which we may ...

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The trajectories are uniquely determined means that the theorem of existence and uniqueness applies (so, the differential equation has to be sufficiently regular). Newton's principle states more: the system is fully determined by the position and the speed, that is, by $2n$ constants, where $n$ is the dimension of the space. As you have $n$ equations (one ...

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In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...

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A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ...

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Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ $$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$ Second possibility : If your box is spherical, By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity. Hence, your ball attains terminal velocity. $$F=6\pi\eta rv$$ $$v=\frac{F}{6\pi\eta ... 2 By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing F_{app} otherwise there will be an acceleration. 2 I'm going to say the same thing as the author but explain more with words. I am also going to take the external force to be zero and ignore it, since the part you're confused about isn't independent of the external force. The force on particle \alpha, due to all the other particles in the system is$$f_{\alpha} = \sum_{\textrm{all other particles ...

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The point is: you can apply Newton's law $dp/dt= F = mg - T$, but you can't assume $dp/dt = m dv/dt$ as $dm/dt$ is nonzero. The fundamental law in Newtonian dynamics is the object's rate of change in momentum equals the net force exerted on it. Only in the special case of a constant mass object can this be translated into a statement about mass times ...

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As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.

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This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i$$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...

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Assume $f$ friction is being applied in direction of F. Direction, now, has no significance as $f$ will come out to be negative if it is opposite direction. $$F+f=ma$$ $$FR-fR=I\alpha$$ symbols have their usual meanings Note that if pure rolling occurs, $f$ is static. Also, $$a=\alpha R$$ You can calculate $f$. If $$|f|> \mu_{static}mg$$ You can ...

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Let the outside pressure be $P_0$. You can find its variation with altitude here. Using Bernoulli Equation, speed of liquid coming out is : $$\rho gh=\frac{1}{2}\rho v^2+P_0$$ $$v(h)=\sqrt{2\rho gh-P_0/\rho}$$ $h$ is the height of water remaining. It is assumed that $r<<R$ and thus $V$ is very small compared to other terms and is neglected. By ...

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There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to ...

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While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...

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The answer to this question depends on whether you're working "in real life" or on a physics problem. On a physics problem, the object doesn't start to move. The net force is exactly 0 at this time, so there is no acceleration. In reality, however, the object would possibly move. There's a variety of reasons for this: The normal force, static ...

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The equation for gravitational force F=Gm1m2/r^2 gives the force of attraction b/n any 2 bodies with point mass m1 and m2 and separated by a distance 'r'..it means both the objects are attracted towards each other by a force F=Gm1m2/r^2..It is also coherent with newtons 3rd law i.e action and reaction forces are equal.. The expression F=ma or a=F/m ...

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The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...

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I'll provide an answer in non-relativistic quantum mechanics. The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously. But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no ...

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Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$. ...

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