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21

Yes, every gravitational force in Newtonian mechanics has an equal and opposing force, and it usually acts on other mass. More specifically, every two pairs of masses feel a gravitational force that's proportional to the product of their masses and inversely proportional to the square of their relative distance, but more important is the fact that both ...


14

Suppose you're standing on a box as shown in (a) below: There are four forces acting. You apply a downward force $mg$ on the top of the box, and by Newton's third law the box applies an upwards force $-mg$ on you. The box transmits your force to the ground, so the box applies a downwards force $mg$ on the ground and the ground applies an upwards force ...


14

Alright I'll throw my hat into the ring with an answer. The idea that it's an unsolved problem is totally bogus. When you start to fall to one side or another if you turn the wheel slightly in the direction you're falling the bicycle starts to follow a curved path. There is a force due to friction that deflects the rider's path into a curve: The ...


10

I do not agree with the angular momentum theory: if you were to hop off your bike at speed and let it go by itself, it would not go very far before falling on its side, even less if you put a ~150 lbs sandbag on your saddle. There is indeed an effect caused by momentum, but this is negligible when compared to the actual contribution of the rider. I agree ...


7

The diagram is misleading. Look at this: At any moment in time, you have the following forces on the particle: Gravity Tension in the string When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string ...


6

The force on rope is equal for both of them at any time. For winning the game the force on ground is responsible.


5

According to Newton's 3rd law of motion, the force exerted is equally distributed between the rope and the ground. Force on the ground can never be greater than force on the rope and vice versa. Therefore both players exert the same force on the rope and ground, that means that the same amount of work is done on the rope and on the ground. The winner ...


5

Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically. These two forces combine to give the ...


4

My knowledge is limited on the subject but matter is typically prevented from collapsing under the weight of extreme gravity by particle degeneracy. This is what keeps neutron stars from collapsing into black holes and is the result of particles resisting occupying the same quantum states. There are also some recent observations that indicate that there is ...


3

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$ \sum \mathbf p_{i}=\sum\mathbf p_f $$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...


3

So far, it is still an open question as to why bicycles are stable at all. There have been a few ideas put forward, but they have been disproved by construction of non-standard bicycles. The most common explanation is that the wheels on a bike act as a gyroscope, preventing the bike from falling over. A bike was constructed with counter-rotating wheels to ...


3

First - Excel is a fine tool for doing this kind of simulation at the level you want to do. Just remember that integrating equations of motion involves certain errors - you want to make sure you minimize these errors. Two things you can do: use small time steps, and take the average acceleration / velocity over the time step to compute the new position. So ...


3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...


2

The answer is that, despite our best efforts, we still can't quite put a finger on it. The gyroscopic forces mentioned by bobie in his answer have been proven not to be sufficient to fully explain why a bike stays upright. It's indeed very surprising that physicists are not able to explain the mechanism behind such an (apparently) simple and ubiquitous ...


2

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...


2

Static friction force arises whenever there is interaction between two bodies by direct contact (touch). There need not be any mutual motion between the bodies. This friction force is necessary to explain why the bodies around us maintain their position so reliably. Without friction forces, there would be nothing opposing their mutual motion and the world ...


2

This looks like a homework problem, so I'll just give you a hint, rather than give you the whole answer. The hint is that according to Archimedes' principle, the buoyant force on a body that is fully or partially submerged in a fluid is equal to the weight of the displaced fluid. So the net force on the barrel at a given time is the weight of the amount of ...


2

You really, really have to be able to draw a diagram to understand many physics problems. Here is my interpretation of what you write above (with apologies for poor lining up of various lines due to the limitations of the drawing package I had to hand). The normal force acts at right angles to the surface and is labelled as $N$. The weight of the ...


2

Notice that $h$ and $r$ are related in the following way: \begin{align} r = R + h \end{align} where $R$ is the radius of the Earth (the distance from the center to the surface) and $h$ is the height above the surface. Then notice that \begin{align} U = -\frac{GmM}{r} = -\frac{GMm}{R+h} = -\frac{GMm}{R}\left[1 -\frac{h}{R} + O(\frac{h}{R})^2\right]. ...


2

The first the expression $U(r) = -\frac {GM} r$ is a potential, but not potential energy. The units are velocity2. This is a widely used potential in solar system astronomy, geology, geophysics, and in aerospace engineering. For example, see ...


2

The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, $r$, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of $r$ only, and not $\mathbf{r}$. Then you can use it to analyze the evolution of $r$ in ...


2

Yes - it really is that simple. There is no "upthrust" from the sand falling out of the car. There is just less mass in the car, and thus less force of gravity on the car. The sand gains momentum because gravity continues to pull on it after it leaves the car - but that no longer affects the car.


2

For calculations of a collision, you look at (kinetic) energy and momentum equations. Momentum is conserved; energy may be dissipated. There is a direct exchange of momentum in the form of $F\Delta t$ - the same impulse that slows one object down accelerates the other, by the same amount. But the force may result in a deformation of the ball such that ...


1

Newton's 2nd law in its most general differential form (ignoring vectors) is $$F_\text{net}=\frac{dp}{dt}=\frac{d(mv)}{dt}. \tag{1}$$ You must use the product rule on that bad boy $mv$. In method 1, it looks like you went straight to $F=ma$ which got you into trouble. You can really only use this simplified form of $m$ is constant or $a$ don't change; ...


1

Coefficient of restitution and hardness are not the same thing. The COR basically tells us how much energy gets lost in the collision process. In the case of a soft/hard ball with identical v1/v2 the COR=1, i.e. there is no energy loss. However, the collision of a perfectly elastic ball with a perfect wall will take a different amount of time, depending on ...


1

Hint. I think the error is in your first equation, adding the forces for mass M1: $\vec{F}_{M_3}+\vec{F}=m_{1}\vec{a}_1\\$. There is an additional force on M1 that you have omitted. Edit: The pulley exerts a force on M1.


1

You are introducing some irrelevant variables, as $F_{M_3}$, $T_1$, $T_2$. Let us make the assumption that $T_1=T_2=T$ (the pulley doesn't rotate and the string is massless). The whole has mass $M=m_1+m_2+m_3$, accelerates with $a$ and the force on $M$ is $$F=Ma.$$ The tension $T$ equals $m_2 a$ and also $m_3g$, so$$a=\frac{m_3}{m_2}g.$$Hence ...


1

The usual integral for the divergence of the velocity field is over a volume. Since $u$ does not depend on $y$ and $v$ does not depend on $x$, we have $$ \begin{align} \int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = \iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\ & = \iint ...


1

Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ...


1

It has not be proven that The Second Law of Thermodynamics is physically derived from other basic physical principles. The H-Theorem is predicated upon some pretty serious, yet plausible, assumptions about how our universe works. To my knowledge, these assumptions have not themselves been explained using other principles and/or experimental verifications of ...



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