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22

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 ...


7

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


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

Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a ...


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

Force on the ground can never be greater than force on the rope Each players individually exerts some force on the rope and on the ground, but that doesn't means that the same amount of work is done on the rope and on the ground, or that one players exerts the same total force as his opponent. The winner overcomes his adversary only if/when because he ...


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 ...


3

Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$ where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere) Can you see the answer ...


2

$F_\phi = -mg\sin\phi$ is valid for any direction. But remember $\sin\phi$ can itself be negative, changing the overall sign of the quantity $F_\phi$. If positive $\phi$ is counterclockwise, then when the bob is to the left (clockwise) of equilibrium, $\phi<0$. This in turn means $\sin\phi<0$ so that $F_\phi=-mg\sin\phi>0$. So even with the ...


2

You correctly identify the residual velocity of the water after bouncing off the bucket as a critical parameter in the calculation. Where you go wrong is in assuming that you can assign any value you want to it. If your bucket's bottom was shaped in such a way as to "turn around" the water jet hitting it, then you would have the maximum possible momentum ...


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

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

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

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

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

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

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 ...


2

Newton's 2nd law in differential form (ignoring vectors) is $$F_\text{net}=\frac{dp}{dt}=\frac{d}{dt}\left(p_\text{train} + p_\text{water in trolley} + p_\text{rain just hitting trolley}\right). \tag{1}$$ You must take into account the change in momentum of the rain that occurs when it falls into the trolley and accelerates up to the speed of the train. ...


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

Remember how work is defined. The key word is displacement. I think that when you wrote $Fd$, you considered $d$ to define a single point, and not displacement. Now back to your problem. The work done by a net force is $$W=Fd=mad=ma{(}\frac{v_{f}^{2}-v_{i}^{2}}{2a})=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}=\Delta E_{c}$$ The single diference is that I ...


1

This is kind of hard to describe without a diagram but anyway i will give it a shot. I am assuming you mean something like the fourth photo on this page: http://www.oldschool.com.sg/index.php/module/PublicAccess/action/Wrapper/sid/9595afb87c8cf767f034c3ae53e74bae/coll_id/4745/recs_ppg/5/desc/wrap-function/pg_id/3 If you consider the simpler case where the ...


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

You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum). In the first approach, you could redo it to add the force that the ...


1

The block is accelerating, but that's due to the pseudo force, not the frictional force, which is zero. You can't just look at the horizontal forces on the block to determine whether or not static friction will hold. You have to consider what's going on with the other surface as well. Here's a simpler situation to consider first in an inertial frame. ...


1

Just to expand on vaaaaaal's answer, let's simplify this very slightly by assuming that the ball falls at it's average fall velocity $v$ for the whole height $h$ over a time $t$. Obviously, $v = \frac{h}{t}$. Then, we know that total momentum is conserved, so the Earth must fall up with speed $v_{e} = \frac{m}{M_{e}}v$. Thus, over the whole time of ...


1

I can't click the comment button so I will post here: there is still friction but unlike tires it doesn't oppose but it holds your feet without which you might slip ay every step. The work is done by your muscles, but if you go deeper it is your muscular system controlled by your nervous system being supported by your skelet system and maintained by your ...


1

The force from the earth on the ball and the force from the ball on the earth are in fact opposite and equal but the amount of work done on each is not the same. The earth is much more massive than the ball so, for an equivalent force, it is going to accelerate much more slowly and move a much shorter distance during the time the ball is falling than the ...



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