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0

Any change in speed is acceleration. Accelerating is equivalent to gravity being different. Braking slows you down, which means you are accelerating backwards, which means you feel gravity pulling you in the same direction as on a downhill slope. As on that downhill slope, more of the your bike's weight is supported by front wheel and less by the back wheel. ...


0

Q1: How does one interpret tensile strength, yield strength, etc.? The answer is to interpret them as the result of a test that tells you what the material can withstand in an engineering application. The type of machine used to measure tensile strength is popularly called an Instron machine (the most famous manufacturer is Instron; kind of like how tissue ...


1

In a system, the total sum of forces when added together equals mass times acceleration: $$ \sum F = \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\mathrm{d}mv}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma $$ Since the sum of the forces on the robots is zero, there is no acceleration. However, the tension of the string is not contingent on the movement. I ...


19

Using the brakes on the front of the bike causes your weight to shift forward. Additional weight allows more force before the tire will slip (skid). If you brake hard enough the back tire of your bike will lift up and at that point all of the mass is distributed on the front tire. Remember the maximum force is $F_{max} = \mu F_{normal}$ and $F_{normal}$ ...


0

He "ignores" the magnetic field due to $I_2$ because he is interested only in the effect of the magnetic field caused by $I_1$ on the loop. So while there are indeed internal forces on the loop from the magnetic field that it creates itself, these are by his own admission not the forces he is interested in.


2

I'll write my comments here as a full answer, as suggested by Floris. I won't use the moment of inertia tensor: it's simpler from pure angular momentum of each point particle. We know that $$\vec{L} = (\vec{r} \times \dot{\vec{r}})\,m .$$ So, for a point particle, $$d\vec{L} = (\vec{r} \times \dot{\vec{r}})\, dm .$$ Noting that $\rho = \frac{dm}{dV}$, ...


-1

Using the method of similarity (see Landau or Arnold), let's assume we can apply the same force to both objects. Therefore, \begin{equation} m\frac{d^2x}{dt^2}=m'\frac{d^2x'}{dt^2}. \end{equation} By reescaling $x'\rightarrow\frac{m}{m'}x$, we get that one can throw the ball $\frac{453}{148}=3,06$ longer than the granade.


2

An olympic shot put shot is 7260 grams and has been thrown 25 yards. 50 yards for a 453 gram grenade is reasonable. The military pentathlon has a 600 gram grenade event that requires precision throwing to 35 meters. Hartmut Nienbar threw a 600 gram grenade 80.3 meters in 1983. A relatively light-weight object will insignificantly slow the release ...


0

I'll attempt to explain the nature of buoyant force with an example. Consider a cube with side $l$ and density $500kg/m^3$, is submerged completely in water. Let the mass of this cube be $m_{cube}=1kg$ We know that density of water, $\rho_{water}=1000kg/m^3$ Now, Water above the cube exerts a force equivalent to its weight. The cube exerts an equal and ...


3

BTW a horsepower, even though an archaic unit retrieved from archaeological digs, is still recognisable to physicists as a unit of power (rate of working) rather than energy. To answer this question, you need an accurate model of your throwing apparatus: i.e. you need a functional characterisation of the force that can be imparted by the hand as a function ...


0

The total force (vector) is the sum of all the forces:$$\vec{F_T}=\sum\limits_i\vec{F_i}$$ Mathematically:$$||\vec{F_T}||=\sqrt{\sum\limits_i||\vec{F_i}||^2}\tag{1}$$ If you have doubt on that you can prove $(1)$ using the mathematical induction.


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Quick answer, the ground. Try to use a lever on unstable ground (like mud) see what happens.


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Quite simply: yes. When you add up the x and y components, you result in a vector whose x and y components have already been broken down. The total force is equal to the magnitude of that vector. As a simple example, if you have $10N$ at $0^o$ and $10N$ at $45^o$ you get a resulting vector $\vec{F}$ whose components are $F_x = 17.07N$ and $F_y = 7.07N$. ...


0

Force is a vector, so it has components in both the $x$ and $y$ directions in your specific example. You have to add the components. You find the components by using trig. $$F_x = F_{x,A}+F_{x,B} + F_{x,C}$$ $$F_y = F_{y,A}+F_{y,B} + F_{y,C}$$ Then the total force is ${\boldsymbol F} = F_x \hat{x} + F_y \hat{y}$


0

Force is a vector which itself has components along different axes, but if you just want the magnitude of the vector, then yes, you use the Pythagorean theorem. There's a good page on vector addition here if you want more info.


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Historically, the use of levers was well understood long before Newtonian mechanics - the scales that merchants used for commerce were all based on levers, and any time accountants are involved, the science will be very well understood. As such, Newtonian Mechanics had to conform to the relationship between the ratio of weights and lengths, not than the ...


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Do you know about conservation of energy? Do you know about conservation of momentum? If you do then here's what you do:: Conserve energy for the entire system (Including velocity of the wedge). At the lower point the potential energy of the smaller block changes (potential energy= mgh) here is one relation between the velocity of the block and the wedge. ...


0

What does the principle of the lever state? it states that there is a compatibility condition $\frac{F_2}{F_1}=\frac{r_1}{r_2}$, related to torques (which, transposing, $F_2 \cdot r_2 = F_1 \cdot r_1$ is another form of the conservation of energy). See a related question for the relation between torque and force. So the lever compatibility condition (for ...


0

You seem to be having quite the fun with this particular diagram over the past few days. The tension in the string is only equal to the weight of the hanging mass if the system is in equilibrium. Other answers have said that if the system is at rest, $T$ and $m_{hanging}g$ are equal, but that's only partially true. The key is that the system has to be in ...


-1

The mass of the body on the table is not directly important - the important thing for the tension in the string is the force of friction $F_f$ in the diagram. So it depends on how smooth the surface of the table is. If $F_f$ is greater than or equal to $mg$ then the tension $T$ is $mg$ and the system is at rest. If $F_f$ is less than $mg$ then the ...


0

It does not depend on the mass but on the friction of the horizontal surface. If friction is high enough so that the masses do not move, the answer is yes. But if the friction is zero the masses will move (regardless of how large the mass on the table is) and the tension will be less than the hanging mass.


1

We call buoyancy a force, but really, what is it ? It's only gravity. This is only a difference in the gravity force applied to the water and the gravity force applied to your object. So buoyancy is not a force applied by the water to the object. It's gravity applied differently to the water and the object by the earth. Here, the real force is gravity.


-1

In a more general and loosely specified manner Newton's Third Law states that one cannot touch without being touched. It seems a very reasonable and obvious statement. However, Quantum Mechanics has shown it is false with such examples as Quantum Non-demolition Measurement. "Quantum non-demolition (QND) measurements improve sensitivity by evading ...


1

I wrote a whole answer, which follows below, but I want to preface by saying this: When you're solving net force problems, it's very helpful to identify your system. If you only want to know the net force and/or acceleration of block 1, then your system is just block 1. In one of my intro physics classes, I was even taught to draw a loop around my system ...


1

Working in the frame of reference of the ground: As the person walks towards the center of the merry-go-round, their tangential velocity decreases: from a high value at the rim to close to zero when they reach the center. Thus, there must be a tangential force acting on the walker to reduce their tangential velocity; it comes from the force of friction ...


0

Without some detailed information this is impossible to answer. In general your inclusion of non-linear hydrodynamic forces probably depends a lot on your righting moment and sea-state conditions. If the dominate forces are the linear ones (large righting moment), then smaller effects from friction, turbulence, etc. on an irregular structure is less ...


1

"Conservation of momentum is probably a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen." Sure it does, at least in Newtonian physics where forces act instantaneously (the link in your first sentence, with the comment that the 3rd law is "definitely not universal", deals with ...


3

Why do physicists strive to make that statement universally applicable? They did do just that from the 17th century to the 19th century, but that is no longer the case. Electromagnetism, relativity theory, and quantum theory put an end to thinking that Newton's mechanics was universal. It's a trivial matter to deduce Newton's third law by assuming that ...


7

What acts here is called impulse (of a Force) Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) they collide. B is at rest (v = 0): Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what steel the balls are made you can calculate the ...


0

Sure. An elastic collision is merely an idealized process which never occurs in nature. On the microscopic level, the two bodies are made up out of atoms. The electron shells of these atoms contain electrons. Since negative charges repel each other, the two bodys will repel each other in a smooth fashion (an inverse square force is acting).


2

No, gripping something with your hands will make you feel tired because that is how muscles work: blood is pumping, cells are working... But imagin the gripped object on a table, and on it a pile of heavy books. The effect on the object is the same (more or less) than when you grip it with your hand, but the books never get tired. You could harvers energy ...


5

Have a look at the answers to Why does holding something up cost energy while no work is being done?. Gripping things takes energy not because a constant, stationary force does any work but because of the way muscles work. A stationary force doesn't do any work so no energy can be harvested from it. The best you could do is capture the heat given off from ...


3

There are two issues at play here. If you lift something upward at constant speed, then the acceleration $\vec a$ is zero. This means that the net force $\vec F_\text{net}$ is zero by Newton's second law ($\vec{F}_\text{net}=m\vec{a}$). As long as something moves with constant velocity, all of the forces add up to zero (i.e., they cancel out). Yes, that is ...


1

$W = x + y$ if the fly is hovering, $W>(x+y)$ when the fly is flying upwards (i.e. accelerating up), $W<(x+y)$ when the fly is accelerating down, and $W = x$ if the fly is free falling (accelerating at g) (assuming there is no air resistance). Air is a fluid. If you replace air with honey and flies with fishes that prefer to swim in honey than water, ...


0

Yes, but note that $U_i = K_f$ because $U_f$ is 0 and $K_i$ is 0. The rest of the working seems fine. $ {1 \over 2 }kx^2 = {1 \over 2}mv^2 $ $ {1 \over 2} (1290.9{N \over M})(.275m)^2 = {1 \over 2}(.435kg)v^2$ but we know $N/m = kg/s^2$ $97.62{kg\cdot{m^2} \over s^2} \over (.435kg)$ = $ v^2$ You may have multiplied wrongly. $N/m * m^2 = N\cdot{m}$ and ...


0

If you assume linear force displacement relationship (small displacements) with stiffness $k$ then $F(t) = k\, x(t)$ where $x(t)$ is the separation of the two objects. Also initially the impact speed is $V = \dot{x}(0)$. The impact time is characterized by the natural frequency of the system $$\omega = \sqrt{\frac{k}{m}}$$ where $m = \left( \frac{1}{m_1} + ...


0

(make comment as an answer) As mentioned in the comment here, the impact forces that are active during the time-frame of the actual impact are 1) unknown 2) difficult to put in analytic form That is why results like the conservation of momentum theorem are used. One can do estimations or approximations of these impact forces but would have to use more ...


1

In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small. The above reasoning is the ...


-2

Ok so first things first. 1) I presume that by "not attached to a ceiling " you mean to say that the entire system is being pulled in the upward direction by a constant force F . 2)The force equations you provided are wrong.If the system is being pulled upwards then we can neglect tension as it is an internal force. 3)acceleration A of the system is F/M ...


2

what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small? Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest: Ball A will exert on b the Impulse of a Force $J$ and its velocity, ...


0

As stated in other answers, $\Delta t$, although small than times you're used to perceiving, is not 0. Because the marbles do not deform very much, the collision does not appear to take much time. However, in the grand scheme of things, the change in momentum for each marble during the collision is also small. Although answers vary, a quick google search ...


1

The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite. Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration ...


3

That's a very nice answer by ACuriousMind. I would like to add something, though. GR is actually not like other gauge theories in some of its aspects (apart from having lots of similarities). For starters, it is background-independent and highly non-linear. In ordinary QFT we usually deal with perturbative expansions, which make sence only for weak-coupled ...


2

In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$ \sum F = \frac{\Delta mv}{\Delta t} $$ Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$ \Delta mv = \int ...


0

Generically, if I have a particle which has potential energy $\phi(x,y,z)$, then the force on that particle will be given by ${\vec F} = - {\vec \nabla}\phi$. So, generically, the motion of particles will "try" to minimze the potential energy. In particular, the only points where the particle will not move will be those points where $\nabla \phi = 0$, or, ...


3

The other forces are also just the result of "spacetime bending", just in a different way. There is no fundamental difference in the description of the other forces through gauge theories and gravity through relativity.1 The reason why it is often said that it is different is that our usual methods of quantizing a theory fail when applied to gravity. But to ...


2

Force is a classical concept that is useful in modeling the mesoscopic world, i.e the world of classical thermodynamics, mechanics and electrodynamics. Exchanged particles are quantum mechanical concepts which mainly work in small atomic size dimensions. There is continuity in physics going from mesoscopic to the microscopic frameworks, and continuity ...


2

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ...


2

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force. The structure of field which causes ...


0

The astronauts do have a potentially deleterious affect on the vibrational environment on the International Space Station. They have to exercise multiple hours a day to keep bone and muscle loss down to a tolerable minimum. The unmitigated vibrations from all that exercising would be harmful to very sensitive micro-g experiments. Before I get to the ...



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