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32

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 $F_{normal}$ ...


24

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...


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 ...


5

The tension is not the same for all parts of the rope. If the tension is $T_1$ between the hands and $T_2$ between the top hand and the ceiling then. $$ T_2 = T_1 + F_2 \\ T_1 = F_1 $$ where $F_1+F_2=W$ are the forces acted upon the arms. You arrive at this if you make two free body diagrams, one at each hand. The result is that $T_2 = W$ and ...


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

Newton law of gravitation is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$ Hence, the equation simplifies to $$F =(9.822) ...


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 ...


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 ...


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 ...


3

What @lemon said is right. While $\bf F = -\nabla \phi\left({\bf r}\right)$ suggests that $\bf F$ depends on $\bf r$, it ultimately depends on time through ${\bf r} = {\bf r}\left(t\right)$. That is, ${\bf F} = -\nabla \phi\left({\bf r}\left(t\right)\right)$. So, there is no inconsistency in writing Newton's second law as (a dot represents a time derivative) ...


3

The perpendicular force is no longer the weight of the box, you need to consider the force by the two workers in the perpendicular direction as well.


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


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 ...


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 ...


3

The braking force acts between the tyre and the road. The centre of mass is above this point so there is a rotational effect which increases the force going down through the front tyre and decreases the force going down through the rear tyre. Because the amount of braking force the tyre is able to produce is limited by the amount of force going down through ...


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 ...


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 $\rho = \frac{dm}{dV}$, ...


2

Lots of excellent answers here, but for fun, lets think about this backwards. Imagine you have the worlds first and only FRONT wheel drive motorcycle, and your rev it up and pop the clutch. What kind of launch do you think you would get with very little weight on the front tire? The reverse is true during braking when the deceleration shifts the weight of ...


2

The logic is simple: a body loses energy because opposing forces cancels themselves out. if you want to move in a direction +x against an opposing force -x the net balance must be in the direction +x. The logic is the one that rules the composition of forces or vector addition If $F_2 > F_1 = -18 N$ then the body will move, accelerate in the -x ...


2

Option 4, none of the above. Your option 1 is wrong because points don't rotate. Your option 2 is closer to correct, but ultimately still wrong. You're overly hung up on points (the origin). It might help to get a handle on what "rotation" is. Points don't rotate. Better said, a rotated point is indistinguishable from the original. What about one ...


2

Assume that the wind hits the panel normally with speed V and is then stopped (or escapes along the panel). That's not the best of assumptions. The air will flow around the plate. Only in the center of the plate will it hit the panel normally and stop. Your analysis does however capture some of the key dynamics in that the drag force is proportional to ...


2

The weight vector, $W=mg$, can be split up into two components, $mg\sin\theta$ and $mg\cos\theta$, where $\theta = 30$ in your case. The $mg\sin\theta$ is the force vector which would generate a torque, $\tau = \bf{L}\times{}mg\sin\theta$.


2

Any force greater than zero can stop the car. Only it will take longer and the distance moved by it by the time it stops also will be greater. If the force is larger these parameters (time to stop and distance traveled before stopping) will decreasing. Theoretically, infinite force is required to stop it instantaneously.


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 ...


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 ...


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

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

I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it. Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground: The spring clearly has work done on it because its kinetic energy increases and ...


2

I think this is a good example, and should be studied and understood carefully. But I don't know where the book goes after making it. Whether or not it is poorly stated depends on the surrounding context. It focuses attention on two things: 1.) the skater is a deformable body. The center of mass is not fixed in the body. 2.) work is defined as a ...


2

An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F⃗ on her from the rail . However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. This is simply confused: This example is nothing but an elastic collision ...



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