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0

$F_n$ is not equal to $mg$. It is equal to $mg + F\cdot \sin{37^\circ}$ because the applied force has also got a vertical component which makes the total friction force bigger in magnitude.


3

For most guns, you can roughly hold them in place while fired. That is, the repulsion will not only "hit" the gun's mass but the astronaut's mass too, not allowing the gun to gain such high speed. With your numbers this leaves at most $$ v = 0.3~\frac{m}{s} = 1.08~\frac{km}{h} $$ for an astronaut plus gun with $m=80~kg$, if no torque is applied. But I ...


15

You've calculated the speed of a remote-triggered gun after it fires the bullet, true. However, there's actually nothing about space in your calculation, as @ACuriousMind noted. In theory, a gun fired on Earth could fly off just as fast, at least for a second. What you should use is not $m_\mathrm{gun}$ but $m_\mathrm{gun} + m_\mathrm{person}$. The gun never ...


0

Based on the calculation in my earlier answer, we were going to try to charge the earth with $10^{12}C$ and put that charge on the moon. Sending all the charge back in a giant lightning strike would then cause such a rapid change in electric field (not to mention that it dumps all the energy of twelve suns for a tenth of a second...) that it would ...


1

I will estimate the potential difference, and later add in the other considerations. I am taking the entire space between the Earth and Moon to be vacuum - so I am totally ignoring the effects of the Earth's atmosphere. The Swinger Limit (http://en.wikipedia.org/wiki/Schwinger_limit) is the largest electric field that can exist before nonlinear effects start ...


1

If we ignore air resistance, between $t_1$ and $t_2$ there are no forces acting on the ball besides gravity so with both the time and distance we can compute the velocity the ball would need to be traveling to be at both points. With $a=-9.81$, integrate twice to get the position function. $$ x(t)= -9.81\frac{t^2}{2}+C_0t+C_1.$$ Plug in the two conditions ...


0

I think what you are trying to get to is $${\rm d}{\bf I}_G = -[{\bf r}\times][{\bf r}\times]{\rm d}m$$ where ${\rm d}{\bf I}_G$ is the contribution to the mass moment of inertia of a mass ${\rm d}m$ located a distance ${\bf r}={\bf R}-{\bf R}_G$ from the center of mass. The 3×3 skew symmetric cross product operator $[{\bf r}\times]$ is defined as ...


3

Why does the mass of the orbitting object have no effect on its revolution at all? It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 ...


1

"So the idea, is that performing experiments himself, inside the train it is possible to detect the value of the acceleration?" You're spot on. Couldn't write it better myself. BTW in GR we have the same situation, as long as you qualify what you mean. All observers will agree on the reading on an accelerometer fixed in a certain frame, and on the readings ...


0

Galilean relativity does not automatically hold in Newtonian mechanics. You are correct that Newton's first two laws appear at first sight to be invariant under galilean transformations, but a galilean transformation only transforms spatial coordinates,they don't actually tell you how forces transform. If your force laws just has forces that depend only on ...


2

You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.] An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or ...


20

But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given ...


0

This is a classic curved bank problem. The assumption that the car doesn't slide is made to simplify analysis, to illustrate how the problem work. There is no a priori reasoining. For more complex problems the assumption will have to be abandoned.


4

Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum). In free-fall, $F = mg$ And we know that, $F = ma$ So we can substitute, $ma = mg$ And divide by $m$, $a = g$ Thus, no matter what mass is, acceleration equals $g$.


10

You certainly know that all things fall at the same rate regardless of their mass (neglecting friction). An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.


3

Well, if you have $2{\pi}r$ and you know the time in which you traveled it, then you could find speed. Nothing prevents this.


0

Here's a mathematical proof to your problem, showing that the polar moment of inertia about the centre of gravity is indeed the minimum, at least for a laminar (2D) rigid body. For a rigid body of mass $m$: The polar moment of inertia taken about a general point P is: $$I_P = \int\limits_m \left( \vec r_P \cdot \vec r_P \right) dm$$ We need to find the ...


1

If you attached two flywheels through a motor to the disk at the positions you show, and the motors start spinning in the same direction, then conservation of angular momentum tells us that as the flywheels spin clockwise, the disk must (and will) rotate counterclockwise. However - if you attach the motors to an external structure, you are preventing ...


1

By definition, for a 2D object comprising a number of distinct point masses $m_i$ rotating about a point $r_0$, the moment of inertia is given as the sum $$I = \sum_i{m_i |\vec r_i - \vec{r_0}|^2}$$ If we write the position of the vector $\vec{r}$ as (x,y) and the point $r_0$ as $(x_0, y_0)$ then we can write this as $$\begin{align} I &= \sum_i{m_i ...


0

I do not have access to this paper which is, I think, at least related to your question: Micronewton electromagnetic thruster by Charrier at Applied Physics Letters (2012) abstract: A low cost and light electromagnetic thruster, consisting in a disc rigidly attached with a coaxial coil, shows steady recoil by losing its linear momentum. The signal applied in ...


0

I'm surprised to see no mention of turbofan jet engines in the answers so far. In fact, the blade tips of most modern turbofan engines do reach supersonic speeds. As predicted by a comment above by tpg, this does produce shockwaves from each blade tip and what you hear is a 'buzzing' sound, which is commonly described as sounding like a buzzsaw. If you ...


0

By saying point(c) (or any of the other points) applies a torque on the disc, it sounds like point(c) is a small physical body (If point(c) exerts a torque on the disc, then disc must exert a torque on point(c)) I'm going to assume that the mechanism by which point(c) causes a torque on the disc is a motor connecting point(c) and the disc. So, if points (a) ...


0

On a free floating body, if a pure torque is applied (with net zero force) then the body is going to rotate about it's center of mass (see http://physics.stackexchange.com/a/81078/392). This is regardless of where the torque is applied, or how many torques are applied. If the net force is zero then the center of mass will not move. Now if C or A or B are ...


4

The equation you quote is an approximation that is only valid if the horizontal force on the string remains the same. In practice that is not the case - and your concern is valid. The increase in length $\Delta \ell =\ell(\frac{1}{\cos\theta}-1)$; how much additional force that generates depends on the unstretched length of the string (or equivalently on ...


0

Firstly, tension force itself depends on the elastic/rigid property of the string. So, the properties are implicit in the mathematical representation of the tension. Let a small part of the string be acted by tension $T_0$ from both sides of the string & hence it remains in static equilibrium as in the first picture. Now, when that part is stretched, ...


0

I think in this example, you have to find the tension "in terms of". To do so, you have to do it using the static equilibrium equation which gives you the solution that you wrote. So, here it works the other way around. You first find the tension in terms of $T_0$ and THEN you find the new length of the wire. Why do you have to do it this way? Well, you ...


0

This answer looks at what happens to your jump height if you actually decrease in mass, while maintaining the same leg jump strength. So, let us assume that the mechanism by which a person jumps can be modelled as a finite impulse that occurs in an instance (i.e. a very very large force applied in a short frame of time). Let this impulse has a magnitude ...


0

I'm assuming that you mean you can use your thrust device further to the thrust you can impart with your own legs. Otherwise, something can lift 75 pound weight won't lift you at all. Also, you need to explain how your thrust device's force changes as you move (a spring, for example, imparts less force the more it relaxes). So I'll assume your device is ...


1

Obviously, I cant fly, but I guarantee if I weighed 75 lbs less, I could jump a lot higher than I currently can. Thank you. I appreciate your guarantee. How high could i jump if I weighed only half as much? Under the assumption that you would be able to impart the same amount of energy with your legs, you would be able to jump twice as high. ...


4

There is a difference - but not exactly why you think. There are prevailing winds around the earth - these used to be called the "Trade Winds" because traders, knowing the direction of the wind, knew how best to navigate the globe. Basically, on the equator (in the tropics) they flow from east to west, and at higher latitudes they flow from west to east: ...


1

No, the baseplate will not move if the two motors are applying torques in such a way that the rotors remain at rest. To analyse this problem, it is best to consider the forces on the three parts (base disk and two rotors) individually, using free body diagrams. There are two motors, each connecting a rotor to the base plate, and the effect that each motor ...


2

F.dr does not determine if a force is conservative or not.All normal forces (conservative or not) produce work equal to F.dr but what determines if they are conservative is the integral of F.dr in a closed loop.If that is equal to zero then it is conservative because no energy is lost in that loop.It is like gravity.You throw something upwards and no energy ...


0

You are assuming that $a$ is positive in the direction of motion. One of the two solutions will be negative because the forces are inconsistent, the correct answer will be the positive one.


0

I'm going to give you another approach. Suppose that you where to perform the following experiments: 1.) a) You happen to find a ball that is not moving, well its going to stay like that if we are in a place where the ball is not interacting with anything (air, gravity, etc). Like this : b) If You happen to find a ball that is moving with constant ...


0

In my opinion, your explanation's weak point lies with the intuitive jump you make by take the limit of $\Delta t$. You may need some mathematics or a diagram to support this intuition. For example, drawing a diagram showing the velocity vector and the small change in velocity vector, showing that in order to keep the magnitude of velocity constant, then the ...


2

When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.


0

If a particle is left alone in an inertial system, it will travel on a straight line, that is $$ \frac{d\vec{v}}{dt} = 0 \tag1 $$ it will not change the magnitude and direction of its velocity. This is an empirical fact. So if a particle does change its velocity the right hand side of (1) is not zero. Lets call it $\vec{A}$: $$ \frac{d\vec{v}}{dt} =: \vec{A} ...


0

Simply put, it's a correlation between a certain action (applying force on a given mass) and reaction (gaining or losing velocity). The mass is a just a factor that was found to give accurate predictions in the world Newton lived in (relativity theory introduced a slight modification). Both force (over time or distance) and velocity represent energy, that's ...


0

Force comes to play when the body is moving with changing velocity. To quote one author's statement is rather noteworthy: Force is the agent that is at the forefront for the initiation of the motion & also responsible for change in the velocity. The famous relation can be deduced intuitively by means of calculus. If $v$ is the velocity of a body ...


2

Let's break this down in two pieces. First - the proportionality of force with mass. Imagine you need a certain force $F$ to accelerate a mass $M$. Now imagine that instead of one mass, you have two - each with mass $\frac{M}{2}$. Attach a piece of string to each. The total force on the two strings must be the same as before, so each string sees half the ...


0

The acceleration is the effect on the objects motion due to the cause(s), which is the vector sum/total of the forces acting on the object, i.e. the total force (composed of many forces) is responsible for the acceleration (which is just one vector). The mass is just the constant of proportionality that links total force and acceleration.


-3

whenever external force is applied on the object automatically a restoring force is developed inside the object to restrict the deformation of the object.The ratio of restoring force perpendicular to the surface to the area is known as stress.The ratio of external force perpendicular to the surface to the area is known as pressure. for example if you press ...


4

Because the equations for linear and angular motion are very symmetrical In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object. In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor. So, the ...


0

I believe your second interpretation is spot on. Whenever anything rolls without slipping, it means that the point of contact of the ball with the frictional surface is instantaneously stationary. Therefore, the frictional force applied to this point does not do any work, and so there is no frictional dissipation. Also, because the ball is still rolling ...


1

$m_1 = m = 10$ kg $m_2 = 2m_1 = 20$ kg $F = 60$ N; $ma = F - \mu mg-T$ $2ma = T - 2\mu m g$ $\Rightarrow ma+\mu mg = F-T$ $ 2(m a+ \mu mg)=T$ $\Rightarrow 1/2=(F-T)/T \Rightarrow T = 2F/3$


11

I want to add to the other answers that when an object is rotating at a supersonic speed, an observer will be hit by a rapid series of sonic shock waves, as the shock wave is an ever-expanding spiral. This is what makes supersonic propellers so terribly loud. The images below depict the process. The red circle in the middle is the trajectory of a propeller ...


3

The total force acting on a raindrop equals $g$ minus air resistance which increases with velocity. In other words, as the raindrop speeds up, air resistance increases which decreases the acceleration (until eventually the acceleration equals zero and terminal velocity has been reached).


2

First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...


0

I don't think you went wrong anywhere, you just didn't find the solution as quickly as the book. Using an integral table, you see that $$ \int\cot\phi= |\ln(\sin\phi)|+C \quad,\quad\int\csc\phi= |\ln(\csc\phi-\cot\phi)|+C.$$ So, integrating $\frac{dv}{v}=(\csc\phi-\cot\phi)d\phi$ and ignoring the absolute values gives $$ v=Ce^{\ln ...


0

$$ \vec{F} = - G m \int \frac{\varrho(\vec{r}\ ') \cdot (\vec{r} - \vec{r}\ ')}{|\vec{r} - \vec{r}\ '|^3} d^3r'$$ where $\varrho$ is the mass distribution. The Integral is taken over the whole volume (or equally: the support of $\varrho$). For mass distributions, that have no 3D-volume, you have to take some Delta-terms, e.g. $$ \varrho(x,y,z) = M\cdot ...



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