New answers tagged forces
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Yes you are correct!
it is a known bug in VMC answers. If you search their forums as well ,some 2 years old questions perhaps, you'll find this.
The friction is static and just opposes the external force. Hence the friction force is just $4.5N$, and body will remain at rest.
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This looks like a simple linear blending problem. It is two-dimensional, but each dimension can be considered independently.
The more to the right the weight is, the larger the fraction of it carried by F2 and F3. Basically, the fraction of the weight carried by F2 and F3 is X/W. Put more mathematically:
(F2 + F3) / (F1 + F2 + F3 + F4) = X ...
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The simple answer is that you can't fully solve this problem--because as you note it is under-constrained--under the assumptions that are made when you first start doing statics (that objects are completely rigid).
The introduction of finite strains bring in additional relationships.
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As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a ...
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Here we need to see that $$f\le N\mu_s= mg \cos\theta\ \mu_s$$
$N\mu_s$is the maximum friction that may be present in the surfaces,whereas here we need equilibrium.
The two forces along the incline must balance each other. So, $$f=ms\sin\theta \le \mu_s mg\cos\theta$$
If this inequality does not hold , ie. $$\tan\theta\le\mu_s$$
then the body will never ...
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Yes, what you wrote is right. Each loop feels force from the magnetic field and the loop cannot exert net force on itself, so your expression is the only option.
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My knowledge of physics does not really extend to these realms, and I apologize if I am wrong or off topic, or talking to far above my head. It is often funny how the mathematization of things can force some level of belief or unification,
even though I am just trusting the mathematician.
I understand that the Kaluza-Klein theory from the twenties provides ...
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You haven't given us enough information to solve the problem. Is there any motion? Do you know the weight of the board? Anything else?
Typically, lever problems have two types of equations that can be useful. The first is just the standard $\sum_i F_i = m\, a$. Note that this is the sum of all the forces. For example, in your picture, the sum of all ...
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I would guess that you're supposed to assume the graph of force versus time is a straight line. If so the equation for $F(t)$ is easy to calculate because you know the force at $t$ = 0 and $t$ = 3s. The change in momentum is equal to the impulse, which is indeed the integral of $F(t)dt$.
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Firstly, as Peter mentioned, your equations implicitly assume that the id fluinside is causing the acceleration. Otherwise you need to specify that the system is being pulled. I'll deal with both situations here.
If the water causes the acceleration
Unfortunately, we can't directly deal with the individual cylinder caps because the cylinder tube exerts a ...
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See the cylinder as a whole object . The second law of motion gives that the cylinder does not accelerate untill it have a net external force.
So, the external pressures on both sides of the cylinder is not equal.The external agent which accelerates the cylinder applies force ($ma$) on left face towards right so that pressure on left fave exeeds by a ...
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A constant net force means:
$$\Sigma\vec{F}=\frac{d\vec{p}}{dt}=C$$
where $C$ is some constant. This means that
$$\int \ dp=p=C\int\ dt=Ct+p_0$$
where $p_0$ is the initial momentum. Now, you can easily verify that
$$p_2-p_1=\Delta p=Ct_2+p_0-Ct_1-p_0=C(t_2-t_1)=C\Delta t$$
In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma ...
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I think I have figured out the answer, hopefully.
Firstly, let us begin by stating the First Law:
$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$
When the net force is constant, we it means that there is no change of momentum, in other words,
$$ \frac{\delta \vec{p}}{\delta t} = 0$$
In this case, we know that the function of momentum is a constant, ...
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The car's engine tries to make the wheels turn. However, the wheels encounter friction against the road so they cannot just spin. As the road has much higher inertia than the car, it will not move when the wheels want to turn. Instead, it is the car that moves.
The end effect is that the engine pushes against the road, just as you do when you push the car: ...
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Each force causes reaction (3rd law). If move a car from the inside the car moves you as well. That's because you are pushing or pulling. However the engine does not push but converts energy in other directions, usually a rotating one (the same as riding a bicycle). This rotating force has its counter-force which is reaction of ground.
Pushing a car you ...
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It explodes with a force of 500N.
This sentence is nonsensical. It can explode and release some energy. It can also explode and impart a very high force onto the fragments (which sill be different for each fragment) for a very short time interval.
Once the explosion takes place, the fragments will not accelerate (they may decelerate due to air drag, ...
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The microscopic description of the normal force is inherently complicated, not simple, and it cannot be explained in terms of just an electrical repulsion or or just the Pauli exclusion principle.
There is a very general and useful concept of a residual interaction. For example, the force between one neutron and another neutron is fiendishly complicated. ...
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For pushing it up, we have to overcome friction(act downwards) as well as the $mg\sin\theta$. So, $$3N=f+mg\sin\theta$$
Now the block is just slipping , so friction is acting upwards, and so does the force applied externally.So,
$$N+f=mg\sin\theta$$
Eliminate $N$ and use $f=\mu mg\cos\theta$.
Solve for $\mu$ you get your answer.
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The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.
Static friction always acts to prevent relative motion. It ...
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$F=ma$. If $F=0$, and $m=0$, $a$ can be anything. Most physical laws are not "A causes B". They usually say that "A and B can coexist in these conditions". So, it is not necessarily "Force causes acceleration". It is "an accelerating body can coexist with a force if $F=ma$"
The net force on a massless string is always 0 -- it has to be (otherwise it will ...
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Non-relativistic mechanics can't. Massless objects travel at the speed of light. The only reason to introduce a massless string is so that you can get some effect from the string without having to worry about the string in calculations. As soon as you start worrying about forces on the string causing it to accelerate you've violated the whole reason for ...
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Quantum Mechanics says force is not physics primitive. It shows the undelying mechanism for them.
What is a force? It is something that changes the velocity of a particle, with the Newton's second law:
$$\vec{F}=m\dfrac{d\vec{v}}{dt}$$
Any other appearances of forces can be reduced to this. For example, when we measure the force with a dynamometer, it is ...
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You've done the two hard parts: (1) relate area to force and intensity and (2) relate distance, acceleration, and time. For connecting these two, I suggest looking at Newton's second law.
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Gravity is pulling us and the ground down. But, unless something bad is happening, neither we nor the ground under us is moving because another force is pushing us and the ground up. That force is electromagnetism.
Think of your foot pushing down on the ground. You can envision a top layer of atoms pushing back up. (Real life is messier than nice flat ...
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Perhaps I did not understand the question correctly, but it seems to me that you cannot use a Gaussian shell in this case, because the intensity of the field $E$ would be different at different points of the shell. If you want the following expression to hold,
$$
\int E\cdot da = E \int da
$$
then $E$ must be equal to the same value all over the Gaussian ...
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Specifying the laws of motion is different from specifying the solutions of the laws of motion for a given force. An example of a law of motion would be $d^2r/dt^2=F/m$. From your given information, and assuming this law of motion, one can infer the function $F_{1,2}(t)$. By assuming a different law of motion, such as $d^3r/dt^3=F/m$, we would infer a ...
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So at atomic level for same charges that repel eachother(electrons etc.), if thought of as a slingshot effect, elastic repulsive collision and buoyancy fields etc. makes repulsion a pseudo or dummy. as for like gravity is doing it all, then it will be more insightful or not?
Not.
To describe all the available data the theory has to be much more ...
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Lennard-Jones potential is a big answer to the confusion
The force is a spacial deriviative of the energy, F = dE/dr. So, positive derivative, where plot goes up you get attractive force, and where it goes down you will have repulsion. You see that there is a huge reuplsion between atoms when the distance between them, r, is very small. So, they cannot be ...
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Calculate Potential energy using the following formula
$PE = mgh = 0.182 \cdot 9.81 \cdot 2\: \mathrm{Joules}$
Average Impact Force = $0.182 \cdot 9.81 \cdot 2 \cdot 0.005\: \mathrm{Newtons}$ (that is the answer)
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The slingshot effect is in fact only a transfer of angular momentum from one body to another. So for example if you're using Jupiter for a swing by manoeuvre with your satellite, you do nothing more than slow down Jupiter (in the Sun-Jupiter-(satellite) system) and speed up your satellite. Of course because of the huge difference in mass, you only see a ...
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For each action of body A on body B, there is a reaction of body B on body A. The two forces do not apply on the same body and thus won't cancel each other when looking at the movement of body B only.
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Gravity is not special at all. It seemed to be special at dawn of the 20th century but now the picture is different.
Fields are more than just forces. Fields can have their intrinsic dynamics, solitons, topological features, nontrivial vacuum.
As of force aspect, electromagnetic field makes a 4-force $qF^{\mu\nu}u_{\nu}$, and gravitational field makes a ...
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Gravity is still viewed as force for better understanding because not all have Einstein's IQ to see curvature of more than 2 dimensions.
Gravity is a force, but not a REAL force. Try pushing rat and elephant over a frictionless surface so that both feel same acceleration. Common sense says that you need more force for elephant.
It means that gravity is a ...
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At the risk of being chided by physicist for vast oversimplification, can I provide an intuitive answer?
In a rotating frame of reference, centrifugal force exists (as does Coriolis force). Observers in rotating frames see freely moving objects travelling curved paths. They conclude that a force exists, and can even generate a formula for it. An observer ...
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You must have to think about the mass of the object. Same force will be applied on two different objects of different mass. Velocity (movement) will depend on the mass. Whose mass is more it will be at greater velocity than other higher mass object.
Let's say, i jump on the earth. Same Force is applied both on earth and me. Since my mass is very less ...
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Newton's third law doesn't imply that things can't move but it does imply conservation of momentum and energy.
Imagine a scenario where an astronaut is in orbit so they don't feel the affects of gravity. If there is an object floating and they push on it (apply a force to it) we know intuitively that the object will start to accelerate in the direction of ...
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The air in the bag will be compressed when a weight is placed on the bag. The change of air pressure in the bag will change according to the bulk modulus of the fluid (air in this case) in the bag.
The equation for bulk modulus is as follows:
$$B=-\frac{\Delta p}{\Delta V/V}$$
In this case, $B$ is the bulk modulus of air, $\Delta p$ is the change of the air ...
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Most of the resonances detected in particle physics scattering experiments are bound states of the strong force, bound for a time interval before decay .
These are created in the interaction and seen in invariant mass combinations of the interaction products, statistically.
The distinction with electromagnetic or weak decays comes from the widths of the ...
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You know Newton's law of gravity with the inverse square distance and two masses $M_1$, $M_2$. Use Coulomb's law and adjust the charges $Q_1$, $Q_2$ to get the same force. Of course, with two variables and only one constraint, you have a bit of freedom. The exercise gives you a second constraint: that the ratio of $Q_1$ to $Q_2$ should be the same as for ...
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My Solution: Newton's 2nd for the moon (little $m$ moon, big $m$ earth):
$$\sum F=ma\rightarrow k\frac{q_mq_e}{r^2}=m\frac{v^2}{r}$$
$$\frac{q_m}{m}=\frac{v^2r}{kq_e}$$
Now calling the ratio we are trying to find $\chi=q_m/m=q_e/M$
$$\chi^2=\frac{v^2r}{kM}$$
you can plug your values in from there. Notice I had to assume the earth and the moon had the ...
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we can actually use $300 - T = 0$ (1) for equilibrium of the lift and $600 -T = 0$ (2) for equilibrium of the man as both systems are at rest.
Adding (1) and (2)
$900- 2T = 0$
$T = 450 N$
Taking into account the system of the man, the man exerts a force downwards and tension pulls him upwards.
Thus total force downward in the man's system is $600 N - ...
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The total force down is 900N. Thus, the total force up must be 900N. For the rope to be in equilibrium, the force up on each side of the pulley is 450N.
Your force down is 600N, the elevator's force down is 300N. If we carry through the math, that means your net force is 150N down and the elevator's net force is 150N up.
Assuming the weigh scale is ...
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Now these two equations can be solved to get the $N$ , by eliminating $T$ . Remember, only the normal force is reading of the elevator.
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Firstly, the centrifugal force is an pseudo force and only appears to exist if you are sitting on the stone and flying along with it. This is because you are in an accelerating (non-inertial) frame of reference. Another common pseudoforce you experience every day is the "force" that throws you into the seat of your car when it accelerates. For further ...
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First of all $mv^2/r $ is a pseudo force that acts in frame of rotating body itself. So, let's work it like that.
The centrifugal force acts outwards radially from the circle of rotation (not along the thread).
First let's see the the first image.
$$T\cos\theta=mg$$
$$T\sin\theta=\dfrac{mv^2}r$$
As we move the body faster. $T\sin\theta$ must increase ...
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Basically no force except tension acts , 1 thing is centrifugal force is a pseudo force .
Secondly length of string is constant , so when you rotate too fast .
Faster than what $\frac{mv^2}{r}$ = $Tcos\theta$(horizontal force) permits for a radii $r$ , then horizontal force has to increase , and T in the string can only attain a max value(after which it ...
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When acted on only by gravity and drag, the ball moves at 10 cm/s downwards.
If the specified upwards force were applied, the ball would be acted on only by g downwards (gravity), 2g upwards (external applied force), and by symmetry to the initial conditions, a drag force of g downwards when the ball travels at 10 cm/s again.
Update:
Symmetry comes from ...
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If you are off by a factor of two, it's probably because the volume of a sphere is
$\frac{4}{3}\pi r^3$ and not $\frac{2}{3}\pi r^3$
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I'll do my best to answer this with the information provided.
In the most general (and hand-wavy) sense possible, something has reached equilibrium if after a while it has stopped changing in some way. For example, if you put a warm person in a cold room the person will warm the room up ever so slightly and the room will cool the person down. Whilst the ...
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There a several notions of equilibrium. These are discussed in an informative online encyclopedia called "Wikipedia", which I recommend. The link to the mechanical equilibrium section is here. (I assume you mean mechanical equilbrium since there is the forces tag.) If you read the article and have trouble understanding anything, please ask a more specific ...
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