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0

It takes zero force to push an object along a frictionless plane with a constant velocity. In fact a force can be defined as the rate of change of linear momentum, and zero force applied to a body yields no change on momentum. When friction is added, only when the applied force exactly matches the friction force the resulting motion is uniform. In ...


0

Some times, we write different equations for force, like $$F=\frac{1}{4\pi\epsilon}.\frac{q}{r^2}.q_{o}$$ $$F=Eq_o$$, $$F=m.\frac{GM}{r^2}$$$$F=ma$$, etc. When the force is due to the property of static charge, we say it as electrostatic force (Coulombs law), when the force is due to mass, we say it as gravitational force (Newton's law of Gravitation). ...


1

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ...


1

Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ $$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$ Second possibility : If your box is spherical, By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity. Hence, your ball attains terminal velocity. $$F=6\pi\eta rv$$ $$v=\frac{F}{6\pi\eta ...


0

If you apply a force on the box, and see no acceleration, then the force you apply is equal to the friction force. Friction is velocity dependent, you cannot say "the friction force is so much" independently of the force you are applying.


1

By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing $F_{app}$ otherwise there will be an acceleration.


1

The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...


2

The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...


1

Assuming isosceles triangle.­­­­­­­­­­­­­­­


1

While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...


2

This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...


1

Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be ...


0

Einstein says: "If, relative to K, K' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K." Apparently, "uniform movement", i.e. one without accelerations, is the assumption behind the whole SR Theory that allows us to compare ...


0

There are a number of wrong assertions in your statement. You say, "surely ... continue accelerating to infinity." Since $ a = \frac {F}{m} $, as long as you apply net force F, you only get constant acceleration a. The acceleration value not only does not go to infinity, but actually does not change, unless the net force changes. When there is no net ...


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Depending on the magnitudes of the separate constant accelerations of the two objects, depending on the angle between their trajectories and depending on their "initial configuration" (initial separation and initial speeds, as determined by members of one suitable inertial system) there are indeed qualitatively distinct outcomes how such two object would ...


-1

The expression "Total accelration" does not fit if the accelrations have different directions. The vector resultant is actually the "net accelration", or the combined effect of these two accelrations, or equivalently, forces. The vector resultant makes sure that only the effective components are added, and the opposing effects cancel out. Maybe an example ...


4

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$, $\vec{F}=\vec{F_1}+\vec{F_2}=m(\vec{a_1}+\vec{a_2})$ When using vectors symbol, its automatically takes care of their directions.


0

Assume you are moving 10 meters in 10 seconds northward. What is the acceleration?. You cannot determine acceleration from this information. This information could represent a constant velocity of 1 meter/second and zero acceleration.


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Newton's law of universal gravitation holds for point like masses. For spatial masses you would have to integrate of infinitely small parts of that mass. However it does turns out that the result of this for a sphere of constant density at a given radius does yield that simple formula, see Gauss's law. To get a better understanding of this, without having ...


0

when your ball is far from a donut, there's a potential energy of the gravitational field. when the ball reaches the donut, that potential energy must be converted into something else, it can't disappear. do it must be that it converted into the kinetic energy of movement, i.e. the ball must be moving at this point, hence, it'll pass the center at some ...


1

1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere ...


-1

The relation is: $\sum F=0$ comes from $\sum F=ma$ If $\sum F=0$, so that either $m$ or $a$ must be $0$. Currently, there is no substance of mass $0$. So, $a=0$. Means that no force is applied to the object. If it is moving, it will continuously move with constant velocity. If the $v=0$ , the object will stay stationary. Hope you understand.


2

There is no law that says the sum of forces on a given object must be $0$, that is simply the condition for mechanical equilibrium. If an object has constant $0$ velocity (or, more generally, any constant velocity), then its acceleration ($\frac{dv}{dt}$) is $0$ and, by Newton's second law as you have it, the net force acting on it is $0$. However, if all ...


3

As Sachin says, there is no limit to acceleration. In fact you can show this by considering an observer hovering at a fixed distance from a black hole. As described in this question, the acceleration required to maintain a fixed distance from a black hole is given by: $$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} $$ where $r_s$ is the radius of the ...


0

I hope this doesn't confuse you, but in one sense, yes, heavier bodies do fall faster than light ones, even in a vacuum. Previous answers are correct in pointing out that if you double the mass of the falling object, the attraction between it and the earth doubles, but since it is twice as massive its acceleration is unchanged. This, however, is true in the ...


1

There's no theoretical limit to acceleration as long as you are not accelerating something to $c$. However, there'll be practical challenges as high acceleration would require high amount of energy.


-1

Although Pranav Hosangadi has explained, I will try to explain where you might be going wrong. I think this will be helpful for you. And I think it will not be waste of time in typing the answer for you. I was going to ask: if mass is an objects tendency to resist acceleration then why do two objects of different masses fall to the Earth at the same ...


-1

Mass is an object's tendency to resist acceleration. This applies when both masses you're testing are subjected to identical forces. From Newton's Law of Gravity, $$F = G \frac{M \cdot m}{r^2}$$ It is fairly obvious that the force the Earth exerts on a heavy body is more that what it exerts on a light body, so you can not compare the accelerations ...


0

Acceleration depends on velocity per unit time, and velocity on distance per unit time. It is clear that acceleration is infinite for us because space is infinite. For a given time, acceleration is infinite.


0

There are two things that limit the maximum traction (F) of a car. One is given by the friction formula, F = μR (Traction = friction coefficient x weight of car), above which the wheels start to spin. The other is the power equation. P = Fv or F = P / v (Traction = power / velocity), which the engine isn't powerful enough to exceed. Note that the max ...


0

The velocity decreases at $(9.81 m/sec)/sec$ Divide that into the starting velocity to find the time it takes the velocity to get to zero.


0

Assume air resistance is 0 and that the bullet goes completely vertically. I don't see any reason to model it as a quadratic. Just normal kinematic equations.


1

Well that's a result using differentiation and derivation. Have you studied calculus? If not, there is a simple way to look at it. $$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$ (Yes there should be a $\Delta t$ in the denominator, too.) Now, what does $\Delta(mv)$ mean? It represents the change in the quantity $mv$. For current situations, ...


1

You are right, there is a $\Delta$ missing in front of the $t$. $\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps. The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like ...


3

1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $ 4v $ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. ...


1

Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a ...


8

There's a simple way to look at this that doesn't involve any maths. Suppose the two cars are parked and are stationary, and you accelerate past them in your car. If you are accelerating forwards then from your perspective it looks as if the two cars are accelerating backwards (at the same rate). But the cars are at rest, so the distance between them can't ...


1

No, it is not correct. Let $a$ be acceleration; $F$ force; $m$ mass; $v_0$ initial velocity; $v_f$ final velocity; $P$ power required; $x$ distance travelled and $t$ time taken. Hence, $$P=\frac{Fs}{t}=\frac{ma\ x}{t}$$ Then $a=\frac{v_f^2-v_0^2}{2x}=\frac{5^2-0^2}{2*10}=1.25\text{ m s}^{-2}$ and $t=\frac{2x}{v_0+v_f}=\frac{2*10}{0+5}=4\text{ s}$ Hence, ...


1

You have a differential equation that says \begin{equation} a(x) = -0.01*w = \frac{d w}{d t} \end{equation} What you did with the change of variables is correct, so $w$ cancels on either side. Otherwise you have a first order differential equation to solve.


2

Lets do a Free Body Diagram of the lift (NOTE: Always do A FBD first). What are the forces acting on the lift? $$\sum \vec{F} = \begin{pmatrix} -T \sin\psi \\ T \cos\psi - W \end{pmatrix} $$ What is the acceleration on the lift? $$\vec{a} = \begin{pmatrix} -a \cos \theta, a \sin \theta \end{pmatrix} $$ Combine them with $\sum \vec{F} = m \vec{a}$ and ...


0

You do know the direction of the tension force, the force of gravity and the direction of the resultant force, so well as the magnitude of the force of gravity. You can create some goniometric equations with that.


1

Here is a link to such a device that one can purchase. The manufacturer claims The Tactical ResQmax™ is capable of deploying a standard line up to 250 feet / 75 meters with the standard projectile depending on the service pressure, weight of line and type of projectile used. Distances of over 400 feet / 120 meters can be achieved using customized line ...


1

Disclaimer: I have no engineering background, so if anything I write is in error, definitely point it out. However, if the 3-axis accelerometer only returns the proper acceleration vector $\mathbf{a}$, then if the object is moving around and physically accelerating, it is impossible to determine the orientation of the object without additional information. ...


0

There is no proportionality constant for the same reason there isn't one in $$\text{distance} = \text{speed}\times\text{time}$$ if we use consistent units. For instance, if distance and time are kilometers and hours, and we have measure speed in kilometers per hour, then there is no conversion factor. The $F = ma$ formula defines a force as being mass, ...


2

In Newtonian mechanics, we have the quantity momentum (I'll get to force a bit later): $$\vec p = m\vec v $$ which is conserved and is thus a quantity of fundamental importance. We can think of the mass as the constant of proportionality between momentum and velocity. But you might ask "why isn't $\vec p = k \cdot m \vec v$ instead?" The answer is that, ...


0

A lot of these comments/posts are suggesting that Newton's law is defining force, but I don't think this is a good way of looking at it, otherwise it is trivial and the statement is vacuous. I look at Newton's law as essentially defining mass, ie. $m=\frac{F}{a}$ and the reason Newton's law is then nontrivial is because it says that $m$ is constant.


3

In the Newton's Second Law, Newton basically defined what Force is. He could have taken that constant as any number that he wanted, he chose 1 for simplicity.


7

I think it is more intuitive to say that (net) force is proportional to acceleration: $F\propto a$. The proportional constant tells us now how easy it is to accelerate an object with a certain force. This proportional constant is called the (inert) mass of said object. Hence $F=m\cdot a$.


5

A lot of the other answers kind of point to this, but the best way of looking at Newton's second law is to think of it as a definition of force -- while it encodes the notion of a push or pull, technically, we have to encode this quantitatively. Newton's second law encodes the fact that pulls create acceleration, and it sets the unit of a pull in such a way ...



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