New answers tagged

0

Just to expand a little bit... The centrifugal force on a particle of mass $m$ is given by $$\vec F_{ce}=-m\vec\omega\times(\vec\omega\times\vec r),$$ where $\vec\omega$ is the Earth's angular acceleration and $\vec r$ is the position of the particle. This force must be taken into account when using Newton's second law in the non inertial frame. Similarly to ...


0

The force exerted on a unit positive charge is $ \dfrac {kq}{r^2} (+\hat r)$ and so an external force $ \dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge. If the step is $d\vec r$ then the work done by the external force in moving that step is $ \dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$. You have moved from $\vec r$ to $\...


1

That question refers to apparent weight, as opposed to the true weight. Your true weight, $W=mg$, is the force of gravity pulling on you. Your apparent weight is how hard the ground has to push on you to keep you from falling. This is the weight you "feel". When you are accelerating up or down in an elevator while standing on a scale, the scale reads ...


1

There are two reasons. One is that the Earth is not perfectly spherically symmetric and so at any point on the Earth's surface, the net gravitational force does not point directly towards the centre of the Earth. Secondly, because the Earth is rotating, there is a pseudo-force, the centrifugal force, that appears to act directly away from the Earth's ...


1

First off, there would have to be a force transfer between the road (blue) into the asphalt (black) beneath it for the sensor to trigger. We need to view the road as not infinitely rigid, it's somewhat "squishy", which allows force transfer at and near to the point(s) of contact. The ability for the force at one point to communicate to the other point ...


1

Velocity is a vector quantity. It has magnitude as well as direction, so if suddenly all gravitational force disappeared then all planets will move tangentially. It can be proved by a simple experiment: Tie a weight to a string and rotate it and suddenly cut that string without disturbing the system. It will move tangentially. In this experiment tension in ...


-2

For a very simple answer, the sun is bigger but the average mass density of the sun from its center to the surface is smaller compared to the average mass density of the moon from its center to its surface, it is the reason that the major causes of tides are from the moon than from the sun.


1

I think that the components must be done in the perpendicular directions $T\cos\theta=mg$ as the bob can't go any further. It is wrong. Because $\Sigma \vec F=\vec 0$ doesn't mean that the bob will be fixed. A body can move because of its inertia even if $\Sigma \vec F=\vec 0$ (first law of Newton).


1

As the force $F$ is constant, and we know that the force of weight $mg$ is constant too; we can say the net force acting on the ball is constant whole the motion. But, from $t=0$ to $t=t_1$ the net force is $F-mg$ and after $t=t_1$ the net force is $-mg$. So, velocity graph will be linear but it has a breaking at $t=t_1$. For $0\le t\lt t_1$ the gradient of ...


2

\begin{equation} T-mg\cos\theta=m\,a_\textrm{centripetal}=m\,\dfrac{v^{2}}{r} \tag{01} \end{equation} At maximum $\:\theta\:$ since $\:v=0\:$ \begin{equation} T-mg\cos\theta=0 \tag{02} \end{equation}


2

The concept of "mediation" depending on a photon comes from the use of Feynman diagrams. Feynman diagrams are the calculational tool of quantum electrodynamics because they give a prescription of how to calculate the integrals in each order of the pertubation series expansion for proton proton or proton electron scattering. Lets make it simple, because the ...


0

First, you have to realize this is the perturbation theory picture: all the "particles" are plane waves. Also, these are virtual particles (they are not eigensolutions of the free space wave equation and are transient, they only exist conditionally on the creation and annihilation event/vertex). Basically, you are developing the $1/r$ potential of a charged ...


0

Hint: By the use of coordinate geometry find the distance between the $2kg$ and other masses. Now first of all use the expression for the gravitation force $F=Gm_1m_2/r^2$ between two particles. And by the use of principle of superposition tackle the particles pair by pair. By this I mean Lets, assume the force between $4kg$ and $2kg$ is $\vec{F_1}$ and ...


1

The sign of the force direction is just given by convention as $q \vec{v} \times \vec{B}$. Whilst the direction of the force on the particles is clearly something that can be measured, the sign of the charge and the direction of the magnetic field lines are man-made constructs. For example you would get the same direction for the force if we decided that ...


0

I can give you an answer using only symmetry considerations, and leave to you the rigorous calculations, since you are probably familiar with differential calculus. The figure below shows a transverse cut of your system of two shells. For any point-like area of the inner shell (red dot) you can draw a diagram like this, which is symmetrical for rotations ...


0

Well, I think that the magnetic force of the moving positive-charge which you describe is directed at the positive z-axis, if I'm not mistaken ... in other words, all 3 axes come into play when one visualizes which way the magnetic force will point .....


0

Regarding the first part, the induced charges can never be more than the charge placed and at best they will be equal in magnitude to the charge placed in the cavity( but opposite in sign). So the fields due to them will cancel out inside the conductor. Outside electric field will induce charges on the surface of the conductor but the direction of fields in ...


1

For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...


1

Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...


2

It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...


0

Actually. the velocity can be determined as follows. V = sq. root of 2gh where: V = velocity in ft./sec. g = acceleration const. 32.2 ft per sec per sec. And h = head in feet of liquid. h = P * 2.31/SG where: P = pressure in psi. SG = liquid specific gravity. With the above information you can now calculate F.


3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...


0

Imagine a body moving with velocity $\vec{v}$ in the $\hat{x}$ direction and now imagine having it losing speed (decelerating, so $\vec{a} \propto -\vec{v}$) the friction force should be opposite the $\vec{a}$ (acceleration) or the $\vec{v}$ (velocity)? if you think carefully about it, you'll convince yourself that it should oppose velocity and not ...


0

Infinite distributions of mass can give rise to some contradictions. The classical well known example is an infinite homogeneous universe. In this case, by symmetry you would say that the the force will be zero everywhere. However, if you chose two arbitrary points in space, one as the center of coordinates and apply gauss law centered at the origin, then ...


1

Using Bernoulli's equation and the momentum conservation equation, we can show that water flowing out of a pipe with cross-section $A$ at speed $v$ exerts a force $F$ on a wall (at 90 degrees), acc.: $$F=\rho Av^2$$ With $\rho$ the density of the water. But your specification of "8" pipe with 500psi stream of water exiting it and hitting a wall at 90 ...


1

In fact, it is not convergent: the horizontal component of the gravitational force can depend arbitrarily on parts of the sheet which are far away, if you are sufficiently malicious in your choice of surfaces which exhaust the infinite plane. To make this explicit, let's suppose we are suspended at the cartesian point $(0,0,D)$, and place some mass of ...


0

When the suction cup presses down it squeezes all the air out, which creates an area of low pressure and sucks the cup in. When you lift the cup the area inside the cup stretches out, and the area of low pressure turns slightly towards high pressure which lets some air out. The air keeps going until the low pressure turns completely to high pressure and cup ...


1

The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ...


0

The first thing first -The concept of friction Friction come in play when there is any tendency of relative motion between 2 surfaces and it is in opposite direction of relative motion (Note- I used word relative motion not simply motion). In simple word friction try to reduce relative motion. in 2 block problem and this problem the difference is T (tension)....


3

As the question is finite versus infinite, I guess we don't need the exact result for a finite disc (though it is not difficult to calculate). The simple intuitive answer is that even though the disk mass is infinite, most of the forces from the bits of disk going out to infinity will cancel out due to symmetry, so the answer is finite. Let us assume we ...


25

If the disk has infinite diameter it is nothing but an infinite plane. For any finite thickness we can consider a layer of mass whose superficial density is $\sigma$. Moreover, if the plane is infinite it does not matter if you are one meter or one kilometer away from the plane. Wherever you look at the plane you will see the same structure. So the ...


7

You can do the integral, and will discover that the answer is "finite" - because not only does the distance to mass increase, but so does the angle. Consider an annulus at radial distance $r$: if you have mass per area $\sigma$, the total mass at that distance is $2\pi r \sigma$; if the vertical distance to the center of the disk is $h$, the vertical ...


1

The torque supplied by $F_m$ results in the torque due to $F_v$, so these torques are equal : $\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either $\vec A \times \vec B = (AB \sin\theta) \hat k$ where $A$, $B$ are magnitudes and $\theta$ is the angle between (b) or $(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ...


0

Although you do not need it you have miscalculated the position of the centre of mass of the arrangement of squares. About corner $O$ the two left-hand squares produce a anticlockwise torque whilst the right-hand square, the normal reaction of the squares due to the ground $N$ and the force $F$ all produce a clockwise torque about corner $O$. Without the ...


1

Due to the fact that the body in equilibrium, all forces must cancel each other. Which forces must cancel each other when a body is in equilibrium? When a body is in equilibrium, resultant of forces acting on it must be zero. In current question, we have three bodies those are in equilibrium (man, pen and the earth). So, net force exerted on each of them ...


0

Your question is similar to No buoyancy inside liquid. This shows that the buoyancy force on Cell5 provided by Cell4 depends on the extent to which the fluid in Cell4 surrounds that in Cell5. In your scenario none of the fluid in Cell4 surrounds Cell5, so there is no buoyancy force. There is only the weight of Cell4 pressing down on Cell5. The following ...


0

Your calculations are right. I think there's a little confusion in your thought, that needs to be clarified: The normal force $N$=$P$ ,which in this case acts horizontally , cannot be compared with the friction force that acts vertically , so your query: "Now I am not sure if this is the minimum horizontal force needed to keep the book from slipping, ...


0

The object will topple about the lower RH corner. Take moments about this corner. The net clockwise moment must be > 0 for the object to topple. There is no need to find the CM. You can calculate moments for each cubical component of the object.


0

You claim the horizontal distance of the CoG ($X$) from the bottom left corner is $\frac{5a}{12}$ (I've not verified this). That is the distance $|OP'|$ in my version of the diagram. The distance $|PP'|$ is therefore: $$|PP'|=\frac{a}{2}-\frac{5a}{12}=\frac{a}{12}$$ When you start exerting the force $F$, a torque about the point P, which is $F\times \frac{...


0

You need to consider the moments of the forces involved. Moment is force times perpendicular distance from the axis. In order to turn the dumper bucket, the clockwise moment of the force from the hydraulic cylinder needs to be at least as much as the anticlockwise moment of the bucket's weight (when loaded). The weight of the bucket acts through its centre ...


0

The body will start to rotate when the torque is barely above zero. As you apply the force, the normal force made by the floor will move to the right to prevent rotation, until it reaches the right corner of the bottom square. After that it cannot keep moving. Thus to solve the problem make force diagram to get $N$ in terms of $F$, and impose a torque equal ...


2

if the body is rolling on a plane, then its degree of freedom is 2: one for rotation about the body's axis and one for translation of its center of gravity in forward and backward direction. if there is no slipping, the translation can be calculated from the rotation and the radius of the block. Thus the degree of freedom is degenerated to 1.


7

You need a better statement of Newton's Law. The one you are using is meaningless, because the word action is not defined. (In today's language of physics the word action is used in an entirely different context, sense, and meaning.) It's based on what Newton wrote, but is only half of what he wrote. Wikipedia gives us the whole thing To every ...


46

The way we are all taught Newton's Laws (by reciting them like mantras as children) is unfortunate because the traditional wording is misleading in many ways. A big problem (though not the only one) with the traditional wording of both Newton's second and third laws is that they incorrectly suggest cause and effect (and hence imply a chain of events, as you ...


1

How about considering a specific force, such as the Newtonian gravitational force between two point masses $m_1$ and $m_2$. We could write the force that mass 1 exerts on mass 2 as follows: $$\vec{F}_{12} =-G\frac{m_1 m_2} {r_{12}^2} \hat{r}_{12}, $$ where $r_{12}$ and $\hat{r}_{12}$ are the distance 2 is from 1 and unit vector from 1 to 2, respectively. ...


3

You're assuming "action" and "reaction" are the same things. Ok, there are lots of reasons why we might consider them to be either similar or exactly the same, but since the third law doesn't say "for every action there is an equal and opposite action", the third law is logically consistent. For the purposes of the third law, reaction is NOT the same as ...


0

It should be "Every action has equal and opposite reaction of SAME TYPE" , the best way to understand these contact force related questions is to draw a large clear free body diagram. Which will eventually lights up your problems. Draw all possible forces on both object(s) and or contact surface. Pls see the image , the force you applied is not belong to ...


9

It does not matter which one you call action and which one reaction, what the law says is that two objects make the same force on each other. Thus if A pushed B with a force of 5 Newtons, B will also push A with the same force.


1

My explanation is as follows Force on the spring in stretching it to length x can be written as $F=k(x-L_0)$ where x is the displacement, $L_0$ is the initial length of the spring and k is spring constant. energy stored is $dE=F.dx$ upon integration we will get $E=\frac{1}{2}.k.(L-L_0)^2$ where $L$ is the final length of the spring. This is I think ...


2

When the volume doubled mass and velocity both get doubled, The force is given by change in momentum, F=(mv-mu)/t since the both mass and velocity get doubled the force becomes four times as initial stage!



Top 50 recent answers are included