New answers tagged

1

If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.


0

For First Case: the rotation produced by the torque at the centre of wheel will rotate the wheel in the clockwise direction, but here friction is present, as friction opposes the motion of particle that's why it acts in anticlockwise direction and helps the body to move. For Second Case: here $mg\sin\theta$ will act along the line of centre of mass, which ...


0

Imagine there are two objects instead of one. One object has a certain mass, and experiences friction. The other object has another mass, and no friction. If those two objects were joined together, you would have no difficulty figuring out the equation of motion. But actually the problem is harder than it looks: the way you have drawn it, the object has ...


1

I agree that friction in the drive mechanism reduces thrust, rather than opposing the motion of the car. However, this is not the case for wheels which are not in the drivetrain - ie where there is front/rear wheel drive instead of 4-wheel drive. Friction in non-drivetrain wheel mechanisms are then sources of resistance to motion. If the car has rear-wheel ...


0

A free body diagram will show a car in motion has air drag force, gravity force and friction force on it. The net force keeps the car accelerated, decelerated, or moving at constant speed. Friction force is due to relative motion between wheel and ground. Engine output spins the wheels (torque from power train system balances the torque produced from the ...


0

PSI does not change with pipe size, only the surface area it is pushing on. Static head pressure is .433 per vertical foot. Water cannot flow over the top of a vessel which is gravity fed. It will only flow as high as the water level in it, because the water in the hose is also pushing back with gravity. A lot of these answers you are reading are ...


1

what's the backward force to balance $F_{fs}$ so as to keep the car moving uniformly? Is that the Force produced by engines? The forwards force comes from the torque produced by the engine of the car and is transferred into the ground via static friction. The retarding force that keeps the car moving at the same speed is mostly air resistance (as well ...


1

I would suggest that the major reason for the ball becoming stationary is its inevitable interaction with the air - i.e. friction, resistance to being parted and eddy currents.


0

The reasoning you have stated is correct, as the ball maintains a constant velocity in those conditions. However, in reality, the ball squishes a little bit in the direction which it is moving, so a greater frictional force acts (opposing linear motion instead of "assisting" rotary motion), causing a deceleration and hence the ball stops. It is also ...


0

You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them.


3

It is due to the viscous nature of any liquid. When you stir, the liquid starts spinning and this causes the liquid (the part which is in contact with the pot) to "drag" the pot(due to friction) along with it in the path of its motion.Hope this answers your question.


1

A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction). As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same ...


1

The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have ...


2

So given a spring with spring constant $k$ can one predict what dissipative force the spring will exert when extended? The answer is "No" because they depend on different things. The stiffness depends on the elasticity of the bonds between the atoms/molecules which make up the spring and the damping depends on the permanent distortion of the bonds which are ...


1

Spring stiffness is not responsible for energy loss. Consider a spring with stiffness $k$ but no damping. The work done in compressing the spring by a distance $x$ will be stored as the potential energy(PE) of the spring. For a spring compressed by a distance $x$, the PE is given by $\frac12 k x^2$. This energy is not lost and can be used by letting the ...


-1

Why do you think that It doesn't make sense to me that the normal force would be smaller than the perpendicular component of mg Yes, it is, because the ground is bending away below the car. If it would bend quicker the car would be flying and there would be zero normal force. No, it's okay. The normal force would be $mg\cos\alpha$ without the ...


2

You are mixing up concepts. The kinetic friction of a car has nothing to do with $F_N$. The concept of friction as taught in school tells you, that friction is independent of velocity. This is clearly not realistic for a car! The friction - let's rather call it "air resistance", for this is the main part - has a very complicated dependence on the velocity ...


2

The resistance to motion in a car is not due to simple friction. At low speeds it is dominated by viscous losses in the engine and transmission while at high speeds it is mostly due to aerodynamic drag. Viscous losses are approximately proportion to the velocity of the car: $$ F_v = Av $$ for some constant $A$. Aerodynamic drag is approximately ...


2

The formula you gave for kinetic friction, $$F_f=\mu_kF_N,$$ is a first-order, simplistic, phenomenological model. It describes systems for which the contact surfaces are uniform in space and time. In the real world, if you want to refer to this model, one should expect the coefficient of friction to be a function of temperature, normal force, velocity, ...


1

Hint: first find the angular velocity of the disc by conservation of energy, $\frac{1}{2}I\omega^2 = mgh$ , $\omega$ = angular velocity of the pulley, I = moment of inertia of the pulley. It is mentioned that the pendulum cover a horizontal distance L in t seconds. And we also know that $$v_{linear-velocity} = R\omega$$ So time t can be found ...


-1

Actually there is a normal reaction equal to $\frac{W}{2}$ on the two edges. One causes sliding friction of $\mu \frac{W}{2}$ and the other doesn't. The amount of material under friction does not matter. You can just lump the mass on the two ends and treat them accordingly. Edit 1 @Floris is correct, the normal reaction is not even because friction causes ...


1

You are correct : to calculate the friction force, you only need to consider the weight of that part of the rod resting on the rough surface, not the whole of it. When the block overlaps the rough area by distance x, the normal reaction on that portion of the block is Wx/L and the friction is F=μWx/L. The work done against friction in moving a short ...


1

Do dimples reduce drag if the flow in question is entirely laminar? No. Do dimples often reduce the total drag on objects which are causing turbulence as they pass through calm air? Yes. I can think of only one conclusive way to answer your question about the frisbee, and that is to get a bunch of similar frisbees, put dimples on some of them, and take ...


0

Your approach looks correct. I didn't check your arithmetic, but I suspect that you got the right answer and that the issue here is that your expectations of Interactive Physics are not realistic. Interactive Physics is a simplified educational version of the Computer Aided Engineering program called Adams. This program is intended to produce results ...


3

I am not sure what you actually want to ask. So I would recommend that you put more effort into your question. Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange: Say the toque applied to the joint of the scissors is $\tau$. What is this force $F$, then? The distance of the rope contact point to ...


1

We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


1

I always wondered the same thing, my guesses are all kinetic energy in tires turns into KE in engine and then lost through heat, as you may notice engine reving up when you down shift , the engine isn't getting any energy from fuel it must from the tires, so its the opposite the tires move the engine.


0

If you did not implant a horizontal impulse on the ruler and there was no friction then the centre of mass of the ruler would fall vertically and so one end of the ruler would move forwards and the other end backwards but the centre of mass of the ruler would stay in the same position. If there was friction between the ruler and the surface then the centre ...


0

Since work done by a force $\vec F$ undergoing a displacement $d\vec r$ is defined as $\vec F \cdot d\vec r$ when this dot product is positive the force and displacement are in the same direction and is negative when they are in opposite directions. The work done by a frictional force does not always have to be negative. Imagine a block $A$ on top of block ...


0

Force is a vector, meaning magnitude and direction. Work done by a force is relative to the direction of a force is the scalar value obtained by performing the vector dot product of the force and the displacement (which is also a vector). If something isn't coming out to what you expect when you compute work, make sure you have the right magnitude and ...


0

The friction force is given by $F_f$ = $\mu$N. In this example, $\mu$ remains the same, while both the normal force and force of friction increase. The normal force is given by $F_n$ = mg. Your confusion arises from what defines m. I like to think of it as "effective mass." Basically, anything that contributes to the interface the friction is acting on. So ...


0

If the pivot is accelerating horizontally (together with the body) at a rate of $a_{pivot}$ then the angular acceleration of the pendulum is $$ \ddot{\theta} = - \frac{m c (a_{pivot} \cos\theta + g \sin \theta)}{I_{zz} + m c^2} $$ where $c$ is the distance from the pivot to the center of mass, $m$ the total swinging mass and $I_{zz}$ the mass moment of ...


0

Although there are excellent answers, I think a more "simplistic" answer is required to correct your thinking. If you start with a piece of lead (1 kg) on the floor, grab and lift it 1 meter, it will gain (1 x 9.8 x 1 =) 9.8 J of energy. If you now open your hand (release it), it will fall by "it self" and hit the ground and "loose" the energy it had ...


0

You can possibly wrap a rope around a pole (trunk of a tree). If you had a tension in the rope of $10N$, you would need exponentially less amount of force to hold the rope from the other end (assuming that you have wrapped the rope around a tree). Basically you are increasing the normal force on the rope, by using the tension in it, which is technically ...


2

I'll assume you meant to say "Rheological definition of friction". Rheology deals with the friction of fluid layers against one another and with any solid boundaries. The factors of fluid viscosity determine the frictional forces exterted within a fluid and with its boundaries. A brief overview of the principles, mathematics can be found here


0

I want to elaborate on MAFIA's correct and important "potential energy is a property of the whole system". The potential energy is not a property of just one of the involved objects, like the lifted ball. That in our common experience all the potential energy somehow seems to be "attached" to the ball is just a consequence of the very different masses.1 ...


0

In the case of the ball on the table, it is in a state of stable equilibrium, where the table is pushing up against the ball to counteract the gravitational force pulling it down. If the table and ball were to be moved to a planet with much stronger gravity, the gravitational force on the ball could be strong enough to break the table and the ball would move ...


3

This answer is only about Where does the stored energy stay in the object, and why does it only convert into vertical motion and not horizontal motion? because I think your other questions have been well-addressed, but this one has only been answered in highly technical terms that may not have clarified anything for you. Think about what happens if ...


0

Because the tension force of a string is a reaction force (like the friction force). This means there is no tension force until the string isn’t stretched. The string will be non-stretched until the force F is lesser than the limiting static friction force. There is a same state for friction force. The friction force is lesser than the limiting static ...


0

If I'm understanding your question properly, the block in that situation is over-constrained (i.e. statically indeterminate - you cannot calculate the unknown forces from the laws of statics alone) - you cannot assume that F equals the limiting friction - both it and the tension are unknown. When you apply a sideways force to the block, it starts to move, ...


2

The first thing is to note that the gravitational potential energy is associated with both the object and the Earth. You may think that only the object has the potential energy because when you drop the object you see it accelerate downwards and gain kinetic energy. At the same time the Earth is accelerating upwards at a rate of $\frac {\text {mass of ...


20

It is wrong to think potential energy is stored in the object. The earth pulls the object down, but the object pulls the earth up. They share the potential energy. The object fails to fall down because the tabletop pushes it up. The earth fails to fall up because the bottom of the table legs push the earth down. The table pushes up and down because it is ...


3

[...] when it already has energy, then why doesn't it fall off from the table top onto the ground by itself? Because it is being held back. It wants to fall straight downwards, but the bookshelf applies a normal force to hold up the book, which is stronger than the downwards force (gravity). Just as the rubber band holds back the spring from elongating, ...


0

In this totally hypothetical situation, everything you said is correct. It's important to remember, however, that there must be a normal force between the bodies for the friction force to exist. In the most intuitive example, this force can be the reaction of the gravitational force (from Earth) from the upper body onto the lower one. The only acceleration ...



Top 50 recent answers are included