Tag Info

New answers tagged

3

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$ F_{\text{slide}} > \mu m g $$ where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...


2

You've stumbled on an interesting idea: how do classical systems that dissipate heat or energy via frictiom arise from quantum systems that perfectly conserve energy in their interactions? Particles in the collision kind of scenario you described don't really exhibit friction. One convenient point is that temperature and heat transfer in quantum physics is ...


0

Whatever you do it produces heat. This has to do with electromagnetic interactions between the two objects which interact. Moving one surface on an other, the surface electrons from both surfaces will be disturbed, they couldn't be at the same place and will be moved slightly. Every displacement has elements of acceleration and any acceleration is escorted ...


0

f(friction) directly proportional to N(normal force acting, perpendicular to the plane of fridge in outward direction here) is the force that opposes motion, now friction will in this case be provided by the atomic attraction, only atomic attraction is taken into account here, because repulsion would not even let it attach to refrigerator, after the force is ...


3

As you may know, the friction is proportional to the normal force of an object or in this case the force of attraction between magnet and refrigerator. If your force is strong enough then the friction will be sufficient and the magnet will not slip (on earth the force of friction must exceed the mass of your magnet multiplied by 9.81 m/s). If we assume ...


-1

I have a 600 gallon tank on a 25 foot tower. The bottom of the tank has a 2 inch outlet and pipe that drops straight to the ground. There is also a 3/4 inch pipe on the other side of the tank that drops straight down the ground. At the bottom of the two inch pipe I reduced it to 3/4 inch pipe. The pressure is amazing out of that pipe. The 3/4 inch pipe that ...


1

The force between the roller and the road is in the vertical direction so there is no effect of friction. Moreover the speed is constant so there is no acceleration. The only forces acting in the vertical direction are the weight, $W=mg$, and the reaction force between the roller and the road, $F$. As the roller is not accelerating vertically, these must ...


-1

A. Friction = force that opposes motion, if you are not applying force to move it, then therefore NO friction is in account. B. Upto limiting friction, mag. Of applied force = friction force, hence 6N. Moreover, after 8N force, the box will start moving, friction(static, max) = (.2)(40). C. Heed to what floris says


0

I am sorry to say this but surface area is not proportional to the force of friction. Friction is given by the coefficient of friction (static or kinetic) multiplied by an objects normal force. What may ne useful is the theory that the coefficient of friction is equal to the tangent of the angle at which an object starts to move on a plane. Also something ...


0

The rotational motion theory, the kinetic motion theory, the friction theory, and added to further too in special conditions


1

Consider two different regimes. First - when the ball is still slipping, the relative motion of the ground and ball causes a force on the ball which (a) slows down the center of mass and (b) increases the angular speed. Once the ball rolls without slipping this force disappears. Second - when the ball does not slip, the contact point must be stationary. ...


1

Yes I think it is propulsion due to rolling, also the tires put a reaction on the road. The engine spins the wheels through transmission as you push the gas pedal and the engine accelerates, then you shift to higher gear and spin the wheels faster with less engine acceleration as you go to higher speed. Hope that answers the question. J


1

Your assumption that the ball will be faster if it rotates faster has a very logical explanation and is completely valid. If your ball has a circumference C it will cover that distance with each rotation. Let us say your C was 1m and the period was 2 rps (revolutions per second) then your ball would cover 2 metres in 1 second. If your period is doubled to be ...


1

The ball rolls because: the friction present between the surface pushes the atoms/particles in contact with surface or ground backwards, hence with centre of mass moving in a direction, the ball's bottom surface(ie. The surface in contact) moves in opposite direction or backwards, causing the ball to roll. The ball slips because there is no friction that ...


1

The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...


1

When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator. The form they give is $$\omega = \omega_0\sqrt{1 - \zeta^2}$$ Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.


5

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...


0

Static friction does not produce or consume work in most of the times. For example for a solid body that rolls without sliding the velocity of the base point $A$ is $\vec v_a = \vec v_{cm} + \vec v_{tangential} \Rightarrow v_a = v_{cm} - \omega R = \omega R - \omega R =0$ which implies that $x_a = 0$. The static friction is a force that acts on $A$ so $W_T = ...


0

It doesn't. The work is done by the active force(ex. A human trying to pull a bull.). This work is converted into frictional energy(ex. Heat generated b/w surfaces)


1

The correct version: $$P \cos\theta-F_{fr}=ma.$$ Here $P\cos\theta$ is a projection of $P$ on the direction of motion. Next we use the expression for the friction force $F=\mu N=\mu mg$ (note that $N=mg$ in this case because the motion is happening on the horizontal plane). Now $$P \cos\theta-\mu mg=ma.$$ From where we find $$P=\frac{m(\mu ...


0

Rolling friction result from for example small changes in the surface or in the wheel material (the rubber in a tire). The surface is not perfectly flat and rigid so there will be some small forces trying to stop the rotating motion: On the contrary, the static friction is not trying to stop the rotation of the wheel. Static and rolling friction are ...


0

Without friction between the crate and the truck bed, the crate would remain at rest in the frame of reference of the road, as the truck accelerates away down the road. The crate moves in the frame of reference of the road, because of the force of friction acting on it. So the work done on the crate, in the frame of reference of the road, is the friction ...


2

I think you are confused about what $d$ is supposed to mean in the equation $W=F\cdot d.$ You seem to be under the impression that $d$ is the distance that the object being acted on moves relative to the object providing the force. But this is not the correct meaning of $d$ in the equation and you know it. Imagine if the car crate were in front of the ...


1

I am reasoning my way to an answer here - I have never seen this problem before so I could be completely wrong. I think the issue is that you need to make sure the normal force is reduced as the tension on the "signal" side of the capstan is reduced. This is presumably why the slightly stiffer wire (as opposed to the rope) gave more predictable results at ...


3

The contact point between the wheel and the car is stationary - there is no "rubbing" there (well there is because the contact point is really a patch but let's keep it simple). To do work you need "force times distance" - and without relative motion there is no "distance". When you apply the brakes in a car you have sliding of the pads relative to the ...


1

As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to ...


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.


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$ $\Rightarrow 1/2=(F-T)/T \Rightarrow T = 2F/3$



Top 50 recent answers are included