Physics problem

[werto165]

I'm doing a physics investigation soon and i was wondering how to work out the coeffiecent of kinectic(dynamic) friction. I'm doing it on braking distances and I'm using a cart which has a break and you can vary the force applied on the wheels. I know how to calculate the value of static friction(file included), but I don't know exactly how to work out kinectic friction. From looking on the internet I've found out that kinectic friction is <= static friction and i dont really get that either. Any help would be appreciated.  

 

 

[T.Vengeance]

You can find out the by pulling the object with a constant velocity. Since the object is at constant velocity, and therefore at equilibrium, both the magnitude of the applied force and the force of friction will be the same. Therefore if you since F_f = u*F_N , you can divide the applied force by the normal force to get the coefficient.

[Ciccioo]

you have a fractal design R4 on the floor and it is not moving

 

you push it a bit: it doesn't move

this is because the static friction is acting, creating a force opposed and equal in intensity to the force you're applying

 

you push it more: it moves

you reached and surpassed the maximum amount of force that static friction can generate, and now kinetic friction is the one that is acting

 

to start moving an object you need to apply more force that when you just need to make it keep on moving, maybe you didn't notice, but if you try pushing something, maybe you will notice the little "bump" after it started moving

 

you can calculate the coefficient just the same way as you do with the static one

[werto165]

you have a fractal design R4 on the floor and it is not moving

 

How did you know???

[Ciccioo]

How did you know???

i'm watching you

 

nice ass

[werto165]

i'm watching you

 

nice ass

Why thankyou!

[helping]

you could apply the concept of inertia to be the same here

[werto165]

you could apply the concept of inertia to be the same here

now I sort of understand why kinectic friction is less but I still don't understand how I would calculate the dynamic friction. Found this on wikipedia:

Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). The coefficient of kinetic friction is typically denoted as μ_k, and is usually less than the coefficient of static friction for the same materials.^[18][19] However, Richard Feynman comments that "with dry metals it is very hard to show any difference."^[20]

New models are beginning to show how kinetic friction can be greater than static friction.^[21] Kinetic friction is now understood, in many cases, to be primarily caused by chemical bonding between the surfaces, rather than interlocking asperities;^[22] however, in many other cases roughness effects are dominant, for example in rubber to road friction.^[21] Surface roughness and contact area, however, do affect kinetic friction for micro- and nano-scale objects where surface area forces dominate inertial forces.^[23]

[werto165]

You can find out the by pulling the object with a constant velocity. Since the object is at constant velocity, and therefore at equilibrium, both the magnitude of the applied force and the force of friction will be the same. Therefore if you since F_f = u*F_N , you can divide the applied force by the normal force to get the coefficient.

So its basically the same as static friction( http://imgur.com/h0mjMBp ), the net force being 0 and being in dynamic equilibrium. So if an object is accelerating down a plane would the coefficient of kinectic friction be changing all the time? Also, would it be possible to detect the change in the friction doing an experiment of the cart going down a ramp? 

[Ciccioo]

the coefficient is constant for every pair of surfaces

 

the acceleration of the object down the plane is constant (because g=9.8=constant and the plane inclination is supposedly constant), and the kinetic friction is constant

[helping]

So its basically the same as static friction( http://imgur.com/h0mjMBp ), the net force being 0 and being in dynamic equilibrium. So if an object is accelerating down a plane would the coefficient of kinectic friction be changing all the time? Also, would it be possible to detect the change in the friction doing an experiment of the cart going down a ramp? 

No, yes and yes. 

 

kF and sF are always constants in regards to two surfaces under ideal conditions. 

 

It is possible to detect friction (not change? but the constant), but since the graph of static and kinetic friction coefficients is a piecewise function it would be necessary to begin collecting data a second or two after the cart begins moving. 

 

you would simply do something like ma = mgsinθ - ucosθg, solve for what you need.

 

to calculate the differences between the change in state from static to kinetic you would just compare the difference of peak or activation energy (N) versus energy required to maintain motion (N)

 

you might be thinking of the actual retarding force of friction, which is indeed a variable (of N), but linear as a constant divisor. So in a system as N increases, retarding force of friction increases, but the friction coefficient is the same. 

[Ciccioo]

you might be thinking of the actual retarding force of friction, which is indeed a variable (of N), but linear as a constant divisor. So in a system as N increases, retarding force of friction increases, but the friction coefficient is the same. 

wouldn't N and the retarding friction be constant, in the given plane scenario?

[werto165]

No, yes and yes. 

 

kF and sF are always constants in regards to two surfaces under ideal conditions. 

 

It is possible to detect friction (not change? but the constant), but since the graph of static and kinetic friction coefficients is a piecewise function it would be necessary to begin collecting data a second or two after the cart begins moving. 

 

you would simply do something like ma = mgsinθ - ucosθg, solve for what you need.

 

to calculate the differences between the change in state from static to kinetic you would just compare the difference of peak or activation energy (N) versus energy required to maintain motion (N)

 

you might be thinking of the actual retarding force of friction, which is indeed a variable (of N), but linear as a constant divisor. So in a system as N increases, retarding force of friction increases, but the friction coefficient is the same. 

With the piecewise function do you mean this http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html. Would using Gravitational potential energy and kinectic energy also work in conjunction with resolving forces work also? And would you happen to know how you'd work out a stopping distance, if the cart wheels were to skid, would the friction be different again or the same due to increased thermal energy.

[helping]

wouldn't N and the retarding friction be constant, in the given plane scenario?

In this case, static and kinetic yes

 

With the piecewise function do you mean this http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html. Would using Gravitational potential energy and kinectic energy also work in conjunction with resolving forces work also? And would you happen to know how you'd work out a stopping distance, if the cart wheels were to skid, would the friction be different again or the same due to increased thermal energy.

Yes that's what I'm talking about. 

I'm not sure about the relationship between kinetic and potential energies on a plane, but I suspect it should hold if you set everything up right, but again I'm not sure. 

 

To work out the stopping distance you can either set up an integral or use standard kinematics equations with negative acceleration (computed based off of retarding force) and a gravity constant at an angle. 

 

if the cart wheels were to skid, would the friction be different again or the same due to increased thermal energy.

not quite sure what you mean here. In most non-precise science energy lost to heat and sound will typically be ignored, wind resistance as well. 

I assume you meant in these calculations that the card wheels were already skidding, because if they're not then there's a lot of different forces at work to factor in and the problem would be moderately difficult to find the friction force of any arbitrary part of the system. 

[CjlabanG]

becos potato