Is Acceleration Due to Friction Constant Regardless of Speed?

[physicsxu]

Is acceleration due to friction constant for a given coefficient of friction? Say you are driving in a car very slowly and you hit the brakes, or you are driving in a car a little bit faster and you hit the breaks, assuming you are driving in the same car on the same road, would the acceleration due to friction be the same in both cases?

My guess is yes, it would, by using the equation Ff = ukFg
ma = ukmg
a = ukg
the uk and g are constant, so a must be constant as well.
Can anyone solidify or correct my understanding of this concept?

 

[physicsxu]

My mistake I don't drive haha. So let's say the car is skidding, acceleration due to friction will always be the same given the same car and same road?

 

[physicsxu]

Thanks : D, this will help me on lab report.

 

[CWatters]

In some situations the coefficient of friction might not be constant. For example during a long skid the temperature of the tyre would increase perhaps melting the rubber. At high speeds road tyres find it hard to displace rain water so it's possible the coefficient is reasonably constant only up to a certain speed beyond which aquaplaning occurs.

 

[Doc Al]

If you apply the brakes, it's friction between the braking mechanism and the wheel that slows you down. Friction between the tyres and the road only applies if you skid (which actually leads to the car taking a lot more time to stop.)

No. The only external force slowing you down is the friction from the road, whether you skid or not. (When you skid, you rely on kinetic friction.)

Yes, the braking mechanism works to slow the wheel, but without friction from the road the car would not slow.

 

[Doc Al]

Is acceleration due to friction constant for a given coefficient of friction? Say you are driving in a car very slowly and you hit the brakes, or you are driving in a car a little bit faster and you hit the breaks, assuming you are driving in the same car on the same road, would the acceleration due to friction be the same in both cases?

Not necessarily. Keeping things simple, as long as you are not skidding, what slows the car is static friction with the road. Recall that static friction can vary, up to a maximum of $\mu_s N = \mu_s mg$.

My guess is yes, it would, by using the equation Ff = ukFg
ma = ukmg
a = ukg
the uk and g are constant, so a must be constant as well.
Can anyone solidify or correct my understanding of this concept?

Once you lock the wheels and enter a skid, then kinetic friction is involved and your analysis applies.

(Of course, real life is a bit more complicated: See CWatters' post.)

 

[Doc Al]

I perhaps didn't phrase it well, but it's how hard you apply the brakes that determines the braking force. If it were simply friction on the road, then how hard you brake would make no difference.

It is always the friction from the road that provides the force that slows the car. Yes, pressing the brake harder affects the wheel, which in turn leads to a greater static friction force with the ground. (If it doesn't, the car's acceleration cannot change.)

An internal force within the car--such as the brake mechanism against the wheel--cannot directly slow the car. An external force is required.

 

[PeroK]

It is always the friction from the road that provides the force that slows the car. Yes, pressing the brake harder affects the wheel, which in turn leads to a greater static friction force with the ground. (If it doesn't, the car's acceleration cannot change.)

An internal force within the car--such as the brake mechanism against the wheel--cannot directly slow the car. An external force is required.

You've emphasised the word "directly", but that's not a word I used. I was talking about the calculation for how fast something will slow. You've given a calculation above based on static fraction. But, frankly, how could you use that formula to determine braking speed? You can't. You've no way of knowing from that calculation how quickly the car will stop. To calculate that you'd have to look at the braking mechanism.

 

[PeroK]

It is always the friction from the road that provides the force that slows the car. Yes, pressing the brake harder affects the wheel, which in turn leads to a greater static friction force with the ground. (If it doesn't, the car's acceleration cannot change.)

An internal force within the car--such as the brake mechanism against the wheel--cannot directly slow the car. An external force is required.

So, what would you say causes a car to move in the first place? The engine or friction on the road?

 

[Doc Al]

You've emphasised the word "directly", but that's not a word I used. I was talking about the calculation for how fast something will slow. You've given a calculation above based on static fraction. But, frankly, how could you use that formula to determine braking speed? You can't. You've no way of knowing from that calculation how quickly the car will stop. To calculate that you'd have to look at the braking mechanism.

I'm not talking about calculating anything. You made statements implying (actually stating) that it was not the road friction that slowed the car; those statements were wrong.

Obviously, the way the car is designed you must step on the brakes to slow the car. But it is the friction from the road that must be present to slow the car. Or are you still denying that?

 

[Doc Al]

So, what would you say causes a car to move in the first place? The engine or friction on the road?

The word "cause" is a bit problematic here. But from a physics point of view, accelerating a car requires an external force. Or do you think you can accelerate a car without friction?

 

[PeroK]

The word "cause" is a bit problematic here. But from a physics point of view, accelerating a car requires an external force. Or do you think you can accelerate a car without friction?

Do you think static friction can accelerate a car without the engine being engaged?

We're arguing about nothing here. I was sloppy in what I wrote. I was only thinking about the calculation:

$F = \mu_k mg$

Which doesn't apply unless the car skids.

 

[PeroK]

I'm not talking about calculating anything. You made statements implying (actually stating) that it was not the road friction that slowed the car; those statements were wrong.

Obviously, the way the car is designed you must step on the brakes to slow the car. But it is the friction from the road that must be present to slow the car. Or are you still denying that?

I've deleted those statements. I've never denied that. Of course, I know that. The original post was about calculating braking speed. I should have said "the coefficient of kinetic friction doesn't apply unless the car skids".

 

[A.T.]

So, what would you say causes a car to move in the first place? The engine or friction on the road?

Depends on the definition of "in the first place". Ultimately the Big Bang causes everything in the "in the first place". But a car will not accelerate, until there is an external force on it.

 

[physicsxu]

Okay I've read your discussion, especially about static and kinetic friction, so now I'm a little worried again. Say I calculate the acceleration caused by friction by pushing a cart on a level surface. I'm using motion sensors and the software capstone, which will graph the acceleration/time graph for me. Obviously the graph is not perfectly straight (a bit zig-zaggy), suggesting that there is a mix of static and kinetic friction at work here (I think?). Say I want to just use the average acceleration caused by friction, I average the points on the graph to get the average acceleration due to friction. I then incline the plane, and measure the acceleration/time graph again, this time not pushing the cart, just letting gravity do the work for me. There, I find the average acceleration again.

What I want to find is the acceleration due to gravity, and I know that I can't just simply calculate it by (average acceleration of cart)/sin(angle of the inclined plane), because friction plays a role here. Can I subtract the average acceleration due to friction I found earlier from the average acceleration due to the parallel-to-the-plane component of gravity I found after, to get the simulated average acceleration if the cart was on a frictionless surface?

My own guess is I can, because I assume the acceleration due to friction does not change whether the cart is rolling on an inclined or level plane, so subtracting that (negative) value from the average acceleration due to the parallel-to-the-plane component of gravity will give me the simulated acceleration due to the parallel-to-the-plane component of gravity on a frictionless surface. Also, if I don't consider the acceleration due to friction, my final calculation for g is about 7m/s^2, whereas if I consider the acceleration due to friction, my g value is about 9m/s^2 (only did the calculations for 1 trial so far).

Sorry if this sounds impractical, my lab instructor didn't give us precises instructions, he gave us the tools and he wanted us to design the experiment to calculate g using a cart and a plane (that can incline) ourselves. Thanks to everybody who contributed so far!

 

[CWatters]

With some qualification what you propose is correct. You could calculate the friction force by applying a known force to the cart and measuring the acceleration on a level surface. The acceleration should depend on the net force (eg applied force - friction force). Then you could assume that the friction force is the same when the cart rolls down an incline and do the maths.

Many of the earlier posts are somewhat irrelevant to this experiment. It doesn't matter what the origin of the friction is as long as the assumption that it's the same is valid. For example it wouldn't work if the two surfaces were very different and that effected the rolling resistance. Nor would it work if the incline was very steep (reduced normal force effecting friction force) or the speeds very different between the two experiments.

 

[physicsxu]

Okay thank you :), now I have confidence in my calculations.

 

[Vibhor]

An internal force within the car--such as the brake mechanism against the wheel--cannot directly slow the car. An external force is required.

Hello Doc Al,

I have a few doubts .

The force provided by the engine turns the wheel . Do we consider this engine force as internal or external ? Can this force change the momentum of the car ? Can this force provide energy to the car ?

My thinking is that the engine force is an internal force . It cannot change the momentum of the car , but can provide energy to the car . It is this energy provided by the engine which is transformed into the mechanical energy of car .

Does this make any sense ?

 

[CWatters]

The force provided by the engine turns the wheel . Do we consider this engine force as internal or external ? Can this force change the momentum of the car ?

What happens on ice (frictionless ice).

 

[Vibhor]

What happens on ice (frictionless ice).

In the absence of friction ,the wheel keeps rotating but the car doesn't move .

 

[Doc Al]

Hello Doc Al,

I have a few doubts .

The force provided by the engine turns the wheel . Do we consider this engine force as internal or external ? Can this force change the momentum of the car ? Can this force provide energy to the car ?

My thinking is that the engine force is an internal force . It cannot change the momentum of the car , but can provide energy to the car . It is this energy provided by the engine which is transformed into the mechanical energy of car .

Does this make any sense ?

I think you've got the right idea.

For the car to accelerate, there must be an external force on it. Any forces exerted by the engine are internal forces. (Obviously, those engine forces are essential in creating the conditions for there to be an external force, but they do not directly accelerate the car.)

The engine converts chemical energy to mechanical energy, and is the source of the car's mechanical energy. An external force (from the ground) is still needed to accelerate the car.

 

[Vibhor]

Ok.

Suppose there is a car moving (rolling without slipping) on a horizontal surface with constant speed 'v' covering a distance 'x' in time 't' . The forces present are static friction $f_s$ , air resistance ( and all other dissipative forces ) $f_D$ , force due to engine $f_E$ . Power supplied by engine is $P$ .

No work is done by friction . Applying law of conservation of energy , Power supplied by engine = Power dissipated by resistive force .

$f_Ev = f_Dv$ . Does that mean $f_E = f_D$ i.e the force due to engine is equal to the resistive force ?

 

[CWatters]

In the absence of friction ,the wheel keeps rotating but the car doesn't move .

So without the external force of friction between the car and the ground the car cannot gain momentum and you were correct when you said..

My thinking is that the engine force is an internal force . It cannot change the momentum of the car..

This what you would expect from Newtons first law.

 

[Vibhor]

What is your opinion about post#22 ?

 

[CWatters]

Im on my mobile and it doesn't display everything correctly but otherwise #22 seem OK.

If there is no acceleration then the forces must sum to zero.