# Acceleration required to cause vehicle skid?

1. Jul 3, 2015

### Ocata

1. The problem statement, all variables and given/known data

Hello, I know how to solve this problem, I just don't understand an aspect of the conclusion.

A car is traveling at some constant velocity forward in a horizontal direction. The ground has a coefficient of static friction of .6. What acceleration in the forward direction would cause the vehicle to skid? Approximate gravity to 10m/s^2.

2. Relevant equations

$[F_{net} = ma] = [F_{friction} = ma] = [μ(F_{N}) = ma] = [μ(F_{G}) = ma] = [μ(mg) = ma] [μ(g) = a]$

3. The attempt at a solution

.6(10) = 6m/s^2

This is fine, but I am not really understanding why m would cancel as though it does not matter.

To me, the heavier the object, the less likely it is to skid. That is, the greater the mass of an object, the greater its weight, and there for the greater the normal force.

So on a particular surface of μ = .6, then:

Force of Friction = $μ(F_{N}) = μmg = .6(10kg)(10m/s^{2}) = 60N$.

However, if I increase the mass from 10kg to 100kg, then:

Force of Friction = $μ(F_{N}) = μmg = .6(100kg)(10m/s^{2}) = 600N$.

So, clearly, mass has an impact on force of friction. If this is in fact true, why does it cancel from the formula above in the part 2 (relevant equations) section?

2. Jul 3, 2015

### FactChecker

You are right about the normal force increasing with mass. But also the force needed to accelerate it increases with mass (F = m*A). And they are in exactly the same proportions. So the two effects of mass on skidding cancel each other.

3. Jul 3, 2015

### Ocata

So would I be correct to say that if there are 4 blocks, 1kg, 10kg, 100kg, and 1000kg, on a surface with coefficient of friction μ, it would, of course, take different amounts of force to move each block. Specifically, for each box to be put into motion from rest (starting from least to greatest mass), I'd have to apply a greater and greater amount of force.

For example, if μ = .10, then:

1kg(10m/s^2) = Fg = Fn = 10N => μ(mg) = .10(10N) = 1N

10kg(10m/s^2) = Fg = Fn = 100N => μ(mg) = .10(100N) = 10N

100kg(10m/s^2) = Fg = Fn = 1000N => μ(mg) = .10(1000N) = 100N

1000kg(10m/s^2) = Fg = Fn = 10000N => μ(mg) = .10(10000N) = 1000N

And so what the equation is saying is that in order for the block to be put into motion, it needs to be pushed with some force that is equal to or greater than 1N, 10N, 100N, and 1000N.

And since F = ma, and we know the force we mush push with and we know the mass we need to push, the only thing that needs to be established is the acceleration, which we can done by calculating:

F = ma

1N = 1kg(a) ==> 1N/1kg = 1m/s^2

10N = 10kg(a) ==> 10N/10kg = 1m/s^2

100N = 100kg(a) ==> 100N/100kg = 1m/s^2

1000N = 1000kg(a) ==> 1000N/1000kg = 1m/s^2

This shows that a mass needs to be pushed with such a force that causes the object to move with some constant acceleration. So for this particular surface type. As long as I can get the object to accelerate to 1m/s^2 or greater, I will have set it into motion, yes?

4. Jul 3, 2015

### UMath1

The force must be greater than 1N, 10N, 100N, and 1000N, not equal to it. If it were equal, the object would remain in equilibrium with no net force.

Yes, but thats the wrong idea. You could set it into motion with .001 m/s^2 acceleration. In order to set it into motion, the applied force must be greater than the force of static friction which you calculated as 1N, 10N, 1000N, or 100N. For example, if the applied force were 1.1 N on the 1kg block. Then the net force would be .1N. And the resulting acceleration would be .1/1 = .1 m/s^2 and the object would be set into motion. Although, it would be set into motion for a very short amount of time.

5. Jul 3, 2015

### Nathanael

Maybe it will help to look at it like this:

Suppose you have a few boxes with different masses in the back of a truck (the back of the truck has a uniform coefficient of friction).
Now the truck starts increasing it's acceleration. All of the boxes will slip at the same time regardless of their mass.

6. Jul 4, 2015

### Ocata

So its not that the object needs to be accelerated to a > 1m/s^2 in order to be set into motion, I just need to push with some force such that the blocks "would" accelerate if no other forces, such as surface friction, were being applied to it.

For instance, suppose the friction is μ = .2. Then,

Normal Force = Fn = 10kg(10m/s^2) = 100N ==> Force Static Friction = fs = μ(Fn) = .2(100N) = 20N , so

20N = 1kg(a) ==> 20N/10kg = 2m/s^2

If I begin to push the box with a very light force and then gradually increase until the box slips, then the acceleration at that moment will not be 2m/s^2?

If I understand correctly, I would need to push with a force such that I would cause an acceleration of 2m/s^2 if the surface was completely frictionless.

On a frictionless surface, a 10kg block causes a Normal Force of Fn = 100N. And if the block were to be pushed in order to generate an acceleration 2m/s^2, then the applied force on the block would be 10kg(2m/s^2) = 20N. So in order to set the 10kg block into motion, a force greater than 20N would need to be applied if it were on a surface of μ = .2. Yes?

7. Jul 4, 2015

### UMath1

Yes, the force would have be greater than 20N. But the acceleration of 2m/s^2 is irrelevant in determining this. The acceleration of 2m/s^2 only tells us the rate at which the object would slow down if it had been previously set in motion and the only force acting on it at the moment was friction.

8. Jul 4, 2015

### Nathanael

No no, I think Octa is assuming that there is static friction but no kinetic friction.

At any rate, you have the right idea. (Although it would be very strange if there was no kinetic friction yet there was static friction.)

Personally I prefer the truck example

9. Jul 4, 2015

### Ocata

Hi Nathanael,

Not sure I understand the truck example. what would be the forces acting on the block?

I made a free body diagram, but not sure if its correct. Could you let me know if the horizontal forces need correction?

#### Attached Files:

• ###### FreeBodyDiagramTruck.png
File size:
10.6 KB
Views:
46
10. Jul 4, 2015

### Nathanael

In the truck example, it is the force of static friction which allows the boxes to accelerate with the truck. Once the acceleration of the truck becomes too large, the static friction isn't big enough to accelerate the boxes that fast and they will slip. Although the maximum force of static friction is larger for the heavier boxes, they will still all slip at the same time.

(In your free-body-diagram static friction should be pointing to the left, and I have no idea what "force of truck bed" is supposed to represent.)

11. Jul 4, 2015

### Ocata

I may understand better after reading your correction. Is this more accurate?

This would suggest that the block only has a force applied to it in one direction. As opposed to a scenario of someone pushing on the block in the direction against the direction of friction.

#### Attached Files:

• ###### FreeBodyDiagramTruckb.png
File size:
8 KB
Views:
52
12. Jul 4, 2015

### Nathanael

Your FBD looks good now. I just mentioned that example because it's quite similar to your original problem you were having confusion about.

You said, "To me, the heavier the object, the less likely it is to skid," which is like saying, "the lighter boxes in the truck will slip first."

I just tried to re-word the question to hopefully help you understand it better. At any rate, the boxes all slip at the same time, which I think you now understand why (and same for the skidding of the car, regardless of it's mass).

13. Jul 4, 2015

### Ocata

I still don't understand it entirely.

If there is a 1kg box on a surface of μ = .2. It needs just over 2N in order to break free from the static friction.

If there is a 10 kg box on a surface of μ = .2. It needs just over 20N in order to break free from the static friction.

As UMath1 indicated, there is be a net force on the object.

(My pushing force) - (Force of Friction) = Fnet

For the 1kg block: 2.001N - 2N = .001N = Fnet

For the 10kg block: 20.001N - 20N = .001N = Fnet

The acceleration of the 1kg block after being set in motion would be: Fnet/m = .001/1kg = .001m/s^2

The acceleration of the 10kg block after being set in motion would be: Fnet/m = .001/10kg = .0001m/s^2

So the heavier box accelerates at a slower rate than the lighter box after breaking free from the static friction.

So if the accelerations are different for different masses, then the equation:

μ(mg) = ma

can not have anything to do with the movement of the boxes because different masses definitely seem to have different accelerations upon or after breaking free of the static friction.

So then what acceleration is referred to when the masses cancel out?

[μ(mg) = ma] = [μg = a] ...acceleration of what?

14. Jul 4, 2015

### UMath1

As nathanael said earlier, I think the frictional acceleration, μg, is used primarily in the case of kinetic friction. That is, friction on an object already in motion. This is the acceleration any moving object wil undergo in the absence of any force other than friction. Just like in your example of the skidding car; once the brakes pressed the wheels stop moving and the only force that acts on the car is kinetic friction. Kinetic friction slows the car down at a fixed accelertion of μg. Which is why if you're going faster, you need more time to come to a stop.

15. Jul 4, 2015

### Ocata

I think in this problem they were talking about static friction. The car needs to accelerate at some rate on order to begin to skid. Isn't it correct that when something begins to slide or skid, you are dealing with static friction?

16. Jul 4, 2015

### UMath1

When something begins to slide or skid, it is kinetic friction. Kinetic friction occurs when an object is in motion.

Static friction occurs when an object is stationary and has not begun to move yet.

17. Jul 4, 2015

### SammyS

Staff Emeritus
Yes, static friction occurs when the two objects in contact with each other are stationary (actually stationary relative to each other - they may both be in motion).

However, the question asks for the acceleration needed for skidding to begin. This occurs when the force required to achieve that acceleration is just a very little greater than the maximum force available from static friction. So essentially you're dealing with static friction for this problem.

Last edited: Jul 4, 2015