Minimum stopping distance so a box on a truck doesn't slip (friction)

[Mohmmad Maaitah]

Homework Statement

A crate is carried in a truck traveling horizontally at 15 m/s. If the coefficient of static friction between the crate and the truck is 0.40, determine the minimum stopping distance for the truck s ch that the crate will not slide on the truck.

Relevant Equations

I get d = 28.6 m
When I use Bing chat and chat gpt they give totally different answers one give 10 and one like 200 or something.
So how to solve it ?

 

[kuruman]

Homework Statement: A crate is carried in a truck traveling horizontally at 15 m/s. If the coefficient of static friction between the crate and the truck is 0.40, determine the minimum stopping distance for the truck s ch that the crate will not slide on the truck.
Relevant Equations: I get d = 28.6 m
When I use Bing chat and chat gpt they give totally different answers one give 10 and one like 200 or something.
So how to solve it ?

View attachment 325391

Why do you doubt your solution? Do you see anything wrong with it?
You have the chat solutions. Do you see anything wrong with them? Why don't you post their solutions here for us to see and help you figure out who is right and who is wrong?

Note: If you must post photos in the future, please make sure they are right side up.

 

[Mohmmad Maaitah]

Why do you doubt your solution? Do you see anything wrong with it?
You have the chat solutions. Do you see anything wrong with them? Why don't you post their solutions here for us to see and help you figure out who is right and who is wrong?

Note: If you must post photos in the future, please make sure they are right side up.

Now chat GPT gives me same answer I got (now I feel confidant with it which is d = 28m)
Chat GPT now:
To determine the minimum stopping distance for the truck such that the crate will not slide on the truck, we need to calculate the maximum deceleration that the truck can have without the crate sliding. The maximum deceleration occurs when the friction force between the crate and the truck is equal to the maximum static friction force, which is given by:

f_static = μ_s * N

where μ_s is the coefficient of static friction, and N is the normal force between the crate and the truck. Since the crate is not sliding on the truck, the friction force is equal to the force required to decelerate the crate and the truck, which is given by:

f_friction = m * a

where m is the mass of the crate, and a is the deceleration of the truck.

Equating these two forces and solving for a, we get:

a = f_static / m = μ_s * N / m

The normal force N is equal to the weight of the crate, which is given by:

N = m * g

where g is the acceleration due to gravity.

Substituting for N in the expression for a, we get:

a = μ_s * g

The stopping distance s is given by:

s = (v_f^2 - v_i^2) / (2 * a)

where v_f is the final velocity (zero in this case), and v_i is the initial velocity (15 m/s).

Substituting for a, we get:

s = v_i^2 / (2 * μ_s * g)

Plugging in the given values, we get:

s = (15 m/s)^2 / (2 * 0.40 * 9.81 m/s^2) = 28.7 meters

Therefore, the minimum stopping distance for the truck such that the crate will not slide on the truck is 28.7 meters.
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Bing chat gave me this:

I doubted my solution because I thought AI will do it better and after trying more than a problem it seems that AI gives different answers in different times so it's not reliable to give you final answer that I need because my textbook only provide final answer for odd numbered problems (physics for scientistic and engineers 1992 edition)

 

[haruspex]

Fascinating. Bing chat turned 2x9.8x0.4 into the mess 29.80.40 and thence into 40, or thereabouts.

 

[kuruman]

The maximum deceleration occurs when the friction force between the crate and the truck is equal to the maximum static friction force, which is given by:

f_static = μ_s * N

where μ_s is the coefficient of static friction, and N is the normal force between the crate and the truck.

Thank you for posting the replies that you got from the bots. Despite giving the correct numerical answer, ChatGPT is wrong and misleading. The segment quoted above is absolutely correct. However the statements immediately following are not. (The text in bold red is mine.)

Since the crate is not sliding on the truck, the friction force is equal to the force required to decelerate the crate and the truck, which is given by:

f_friction = m * a

where m is the mass of the crate, and a is the deceleration of the truck.

Here is why. It is true that the crate and the truck move as one and therefore have the same acceleration. It is true that the force of static friction between the crate and the truck decelerates the crate. It is not true that the same force of static friction decelerates the truck as ChatGPT wants us to believe.

If static friction were responsible for the truck's deceleration, then there would be no slipping and the truck would stop instantly without moving any distance before stopping. That is not what happens. The truck decelerates because of kinetic friction at the brakes and between the tires and the road. The net force decelerating the truck is $~F_{\text{truck}}=m_{\text{truck}}a~$ where $a=-3.92~$m/s². Since we don't know the mass of the truck, we cannot find a number for that force.

 

[Merlin3189]

I think you are being a bit hard on the poor old (new?) bot. They give a correct formula that enables a correct solution. Their slip of the pen in their incidental remarks doesn't really impact the result, provided you read the "where m= ... " attached to the formula. Would we nit-pick a human solution that gave a correct solution using correct relevant equations, but made such a remark?
If it were someone whose first language was not english, we might just assume they meant to say, something like"... friction force is equal to the force required to decelerate it at the rate of crate and truck ...", since they followed it up with the correct maths - the touchstone for PF.

 

[TonyStewart]

On good road and tires, the μ_s=0.8 for trucks. For this load μs=0.4, so we can assume good tires and road will not limit load slip and d = (v²)/(2gμ)

The net force decelerating the truck is $~F_{\text{truck}}=m_{\text{truck}}a~$ where $a=-3.92~$m/s². Since we don't know the mass of the truck, we cannot find a number for that force.

I disagree, the truck mass does not count, rather it's static friction must be greater. ( and have good tires and road surface)

The μ_s=0.8 for good trucks. = good tires + road will not limit load slip

d = (v²)/(2*g*μ_s) appears to be the correct formula
d= 28.7 m (you are correct with 0.1 error due to round off 9.81 to 2 sig figs.
so 29 m is better for you using2 sigfigs.

ChatGPT seems to find false results with compound division frequently , I found.

e.g. 2x error
a = 0.40 x 9.8 = 3.92 m/s^2
d = (15^2)/(2 x 3.92) ≈ 57.4 m (this has calc. error)

e.g. confused between start and stop velocity.
Deceleration = Final velocity^2 / (2 * stopping distance)

Since the truck comes to a stop, the final velocity is 0 m/s. Substituting this into the equation, we get:

Deceleration = 0^2 / (2 * stopping distance) = 0

Now, equating the frictional force with the force required to decelerate the truck, we get:

μ * m * g = 0

Solving for stopping distance, we get:

stopping distance = μ * m * g / 0

Note that dividing by 0 is undefined, so there is no stopping distance required for the truck to bring the crate to a stop without sliding on the truck. This means that the crate will start sliding on the truck as soon as the truck begins to decelerate, regardless of the stopping distance.


ChatGPT3.5 is not perfect and does not use a calculator, so trust but verify!
Nor am I. ( it was programmed by humans ya know)

 

[kuruman]

I disagree, the truck mass does not count, rather it's static friction must be greater. ( and have good tires and road surface)

I don't understand your objection. Which of the two statements do you disagree with and why?

The acceleration of the truck is the same as that of the crate and equal to -3.92 m/s ².
The net force on the truck is equal to its mass (unknown) multiplied by its acceleration (known).

 

[TonyStewart]

I don't understand your objection. Which of the two statements do you disagree with and why?

The acceleration of the truck is the same as that of the crate and equal to -3.92 m/s ².
The net force on the truck is equal to its mass (unknown) multiplied by its acceleration (known).

Pls allow me to re-quote the statement
"Since we don't know the mass of the truck, we cannot find a number for that force. "
We can make an assumption that the truck can brake fast enough to cause slippage on the load.
But if the road is icy then we cannot guarantee the tires will slip before the load. I looked up the typical static friction = 80% g for a good truck on good tires and road. Thus, if the pallet load was equal to the weight of the truck ( like a Betz foil blade ) then the acceleration force must be reduced 50% then distance doubles and then the tires could slip or the load.

But in reality, one must consider worst case conditions with road vibration increasing random acceleration motion into its lower dynamic % friction of gravity. Thus a 2x stopping distance safety factor is wise but depends on actual vibration g levels in the horizontal direction. For ice the safety factor required is 10x and I have experienced 50x.

 

[kuruman]

Pls allow me to re-quote the statement
"Since we don't know the mass of the truck, we cannot find a number for that force. "
We can make an assumption that the truck can brake fast enough to cause slippage on the load.
But if the road is icy then we cannot guarantee the tires will slip before the load. I looked up the typical static friction = 80% g for a good truck on good tires and road. Thus, if the pallet load was equal to the weight of the truck ( like a Betz foil blade ) then the acceleration force must be reduced 50% then distance doubles and then the tires could slip or the load.

But in reality, one must consider worst case conditions with road vibration increasing random acceleration motion into its lower dynamic % friction of gravity. Thus a 2x stopping distance safety factor is wise but depends on actual vibration g levels in the horizontal direction. For ice the safety factor required is 10x and I have experienced 50x.

You did not answer what I asked in post #8. Instead, you changed the physical situation to something different from the original. So be it.

 

[TonyStewart]

You did not answer what I asked in post #8. Instead, you changed the physical situation to something different from the original. So be it.

You are correct because I did not disagree with either statement. Neither mass needs to be known for the reasons given. Did you want to correct your feedback? or change my assumptions? or say something which is not entirely favorable.

 

[haruspex]

I disagree, the truck mass does not count, rather it's static friction must be greater. ( and have good tires and road surface)

You appear to have misunderstood @kuruman's point in post #5.

 

[TonyStewart]

You appear to have misunderstood @kuruman's point in post #5.

Thankyou for giving me the opportunity to review #5 and my answer in #9 and clarification in #11.

Let me attempt to clarify your misunderstanding. It is correct that the deceleration of the truck is not solely due to the force of static friction between the crate and the truck, but rather due to a combination of kinetic friction between the brakes and the wheels and the friction between the tires and the road.

Since I assumed a good truck in a nominal condition and road, the truck will not slip but rather both the truck and the pallet will experience the same acceleration with the truck masses to include the load but NEVER SLIP since the truck has twice the traction as the load. So my answer and the explanation are not in conflict with the correct answer and #5. I hope you now understand and might accept my answer.

 

[haruspex]

Thankyou for giving me the opportunity to review #5 and my answer in #9 and clarification in #11.

Let me attempt to clarify your misunderstanding. It is correct that the deceleration of the truck is not solely due to the force of static friction between the crate and the truck, but rather due to a combination of kinetic friction between the brakes and the wheels and the friction between the tires and the road.

Since I assumed a good truck in a nominal condition and road, the truck will not slip but rather both the truck and the pallet will experience the same acceleration with the truck masses to include the load but NEVER SLIP since the truck has twice the traction as the load. So my answer and the explanation are not in conflict with the correct answer and #5. I hope you now understand and might accept my answer.

I suggested you had misunderstood because you wrote that you disagreed with something in post #5. If I understand the above, you no longer disagree with post #5.
If so, let that be an end to this digression.

 

[TonyStewart]

The only part I disagreed with is that you do not need to know the mass of the truck to answer the question. I only had to look up what is a "nominal" truck coefficient of friction and that it had to be greater than the "break point" for constant de-acceleration braking and minimum stop distance.

or the skid inflection point of the "skid". ;)
I hope did not sound too much like an annoying ChatGPT bot in a previous comment ;( it is rather supercilious.

 

[haruspex]

The only part I disagreed with is that you do not need to know the mass of the truck to answer the question. I only had to look up what is a "nominal" truck coefficient of friction and that it had to be greater than the "break point" for constant de-acceleration braking and minimum stop distance.

But that is all irrelevant.
We are tasked to "determine the minimum stopping distance for the truck such that the crate will not slide on the truck." If there is some combination of road/truck/weather/braking pattern that allows the truck to stop in distance x without the box sliding then the answer is $\leq x$.

 

[TonyStewart]

That was my point too. The truck mass was irrelevant.

 

[haruspex]

That was my point too. The truck mass was irrelevant.

But you argued that that is only because, in practice, the truck tyres' coefficient of friction is adequate. That is wrong on two counts:
1. Even if the coefficient were much lower, the masses would still be irrelevant. They cancel out in the equations.
2. The coefficient is not relevant either, for the reason I gave in post #16.

I really do not like continuing to pollute this thread with an irrelevant discussion, but I cannot let your misleading comments go unchallenged. If you wish to continue, please do so by messaging me directly.

 

[TonyStewart]

I understand that the masses cancel out in those equations, but the coefficient of friction is still relevant as it determines the maximum force that can be applied before sliding occurs, which is the root of this question.

If the truck's coefficient of friction is not adequate, then the question is unanswerable. the load will not slide.

In addition, I also point that the coefficient of static friction between the truck's tires and the road is not directly relevant to the static coefficient of friction between the box and the truck bed. But it must be greater to solve the problem.

I started this objection saying the truck's mass does not need to be known, only that it's coefficient of friction is greater.
I hope this clarifies my perspective on the matter

There were criticisms of this fact, yet unfounded. I do not appreciate your forum comments of pollution supported with poor logic.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

The OP has moved on several days ago, so the current debate is not helpful. Thread will remain closed.

If anybody has a problem with the thread closure, please send me a PM. Thanks.