Calculating the energy of a stopping vehicle

[Boowee]

I'm looking to calculate the horizontal force produced on a plate covering a pit from a fork lift truck coming to a studden stop whilst traveling across the plate. The information known is:

Mass of truck: 9800kg
Velocity : 0.75 m/sec

Any ideas?

 

[sophiecentaur]

It depends entirely on the time for which the braking is needed.
Change of Momentum = Impulse = Force times Time
This make sense when you think about how it feels when braking hard and when braking gently.

 

[john.phillip]

Ignoring the rotative inertia of the wheels and drivetrain, the maximum breaking force will be dictated by the maximum friction coefficient.

Asssuming a maximum friction coefficient of mi=1, the maximum possible force would be Bf=weight x mi=96138N

The deceleration will be a = -Bf/mass = 96138/9800 = -9.81m/s2 (not surprising as by definition mi=a/g)

Minimum stopping distance will be d=v^2/(2*a)=0.75^2/(2*9.81)=0.0287m or 28.7mm

If coefficient is as low as 0.4, you can recalculate the maximum expected force based on the above.

If the plate is not restrained horizontally and neglecting the plate self weight, the friction coefficient between the plate and the ground must be higher than the friction between the tires and the plate, for it not to move during breaking.

 

[Boowee]

Many thanks for the info guys. I've been thinking about your comments and our estimated stopping time is around 0.2 seconds and a reasonable friction coefficient would be around 0.2 - 0.3 for a steel plate and rubber tired wheel.

Acceleration would be V/t as there is no final velocity, 0.75/0.2 = 3.75m/s²

I'm sort of thinking out loud and also feeling a little thick on the maximum force formula, would the value Bf be the friction force resisting movement of a non wheeled body. If we taken the force of the decelerating body as mass x acceleration the force would be 9800 x 3.75 = 36750kg, this is produced by the braking mechanism?

Then would the force on the plate be the difference between the two

[(9800 x 9.81) x 0.2] – [(9800 x 9.81) x 3.75] = 341289.9N

Please educate a keen cave man.

 

[john.phillip]

The correct unit is [N]: 9800 x 3.75 = 36750N (3746kgf)

I'm not sure what you are trying here, but this equation is unbalanced: [(9800 x 9.81) x 0.2] – [(9800 x 9.81) x 3.75] = 341289.9N. The first term will give [N] and the second term [N x m/s²], so you cannot add them together because they have different units.

Maybe you were looking for: (weight x mi - m x a) = 9800 x 9.81 x 0.2 - 9800 x 3.75 = -17522.4N. This equation does not give the "net" force on the plate, what it tells is when the friction coefficient is 0.2 the maximum acceleration cannot be 3.75m/s², it must be lower as it is missing a friction force of 17522.4N.

Working with provided time of 0.2s:

Acceleration will be a = v / t = 0.75 / 0.2 = 3.75m/s²

For this acceleration to be feasible the friction coefficient must be mi = a / g = 3.75 / 9.81 = 0.38 which is beyond the expect values of 0.2 and 0.3.

With friction coefficient of 0.38, the maximum breaking force is Bf = weight x mi = 9800 x 9.81 x 0.38 = 36532.4N (3724kgf)

Coming back to the equation (weight x mi - m x a) = 9800 x 9.81 x 0.38 - 9800 x 3.75 = 217.56N. The non zero difference is due to a roundoff error and based on this equation result you can relate friction coefficient 0.38 to acceleration 3.75m/s² as they will balance out.

A wheeled body can generate three different friction coefficients: One static when parking brakes are engaged and the vehicle is not moving; One dynamic when the wheels are blocked and the vehicle is sliding; And another one dynamic when the wheels are turning. For low speeds static friction is higher than the other two, so the maximum possible force transmitted to the ground will happen at the instant the vehicle is about to stop and is not slipping.

A non wheeled body have two friction coefficients, one static and one dynamic, the dynamic friction will occur when the body is sliding and is less than the static friction coefficient. A parked wheeled body and a non wheeled body will have the same static friction when both are in contact with the same ground with the same materials. A sliding wheeled body and a sliding non wheeled one will have the same dynamic friction under the same conditions.

Rotative inertia will not influence the maximum breaking force as it is dictated by the friction coefficient only, but time and stopping distance will be higher to dissipated the energy stored in the rotative mass of the wheels and drivetrain. That is the main difference between wheeled and non wheeled bodies during breaking, the stopping distance is higher for wheels.

The breaking mechanism will produce a greater force in the breaking drums than onto the ground, as the drums are smaller than the tires. The tires will produce a force on the ground that is related with the friction coefficient and the drum force.

 

[Boowee]

John

Many thanks, that makes a lot more sense than my own rambling nonsense.

 

[sophiecentaur]

I imagine you are basically wanting to know just how strong the fixings need to be to stop the plate from shifting if you put the brakes on hard. This, I should think, is information which is used by every 'manhole' designer / supplier when producing such things for normal roadbuilding.
There is almost certain to exist some well established code of practice for such things. You might be best to look at that and buy in whatever fixing is normally used.
Sorry if this covers something you've already done.

 

[john.phillip]

I second sophiecentaur, if you are required to apply a code, stick to the code. I'm considering your question as being more to the curiosity side.

 

[Boowee]

John

Thanks for the interest. To give you a better idea of what I'm looking for, the situation is as follows:

I am supplying avery large rail company with some simple heavy duty cover plates to bridge over some pits, the clients structural consultants are asking questions to which they don't know the answer i.e. if the fork lift truck comes to an abrupt stop, fully loaded on the cover plates; What load is imparted horizontally from the pit board into the concrete.
In an ideal work the answer would be simple; Find out yourself. However we are looking to try an make a good impression on the client and i would also like to know the answer myself.

Not sure if that's of any more assistance.

 

[john.phillip]

As they are structural consultants, either they are trying to figure out what will be the size of the restraints, or they might be questioning the plate thickness you provided. During break, most of the fork lift weight will be directed to the front axle and depending on the pit size, restraints, plate material and thickness, it might bend under the dynamic load.

 

[sophiecentaur]

Still sounds to me that you should be using 'standard practice'. There's bound to be some dusty document somewhere with all the necessary (albeit generous) spec in it. If the consultants are trying to cut costs in some way rather than do the 'normal' thing then I'd not want to be involved.