Stability Triangle of Forklift

[Red_CCF]

Hi

I came across this:

http://www.free-training.com/osha/forklift/Physics/54.htm

Here I don't understand why the forklift will remain stable only when the CG is inside the triangle. If the CG is outside the triangle but still within the square formed by the four wheels of the forklift why would it tip over?

Also, it was mentioned in this that tight turning causes tipovers, why is this the case? I read in another forklift training (can't find the link) that the centrifugal force throws the CG outside the stability triangle but I don't really understands this.

Thank you

 

[tiny-tim]

Hi Red_CCF!

If the CG is outside the triangle but still within the square formed by the four wheels of the forklift why would it tip over?

The two steering wheels are there just to confuse you: the load rests on a pivot pin halfway between them.

From the good ol' United States Department of Labor at http://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=STANDARDS&p_id=9829" …

A-4. The Stability Triangle.

A-4.1. Almost all counterbalanced powered industrial trucks have a three-point suspension system, that is, the vehicle is supported at three points. This is true even if the vehicle has four wheels. The truck's steer axle is attached to the truck by a pivot pin in the axle's center.​

Also, it was mentioned in this that tight turning causes tipovers, why is this the case? I read in another forklift training (can't find the link) that the centrifugal force throws the CG outside the stability triangle but I don't really understands this.

Essentially, in the non-inertial frame of the driver, the centrifugal force is a horizontal extra force of gravity …

you add it to the usual down-to-earth vertical gravity, and that gives you a total effective gravity at an angle …

(you can do the same thing with linear acceleration, to find the angle at whch your furry dice will hang! )

the weight of the load acts along a line through the centre of mass parallel to the effective gravity, and if that line falls outside the stability triangle, the truck tips over.

(if you're turning sharp left, the centrifugal force is to the right, so the total effective gravity is down and to the right, which is why the truck tips over to the right )

 

[jack action]

Also, it was mentioned in this that tight turning causes tipovers, why is this the case? I read in another forklift training (can't find the link) that the centrifugal force throws the CG outside the stability triangle but I don't really understands this.

From http://hpwizard.com/car-performance.html" :

Rollover limit

Because the friction force acts at ground level and the inertia force acts at the center of gravity, there will be a weight transfer from the inside tires to the outside tires. Once the vertical force acting on the inside tires is equal to 0, then the inside tires are off the ground, leaving the total weight of the vehicle on the outside tires. This is the rollover limit. The maximum load that can be transferred from the inside tires will be half of the total vertical force.

The maximum lateral acceleration is:

$$a_{L_{max}} = 0.5 \frac{T}{h}$$

 

[Red_CCF]

Thanks for your help!

Hi Red_CCF!

A-4. The Stability Triangle.

A-4.1. Almost all counterbalanced powered industrial trucks have a three-point suspension system, that is, the vehicle is supported at three points. This is true even if the vehicle has four wheels. The truck's steer axle is attached to the truck by a pivot pin in the axle's center.​

So if the CG shifts outside the stability triangle but still inside the four wheels, would it tip over or just lean so badly on one side that it cannot operate?

Because the friction force acts at ground level and the inertia force acts at the center of gravity, there will be a weight transfer from the inside tires to the outside tires. Once the vertical force acting on the inside tires is equal to 0, then the inside tires are off the ground, leaving the total weight of the vehicle on the outside tires. This is the rollover limit. The maximum load that can be transferred from the inside tires will be half of the total vertical force.

aLmax=0.5Th

With regards to the inertia force, is it same as inertial/centrifugal force in a non-inertial frame? If that's the case, I don't understand how the friction/centripetal force can co-exist with the inertia force when viewed from a specific frame of reference?

Thank you very much

 

[Quinzio]

Hi

I came across this:

http://www.free-training.com/osha/forklift/Physics/54.htm

Here I don't understand why the forklift will remain stable only when the CG is inside the triangle. If the CG is outside the triangle but still within the square formed by the four wheels of the forklift why would it tip over?

Also, it was mentioned in this that tight turning causes tipovers, why is this the case? I read in another forklift training (can't find the link) that the centrifugal force throws the CG outside the stability triangle but I don't really understands this.

Thank you

Ok, but words have a meaning.
The way the problem is explained is confusing and poor.
CG is a fixed point in the reference frame of the lifter. It depends on the masses distribution. Once the masses are fixed, the CG is fixed.
IF the CG projection along a vector parallel to gravity falls out of the wheels square, the lifter topples.
IF the 3 points suspension system makes the lifter roll (on which axis ?) then the CG point moves and may fall out of the wheels square.
IF the lifter is accelerating, then addition torques are trying to make the lifter topple.
The torque basically depends upon the height of the load and the acceleration itself. That's why fork drivers always bring the load to ground level when they move.

 

[tiny-tim]

Hi Red_CCF!

So if the CG shifts outside the stability triangle but still inside the four wheels, would it tip over or just lean so badly on one side that it cannot operate?

Tip.

Weight is a force. Every force has a line of application. Weight's line of application goes through the centre of mass (CG).

If that line passes inside the stability triangle, there is no problem.

But if it passes outside the stability triangle, the torque of the weight will make the body of the forklift rotate … that makes the line even further outside, which makes the torque more, and the body will keep going until it hits the chassis.

At that point, the question is whether the CG is now outside the wheelbase … if it is, the whole forklift (body and chassis together) will tip.

With regards to the inertia force, is it same as inertial/centrifugal force in a non-inertial frame? If that's the case, I don't understand how the friction/centripetal force can co-exist with the inertia force when viewed from a specific frame of reference?

If the forklift isn't changing speed or direction, then the line of application of the weight is vertical.

If the forklift has acceleration a forward, then from the point of view of the driver (ie in his non-inertial frame of reference), there is an extra "fictitious gravity" …

this is horizontal, equal to a backward.

If the forklift is turning with a turning circle of radius r at speed v, again there is an extra "fictitious gravity" …

this is horizontal, equal to v²/r sideways and outward.

(multiply it by m, and you have the centrifugal force mv²/r)

To find the line of application of the weight, you simply add all the gravity together, ie g down, a backward, and v²/r sideways …

if that line falls outside the stability triangle, the forklift tips.

(and, for a particular speed or turn, as you raise the CG, the angle of the line stays the same, but since it now starts higher, it gets closer to the edge of the stability triangle)

(btw, "inertial" means proportional to the mass, so both real gravity and "fictitious gravity" produce inertial forces)