WDH and airbag setup Question

Ron Gratz wrote:
Let's say the length of your beam is "WB". Since it's a uniform beam, the distance from the front axle to the CG is WB/2 and the distance from the CG to rear axle is WB/2. With the beam at an angle, "theta", the horizontal distances from front axle to CG and from CG to rear axle are cosine(theta)*WB/2.

To maintain rotational equilibrium, the sum of the moments about any point must equal zero. For your example below, the 25# force at the front generates a CW moment equal to 25*cosine(45 degrees)*WB/2. The 75# force at the rear generates a CW moment equal to 75*cosine(45 degrees)*WB/2. Clearly, your free body cannot be in rotational equilibrium unless the upward force at the front and the upward force at the rear are equal. The end reactions shown in the diagram below must both be equal to 50#.

In general the sum of moments about the CG must give:
Ff*cosine(theta)*D1 - Fr*cosine(theta)*D2 = 0
where Ff and Fr are the vertical upward forces acting at front and rear, D1 is distance from front axle to GC, and D2 is distance from CG to rear axle.
Therefore, we have: Ff*D1 = Fr*D2
And we know: Ff+Fr = W = weight of beam
These two equations give: Ff = W*D2/(D1+D2) and Fr = W*D1/(D1+D2)

The distribution of weight is not dependent on the value of "theta".

When you stand the beam vertically on end, as you did in your first diagram, the front reaction force, the rear reaction force, and the weight force are all collinear. The values of D1, D2, and D1+D2 are all zero. There are no moments acting about the CG. You have a case of unstable equilibrium. Ff and Fr could be any value as long as their sum is equal to the weight. However, in the limiting process of approaching 90 degree slope, we know that the above equations for Ff and Fr apply.

Ron

Following diagram provided by 2500HDee

Ron, you are right. Thank you for the correction. It was the center of gravity's distance from the fulcrum that I was thinking of as you pointed out in the wheelbarrow post. I am going to delete my other posts as if to not leave incorrect information posted.

I am going to run some numbers on the COG of the truck and load in the bed though, as I do think that it is significant in the weight distribution of the truck. I am referencing a specific time that a kid I was roofing with loaded an entire pallet of shingles in an F150 and the front suspension was at full droop. And yes, he did have them loaded over the axle, not behind it.

Also, when setting up race cars we used 4 wheel scales and adjusted the coil overs up or down to balance the load side to side and or front to rear but I realize I am providing personal examples here and not numbers so I will get back with everyone soon.

Thanks for the discussion.