WDH and airbag setup Question
So here is my setup: 2010 F250 Diesel 2WD CC Tow vehicle 11.5K trailer fully loaded 1200 to 1250 Pound Tongue weigh once loaded Hensley Arrow hitch 1400 pound Springbars Truck equiped with air...
Forum Discussion
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
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
