Statics - Truck's max load in center of mass

[Femme_physics]

Homework Statement

During the weight-in of a truck it's been determined that the load of the weight acting on the front axis and back axis are 18 [kN] and 12 [kN] correspondingly, when the truck is empty as shown in the diagram.
The max load allowed on every axis is 40.Calculate the size of the max load that's allowed to load on the truck, presuming it's centered in the center of gravity of the truck. Xc.

(The first question was to find Xc, I found it correctly)

http://img717.imageshack.us/img717/9959/truckmy.jpg

The Attempt at a Solution

Attached

 

[gneill]

The total weight of the unloaded truck is (12 + 18)kN = 30 kN. This weight, if taken to be concentrated at the center of gravity of the truck, ends up distributed to the axles as shown. It follows that any additional weight added at the center of gravity would be distributed likewise.

Suppose weight w is added at the center of gravity. How does it divide between the axles?

 

[Filip Larsen]

Firstly, for calculating the position of CG you can just use one of the two axles as reference point so you don't have to "invent" distances on the truck that is not given. For instance, using the front axle as reference you'd get x_c = (5m*12kN + 0m*18kN)/(12kN+18kN) = 2m.

Secondly, if you place a load L in CG, the total weight is then L+30 and you can write up two equation that via torques around either axle calculate the load on the other axle and set that equal to 40 and solve for L. The equations with the smallest L wins.

 

[Femme_physics]

The total weight of the unloaded truck is (12 + 18)kN = 30 kN. This weight, if taken to be concentrated at the center of gravity of the truck, ends up distributed to the axles as shown. It follows that any additional weight added at the center of gravity would be distributed likewise.

I see.

Suppose weight w is added at the center of gravity. How does it divide between the axles?

In terms of percentages:

12/30
18/30

Firstly, for calculating the position of CG you can just use one of the two axles as reference point so you don't have to "invent" distances on the truck that is not given. For instance, using the front axle as reference you'd get xc = (5m*12kN + 0m*18kN)/(12kN+18kN) = 2m.

Dooly noted :)

Secondly, if you place a load L in CG, the total weight is then L+30 and you can write up two equation that via torques around either axle calculate the load on the other axle and set that equal to 40 and solve for L. The equations with the smallest L wins.

I see what you mean. Let me try that...

 

[Femme_physics]

Got it! Clearly it's the LOWEST weight that's the correct result, because that's when the first axle reaches 40. Thank you! :)