Center of gravity question.?

Question

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4m high and 2.4 m wide and its CG is 2.2m above the ground how steep a slope can the truck be parked on without tipping?

Answer 1

This is a moment or torque, and equilibrium problem. Draw a fairly big diagram of a rectangle representing the truck parked on an inclined parking lot. If the center of gravity is located in the middle of its width, then it's at 1.2m from either side of the truck , and as given in the problem is 2.2m from the ground. Now if the angle of incline from the horizontal is theta, then the component of the weight of the truck parallel to the incline is wsintheta, and the component normal to the incline is wcostheta.

Now take moments about the left corner of the truck. Let's call that point A. The moment about A: wsintheta*2.2=wcostheta*1.2. Divide both sides by wcostheta*2.2:

sintheta/costheta=1.2/2.2
=0.54
tantheta=0.54

What's the corresponding angle? Go to your trig table to get that. Or compute the reciprocal of 0.54 and you get 1.85, meaning the slope of the incline is 1:1.85.

Answer 2

On small airplanes, gasoline is saved in tanks close to the plane's midsection of gravity. As gasoline is burned, little replace contained in the midsection of gravity takes position, in spite of the indisputable fact that it does ensue. plane trim can atone for that small replace. On more effective plane, trim will preserve the small differences alongside with autopilot. more effective differences in gasoline intake are managed by technique of move pumps to flow gasoline between the tanks to save the CG interior of limits or by technique of eating gasoline from quite some tanks in certain sequences and lengths of time..

Answer 3

It is parked sideways or up/down the hill?

28.6 Degrees for sideways.