- #1

Ashferico

- 9

- 0

- Homework Statement
- A truck weighing 80kN is supported at its two axles. The truck also carries a load on a lifting arm of 45kN. It has stabilizer legs at the rear of the vehicle. What is the required force of the linear actuators that are driving the stabilizer leg to jack the rear tyres so they are not in contact with the ground?

- Relevant Equations
- Resultant Torque = Moment of Intertia I x Angular Acceleration

Moment of Inertia I = Mass x (Radius of Gyration)^2

Parallel Axis Theorem: I = I(cm) + md^2

Angular Displacement = 0.5 x Angular Acceleration x (time^2)

Resultant Torque = Sum of Clockwise Moments - Sum of Anticlockwise Moments

Moment = Force x Perpendicular Distance from Pivot

Hi,

I previously posted about the statically indeterminate truck problem. Thank you to everyone who helped me. However, I now realized that isn't the problem I need to solve. I need to know the force of the linear actuators to lift the rear tyres off the ground.

Since the tyres will be lifting off the ground, it can't be in static equilibrium.

If it rotating about the front tyres, then there must be a resultant torque:

Resultant Torque = Moment of Intertia I x Angular Acceleration

If I make the assumption that rear tyres are raised 1cm from the ground:

The angular displacement is 0.15 degrees.

If I make the assumption this happens over 5 seconds:

Angular Displacement = 0.5 x Angular Acceleration x (t^2)

Angular Acceleration = 0.012°/s^2 =0.0002 radians/s^2

Now if we use:

Resultant Torque = Moment of Intertia I x Angular Acceleration

&

Resultant Torque = Sum of Clockwise Moments - Sum of Anticlockwise Moments

Moment of Intertia I x (-0.0002) = (80 x 0.46) + (45 x 1.85) - (Rcy x 4.8)

If I calculated the Moment of Inertia of the entire truck and box combined, then I can calculate Rcy.

Is this Rcy, the linear actuator force required by the stabilizer leg to jack the vehicle?

Likewise, how can I calculate the Moment of Inertia of the truck about the front wheel's tyre?

I previously posted about the statically indeterminate truck problem. Thank you to everyone who helped me. However, I now realized that isn't the problem I need to solve. I need to know the force of the linear actuators to lift the rear tyres off the ground.

Since the tyres will be lifting off the ground, it can't be in static equilibrium.

If it rotating about the front tyres, then there must be a resultant torque:

Resultant Torque = Moment of Intertia I x Angular Acceleration

If I make the assumption that rear tyres are raised 1cm from the ground:

The angular displacement is 0.15 degrees.

If I make the assumption this happens over 5 seconds:

Angular Displacement = 0.5 x Angular Acceleration x (t^2)

Angular Acceleration = 0.012°/s^2 =0.0002 radians/s^2

Now if we use:

Resultant Torque = Moment of Intertia I x Angular Acceleration

&

Resultant Torque = Sum of Clockwise Moments - Sum of Anticlockwise Moments

Moment of Intertia I x (-0.0002) = (80 x 0.46) + (45 x 1.85) - (Rcy x 4.8)

If I calculated the Moment of Inertia of the entire truck and box combined, then I can calculate Rcy.

Is this Rcy, the linear actuator force required by the stabilizer leg to jack the vehicle?

Likewise, how can I calculate the Moment of Inertia of the truck about the front wheel's tyre?