Equations of Motion: How Many Independent Equations?

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SUMMARY

The discussion focuses on determining the number of independent equations of motion for a rigid body in equilibrium, specifically in 2D and 3D systems. It establishes that in 3D, there are three degrees of freedom for linear accelerations and two for rotational acceleration, leading to a total of five independent equations. The conversation clarifies that redundancy occurs not merely from parallel torques but from the alignment of axes used in force and torque equations. The importance of ensuring that the axes of torque do not lie parallel to the linear force equations is emphasized.

PREREQUISITES
  • Understanding of rigid body dynamics
  • Knowledge of linear and rotational motion equations
  • Familiarity with torque and force balance concepts
  • Basic grasp of 2D and 3D coordinate systems
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  • Study the principles of rigid body equilibrium in 3D systems
  • Learn about the relationship between torque and force equations
  • Explore examples of independent equations in 2D and 3D dynamics
  • Investigate the implications of axis alignment on redundancy in equations
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Students and professionals in physics, mechanical engineering, and robotics who are analyzing motion dynamics and seeking to understand the formulation of independent equations of motion.

Thejas15101998
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Say, I have a system at rest. I was wondering - how many equations of motion can the system have (without redundancy)? Well, I thought that equating the forces along 2 or 3 different axes would give 3 independent equations. Also equating torques would give some equations, but how many of them (independent) can I formulate? Kindly help me.
 
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Thejas15101998 said:
Say, I have a system at rest. I was wondering - how many equations of motion can the system have (without redundancy)? Well, I thought that equating the forces along 2 or 3 different axes would give 3 independent equations. Also equating torques would give some equations, but how many of them (independent) can I formulate? Kindly help me.
Assuming you mean a rigid body in equilibrium, count the potential accelerations. In 3D, the mass centre has three degrees of freedom for linear accelerations. That leaves rotational acceleration about the mass centre. In 3D, there are two degrees of freedom for the orientation of the net torque.

But it does not have to be three linear force equations and two torque. There are other ways of getting five independent equations in 3D. E.g. in 2D, instead of two linear and one torque you could have one linear and two torque, provided the two torque axes do not lie on a line parallel to the linear force equation.

Edit: I should have written "... do not lie on a line normal to the linear force equation".
 
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haruspex said:
But it does not have to be three linear force equations and two torque. There are other ways of getting five independent equations in 3D. E.g. in 2D, instead of two linear and one torque you could have one linear and two torque, provided the two torque axes do not lie on a line parallel to the linear force equation.

Could you please give me an example? And moreover, If two torque equations (considered from two different points on the plane say in case of 2D) are such that the torques are parallel to each other then does that always imply redundancy or not?
Thank You
 
Thejas15101998 said:
If two torque equations (considered from two different points on the plane say in case of 2D) are such that the torques are parallel to each other then does that always imply redundancy
It's not whether the torques are parallel. In a 2D set-up, torques are all normal to the plane, so are all parallel. But what I wrote before is not correct either.

The issue is whether the vector displacement of the two axes is normal to the direction used for the linear equation. E.g. suppose you consider force balance in the X direction and torque balance about the origin. For the third equation you can use a force balance equation in any direction except parallel to the X axis, or a torque balance about any point not on the Y axis.
To see this, suppose the system of forces sums to a force FX in the X direction, FY in the Y direction, and a torque τo about the origin. The torque about a point at (0,y) is τo+yFX. So if we write equations for forces in the X direction and torque about the origin, an equation for torque about (0,y) woukd be redundant.
 

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