21st November 2006 - 03:49 AM
My question is:
A)What is this equilibrium position? Is it the position when the body stops deforming and comes to rest?
B)Can I say, that, even in a static problem from the instant the body starts deforming/displacing to the instant the body stops deforming, it will be accelerating?
However, this acceleration is very small in magnitude and hence could be neglected?
Also,in a static problem,from the instant the body starts deforming/displacing to the instant the body stops deforming/displacing it will have the effect of inertia.
Is this ok?
Equilibrium is the position where the net force is zero. If the object is PLACED at rest at the equilibrium position, it will remain there until it is disturbed from outside. If the object is not placed at rest at the equilbrium position, it will oscillate around the equilibrium position. The equilibrium position will be the center of the oscillation. In that case, the positions where the body comes to rest will be at the extremes of the vibration, not at the equilibrium position. At these extreme positions, the velocity will be zero just for an instant, so the body does come to rest just for an instant, but it will not stay at these extreme points because it will have an acceleration toward the equilibrium position. At the equilibrium position, it will have a velocity (as it passes through the center position of its vibration) but NO ACCELERATION, because the net force is zero at the equilibrium position.
The short answer is: The equilibrium position is the location where the force and acceleration are zero, NOT where the velocity is zero. Velocity may or may not be zero at the equilibrium position. If it is zero, the object will remain there at rest (Newton's first law), and if it is not zero, the object will vibrate around the equilibrium position.
Everything you say in this part is correct. When the object is moving from its initial static position to its final static position, it will in general be accelerating and inertia will have an effect. To this extent, it is now a dynamic problem rather than a static one. However, a static problem does not study this motion, it studies ONLY the final position. The acceleration may or may not be very small, but that is not the main reason why it is ignored. The reason is that a static problem only studies the final position after the motion has stopped, so the acceleration is no longer relevant. For a dynamic problem, you observe the motion and try to relate force to motion and deformation. For a static problem, you close your eyes and wait until all the motion stops. Then you open your eyes and look at the final position, and pretend that the object was always there; in other words, you ignore any changes that led to the current configuration and only ask about the equilibrium of forces and torques in the current configuration.
Hope this helps!