kajal
According to Newton's second Law:

"If the resultant force acting on a body is non zero the body moves with an acceleration and:

1) The acceleration is directly proportional to the force.
2)The acceleration is inversely proportional to the mass."

Now, let us say, we have a time varying force p(t) applied to a body with a single degree of freedom:

Then according to Newton's second law, at any instant of time "t":

p - k*u = ma

-> p (force at "t" seconds)
-> k*u (u is the displacement at "t" and "k" is the stiffness)
-> a (acceleration at "t" seconds)

"minus" sign above is because the force due t stiffness is opposite to the applied force.Right?

My question is ,

1.)If the external force varies with time can we say that the resultant force (resultant force means external force plus force due to stiffness) will always be non zero?

2.)Can we say that the body will accelerate only if the external force varies with time?

3.)If the external force does not vary with time (as in a static problem) the body will not accelerate and then the above equation will be:

p - k*u = 0 (as in a static problem)

4.)The internal forces in the body (may be axial forces , shears and moments) will not depend on the total displacement because out of the total displacement a part of the displacement will be a rigid body displacement and a part of the total displacenment will be a deformation.Right?

mr_homm
QUOTE
1.)If the external force varies with time can we say that the resultant force (resultant force means external force plus force due to stiffness) will always be non zero?

No. It is possible for it to be zero sometimes. If fact, if the object is vibrating then it MUST be zero at some times. This is because vibrating motion on one degree of freedom must at some time reach a maximum speed (otherwise, the motion could not repeat). At this time, the acceleration is zero, hence the net force must be zero.
QUOTE (->
 QUOTE 1.)If the external force varies with time can we say that the resultant force (resultant force means external force plus force due to stiffness) will always be non zero?

No. It is possible for it to be zero sometimes. If fact, if the object is vibrating then it MUST be zero at some times. This is because vibrating motion on one degree of freedom must at some time reach a maximum speed (otherwise, the motion could not repeat). At this time, the acceleration is zero, hence the net force must be zero.

2.)Can we say that the body will accelerate only if the external force varies with time?

No. A constant external force will shift the equilibrium location to a point where p - k*u = 0. However, the mass may vibrate around this point, and the vibrating motion will have acceleration.
QUOTE

3.)If the external force does not vary with time (as in a static problem) the body will not accelerate and then the above equation will be:

p - k*u = 0 (as in a static problem)

No. You are assuming that the object is placed EXACTLY AT the equilibrium location. In that case, the net force is zero and your statement is correct. However, if the mass is not placed at the equilibrium location, it will vibrate around that location, and then since the vibration involves acceleration, it will NOT be true that p - k*u = 0.
QUOTE (->
 QUOTE 3.)If the external force does not vary with time (as in a static problem) the body will not accelerate and then the above equation will be:p - k*u = 0 (as in a static problem)

No. You are assuming that the object is placed EXACTLY AT the equilibrium location. In that case, the net force is zero and your statement is correct. However, if the mass is not placed at the equilibrium location, it will vibrate around that location, and then since the vibration involves acceleration, it will NOT be true that p - k*u = 0.

4.)The internal forces in the body (may be axial forces , shears and moments) will not depend on the total displacement because out of the total displacement a part of the displacement will be a rigid body displacement and a part of the total displacement will be a deformation.Right?

Yes. Rigid body displacement will not affect or cause internal forces. Only the net external force interacts with rigid body displacement. Deformation is related to internal forces. Basically, external forces affect rigid body displacement and internal forces cause deformation.

Hope this helps!

--Stuart Anderson
kajal
Dear Sir:

Thanks for the response!!
To my questions 2 and 3 posed,

QUOTE
2.)Can we say that the body will accelerate only if the external force varies with time?
3.)If the external force does not vary with time (as in a static problem) the body will not accelerate and then the above equation will be:

p - k*u = 0 (as in a static problem)

You said:

"No. You are assuming that the object is placed EXACTLY AT the equilibrium location. In that case, the net force is zero and your statement is correct. However, if the mass is not placed at the equilibrium location, it will vibrate around that location, and then since the vibration involves acceleration, it will NOT be true that p - k*u = 0."

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?

mr_homm
QUOTE
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?

A)
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.

B)
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!

--Stuart Anderson
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