I Dynamics of systems of material points

  • I
  • Thread starter Thread starter Hak
  • Start date Start date
  • Tags Tags
    Forces Systems
AI Thread Summary
The discussion focuses on the complexities of calculating resultant moments and work in systems of material points. When two forces act on a system, the resultant force can be zero while the resultant moment can still be non-zero, highlighting the importance of reference points in moment calculations. The total work done by forces in a system is derived by summing individual works, which can yield a non-zero result even when the net force is zero. This is because work is calculated based on the displacement of the particles affected by the forces. Understanding these dynamics is crucial for accurately applying the fundamental laws of material point dynamics to systems.
Hak
Messages
709
Reaction score
56
I have difficulty understanding the extension of the fundamental laws of material point dynamics to systems.

Example 1:
Consider a system consisting of two material points. Suppose that the two forces acting on the two constitute a pair of forces of nonzero arm. The resultant of the forces acting on the system is zero. The resultant moment is not! Why is it that to calculate the resultant moment we add up the moments of the two forces calculated separately, thus obtaining a different result than if we calculated the moment of the resultant of the forces (which would result in a null moment)?

Example 2:
Same system, but this time the pair of forces has zero arm. Assuming that due to the effect of the two (constant) forces, the two points move by a stretch ##s##, why is the total work of the two forces derived by summing the work of the individual forces, resulting in ##2Fs##, instead of calculating the work of the resultant of the forces (which would give work equal to ##0##)?
 
Physics news on Phys.org
Hak said:
Why is it that to calculate the resultant moment we add up the moments of the two forces calculated separately, thus obtaining a different result than if we calculated the moment of the resultant of the forces (which would result in a null moment)?
Because we want the resultant moment to represent the rate of change of angular momentum, which can be non-zero, even if the net force is zero.

Hak said:
why is the total work of the two forces derived by summing the work of the individual forces, resulting in ##2Fs##, instead of calculating the work of the resultant of the forces (which would give work equal to ##0##)?
Because it takes a non-zero amount of energy to do this, so your approach would not make sense.
 
Hak said:
Consider a system consisting of two material points. Suppose that the two forces acting on the two constitute a pair of forces of nonzero arm. The resultant of the forces acting on the system is zero. The resultant moment is not! Why is it that to calculate the resultant moment we add up the moments of the two forces calculated separately, thus obtaining a different result than if we calculated the moment of the resultant of the forces (which would result in a null moment)?
A force by itself does not have a moment. Not even a zero one. In order to compute a moment you need the force and a reference point with respect to which to compute the moment. For a single particle it would be relatively natural to pick the particle itself as the reference point - thereby obtaining zero moment for any force with a line of action through that particle.

Now here is the thing: In order to add moments you must compute them relative to the same point, so once you picked a reference point you need to stick with it for all forces. Hence, in your two particle case, at least one of the forces will provide non-zero moment.

Note: The total moment generally depends on the point of reference. However, this is not the case if the net force is zero as in your case.

Hak said:
Same system, but this time the pair of forces has zero arm. Assuming that due to the effect of the two (constant) forces, the two points move by a stretch s, why is the total work of the two forces derived by summing the work of the individual forces, resulting in 2Fs, instead of calculating the work of the resultant of the forces (which would give work equal to 0)?
Because the work done by a force is the force multiplied by the displacement of what it is acting on - the particles in this case. The forces are opposite but so are the displacements of the particles. Hence 2Fs.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top