Confused about which forces are external when Newton's Second Law is used

AI Thread Summary
In analyzing the forces acting on a trolley and an oscillating sphere, it's crucial to define the system boundaries to distinguish between external and internal forces. When considering the trolley, string, and sphere as a system, the weights and reaction forces from the ground are external, while the tension in the string is internal. If the system is defined solely as the trolley, the string exerts an external force on it. The choice of system boundary significantly impacts the analysis, as any force crossing this boundary is deemed external. Understanding these dynamics is essential for accurately calculating torque and angular momentum.
curiousPep
Messages
15
Reaction score
1
Thread moved from the technical forums, so no Homework Template is shown.
So I have a trolley of mass m that moves on a straight line.
A sphere of mass m, is attached on the trolley with a light string of length a and it is left to oscillate.

Just to give some idea of their positions:
r_trolley = xi
r_sphere = (x-asinθ)i - acosθj (θ is the angle between the string and the vertical - j axis)

Are the external forces the weight of the trolley, mass and the reaction betweeen the trolley and the rack/ground?
 
Physics news on Phys.org
Welcome!
Is the sphere oscillating under the trolley?
You determine the limits of the system and which forces are external or internal.
If the system to study is trolley-string-sphere, the weights and the reaction forces are external, while the tension of the string is internal.
If the system is only the trolley, then the string pulls externally from it.
 
Lnewqban said:
Welcome!
Is the sphere oscillating under the trolley?
You determine the limits of the system and which forces are external or internal.
If the system to study is trolley-string-sphere, the weights and the reaction forces are external, while the tension of the string is internal.
If the system is only the trolley, then the string pulls externally from it.
Yes the sphere is hanging below the trolley.
I see, I have been trying to calcualte the rate of change of angular momentum (torque) using:
##\dot{\vec{h_{a}} + \dot{\vec{r_{a}}x p = Q,####\Q = sum_{n=i}^\n (r_{i} - r_{p}) x F_{i} ,##
PS:sorry for the equations, if someone can fix it , that wouldb ereally helpful (thank you n advance!)

F_i are the external forces of the system.
In the case that I take p as the combined momentum of the trolley and the sphere, does it mean that the exetrnal forces that I consider are the weights and the reaction force of the trolley?

Thank you!
 
  • Like
Likes Lnewqban
curiousPep said:
In the case that I take p as the combined momentum of the trolley and the sphere, does it mean that the exetrnal forces that I consider are the weights and the reaction force of the trolley?

Thank you!
Weights and reaction force from the ground, yes. The latter may include a horizontal component.
 
  • Like
Likes curiousPep and Lnewqban
We really need a system diagram and a Free Body Diagram (FBD) to see what is happening.
 
  • Like
Likes Lnewqban
curiousPep said:
Are the external forces the weight of the trolley, mass and the reaction betweeen the trolley and the rack/ground?
Which forces are external/internal depends on how you define the bodies for analysis. Which definition of bodies is sensible depends on the specific question being asked.
 
  • Like
Likes jack action, curiousPep, Lnewqban and 1 other person
By considering the trolley, string and mass, the external forces are the weights and the reaction force but the weight and reaction force for the trolley have no effect in the cross product to callculate Q, since r_a - r_p (p is the centre of mass of trolley) is parallel to the direction of F_i (external forces)
 
curiousPep said:
By considering the trolley, string and mass, the external forces are the weights and the reaction force but the weight and reaction force for the trolley have no effect in the cross product to callculate Q, since r_a - r_p (p is the centre of mass of trolley) is parallel to the direction of F_i (external forces)
As I mentioned in post #4, the reaction from the ground may have a horizontal component.
What is ##r_a##, and how are you defining Q?
 
  • Like
Likes curiousPep
To determine which forces are external requires the definition of a system boundary. Often it is useful to include all parts in the system within the boundary, but this is not always true. The choice of system boundary is a matter of experience, but a good choice makes life much simpler.

One the system boundary is defined, then any force that crosses the boundary is an external force, and any force that with its reaction is internal to the system boundary is not an external force. This is where looking for reactions can be very important and the place where d'Alembert's Principle can lead the user astray.
 
Back
Top