Prove the average external force is Zero

Click For Summary
The discussion centers around proving that the average external force on a system of particles, which starts and ends at rest, is zero. Participants express confusion about defining average force and how to incorporate both internal and external forces in their calculations. The key insight is that internal forces do not affect the motion of the system's center of mass due to Newton's third law, leading to the conclusion that the total change in momentum must also be zero. This implies that the average external force over the period is indeed zero. Understanding the role of internal forces is crucial in solving the problem effectively.
Manolisjam
Messages
49
Reaction score
0
1.Problem statement
Prove the average external force of a system of particles N starting from rest and ending at rest is zero.

Homework Equations


If the system moves periodically prove the av. external force is zero in a period

The Attempt at a Solution


I don't quite understand what i am asked .Since i don't have a definition for a average force of a system of particles. Yet i tried to derive the eq for momentum of a system.So this is the momentum of the system i found. Now i know that going from rest to rest change of momentum must be zero . But i have internal and external forces. So don't know how to proceed
12.jpg
 

Attachments

  • 12.jpg
    12.jpg
    14.9 KB · Views: 385
Last edited by a moderator:
Physics news on Phys.org
There is someething with internal forces you need to exploit here. has to do with one of the Newton laws...:
what is the effect of an internal force on the motion of the system ?
 
BvU said:
There is someething with internal forces you need to exploit here. has to do with one of the Newton laws...:
what is the effect of an internal force on the motion of the system ?
Its zero the rest of the math is ok? SO the double sum is zero.
 
Can't read the small writing. What I hinted at is action = - reaction so the internal forces can't influence the center of mass of the system
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 2 ·
Replies
2
Views
827
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 25 ·
Replies
25
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
3K