Will a Particle Remain in Equilibrium if Three Reversed Forces are Removed?

  • Thread starter Thread starter Final_HB
  • Start date Start date
  • Tags Tags
    Forces Particle
AI Thread Summary
A particle influenced by five forces remains in equilibrium even when three reversed forces are removed. The remaining two forces must balance each other to maintain equilibrium. If the three forces cancel each other out, their removal does not affect the state of the particle. This can be mathematically expressed by showing that the sum of the forces equals zero in both scenarios. Therefore, the particle will continue to be in equilibrium without the three forces.
Final_HB
Messages
35
Reaction score
0

Homework Statement


A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether.

The Attempt at a Solution


My thinking for this is:
With three of the forces gone, the 2 forces (F1 and F2 ) left must balance the system if its to remain at equilibrium.
If change of forces occurs, the system will react by moving to a new state of equilibrium.
No reaction means that the 3 forces in the question cancel out each other, and no shift occurs.
If the three forces balance to 0, there is no need to have them as acting on the particle.

Right?

If I am right, Is there any maths way to say this, or a better way to explain what I am trying to say. Thank you in advance.
 
Physics news on Phys.org
Final_HB said:

Homework Statement


A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether.

The Attempt at a Solution


My thinking for this is:
With three of the forces gone, the 2 forces (F1 and F2 ) left must balance the system if its to remain at equilibrium.
If change of forces occurs, the system will react by moving to a new state of equilibrium.
No reaction means that the 3 forces in the question cancel out each other, and no shift occurs.
If the three forces balance to 0, there is no need to have them as acting on the particle.

Right?

If I am right, Is there any maths way to say this, or a better way to explain what I am trying to say. Thank you in advance.

You are right, but it can be formulated mathematically. Write up the sum of forces for both cases, and manipulate the equations.

ehild
 
Well...

F1+F2+F3+F4+F5=0

And for the second:
F1+F2-F3-F4-F5=0

taking one away from the other gives us:
2F1+2F2=0
So,
F1+F2=0
 
Final_HB said:
Well...

F1+F2+F3+F4+F5=0

And for the second:
F1+F2-F3-F4-F5=0

taking one away from the other gives us:
2F1+2F2=0
So,
F1+F2=0

If you mean subtract " by taking one away from the other" the result is 2(F3+F4+F5)=0, so the resultant of these three forces is zero, they can be omitted. And then F1+F2=0, too.

ehild
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top