2
$\begingroup$

Is it possible to vectorially add and find the resultant of several forces acting on different points of an extended body? For example, if I apply a couple (equal and opposite forces) to the two ends of a rigid rod, can I claim that the net force acting on the rod is zero? All I know form vector addition that vectors can be added which act at the same point/particle.

EDIT: I think, we cannot add forces on a rigid body acting at different points. If we could, then the net force on the rod would be zero, and assuming Newton's 2nd law is applicable, then it says the acceleration of the rod it zero. But the rod rotates and therefore, various points on the rod do accelerate. One might say that the centre-of-mass doesn't accelerate. However, if we apply Newton's second law directly to a rod, one has no clue whether the acceleration $\textbf{a}$ sitting on the RHS of $\textbf{F}=m\textbf{a}$ is really that of the centre-of-mass. And to arrive at the concept of centre-of-mass one really have to use Newton's third law independent of Newton's second law (as it is done for a system of N-particles).

$\endgroup$
  • $\begingroup$ I don't know when you posted your EDIT, but I hope the answers that have been given convince you that you can add the forces. The $\vec{a}$ on the right hand side is the acceleration of the center of mass. This is shown in the development of newton's laws for extended bodies. $\endgroup$ – garyp Sep 26 '16 at 19:13
  • $\begingroup$ The center of mass does not accelerate when the net force is zero, but the body can rotate about the center of mass when the net torque is non zero. $\endgroup$ – ja72 Sep 26 '16 at 20:00
  • 1
    $\begingroup$ Possible duplicate of Torque in a non-inertial frame $\endgroup$ – ja72 Sep 26 '16 at 20:03
  • $\begingroup$ "However, if we apply Newton's second law directly to a rod, one has no clue whether the acceleration $a$ sitting on the RHS of $F=ma$ is really that of the centre-of-mass" Actually, $a$ means precisely the acceleration of the center of mass. $\endgroup$ – DanielSank Sep 27 '16 at 6:59
  • $\begingroup$ I'm voting to close this question as off-topic because it makes a critical assumption which is incorrect. $\endgroup$ – DanielSank Sep 27 '16 at 7:00
3
$\begingroup$

Newton's second law, when applied to point particles, states that there will be no motion if the sum of applied forces equals zero. Since a rigid body is composed of infinitely many points particles, there will be no motion if and only if the sum of applied forces on each and every point particle the body is composed of is zero.

In the context of your question, saying the resultant force on your rod is zero would be pretty misleading. Mathematically, if you sum those vectors the sum will be zero, but you are considering that you can move your vectors freely and nothing changes. Since you now have a set of particles, each of them acts differently, so you cannot detach those force vectors from its associated points. Indeed, if you apply equal and opposite forces to opposite ends of a rod, even though summing them would give you zero, you will create a torque that will rotate the rigid body around its centre of mass. The total force applied to the system is zero, but this doesn't mean internal forces play no role, because the system is composed of smaller systems that are not isolated (namely, the particles).

$\endgroup$
1
$\begingroup$

 if I apply a couple (equal and opposite forces) to the two ends of a rigid rod, can I claim that the net force acting on the rod is zero?

Yes you can. You are actually asking here, if Newton's laws still work, and yes they do:

$$\sum \vec F=0 \\ \sum \vec F=m\vec a$$

The rod experiences equal but opposite forces that balance out, so in accordance with Newton's 1st law it doesn't accelerate.

Acceleration $\vec a$ is the key thing here. Because, no it doesn't accelerate, but that doesn't mean that it doesn't move! It can still rotate on the spot.

Let's have a look at the rotational equivalents of Newton's laws:

$$\sum \vec \tau=0 \\ \sum \vec \tau =m\vec \alpha$$

If net force is zero (no matter where the forces attack), the object doesn't accelerate, $\vec a=0$, but the net torque must be zero in order for it to not rotate, $\vec \alpha =0$.

So, to be completely still, both $\sum \vec F=0$ and $\sum \vec \tau=0$ must be the case. But if you are only concerned with forces, then you just chose not to care if there is rotation.

$\endgroup$
1
$\begingroup$

See answer in Torque in a non-inertial frame.

Follow the rules of motion:

  1. Sum of forces equals mass times acceleration of the center of mass: $$ \sum_i \vec{F}_i = m \vec{a}_{cm} $$
  2. Sum of torques about the center of mass equals change in angular momentum: $$ \sum_i (M_i + \vec{r}_i \times \vec{F}_i) = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$ where $\vec{r}_i$ is the relative location of force $\vec{F}_i$ to the center of mass.

The key here is that the net forces only describe the motion of the center of mass and not of the entire body. In your case a pure torque will allow the body to rotate about the center of mass.

The rules from the equations of motion are:

  1. If the net torque about the center of mass is zero then the body will purely translate
  2. If the sum of the forces on a body are zero (but not the net torque) then the body will purely rotate about its center of mass.
$\endgroup$
0
$\begingroup$

The rod's center of gravity will not experience any change in momentum if two equal forces are applies simultaneously to different (but opposite) points on the rod.

$\endgroup$
0
$\begingroup$

Answer to your question is -Yes, we can vary well add and find the resultant of several forces acting on different points of an extended body. But this is a complex process let me explain, but before that i want to come at your Question - in this particular question the resultant will work at infinity and you know if an almost zero force is working at infinity then by applying mathematical limit we get finite torque.

Now come to how i know that i.e. how to add - let first take 2 forces in a plane, BY PRINCIPLE OF TRANSMISSIBILITY of forces we know we can move point of application of a force on any point on its line of action. Now these 2 forces have line of action which will meet some where at any point if they are not parallel, in that case addition of two force is nothing but resultant of these 2 forces (as calculated by parallelogram law's mathematical formula) which will act at point where line of action of these 2 meet.

For 3 force, we first find out resultant of 2 (as i explained in 2nd para) then it is problem of 2 forces.
Special cases - (a) if 2 forces P and Q are like parallel which work at A and B then their resultant is parallel to them working at point C on line joining AB where (AC)P=(CB)Q [internal division of AB in ratio P:Q] (a) if 2 forces P and Q are unlike parallel which work at A and B then their resultant is parallel to them working at point C on extension of line joining AB where (AC)P=(CB)Q [i.e. external division of AB in ratio P:Q]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.