1
$\begingroup$

When an object falls down in a liquid it has two forces working on it one gravity and another is viscous drag . Now when these forces are same the object attains the terminal velocity in liquid. Now if these forces have been neutralized then why shouldn't the viscous drag (which is proportional to instantaneous velocity of the ball so it isn't zero even after attaining the terminal velocity ) slow the ball down or accelerate it another way or upwards?

$\endgroup$
  • 1
    $\begingroup$ Didn't you just say that these forces have been neutralized? Why is the viscous drag considered then? $\endgroup$ – Steeven Aug 22 '18 at 4:58
  • $\begingroup$ @Steeven my concern is that the viscous drag is proportional to instantaneous velocity. Then it should work even after attaining the terminal velocity as terminal velocity isn't zero. $\endgroup$ – Nobody recognizeable Aug 22 '18 at 5:18
  • 3
    $\begingroup$ Sure. But gravity also still works after reaching terminal velocity. $\endgroup$ – Steeven Aug 22 '18 at 5:58
3
$\begingroup$

Because gravity is still acting on the object. When the body is moving at terminal velocity, it's not experiencing no forces. It's just not accelerating because the vector sum of all forces on it is zero.

In the case of this object, we can say that $F_{gravity}=mg$, and since $F_{total}=0$, we must have $F_{viscous}=F_{down}=mg$. Now the total force on the object is zero, so it doesn't accelerate (or decelerate) anymore.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ my concern is that the viscous drag is proportional to instantaneous velocity. Then it should work even after attaining the terminal velocity as terminal velocity isn't zero. $\endgroup$ – Nobody recognizeable Aug 22 '18 at 5:18
  • $\begingroup$ @Nobodyrecognizeable Yes, you still have viscous drag; however you also have gravity, and they cancel. $\endgroup$ – Allure Aug 22 '18 at 5:19
  • $\begingroup$ as you've said the forces have already neutralized. I think the gravitational force is constant. But the viscous drag is accumulative as your velocity increases in the first part of falling. Then it is neutralized by viscous drag. But now have just viscous drag as you've neutralized gravitational force. Where am I wrong ? $\endgroup$ – Nobody recognizeable Aug 22 '18 at 5:24
  • $\begingroup$ It is not a cumulative force. At any point in time the weight is always there, with the the drag that is proportional to velocity squared. $\endgroup$ – QuIcKmAtHs Aug 22 '18 at 5:40
  • $\begingroup$ @Nobodyrecognizeable the viscous force is not accumulative. At any point in time there's a single drag force, it's just that the drag force is changing. When the body is moving slowly, the viscous force is small while the gravitational force is large, which makes the body accelerate. If the body is forced to move faster than terminal velocity, then the viscous force is larger than gravitational force and the body slows down. $\endgroup$ – Allure Aug 22 '18 at 5:42
2
$\begingroup$

When an object has achieved terminal velocity, it experiences no acceleration. From this, we can deduce that $F_{net}=0$.

Well, what is $F_{net}$? It is the net force on a object; after all forces have cancelled out, what we are left with is $F_{net}$.

In our case, $F_{net}=F_{drag}-F_{weight}=0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ my concern is that the viscous drag is proportional to instantaneous velocity. Then it should work even after attaining the terminal velocity as terminal velocity isn't zero $\endgroup$ – Nobody recognizeable Aug 22 '18 at 5:18
  • $\begingroup$ At terminal velocity, drag is equal to weight. $\endgroup$ – QuIcKmAtHs Aug 22 '18 at 5:43
-1
$\begingroup$

there is more of a theoretical explanation available for this question than a mathematical one. forces do not have an impact on the velocity of an object, rather they have the impact on rate of change of velocity of an object(acceleration). establishing this, we can say that velocity is more about the nature of that object in our universe. this can be very well understood by considering newton's first law of motion.

"If an object experiences no external forces, it will either stay in rest provided it was already not moving, else it will move with a constant velocity provided it was moving before." hope it was helpful

| cite | improve this answer | |
$\endgroup$

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.