0
$\begingroup$

enter image description here

Why does the normal force on bottom of the track have anything to do with the normal force on top of the track? Why isn't the normal force at the bottom simply $mg$?

$\endgroup$
  • 3
    $\begingroup$ If the normal force at the bottom was exactly $mg$, then the normal force would cancel the gravitational force, and the net force on the car would be zero. This would mean that the car is traveling at constant velocity at this point, but it's clearly not, because it's changing its direction by moving in a circle. $\endgroup$ – march Apr 9 at 16:20
  • $\begingroup$ @march but $N_y$ cancels, correct? $\endgroup$ – user644361 Apr 9 at 16:34
  • 2
    $\begingroup$ No. I just explained why that intuition is wrong (this is a common misconception among students first taking an intro physics class!). $N_y > mg$ and is pointed upwards so that there can be a net force upward leading to the car moving in a circle. At the top, $N_y$ is pointed downward and helps the gravitational force move the car in a circle. $\endgroup$ – march Apr 9 at 16:36
  • $\begingroup$ @march oh, I get it. $ma_y \neq 0$ because there is a change in velocity. $\endgroup$ – user644361 Apr 9 at 16:43
  • 1
    $\begingroup$ To be specific, you use Newton's 2nd Law (which involves the net force and the instantaneous acceleration) along with knowledge about the situation (e.g. that it's moving in a circle) to infer/derive what the forces should be. Your last sentence about $V_y$ is correct. $\endgroup$ – march Apr 9 at 18:21
0
$\begingroup$

Normal force is a special force. According to the situation, it can be greater or lesser than $mg$. More precisely, if an object $A$ has mass $m$, and it's exerting $F$ force on the surface normal on it, then $F$ is also the normal force experienced by the object (Newton's III Law)

So, if the object is freely falling in a lift, it is not in contact of anything, and hence normal force on it is zero.

If the object is going around a vertical loop, it has to push the ground hard enough to get enough centripetal acceleration.

In your example, it is indeed a vertical loop-the-loop problem. For the car to go in a circular motion, the force should be radially inwards. At the bottom, only the track can provide such a force (which means the car would have to really press against the track).

At the top however, gravity is going to push the car downwards towards the centre, so the car can escape at sufficiently high speeds without even touching the track. (i.e. $F_N = 0$ is possible at the top; gravity would maintain the centripetal acceleration)

$\endgroup$
0
$\begingroup$

This really is a centripetal acceleration equation, compounded by gravity. You know when you get to a1g acceleration, the normal force at the bottom is double it's resting Force and would be a normal force of zero at the top. To get to point six at the top you would have to add 75% more centripetal acceleration, and I believe you would have a normal force of 2.2 kg at the bottom. If my math is correct your velocity would have to be 9.3 meters per second.

$\endgroup$

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.