# Why is normal force at the bottom dependent on normal force on top? 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 $$mg$$?

• 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. Apr 9 '19 at 16:20
• @march but $N_y$ cancels, correct? Apr 9 '19 at 16:34
• 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. Apr 9 '19 at 16:36
• @march oh, I get it. $ma_y \neq 0$ because there is a change in velocity. Apr 9 '19 at 16:43
• 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. Apr 9 '19 at 18:21

It is not equal to $$mg$$, because the body is not traversing in a straight line. Here it is circular motion, and normal reaction is always directed towards center, so here normal reaction is acting as centripetal force.

But here $$mg$$ is also acting, and for upper and lower semicircular tracks, it has different directions. So here we have to consider net force acting as centripetal force.

Bottom scenario:

$$mg$$ is acting downwards, and $$N$$ upwards. Net force must be upwards which acts as centripetal force.

$$N-mg=\frac{mv^2}{R_{track}}$$ $$N=mg+\frac{mv^2}{R_{track}}$$

Top scenario:

$$mg,\,N$$ both act downward. Net force must be downward which act as centripetal force.

$$N+mg=\frac{mv^2}{R_{track}}$$ $$N=\frac{mv^2}{R_{track}}-mg$$

Note*: the centripetal force must act always towards center. It is the force responsible for keeping the object bound to circular motion or any curvilinear motion

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)

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.

You need to know that a motion can be accelerated in two independent ways or any combination of these; either when there is a change in velocity, or when there is a change in direction of motion.

The above problem has a constant velocity but the direction of car at every point of the loop changes, which mean the motion is accelerated. Newton's laws tell us that for a motion to be accelerated, a net force must act on the body.

Enters centripetal force:

If $$mg$$ was the only normal, it would mean there is no force to keep the car in circular motion and it would continue in a straight line which is not the case, so there must exist some mysterious force that is keeping the car from going straight, which is indeed, the centripetal force.

Imagine this: from the car's frame of reference, it would be exerting more than just its weight on the loop; if you were inside, you'd be pushed to the floor of the car.

This is why you have another force in play. I hope you can take over from here.

P.S. you intuition isn't wrong but it would apply when the car has not to move in a loop in the next lap, meaning, if it leaves the circular motion at that very precise moment when it reaches the bottom. Then, there would be no force required to keep it in circular motion and thus only mg would act in it in vertical direction.

The connection is that the centripetal force that keeps the car going in a circle is the same at the top of the track as at the bottom (because the car is travelling at a constant speed). If this centripetal force is $$F$$ then at the top of the track we have

$$F = N_{top} + mg$$

where $$N_{top}$$ is the normal force at the top of the track. At the bottom of the track we have

$$F = N_{bot} - mg$$

where $$N_{bot}$$ is the normal force at the bottom of the track. We can elimiate $$F$$ from these two equations to get

$$N_{top}+mg = N_{bot}-mg \\ \Rightarrow N_{bot} = N_{top}+2mg$$