# What force causes the ball to move up?

Let's consider this super-simple example. The ball (mass $m$) is moving from position 1 to position 2 at constant horizontal velocity $v$ until the ball reaches the red arrow (in position 1, it has vertical velocity $0$). Gravity force is equal to $mg$.

When the ball reaches the point marked with the red arrow, it starts moving up, therefore its vertical speed is not $0$ anymore.

Its vertical speed has changed = there was vertical acceleration, so there had to be a non-zero net force in the vertical direction that acted on the ball, right? This is exactly what Newton's first law says.

My question is - what force caused the ball to accelerate vertically so it could reach point 2? What source caused that additional force that allowed it to move upward, up the hill?

In position 1, the gravity force acting on the ball is cancelled out by the ground reaction force, so the net force is zero. In position 2, the net force isn't zero anymore, directed 'down' (because gravity force is greater than ground reaction force acting on the ball). So we have two cases - in both the net vertical force is either zero or it is directed down. But it had to be directed up, because the ball moved up!

I know it can be explained with conservation of energy law, but I think there has to be an explanation in terms of forces and momentum only as well (the laws of dynamics are always true).

• Have you thought through what happens when you run your experiment in reverse? What must the ground reaction force be at "2" for the acceleration to be parallel to the slope? Jan 2, 2016 at 9:17
• @EricTowers The direction doesn't matter (ball moving from 1 to 2 and reverse - forces are the same in both cases). I've answered this in the first comment in dmckee's answer. Jan 2, 2016 at 9:32
• The slope is constant at 2, so the normal force will be $mg \cos\alpha$ (normal force is always perpendicular to the ground surface) Jan 2, 2016 at 10:03

Floris has given you a correct answer but he hasn't addressed the misconception that is preventing you from seeing it as correct.

In the question you write

In position 2, [...] (because gravity force is greater than ground reaction force acting on the ball).

the normal force (the force exerted on the ball by surface it's situated on) is 'at most' equal to $mg$ - its weight (it's obviously smaller when the surface isn't parallel to the ground)

but these statements are wrong.

The key is understanding the normal (or solid reaction) force better.

In the simplest first-course understanding the normal force has two properties

1. perpendicular to the region of contact and
2. takes on whatever magnitude is required to keep the objects from inter-pentrating.

You seem to be very clear on (1) but have misunderstood (2).

So, what happens when the ball reaches the point where the slope begins? Well, the normal force begins to point a little bit backward instead of completely up due to property (1), but it also grows to have larger magnitude because of property (2) and the fact that the only way for the ball to continue forward without interpenetration is to go upward as well.

The result then is a net force on the ball that is upward and backward, causing the ball to rise and slow.

Some interesting question to ask yourself are:

• What happens to the normal force if the slope becomes constant instead of continuing to change?

• What happens to the normal force as the ball slows toward a stop at it's maximum height? What value does it have at the maximum height?

• What happens to the normal force as the ball descends the ramp again?

The answers are slightly subtle, which is one of several reasons we don't do changing slopes in the introductory course.

• 1. The normal force becomes $mg \cos\alpha$ ($\alpha$ is the slope angle). 2. Again, $mg \cos\alpha$. 3. It will increase, just like in the case when the ball was going in the opposite direction, up the slope. Technically, is it caused by the centrifugal force? (centrifugal force appears in non-inertial frame reference, and the frame reference here would be a piece of slope with the ball). Jan 1, 2016 at 18:17
• More or less? I mean the 3 questions. Jan 1, 2016 at 18:32
• Note that vertical upward acceleration only takes place in the bend, that's why it didn't show up in my answer. What happens in the bend can be shown by considering it a rotation but not w/o calculus.
– Gert
Jan 1, 2016 at 19:38
• @Gert yes, I'm aware of that in only happens in the bend. What about my answers, do they make any sense? Jan 1, 2016 at 19:52
• @user5539357: from a rotational POV, the normal force provides the centripetal force, while $mg\sin\theta$ provides the decelerating force.
– Gert
Jan 1, 2016 at 20:22

You answered this question yourself. You state: "In position 1, the gravity force acting on the ball is cancelled out by the ground reaction force, so the net force is zero.". The ground reaction force you mentioned keeps the ball from falling through the surface. It can be considered as the repulsion from the electron cloud surrounding the ball by the electron cloud of the surface. As the ball reaches the slope, the direction of this repulsive force changes from perfectly vertical to slightly tilted, causing a horizontal force that will slow the ball down. Due to the velocity of the ball hitting the slope the electron cloud are pushed closer, increasing the upward force (I guess).

• The ground reaction force can't at all be thought of a repulsion force between the electrons. Even if the objects were made of neutral particles (which in fact is the case, at a macroscopic level), there would still be a reaction force. Moreover, it completely disappears when the objects are not in contact. A reaction force is actually a fictitious force, in the sense that it is not produced by elementary interactions (if one doesn't take into account the internal interactions thank to which the the objects are solid), it really is a force that constrains a system to follow some fixed path. Jan 1, 2016 at 17:39
• +1 The electron clouds will distort, and the inter-atomic bonds will compress. Both effects lead to a restoring force that has vertical and horizontal components. Jan 1, 2016 at 17:40
• Other than that, and the fact that electrons don't need to be pushed closer to cause the variation in acceleration, your explanation is correct. Jan 1, 2016 at 17:40
• @GiorgioComitini Reaction force fictitious? What do you mean? Normal forces are forces of constraint, whose magnitude and direction are variable and dependent on the motion, but they are perfectly real. In this case, it is a inter-atomic (or inter-electron) restoring force. Jan 1, 2016 at 17:43
• As I said, they are fictitious in the sense that they are not produced by elementary interactions, but this is not the only property that makes them fictitious. They are, as you said, forces of constraint. Of course, they are completely real: they produce real effects on the dynamics of the system. However, if one is willing to study the system using a formalism different from (but completely equivalent to) Newton's dynamics, one can describe a constrained motion without using reaction forces. This is what is meant by "reaction forces are fictitious forces". Jan 1, 2016 at 17:50

There is a normal force from the surface on the ball. As long as the surface is straight, this force is equal to the force of gravity (the weight of the ball). Once you reach the curve, the normal force increases and the ball is accelerated upwards. This is the same thing that happens when a ball follows a circular track: there is an additional centripetal force that changes the direction of the ball (in the case of a circular track, the change is perpendicular to the velocity and the total speed doesn't change).

In this case, as the velocity changes, there will appear a component of gravity along the surface of the track - this slows the ball down. Your premise that the horizontal component of the velocity of the ball doesn't change ("at constant horizontal velocity $v$") is wrong unless there is another external force to offset the pull of gravity.

• The speed doesn't change until the ball reaches the red arrow (I should have added that). Jan 1, 2016 at 16:52
• Yes you probably should edit that into your question Jan 1, 2016 at 16:53
• When the ball is at rest, the normal force (the force exerted on the ball by surface it's situated on) is 'at most' equal to $mg$ - its weight (it's obviously smaller when the surface isn't parallel to the ground). The interesting question is why the normal force is actually greater in this case. Jan 1, 2016 at 17:05
• For the same reason that when a ball hits a surface it exerts a force greater than its weight - the surface is moving towards the ball. Jan 1, 2016 at 17:44
• To make the ground exert greater force on the ball, the ball itself needs to exert a greater force on the ground, right? And the only such force I can think of would be associated with inertia (centrifugal force acting on the ground). The problem is, centrifugal force appears only in non-inertial reference frame. Jan 1, 2016 at 19:10

You can also look at what happens in the first part of the incline as a rotation:

The ball rotates around the point $O$, with radius $R$.

The centripetal force $F_c$ is provided by the Normal force ($mg\cos\theta$), while $F_t=mg\sin\theta$ provides a decelerating torque, so that:

$$-Rmg\sin\theta=I\frac{d\omega}{dt}$$

$$-Rmg\sin\theta=mR^2\omega \frac{d\omega}{d \theta}$$

$$-R\omega d\omega=g\sin\theta d\theta$$

Note that the ball enters the incline at $v_0$, so $v_0=\omega_0 R$.

Integrating between $0$ and $\theta$ we get:

$$\omega^2-\omega_0^2=\frac{2g}{R}(\cos\theta -1)$$

Re-worked with $\omega=\frac{v}{R}$, we get:

$$v^2=v_0^2-2gR(1-\cos\theta)$$

So $v$ is reduced and the vector now also has a vertical component:

$$v_y=v\sin\theta$$

Once the ball leaves the arc the only net force acting on it is $mg\sin\theta$.

Edit: Centripetal Force

Firstly, $v^2=v_0^2-2gR(1-\cos\theta)$ can be easily verified by multiplying both sides with $\frac{m}{2}$ and reworking:

$$\frac{mv_0^2}{2}-\frac{mv^2}{2}=mgR(1-\cos\theta)$$

This is the energy conservation equation, e.g. for $\theta=\frac{\pi}{2}$ then:

$$\frac{mv_0^2}{2}-\frac{mv^2}{2}=mgR,$$

which is what we expect.

We know that $F_c=\frac{mv^2}{R}$, so that:

$$F_c=\frac{mv_0^2}{R}-2mg(1-\cos\theta)$$

• I think the centripetal force isnt equal to normal force. Instead it is the net force of normal force and gravity force component (normal force minus $\cos mg$). Jan 1, 2016 at 22:50
• @user5539357: in the radial direction $F_N=F_t=mg\cos\theta$. This balance of forces keeps the ball on its circular trajectory. That's the meaning of centripetal force here: a force needed to prevent the ball from penetrating the arc ramp but not larger either, which would make the ball move inwards to the centre-point of the arc (and leave the arc ramp).
– Gert
Jan 1, 2016 at 23:04
• See here: imgur.com/9Hzlir3 Centripetal force is not equal to $F_N$. (you can also look here at solution a) physics.info/centripetal/practice.shtml) If $F_N$ was equal to $Fg'$, then the net force would be $0$ (and we would have no centripetal force). dmckee explained this in his answer. Jan 2, 2016 at 9:27
• @user5539357: it appears a bit more complicated than that because $\omega_0 \neq 0$. See my edit. Thanks.
– Gert
Jan 2, 2016 at 14:40
• – Gert
Sep 8, 2017 at 13:05

If the vertical momentum does not change as the ball pass through this horizon, what happens?

The answer to this question is: the ball would cross the floor.

The key here is that exist a source of force that is the floor. This source is such that the ball would never cross the floor. The origin of this force is the geometric constraint of the particular system. Fundamentally, this force is only a idealization. Is actually a complicated system very close to electrostatic equilibria that resides in the structure of matter of the ball and the floor.

Ok. Let's answer your question now. When the ball are with some horizontal momentum in the flat floor the vertical momentum that this ball receives from the gravitational field is matched with the momentum of the floor necessary for obey the condition the ball can't cross the floor .As the floor start to be non-flat and get some curvature upwards, the setup of the floor's forces start to change in a way that the condition the ball would can't the floor be appreciated.

Let's understand this more closer. If we pick the starting point when the floor start to warp and take a small interval around this point we can approximate this warp to a tiny straight line with different angle. To the above condition be satisfied the momentum of the ball at this point needs to be aligned with the this straight line, and then, the floor need to supply an extra momentum in the vertical such that the net momentum, horizontal plus vertical, is aligned with the tiny warp. This angle is very tiny too in the initial of the warp, but hey increase as we pass to other points across the warp and the same analysis is the same for other points.

The floor do not delivers energy to the ball, so the kinetic energy can't be changed by this process. So, the magnitude of the momentum need to be preserved by this process. But, now, as this particle are getting vertical momentum upward, they start to getting upper. The gravitational field start to reduce his kinetic energy because of that and then, this ball can't getting upper indefinitely. If the gravitational field is not there, the above explanation about warp is yet valid.

Other way to see why the floor do not change the magnitude of the ball's momentum is that, by the warp discussion above, the instantaneous momentum is always perpendicular to the floor's force (the delivery of vertical momentum on the tiny straight wap), only changing the direction of the velocity. You can see that by thinking that at each point you can think the floor's force as a stretched rope tight to the ball fixed at some point.

Conclusion: is the floor that pushes the ball upward by conservation of your shape and not ghost property (the ball can't cross the floor) of this objects.