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I'm studying the motion and forces involved in a ball (bidimensional) rolling down a ramp inclined by an angle $\theta$ from the x-axis.

This is the body diagram (I didn't draw forces but there is only gravitational force acting on the center of mass and friction acting on the lowest point of the ball).

Wheel rolling down a ramp

The ball is not moving at $t = 0$ and starts rolling down due to the $sin(\theta)$ component of the gravitational force.

From what I learned, to let the ball roll, we need a static frictional force acting on the base of the ball, but there is something I really don't get.

I have seen that the lowest point of the ball is always stationary when rolling, so I assume that the gravitational force acting on that particle should be the same as the static frictional force acting on that point.

So, summing up we have for Newton 2nd law that: $F_g \sin(\theta) - F_s = ma$ . But even here things are getting confusing for me. Gravitation force is acting on the center of mass (at least I can simplify it like that) but frictional force is acting only on the lowest point of the ball.

So I think this should be more precise: $dm \ g \sin(\theta) - f_s = ma$ . So only an infinitesimal gravitational force is acting on that point.

But that means that the only actor causing the acceleration is the frictional force, because the gravitational force is negligible . But this is impossible, because the body would accelerate upwards, breaking all rules of physics. What is wrong with my way of thinking it?

Also, even if I consider all of the gravitational force acting on the point, like $F_g \sin(\theta) - f_s = ma $ , if the point is not moving, then the point should not accelerate, but that means that $f_g \sin(\theta) = f_s $, but that is impossible by definition, because frictional force in this case is defined as $f_s = N \cos(\theta) = mg \cos(\theta)$ (unless $\theta = 45°$, but this is not the case).

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  • $\begingroup$ Can you draw the free body diagram? $\endgroup$
    – Eli
    Jul 29, 2022 at 7:34
  • $\begingroup$ @Eli Yep, Done! $\endgroup$
    – TechMatt
    Jul 29, 2022 at 11:20
  • $\begingroup$ Where are the forces ? $\endgroup$
    – Eli
    Jul 29, 2022 at 12:46
  • $\begingroup$ @Eli There is only gravitational force acting at the center of the wheel and static friction acting on the lowest point of the axle, the one making contact with the surface. (the wheel itself is not making contact, only the axle). $\endgroup$
    – TechMatt
    Jul 29, 2022 at 13:38

1 Answer 1

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and starts rolling down due to the $\sin(\theta)$ component of the gravitational force.

I would say that because the total forces on the ball are unbalanced (gravity, ramp normal, friction), the ball accelerates. It's not just due to gravity. I'm sure you know that, but it helps to be explicit.

I have seen that the lowest point of the ball is always stationary when rolling, so I assume that the gravitational force acting on that particle should be the same as the static frictional force acting on that point.

Not necessarily. A stone thrown in the air is stopped for a moment. That doesn't mean that gravity is balanced out by something. In fact it was accelerating the whole time and was only stopped for an infinitesimal point of time. The same is true for the point at the bottom of your rolling ball. It does stop for a moment, but that doesn't mean that the forces on it are in balance.

But we're usually not going to focus on specific parts of the ball. If we did that we'd have to consider internal forces as well (the bottom of the ball has forces on it from the rest of the ball). If you don't consider those, then you're missing things.

  • An object with $v = 0$ does not immediately imply that $a = 0$ and therefore that $F_{\text{net}} = 0$.
  • A tiny part of an object has more than just the external forces acting on it. Internal forces from the solidity of the object are also present. You may ignore those forces when your system is the ball. You may not ignore them when your system is this spot on the ball.

For this type of problem you should focus on the ball as a single (accelerating, translating, rotating) mass.


The forces are acting on two different parts of body.

That's fine. You can note that and keep track. Just don't split the ball into multiple parts.

  • For translation questions ($F=ma$), you don't care where the force acts. Top, middle, bottom, doesn't matter. The acceleration is still in the same direction.
  • For angular momentum, rotation, you do have to keep track. For gravity, we pretend it all acts at the center of mass, so it creates no torque. For friction, it acts at the contact point so has a lever arm equal to the radius.

This lets us ignore the internal forces between the different infinitesimal parts of the ball, but still be able to sum useful forces to analyze the translational and rotational effects.

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  • $\begingroup$ ok but, I still don't get the fact that gravitational force is acting on the center of mass, when friction is acting on the lowest point. I have seen a lot of exercises where they use the Newton 2nd law like I mentioned before but, how can I apply that here? The forces are acting on two different parts of body. $\endgroup$
    – TechMatt
    Jul 29, 2022 at 11:15
  • $\begingroup$ Gravity acts on every point on the ball. In some situations you can pretend that it all acts on the center of mass. But that is a simplification for the math, not what is really happening. For a rigid body translating, you don't care where it acts. The direction and the total force is all that matters. For angular momentum, you do care where it acts. $\endgroup$
    – BowlOfRed
    Jul 29, 2022 at 17:10
  • $\begingroup$ Yep that's true, But The problem is not with gravitational force, it is with friction. I know that gravity acts on every particle of the body, the problem is that in this case friction is not, it's only acting on one point. $\endgroup$
    – TechMatt
    Jul 29, 2022 at 21:31
  • $\begingroup$ Is that a problem? The location doesn't matter for F=ma. Why does it matter for this case? $\endgroup$
    – BowlOfRed
    Jul 29, 2022 at 21:33
  • $\begingroup$ Ok, so, I've always thought of Newton 2nd law in vector notation. So, does that mean that even if x components of the two forces are applied in different positions, I can still treat them as if they are both lying on the x-axis, thus doing any operation on them without caring on their position but just on the fact that they have an x-component? $\endgroup$
    – TechMatt
    Jul 30, 2022 at 16:43

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