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Suppose there is a ball on an inclined plane. So, there are three forces acting on it: normal force from the plane, gravity (which can be decomposed in the component of gravity perpendicular to the plane and the component of gravity parallel to the plane), and the friction force. Normal and gravity acts on the ball's CG; friction acts on the contact point between the ball and the plane.

I'm wondering: if friction force is strong enough to counteract the component of gravity force parallel to the plane, will the ball even start to roll/slide?

Well, since the net force acting on the ball is zero, I think the ball would not roll nor slide down. It won't be totally still, though. There is the torque from the friction force, which would make the ball skid in its place. Does that make any sense?

To me it is the same case of placing a spool on an inclined plane while holding its string. As the gravity force tries to push the spool downwards the plane, I pull the string to counteract the gravity force. The spool stays on its place, just rotating (on this case, I mimic the friction force by pulling the spool string upwards).

Even if my analysis is correct, I think no practical surface would have a friction coefficient high enough to keep a ball from rolling and/or sliding.

Best regards

ri_ri
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  • 3 forces - gravity, friction and normal. That they are vectors (and have components) means that you could write it with 6 components (9 components!?!), but there are still only 3 forces. – user121330 Sep 15 '14 at 19:55
  • High friction between ball and plane means rolling, low friction means sliding. If the plane is tipped, the ball moves unless you take into account imperfections in either the ball or the plane. – user121330 Sep 15 '14 at 19:58
  • Couple of corrections: the friction force acts along the plane of the surface, not upwards, and you can't be sure the net force on the ball is zero since the friction force and the reaction force are initially unknown. In fact there will be a net force on the ball since it will either roll (if friction is high) or slip (if friction is low). To work out the friction force and the reaction force you must solve the system of equations for centre of mass motion and rotation. – MartinG Sep 15 '14 at 20:03
  • Thanks, user121330 and MartinG. I'm gonna fix the mistakes you pointed in my description. – ri_ri Sep 15 '14 at 20:08
  • Yes, @MartinG, I've watched the following video where the ball acceleration (and friction force) are worked out in terms of the ball's mass, radius, moment of inertia and plane inclination: https://www.youtube.com/watch?v=7abmkvIpfE8 – ri_ri Sep 15 '14 at 20:19
  • @MartinG: what bothers me is WHY can't we know in advance what value the friction force is? Just as a block laying on a horizontal plane: if I apply 100N to it, the surface will apply 100N of static friction in the opposite direction, preventing the box from moving. In the "ball on inclined plane" problem, suppose the parallel component of the gravity is 100N. I thought the surface would respond immediately by applying 100N of friction on the ball, in the opposite direction. Why can't we know the friction force in advance? – ri_ri Sep 15 '14 at 20:26
  • The "contact point" of a real compressible ball with a real compressible surface is a tiny indentation. Even a frictionless compressible ball would require a minimal plane angle, before it could start rolling. A good real life system to show that is a train car on tracks. – CuriousOne Sep 15 '14 at 20:29
  • @MartinG: Do you mean even if I have pre-defined values for all variables: ball mass, radius, moment of inertia, plane inclination and friction coefficient between plane and ball, I'm not able to predict the friction force and if the ball is going to roll and/or slip? – ri_ri Sep 15 '14 at 20:44
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    @ri_ri: that's the difference between statics and dynamics. In general $F=ma$. When you can be sure everything's at rest (statics), $a=0$ and therefore $F=0$, where $F$ is the net force. But when things are in motion (dynamics) the forces don't in general balance and you have to solve for acceleration as well (as well as torque and angular acceleration, which satisfy analogous relations). – MartinG Sep 15 '14 at 20:49
  • @ri_ri (in answer to your last question): you can't be sure it'll roll or slip, but you must be open to that possibility. If the object in question is not round it may have a stable equilibrium, and you can test for that possibility by postulating a static solution. If the equations are solvable under this assumption that's the solution, otherwise you must discard the assumption and bring acceleration and angular acceleration into the equations. – MartinG Sep 15 '14 at 20:53
  • @MartinG: if (in real life) I put a ball to move on a inclined plane, always giving the same initial conditions, it will always describe the same movement (rolling and/or slipping), so I don't understand why I can't predict what kind of movement it will describe. It is deterministic, right? Did you mean there are more advanced methods to describe its movement (through integration maybe)? Thank you for helping me to understand this :) – ri_ri Sep 15 '14 at 21:18
  • Yes, it's deterministic, but you still have to work it out. No integration necessary - just write down all the equations and solve them. – MartinG Sep 15 '14 at 21:20
  • @MartinG: Yes, I know about the equations, but solving the equations starts from the assumption that I have to choose if there will be rolling with slipping or without slipping. That is something I need to decide beforehand. If I decide there will be pure rolling, I make the constraint (velocity of center of mass = linear velocity of a particle in the disc rim). I would like to know how to calculate the ball movement (and friction force) without having to apply this constraint. I would like to discover if the ball would roll with or without slipping given the initial variables – ri_ri Sep 15 '14 at 21:38
  • @ri_ri You don't get to choose about the slipping or rolling. If the static friction is greater than the coplanar component of the gravity, the ball rolls, and if not, it slips. – user121330 Sep 15 '14 at 21:43
  • @user121330: but, if I understood correctly, I can't know the static friction without choosing and applying a slip/no-slip constraint to the system of equations first. I can't see what I'm missing here. – ri_ri Sep 15 '14 at 21:46
  • The ball doesn't accelerate in the normal direction. Static friction (insofar as we calculate it in freshmen physics) is a number multiplied by the normal force. – user121330 Sep 15 '14 at 21:55
  • @ri_ri: sometimes you have to guess. If you guess wrong you will either get no solution or no physically meaningful solution. In that case you revise your guess. – MartinG Sep 15 '14 at 22:05
  • @MartinG, There are definitely times to guess, but I'm not sure this is one... – user121330 Sep 15 '14 at 22:12
  • For example, you might start by guessing that the system is stationary, and revise your guess (and the coefficient of friction) when you discover that it isn't. – MartinG Sep 15 '14 at 22:45
  • @MartinG: Static friction appears in an inequality, not an equation, right? – user121330 Sep 15 '14 at 22:48

1 Answers1

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Can a ball stay still while laying on a inclined plane?

In freshmen physics, the inclined plane and ball are perfect and the ball moves, so for your purposes, no.

If either the surface or the ball have imperfections, we can tip the plane and the ball won't move until gravity exceeds the sum of the normal forces. To imagine those normal forces, we look very closely at the interface between the ball and the surface to see that the ball rests on high (3+ non-colinear) points on the surface. So long as the high points (of the interface) surrounding these three points are similarly high, the ball rolls when the gravity vector points outside that triangle.

I'm wondering: if friction force is strong enough to counteract the component of gravity force parallel to the plane, will the ball even start to roll/slide?

The ball is either moving or the imperfections in the ball and plane are keeping the static condition outlined above. In the latter case, the normal force is stopping the ball, not the friction force. Imagine a bicycle with one wheel on a flat stair and another on a higher or lower flat stair - will the bicycle accelerate?

Well, since the net force acting on the ball is zero, I think the ball would not roll nor slide down. It won't be totally still, though. There is the torque from the friction force, which would make the ball skid in its place. Does that make any sense?

It sounds like you want the ball to start spinning without translating. No, that doesn't make any sense.

To me it is the same case of placing a spool on an inclined plane while holding its string.

Holding the string gives you a no slip condition on one side and either static (no movement) or kinetic (it translates and rotates) friction on the other.

As the gravity force tries to push the spool downwards the plane, I pull the string to counteract the gravity force. The spool stays on its place, just rotating (on this case, I mimic the friction force by pulling the spool string upwards).

If you pull on the string, you're doing something more complicated and un-sustainable as your arm isn't as long as the string.

Even if my analysis is correct, I think no practical surface would have a friction coefficient high enough to keep a ball from rolling and/or sliding.

Your analysis is not correct. Velcro and glue do a nice job of arresting movement.

To solve problems like this, check to see if the ball rolls by using the static friction inequality and then use either a no-slip condition or a kinetic friction force that appears in both the the sum of the forces and the sum of the torques. One might also choose to use the energy equation for the no-slip case.

user121330
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  • Such a detailed answer. Thank you very much. So, I drew what I think is an example of the imperfections on the plane you are talking about: – ri_ri Sep 17 '14 at 00:10
  • It is here. The green triangle is the imperfection in the plane, formed by three non-colinear points. – ri_ri Sep 17 '14 at 00:17
  • If I understood correctly, the more you incline the plane, the bigger the imperfection should be, in order to the gravity force to point inside it and the ball stay at rest, right? – ri_ri Sep 17 '14 at 00:20
  • Also, I didn't get what you meant by "So long as the high points (of the interface) surrounding these three points are similarly high". Where are these high points surrounding the three points which forms the green triangle on my image, for instance, and what they are for? – ri_ri Sep 17 '14 at 00:22
  • @ri_ri, it would be difficult to draw the interface imperfections as there is a ball in the way and our computers display in 2 dimensions, so while your green triangle could account for one of the points (and the plane the other), the third point is necessarily deeper or shallower than the screen's plane. – user121330 Sep 17 '14 at 14:42
  • @ri_ri if the imperfections further from the interface are larger, the ball could simply move to a new equilibrium. Instead of picturing the interface, simply balance the ball in as few fingers as possible. The interface is doing the same thing on a smaller scale. If you put a fourth finger higher than the others in the direction that you tip your hand, the ball may roll to that finger and find a new equilibrium. – user121330 Sep 17 '14 at 14:45
  • now I think I understand what you mean. I drew another diagram: it is here. This diagram shows the ball at a lateral view, and at a view from below the surface (coordinate systems are denoted for clarity). The imperfection is the blue triangle. So, as long as the inclination is so that the gravity points downwards inside the blue triangle, the ball is at rest. But, tilt the plane a bit more, and the gravity will point inside the green triangle, formed by the green point and two blue points. – ri_ri Sep 17 '14 at 16:16
  • If the green point is higher than the other two red points, the ball is at rest again. Otherwise, it will roll. As the plane gets more inclined, I presume that the area of the triangle should be large enough for the gravity to point inside it (it would not be an "imperfection" anymore, as it would considerably change the shape of the plane). Can you please confirm all that? – ri_ri Sep 17 '14 at 16:21
  • sorry, on my penultimate comment, I meant "two red points", not "blue points". – ri_ri Sep 17 '14 at 16:24
  • Sounds close enough. The center of gravity will be in the triangle if the ball is stationary. I'm less comfortable saying that the ball will remain at rest if the green point is higher, but there's a similar equilibrium condition there along with the energy from moving from the initial position to the new possible equilibrium. One thing to note: planes are perfectly flat constructions whereas real surfaces all have imperfections (similarly for spheres and balls). – user121330 Sep 17 '14 at 16:41