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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

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  • $\begingroup$ 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. $\endgroup$ – user121330 Sep 15 '14 at 19:55
  • $\begingroup$ 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. $\endgroup$ – user121330 Sep 15 '14 at 19:58
  • $\begingroup$ 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. $\endgroup$ – MartinG Sep 15 '14 at 20:03
  • $\begingroup$ Thanks, user121330 and MartinG. I'm gonna fix the mistakes you pointed in my description. $\endgroup$ – ri_ri Sep 15 '14 at 20:08
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    $\begingroup$ @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). $\endgroup$ – MartinG Sep 15 '14 at 20:49
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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.

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  • $\begingroup$ 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: $\endgroup$ – ri_ri Sep 17 '14 at 0:10
  • $\begingroup$ It is here. The green triangle is the imperfection in the plane, formed by three non-colinear points. $\endgroup$ – ri_ri Sep 17 '14 at 0:17
  • $\begingroup$ 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? $\endgroup$ – ri_ri Sep 17 '14 at 0:20
  • $\begingroup$ 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? $\endgroup$ – ri_ri Sep 17 '14 at 0:22
  • $\begingroup$ @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. $\endgroup$ – user121330 Sep 17 '14 at 14:42

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