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I can't seem to understand in which direction static friction faces for inclined plane motion with rolling motion. This considers rolling motion without slipping, how do i find the direction of the static frictional force?

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The way i ended thinking about it is as follows: I convinced myself friction is forcing opposing motion. Now imagining a ball rolling down an incline, I considered the point of contact of the ball with the incline, call it x. This point intends to move in a direction that is opposed to the general translational motion of the ball as a whole (to be more precise, you have to consider infinitesimally small change in time from the moment x is in contact with the incline).

Hence the force of friction is opposed in direction to x's motion thus, (finally) the force of static friction is parallel to the general direction of translational motion of the ball.

Note that it is "static friction" because otherwise x would remained in contact with the incline as the ball translates downward, which we refer to as "slipping".

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  • $\begingroup$ This is the same way I've been trying to think of it, parallel is abit ambiguous but I think what you mean is that static friction is in the same direction as the translational motion. I seem to understand it thinking of that point x thank you. @Patzerook $\endgroup$ – Teyash Arjun Jan 9 '16 at 20:37
  • $\begingroup$ This is incorrect (please correct me of I misunderstood the text). The direction of the static friction force is indeed parallel to the translational motion (of the centre of mass) but is not necessarily in the same direction as this motion. If a wheel rolls up a hill, static friction is upwards along the hill; if the wheel rolls down the hill, static friction is still directed upwards along the hill. Static friction does not depend on direction of velocity. $\endgroup$ – Steeven Jan 10 '16 at 14:43
  • $\begingroup$ I agree. My logic was flawed. I think the way to think about is that static friction acts as to prevent slipping rather then opposing the motion. So down a ramp, it ll be indeed in an opposite direction to the acceleration of COM. Sorry. $\endgroup$ – Patzerook Jan 10 '16 at 17:46
  • $\begingroup$ Yes. Friction tries to prevent motion. That might be a more general statement. If the object is moving, (kinetic) frictino will try to stop it. If the object is standing still, (static) friction will work against forces that try to make it move. $\endgroup$ – Steeven Jan 10 '16 at 19:06
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There seems to arrive much confusion on this topic. I've updated the answer here to give a clearer picture of what goes on.

Consider a star instead of a ball rolling down the incline: enter image description here

For it to roll without slipping, whenever a leg is touching the ground it must stand still (it must not slip or slide). That means that during the time of contact for one leg, all forces must balance since the net force must be zero when there is no acceleration. (During this contact, the situation can therefore be considered a leg / cone / box or another object standing still, that is motionless, so that Newton's first law simply applies, if this helps the intuitive view).

There is a normal force $F_N$ and of course the weight $W=mg$. They pull as shown on the drawing, but those two alone do not cancel out but will result in a net force downwards along the incline.

So, to avoid this acceleration down the incline, a static friction force $f_s$ must be present, and it must be directed upwards along the incline.

We could add more legs to the star. If we add more and more and at some point infinitely many legs, the star becomes a complete circular wheel. Each "leg" (each point on the circle) now touches for an infinitely short period of time.

The diagram drawn above applies for each "leg" (point). The diagram below (from this source) illustrates the idea again. And note here that the velocity direction is not drawn:

enter image description here

The important note! These drawings are independent of the velocity of the ball. It doesn't matter, in which direction the ball rolls - up or down the hill - the drawings will be the same for the star and for the wheel. If the star rolled upwards, I would have drawn the $v_{cm}$ arrow and the $\omega$ arrow oppositely, but normal force and weight would remain the same! So static friction would also have to remain the same to keep avoiding any acceleration downwards.

Static friction has got nothing to do with the direction of rolling.


In addition:

In a case where the inclined surface is not very rough, the point of contact might be slipping. Static friction is experimentally defined as:

$$f_s\leq\mu_s F_N$$

which as shown gives the limit of static friction. That is, the upper limit of static friction that the surface is able to exert. Larger normal force $F_N$ and rougher surfaces of incline and ball $\mu_s$ will increase the maximum static friction possible. If (from the drawings and force diagrams above) the net force from weight $W$ and normal force is so large that the static friction needed to balance it would exceed what is possible from this formula, then the ball will slip and start sliding instead of rolling.

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  • $\begingroup$ So can I think of this as temporarily at an instant of time to be motionless in the middle of the slope, use a force diagram from there and work out everything? $\endgroup$ – Teyash Arjun Jan 9 '16 at 20:25
  • $\begingroup$ I understand what you mean by prevents slipping, where im confused is that if it starts from the bottom of an incline, that just makes me abit more confused $\endgroup$ – Teyash Arjun Jan 9 '16 at 20:26
  • $\begingroup$ Yes, you can think of taking a snapshot of the motion. For this snapshot the same forces still work. $\endgroup$ – Steeven Jan 9 '16 at 20:50
  • $\begingroup$ The direction of motion doesn't matter. If it rolls up or down the incline doesn't matter. The static friction will be there we prevent the wheel from sliding since gravity pulls down. So it will always pull upwards along the incline in such case. $\endgroup$ – Steeven Jan 9 '16 at 20:52
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    $\begingroup$ @Steeven NIce answer. But I got confused as you have stated "to avoid this acceleration down the incline, a static friction force fs must be present, and it must be directed upwards along the incline." Does that mean that the net acceleration must be zero for a wheel rolling without slipping down the inclined plane? $\endgroup$ – suiz Apr 30 '18 at 6:46

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