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Suppose two bodies B1 and B2 are in contact with each other in a static situation. A torque M1 is applied to the first body B2. The torque M1 is transmitted to the second body B1. Both bodies won't move because they hinder each others movement due to the inclined contact surfaces.

Now my first question is, what force is transmitted from B2 to B1 directly originating from M1. More particular, in which direction does the force point?

My guess is that the force points orthogonally away from the contact contour of B2 (dotted line). This force would the be split into a transversal component t and an orthogonal component o. In this case, t is always smaller then or equal to the maximum static friction (t <= normal force o * (static friction coefficient), meaning t <= Frmax = o * mu).

Alternatively, the force could point orthogonally away from the surface of B1, which would then have only a component in direction o with t = 0.

Another alternative, which I think I have excluded, is that the force points in a direction tangential to the movement path of the contact point, which is circular around the pivot axis of B2.

The second question is, how big does a second torque M2 have to be to be able to move the second body B1 into clockwise direction (M2 results from F2, but lets ignore the lever arm).

I have three possible equations. Basically the question is, can the force t be transmitted by friction and work against M2, making it necessary to overcome t (muliplied with the lever arm x to get a torque)? I could not figure out if static friction is overcome by the difference of forces (M2 - r * o) or the maximum of forces (max(M2,r + o)). (I think Frmax is technically defined incorectly because there is also a lever arm between M2 and Frmax, but if that one doesn't matter in principle, lets ignore that)

The first two equations are based on the assumption that the normal part of the transmitted force o mulitplied by the lever arm r will be added to M2, making the necessary M2 less big. Or is that assumption wrong, making the third equation correct?

P.s. This question losely expands my previous question.

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  • $\begingroup$ It is confusing to have two open questions at the same time with the same title. $\endgroup$
    – JAlex
    Oct 27 '20 at 16:52
  • $\begingroup$ I'm confused by the naming of the torques. E..e, you refer to a torque M1 on B1 but the drawing shows a torque M2. $\endgroup$
    – Bob D
    Oct 27 '20 at 20:37
  • $\begingroup$ @BobD There was a mistake, but the edit corrected it. Should be clear now. $\endgroup$
    – Modulus
    Oct 28 '20 at 8:20
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When you have two surfaces contact (lets ignore friction for now), the contact force develops along the contact normal.

In your case you have a surface (edge) in contact with a point (corner). Think of the corner as small circle and realize that the circle can develop a contacting force along any direction, but in the surface on the other body cannot. The result is the contact normal has to be along the perpendicular line to the edge. That is the blue direction in the figure below.

fig1

This contact can only supply forces of equal and opposite magnitudes on the two bodies. No torque is transferred through the contact, because the contact normal is the line of action of the force. Typically the normal force $N$ is calculated based on the fact the contact cannot interpenetrate, and thus the speed of the contact point on each body must match along the contact normal.

In the tangent direction relative speed is allowed (slipping), and this may result in sliding friction or not. Sliding friction opposes motion and has the magnitude of $F = \mu N$

Now a special case exists when the friction coefficient is high where the body jams in place. This can happen if there is such friction force $|F| < \mu N$ that can cause the bodies to have no slip.

So you can calculate the normal force without friction, and then find the friction needed to have no-slip. If friction exceeds traction you have slipping and a know friction value, but unknown slip amount. If friction is less than traction then you have sticktion with an unknown friction force, but known motion (tangential speed match).

Use the force balance to find the maximum torque such that $|F| < \mu N$ for whichever direction $F$ might be at.

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  • $\begingroup$ I understand the argument. But in this case, slip can never occur. M1 may be huge, the bodies will still jam. So shouldn't there be force transmission along this sort of form fit? Or in other words, if F is = μN for a given M1, where does the rest of the force go if I double M1? And further, N is not directed onto the axis of B1, so there would be torque in clockwise direction from N. But friction can transmit torque (see force limiters / friction overload clutches). So why would it not in this case? (I don't know if it does, but I don't understand why it wouldn't). $\endgroup$
    – Modulus
    Oct 30 '20 at 8:13
  • $\begingroup$ @Modulus in the case where the parts an jam without friction you end up with an redundant constraint or as commonly known a statically indeterminate system. You just can't determine the forces unless flexibility is considered. $\endgroup$
    – JAlex
    Oct 30 '20 at 18:34
  • $\begingroup$ I think I got it now. The parts only jam if |F|<μN and the the split between F and N is only determined by the angle (because M1 = F * r * sin(angle(r,F)) - F * r * sin(angle(r,N))). And μN will always be bigger than F if the angle is chosen such that the bodies jam. $\endgroup$
    – Modulus
    Nov 2 '20 at 13:36

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