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In the picture below, in a), a body K1 is pivotably attached to a bearing. My question is about the torque that results from a force exerted onto a surface of the body K1.

A first force F1 applied orthogonally onto the surface should result in a torque M1 in clockwise direction.

Is it correct that a second force, F2, applied almost parallel to the surface will result in a torque M2 in counterclockwise direction?

My thoughts are, F2 is split into F2t and F2o (transversal and orthogonal components) by the surface of the body K1. To get a torque, F2o is multiplied by the lever b and F2t is multiplied by the lever a (M2 = F2t * a - F2o * b > 0). As a>b and F2t>F2o, the torque from the force F2 results in counterclockwise direction.

Applying these thought to the two bodies K1, K2 in b), a torque of M3 applied to the body K2 will result in a torque M4 in the body K1. (The bodies won't move because they are in each others movement path)

Is this correct or am I forgetting something? What is the job of friction in this case? From looking at b), K2 should push K1 away by applying a clockwise torque, but that is wrong then, right?

Suppose there is enough friction so that no slippage occurs.

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  • $\begingroup$ Related wiki on contacting bodies and their instant centre of rotation. Hin't torque about the instant center is an important quantity. $\endgroup$
    – jalex
    Commented Oct 27, 2020 at 17:40
  • $\begingroup$ @JAlex that assumes concurrent movement. In this case, the two bodies are plainly stuck and don't move together, so I don't know how to apply this. $\endgroup$
    – Modulus
    Commented Oct 28, 2020 at 8:20
  • $\begingroup$ Sorry, I confused this problem as a dynamics problem, when it is in fact a statics problem. The dynamics problem is super hard to tackle, and the statics problem with friction is only slightly easier as no kinematics need to be considered. $\endgroup$
    – jalex
    Commented Oct 28, 2020 at 13:14

2 Answers 2

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I think friction is required for any torque to be applied CCW, which be definition works against torque being applied CW. So with no friction it would be net torque CW, but with "infinite" friction (i.e. no slipping) it would be net torque CCW (and also locked up and not spinning). I don't think this question can be solved without some assumption or knowledge about the friction involved.

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  • $\begingroup$ lets suppose infinite friction (or enough friction for no slippage to occur). In the case that F2 is a linear force, would there be ccw "spinning"? I edited that into the question. $\endgroup$
    – Modulus
    Commented Oct 26, 2020 at 14:00
  • $\begingroup$ There can't ever be spinning in the CCW direction because the two objects will bind. All it can do is reduce the net torque in the CW direction up until there is no slippage and all the CW net torque is completely nullified. $\endgroup$
    – DKNguyen
    Commented Oct 26, 2020 at 14:01
  • $\begingroup$ Yes. But there would be a resulting torque ccw (which would make the object spin if K2 would allow it)?. $\endgroup$
    – Modulus
    Commented Oct 26, 2020 at 14:03
  • $\begingroup$ Are you now asking about a free force being applied along F2 to K1, with K2 not being present? With K2 present, there would be an something akin to an internal force since the stress is there but nothing is moving because of binding. There can't be a CCW net torque in that case, by definition. $\endgroup$
    – DKNguyen
    Commented Oct 26, 2020 at 14:06
  • $\begingroup$ Or differently phrased, suppose I would be excerting a torque in clockwise direction to K1. This torque would have to be F2t * a + F2o*b*(friction coefficient) to be able to move K1, right? So the F2t part is transmitted to K1 and holds K1 in place? $\endgroup$
    – Modulus
    Commented Oct 26, 2020 at 14:08
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Don't wrap yourself around in pretzels. Even for planar cases, assume they are defined in 3D (with z-axis out of plane) and use the cross product to define torque

$$ \vec{\tau} = \vec{r} \times \vec{F} $$

which expands to

$$ \pmatrix{ 0 \\ 0 \\ \tau_z} = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \pmatrix{F_x \\ F_y \\ 0} $$

and projects into 2D as

$$ \tau_z = -y F_x + x F_y $$

The nature of force does not matter. Use the combined normal and frictional forces to find the net torque, or just an individual component to gauge the effect of on the body.

The same goes for moment of impulse in case you have contacts.

$$ \vec{\gamma} = \vec{r} \times \vec{J} $$

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  • $\begingroup$ That does seem to lead to the same conclusion. I am now pretty confident that F2 is transmitted into a transversal and an orthogonal component. However, I noticed that I don't know the direction of F2 (see physics.stackexchange.com/q/589933/278169 ). $\endgroup$
    – Modulus
    Commented Oct 28, 2020 at 8:18
  • $\begingroup$ You have an corner contacting an edge. The contact force is perpendicular to the edge only, unless friction is considered. With friction the maximum angle away from perpendicular is $\theta \leq \mathrm{atan}(\mu)$. You find the angle, or the tangential component by solving the problem of no-slip. $\endgroup$
    – jalex
    Commented Oct 28, 2020 at 20:52
  • $\begingroup$ I've considered that possibility. However, it leads to two questions. One, the bodies in this case are stuck, because of the pivot axes. Therefore, in addition to the friction being limited by the maximum friction before slip occurs, there is a form fit. Does this form fit transmit forces, making the tangential component potentially bigger than atan(mu)? If not, where does the force go, if slip can't occur? And two, how much force does K2 even transmit in that direction? Or is the force that K2 transmits safe limiting by atan(mu) even though there is a (sort of?) form fit? If yes, why? $\endgroup$
    – Modulus
    Commented Oct 29, 2020 at 10:18
  • $\begingroup$ If this is for rigid bodies you have a statically indeterminate problem because forces generated by one pivot can be directly reacted up another pivot. To bypass this issue, you need to consider the flexibility of the bodies and the contact geometry. Not an easy problem to do in general. Ideally you want to use some FEA software to establish the reaction forces on the pins. $\endgroup$
    – jalex
    Commented Oct 29, 2020 at 12:23
  • $\begingroup$ Yes, rigid bodies. I don't need to know the exact values or anything, but I need to understand the physics. Is it possible at all for M3 to result in a positive M4 or will M4 always be negative because of the contact surface of K1? I tried formulating a different question but with regards to the same problem after I noticed there were even more problems behind the first problem physics.stackexchange.com/q/589933/278169 $\endgroup$
    – Modulus
    Commented Oct 29, 2020 at 13:06

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