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I have a locking mechanism in mind, that is made with two racks. Rack 1 is fixed and then we push down rack 2 to lock something. After that a horizontal force is applied to rack 2. I calculated the resulting vertical force as shown in the sketch. I have two questions: Is the model accurate enough or do we over estimate the vertical force too much or with other words how much does the friction affect the vertical force? I summed all the forces on one tooth of rack contact. Does taking into consideration more teeth change things a lot?

Sketch of mechanism

UPDATE

I managed to analyze the forces on the slope. Everything is on the sketch. I also calculated that for the coefficient of friction 0,36, slope is self locking at an angle 20°. Steel on steel is friction 0,12 and we get self locking at around 7°. With self locking there is also no vertical force.

On thing is that I can't answer to myself. Why is my first equation with tangens inncorect? Triangle of horizontal, normal and vertical force there is different than with horizontal, normal and shear force on this sketch. Maybe someone can help me with this.

Sketch

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    $\begingroup$ to estimate friction, start with the materials. engineeringtoolbox can give a ballpark value. $\endgroup$
    – Abel
    Jan 14, 2022 at 22:20
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    $\begingroup$ I have a feeling that your horizontal force should be the hypotenuse. The component forces should be the Fn and the Ft (along the face perpendicular to Fn). But its been a long time since I have looked at this. $\endgroup$
    – Forward Ed
    Jan 15, 2022 at 4:07
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    $\begingroup$ Now that my vague recollection have been confirmed. I will give you a tip. "Normally" applied forces are applied are occurring at some angle and its the applied force that needs to be broken down into components. Quite often those components are horizontal and vertical. So when learning it is quite often easy to falsely assume that all force component triangles will have Fh as a leg (side) rather than as the triangle's hypotenuse. $\endgroup$
    – Forward Ed
    Jan 15, 2022 at 17:27
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    $\begingroup$ Another way of thinking about it, and I am not always sure its always true, Is if you have an applied force, then the components making it up have to be less than or equal to the applied force. So in this case, you initially took Fn as being larger than your applied force. $\endgroup$
    – Forward Ed
    Jan 15, 2022 at 17:32
  • $\begingroup$ @Forward Ed, I agree with you, it was a mistake from my side and I do it quite often, when I don't put the effort in force triangles. That means when I do quick checks. This quick checks usually become this kind of thing we have here now ;) $\endgroup$
    – Grega
    Jan 16, 2022 at 7:58

3 Answers 3

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Let's review the concept of " shear" and "shear friction" between two objects than undergo movement in opposite directions:

  • "Shear Force" - A force acting parallel to the surface in contact, denotes as "$S$" hereafter.

  • "Shear Friction" - A resistant force on the contact surface that is caused by the normal force ($N$) exerted by the object in motion; it is a reactive force, and its magnitude is depending on the relative roughness ($\mu$") of the contact surface and the magnitude of the normal force. Shear friction acts on the contact surface in the direction opposite to the shear force $S$ with a magnitude of "$\mu N$". The motion/sliding of the object on a plane is possible only if $S > \mu N$

enter image description here

Comment:

For the system, the applied force $F_H$ must exceed the shear friction ($\sum \mu N$), induced by the weight of the upper rack ($N = W/2$), on the horizontal contact faces. In structural design, the upper corners of contact are assumed to resist the entire horizontal force.

enter image description here

ADD: Regarding the latest update

Assume the surface has a friction coefficient "$\mu$" and the system is in equilibrium (stay stationary) after exerting $F_H$:

enter image description here

If $\mu N < S$, the reaction side forces are no longer in static equilibrium without additional force/effect:

enter image description here

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    $\begingroup$ ok, something about this was bugging me and I was running around in circles buying into your description then turning around and not accepting it. I repeated that cycle all day long. Tonight I had a game night with a mechanical engineering, a civil engineer and an astrophysicist. At the end of the night I presented this solution and after a bit of mulling over we finally out what we believe is not quite how we would display the solution. In school we were taught to draw your component vectors tip to tale. Meaning the arrow of one component vector will be at the non arrow head of the next. $\endgroup$
    – Forward Ed
    Jan 16, 2022 at 4:54
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    $\begingroup$ Which you basically have but they are moved to a PARRALLEL location as it does not essentially matter what side you draw them on/what order you draw them in. That being the case the individual component vectors never change direction. As a result, your blue arrows MUST be in the same direction. Your Fsv and Fnv are the same force in magnitude and direction. Fsv=Fnv. Fsv = Force along the face * COS Alpha or Fsv = Force normal * COS Alpha. They are not additive, its one or the other. Due this all with the assumption that there is no friction just to simply the forces then come back. $\endgroup$
    – Forward Ed
    Jan 16, 2022 at 5:00
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    $\begingroup$ When you want to work with friction, First find the Normal force. Then find the Friction force. last find the Force along the face. If the friction force is greater than the force along the face, nothing moves. If the friction for is less than the force along the face, Then subtract the Friction Force from the Force along Face to get your Resulting Sliding Force. Multiply the Resulting Sliding force by COS Alpha and you will have your vertical force. $\endgroup$
    – Forward Ed
    Jan 16, 2022 at 5:03
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    $\begingroup$ Yes, when shear friction capability is less than the shear force demand, the driving block slides up. I don't know what you are planning, but in reality, don't forget the weight of the rack. $\endgroup$
    – r13
    Jan 16, 2022 at 14:37
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    $\begingroup$ I don't think you have a full understanding on the sketch and the equation. Allow me to point out - in the sketch, the component of F_H has components of "N" (normal force) and "S" (shear force demand) on the right of the incline (driving side), on which, the force F_H and the component forces must remain equilibrium, sumFy = 0, & sumFx = 0. I have shown sumFy = 0, and leave you to check whether sumFx = F_H+Ncos(a)+Ssin(a) = 0. Then compare S with muN on the reaction side, if muN < S, the upper element will move that breaks the equilibrium on the reaction side unless a vertical force is added. $\endgroup$
    – r13
    Jan 16, 2022 at 18:55
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Your calculation seems to be right except $\ F_v$ should be pointing up. Because $F_N \ $ should attach to the end of $\ F_H$ pointing to right and up.

The more gears you have, if you have the same horizontal force, the vertical force remains the same but the strasses on gears decrease.

In reality, too many gears will lock in the mechanism, making it harder to move.

f_v

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  • $\begingroup$ I agree with force up, but the equation is the same. Rack 2 will move "uphill" with horizontal force applied. I also agree about harder moving, but how much is that. Is it possible to evaluate? $\endgroup$
    – Grega
    Jan 14, 2022 at 21:25
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Assumptions:

  • neglecting forces from gravity (weight of top part)
  • neglecting possible effects from offset load

(1) Decompose the external force into orthogonal components:

$\vec{F_H} = \vec{F_N} + \vec{F_T}$

(2) The geometry gives us the magnitudes of the decomposed components

$F_N = F_Hcos(\alpha)$

$F_T = F_Hsin(\alpha)$

(3) Write the stipulated equilibrium condition: that the parts "stick".

$$F_T = {\mu}F_N$$

(4) Solve the above for limiting value of $\mu$

$$F_Hsin(\alpha) = {\mu}F_Hcos(\alpha)$$

$$\mu = \frac{sin(\alpha)}{cos(\alpha)} = tan(\alpha)$$

(5) The "vertical force" on the bottom part is just the vertical component of $F_T$

$$F_Tcos(\alpha) = F_Hsin(\alpha)cos(\alpha)$$

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