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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

Forward Ed
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Grega
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    to estimate friction, start with the materials. engineeringtoolbox can give a ballpark value. – Abel Jan 14 '22 at 22:20
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    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. – Forward Ed Jan 15 '22 at 04:07
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    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. – Forward Ed Jan 15 '22 at 17:27
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    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. – Forward Ed Jan 15 '22 at 17:32
  • @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 ;) – Grega Jan 16 '22 at 07:58
  • But for me it is hard to do that intuitively. Now I will try to remember that I always have normal and shear force as a components and not horizontal force as a component of a normal force. – Grega Jan 16 '22 at 08:04

3 Answers3

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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

kamran
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  • 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? – Grega Jan 14 '22 at 21:25
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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

r13
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    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. – Forward Ed Jan 16 '22 at 04:54
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    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. – Forward Ed Jan 16 '22 at 05:00
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    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. – Forward Ed Jan 16 '22 at 05:03
  • @r13 regarding your comment of horizontal contact surfaces. That was my mistake at sketching. I should drawn a gap there, because I'm only interested in forces on the slope. This is also what happens in reality. There is no contact on horizontal forces. – Grega Jan 16 '22 at 07:40
  • @r13 regarding your ADD update. I think I did the same in my latest update. First I put shear force and friction force in equilibrium to see when the ramp is self locking. After that I calculated the option when friction force is smaller than shear force. In that option sum of forces is not zero and rack 2 wants to move up. Can we check that equation? – Grega Jan 16 '22 at 07:50
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    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. – r13 Jan 16 '22 at 14:37
  • @r13 I think that your equation of equilibrium on sketch does not take friction into account, because there is no coefficient of friction in equation. You should write equilibrium for vertical component of shear force and friction force and equilibrium for horizontal component of shear force and friction force. If there is no friction force, rack 2 will move in direction of shear force. I agree about the weight. It also counts in the equilibrium, but that is more straigh forward to do after this slope is solved. – Grega Jan 16 '22 at 17:02
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    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. – r13 Jan 16 '22 at 18:55
  • @r13, so the equlibrium you have written on the sketch for y axis, is just for driving side? Out of force N we get friction force and we compare the equilibrium on the reaction side? I think I already did that, but I just passed few steps in between. But no one wants to comment my equations 1, 2 and 3 on the sketch in my last question update. – Grega Jan 16 '22 at 19:36
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    Yes. My equation is representing the driving side. On the reaction side, you will have F_H' and N' opposite to the driving side forces F_H and N, and "muN" (shear friction) opposites to S, which is the pair of forces you need to compare. If muN = S, the entire system is in static equilibrium, if muN < S, sliding will occur, which cannot be prevented given the premisses in your original sketch - weightless and lack of additional stopping force = S - muN. – r13 Jan 16 '22 at 20:05
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    Your question really concerns the stability of objects on a slope, which you shall review further. In reality, your design won't have that complication. At least, you shall add weight to the calculation unless it is truly weightless, and the geometry is flawless as assumed - no other surfaces in contact other than the sloping faces. – r13 Jan 16 '22 at 20:09
  • Ok. I think we are talking about the same thing here. I will try to clean up my sketch and post it again tomorrow. – Grega Jan 16 '22 at 21:56
  • Let us continue this discussion in chat. – Grega Jan 17 '22 at 13:45
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    We have already dragged this simple problem too far. You can modify your question and post it as a new one to draw comments from the others. – r13 Jan 17 '22 at 15:39
  • Ok. @r13, I understand. Sorry for this and thanks for the debate. I used this problem and one other to get some things straight in my understanding. I now know how to approach the whole analysis, also with weight added. – Grega Jan 18 '22 at 13:12
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    I discontinued this conservation because 1) the problem is non-realistic, 2) thus it reduces to a lesson we learned earlier in school - kinetic motion, which has Newton on its back that leaves no room for "debate". As long as you realize N & S are component forces, and the concept of shear demand and shear (friction) capability, you are already in a good direction. One piece of advice though, in structural engineering, shear friction is rarely been relied upon due to its high uncertainty, so it has never been an important topic in statics but dynamics. – r13 Jan 18 '22 at 15:24
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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 $F_T$

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

Pete W
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