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What is the formula to calculate force F2 given L1, L2, L3 and F1? (updated: wrong question asked. originally it said: "What is the formula to calculate force F1 given L1, L2, L3 and F2? Now question is correct...bahhh, question was correct in the attached sketch though)

The angle in red lever arm is 90 degree. Please see attached image.Image showing bent second class lever arm

update

Thanks to r13 for an updated answer. I still do not how that formula is correct (I do not say I know it is wrong, but I do believe it is wrong. According to r13 last formula I end up with this: F2=2(F1*L1)/L3, which again is not using distance L2 - which I definitely belive will affect F2). I need to understand better how the correct formula is beeing set up, so I know what I'm doing. Please look at my second sketch added in this comment, and also see the text and question(s) in the sketch. I believe it is possible to simplify the the bent second class red lever to another lever (see new sketch), and that it is easier to set up (and understand - for me) the formula(s) used, to calculate F2. enter image description here

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  • $\begingroup$ Show what you have done to solve this. Perhaps start here: engineeringtoolbox.com/levers-d_1304.html#gsc.tab=0 $\endgroup$
    – Solar Mike
    Commented Jan 21 at 10:39
  • $\begingroup$ Thanks for that, but I have allready looked at formulas for those setups... but do not know how to make a formula for the kind of lever arm in the figure I posted. So I was hoping that someone with a better understanding of this, could help me out. $\endgroup$
    – bongobongo
    Commented Jan 21 at 13:09
  • $\begingroup$ I have allready looked at formula for a second class lever arm (see link below)... but do not know how to make a formula for the the bent second class lever arm in the figure I posted. So I was hoping that someone with a better understanding of this, could help me out. firgelliauto.com/blogs/news/… I do not want to guess the formula to use, so I was hoping for some help for the formula. $\endgroup$
    – bongobongo
    Commented Jan 21 at 13:22

2 Answers 2

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The tendency to rotate about the hinge point must be kept at "zero".

$\sum M_{hinge} = 0$

$F_1*L_1 - F_2*L_3 = 0$

$F_2 = F_1*(L_1/L_3)$

Moment (M):

enter image description here

enter image description here

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  • $\begingroup$ Thank you for that answer. But how can you skip using L2 (ref. my figure) in the formula? Could you please explain? $\endgroup$
    – bongobongo
    Commented Jan 21 at 14:54
  • $\begingroup$ The tendency to rotate (called the "moment") about the support is the product of force and its lateral distance measured to the support point. Here, you have two tendencies - F1*L1 and F2*L3, and L1 & L3 are the lever arms between the forces and the support hinge. If you have a vertical downward force on the vertical leg, then you will have one more tendency to rotate - F*(L1+L2), in the same direction as F1*L1. $\endgroup$
    – r13
    Commented Jan 21 at 15:15
  • $\begingroup$ I still believe one must use the L2 in the formula Example a: L1 = 1, L2 = 1. Gives the horisontal lever part is 2. Which again mean that at the right point of horsontal lever part has a force down on: MA = L/E = 1/2. If F1 = 100kg then that force is 50kg. B) If you change L2 to say 9, then the downward force on the bottom right part of the lever is: MA = 1/10 which gives 10 kg. If lever has no mass and hinge has no friction, it looks to me that the force F2 in example 2 here, is much less than in the first example, if L3 is same for both examples. How can L2 be removed in the formula? $\endgroup$
    – bongobongo
    Commented Jan 21 at 16:01
  • $\begingroup$ You need to review how the lever arm is measured - from the line of action (the force) measured perpendicularly to the support point. L2 will be in the equation only if there is a force acting up/down the vertical leg. $\endgroup$
    – r13
    Commented Jan 21 at 16:10
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    $\begingroup$ Added one more sketch for you to review. If you can't get it, sorry, I gave up. $\endgroup$
    – r13
    Commented Jan 21 at 16:33
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For the benefit of other future users. r13's answer is correct and I upvoted it.

Trying to answer why L2 does not play a roll in this lever is that there is no orthagonal force acted on its end, the corner of the lever, in other wrds zero orthagonal force is acting at the right end of L2.

L2 could be 10 times bigger or 10 times smaller. The balance of moments is still the same.

$$\Sigma M_{hinge}=0 \quad F_1*L_1=F_2*L_3$$

It is the definition of momen of a force about a point that the moment, M,is the force vector multiplied by its distance from the point. It is unfortunate that OP has never understood this foundamental definition. However regardless of OP's belief the fact remains that $$F_2= F_1*L_1/L_3$$

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  • $\begingroup$ Thank you for taking the time to explain this. Also thanks to r13 - for not giving up. $\endgroup$
    – bongobongo
    Commented Jan 22 at 9:25

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