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Forces in scissar lift

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Agunia

Agunia

Student
Joined
Dec 29, 2020
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4
Location
PL
jack_n6fkit.png



Hi

I need help, how can i calculate force in screw and in rods ?
 
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A clue, draw some free body diagrams, then post what you have done and then we might help

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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IMG_20201230_133029_1_chw062.jpg


I think direction of Rb force is correct but Ra is not. Only 2 unkown forces Ra , Rb so i didnt use sumM=0. I think im doing huge mistake.
 
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Hi

Where are the dimensions of the scissor lift ? There are several ways to do this but you need some dimensions and angles to work with, is this a student post?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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This thing has a single pinned support. What keeps it from rotating counter-clockwise under its own weight until it hits the floor?

============
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill
 
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podn_vulsbz.jpg


Force in screw depend on angel alfa, Qmax is when alfa=25deg, for alfa>25deg Q=const=Qmax. How can i calculate forces in screw and arms correct ? This structure work as lattice i think.
 
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Resolution of forces start with the load Q.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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PODN1_rwd41n.jpg


This is still bad because i got 2 unkown forces Ra and Rb and 3 equation (x=0 y=0 M=0).
 
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Take moments from Ra so Ra*300 sin α = Rb * 150 sin 20°, Transpose and find Rb


“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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Add a roller support for stability, draw to scale, calculate a few dimensions and it's a piece of cake. You wouldn't want us to do the calculation for you, would you?

Note that 169/tan25 = 362.4, which I rounded to 362.

image_mbw3mw.png



BA
 
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Think of this apparatus as a giant nut cracker where the vertical force Q is applied at the end of each handle. The two handles are joined together by a pin at the roller support. The 'nut' is located 135mm to the right of the left end.

image_yllcew.png


What is the force required to crack the nut? It is a vertical force caused by a beam hinged at the left with a 227mm cantilever and a 135mm back span. Say that force is F. The 'nut' in our case is a diamond shaped frame of length 270mm and height 270*tan25o. Two corners are tied together by a horizontal member to prevent them from spreading.

You should be able to calculate all of the forces using statics.

BA
 
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Hey, Agunia, keep us in the loop. Have you solved the problem yet? If not, why not? If Eng-Tips solves it for you, your professor might not take kindly to our interference. I gave you a suggestion; it is up to you to finish it. After you calculate the force acting at the top and bottom of the diamond frame (or walnut), the rest is simple statics. Is there a problem?

I didn't go through your calculation sheet, but at a glance, it appears far too complicated. Let E-T know if we can help.

BA
 
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Greg...

Thanks for the explanation. However, is friction in the foot pin enough before engaging the load and is it enough if the load and foot are not precisely aligned? I have only used symmetrical scissor jacks. This type doesn't give me a lot of confidence.

Fred

============
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill
 
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fel3,

I'd say you were correct the first time. The structure is unstable as shown. It needs more support to prevent it from rotating about the bottom pin. That is why I added a roller support at the left end in my model.

Friction should not be relied upon to prevent rotation. The "foot" on the ground can't help because it's below the pin. A socket joint at the top could help, provided the vehicle being jacked is capable of providing adequate lateral support. That is not specifically stated, and should not be relied upon.

I share your lack of confidence in the jack as drawn.

BA
 
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Your left hand bottom sketch is unstable, Greg.
The models shown below are much better, although I wouldn't trust any of them on a steep gradient.

image_ane4b0.png


The main difference between your model and these is the fact there are two pins instead of one, above the base plate providing a degree of stability. The assembly could collapse in an orthogonal direction too. For that, there is the width of the base to provide some stability.

image_j8zz78.png


BA
 
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GregLocock said:
It is stabilised by the wheels on the other side of the car, although I agree it isn't ideal.
Click to expand...

So it relies on the wheels on the other side of the car being locked against rotation. Otherwise, it is not stabilized. I agree it isn't ideal.

BA
 
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