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Forces in scissar lift

Forces in scissar lift

(OP)

Hi

I need help, how can i calculate force in screw and in rods ?

RE: Forces in scissar lift

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

RE: Forces in scissar lift

(OP)

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.

RE: Forces in scissar lift

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

RE: Forces in scissar lift

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

RE: Forces in scissar lift

(OP)

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.

RE: Forces in scissar lift

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

RE: Forces in scissar lift

(OP)

This is still bad because i got 2 unkown forces Ra and Rb and 3 equation (x=0 y=0 M=0).

RE: Forces in scissar lift

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

RE: Forces in scissar lift

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.

BA

RE: Forces in scissar lift

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.

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

RE: Forces in scissar lift

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

RE: Forces in scissar lift

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

RE: Forces in scissar lift

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

RE: Forces in scissar lift

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.

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.

BA

RE: Forces in scissar lift

Quote (GregLocock)

It is stabilised by the wheels on the other side of the car, although I agree it isn't ideal.

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

RE: Forces in scissar lift

To better represent real-life use, the rolling support should be on the top arm rolling in the vertical direction. Rotate the left support 90* so the arrow points left and put it at the top arm. Still, I don't believe the location used to get it to work in modelling software changes the force calculations in the jack.

To stick as close to reality as possible, the rolling support plane should be an arc so the support moves to the right as the jack is raised. But that still shouldn't change the forces seen by the members in the jack.

RE: Forces in scissar lift

Quote (BAretired)

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.

Even the full scissor jacks you posted rely almost completely on the car being (relatively) rigid and at least two of its wheels remaining in contact with the ground for stability under load. The only real benefit of the stability added by their symmetric design and rectangular base is so you don't have to hold it up with one hand while you try to get it engaged with the car and the ground. In an unconstrained load situation, I wouldn't trust a scissor jack with a cinderblock.

RE: Forces in scissar lift

Quote (LionelHutz)

To better represent real-life use, the rolling support should be on the top arm rolling in the vertical direction. Rotate the left support 90* so the arrow points left and put it at the top arm. Still, I don't believe the location used to get it to work in modelling software changes the force calculations in the jack.

One extra roller support is needed somewhere to make the structure stable. It carries no load provided the upper pin is directly above the lower pin. It could be in the position shown and, as noted by by LionelHutz, does not change the internal forces or moments in the members.

Quote (handleman)

Even the full scissor jacks you posted rely almost completely on the car being (relatively) rigid and at least two of its wheels remaining in contact with the ground for stability under load. The only real benefit of the stability added by their symmetric design and rectangular base is so you don't have to hold it up with one hand while you try to get it engaged with the car and the ground. In an unconstrained load situation, I wouldn't trust a scissor jack with a cinderblock.

Almost completely is not completely, handleman. The full scissor jacks I posted do not rely on wheels remaining in contact with the ground or the rigidity of the car. The applied load could be a sack of mashed potatoes; as long as the centre of gravity of the load remains inside the rectangular base plate, the structure is stable.

However, a slight lateral force could topple the scissor jacks if the resultant of the applied forces lies outside the edge of the base plate. If I were planning to work under the car, I too would want the additional support of some wood blocking or possibly a concrete block (not sure I would trust a cinder block).

The discussion of stability of the structure, while important, was probably not the concern of the OP.

BA

RE: Forces in scissar lift

Quote (handleman)

Wait... Are you an engineer or a theoretical physicist? DURING REAL WORLD USE, almost every item described as a "jack" relies entirely on external guidance of its load.
I am a retired structural engineer. What the hell does "external guidance of its load" mean?

Quote (handleman)

The addition of a horizontal roller support in the location you have shown is idiotic.

The vertical roller support I have shown in my latest diagram prevents horizontal movement but permits vertical movement. Nothing idiotic about it.

BA

RE: Forces in scissar lift

No "jack" during use is designed to functionally support a load that is not in the direction of its action. A jack acting vertically is totally reliant on outside factors to laterally stabilize the load. It essentially acts as a pinned 2-force member.

Sorry.... Your first roller support was idiotic. Your second is a graphical representation of how this device actually works. I.E. this jack mechanism relies on external guidance of its load.

RE: Forces in scissar lift

Quote (handleman)

No "jack" during use is designed to functionally support a load that is not in the direction of its action. A jack acting vertically is totally reliant on outside factors to laterally stabilize the load. It essentially acts as a pinned 2-force member.

The provision of outside supports to laterally stabilize the load is a wise precaution with any jack, on that we agree. However, the bases in the photos below are not pins, and they do offer some stability, albeit small. They are decidedly better than a pinned base.

Quote (handleman)

Sorry.... Your first roller support was idiotic. Your second is a graphical representation of how this device actually works. I.E. this jack mechanism relies on external guidance of its load.

Once again, you are mistaken, handleman. (No, you are not!!) You seem to be batting 1000 today. The horizontal roller in the first diagram is just as effective as the vertical roller in the most recent diagram. In both cases, they prevent rotation of the assembly about the lower pin.

EDIT: The horizontal roller in the first diagram prevents Node A from moving up or down, so it was a mistake. Following is the corrected diagram.

Again, while very important to make the structure stable, none of this responds to the OP's question, which addressed forces in the rods, an exercise in elementary statics.

EDIT: My apologies, handleman. You were right!!! (embarrassed murmuring in background)

BA

RE: Forces in scissar lift

I'm pretty sure the second pin at A would work fine as long as Q is kept pointed towards the bottom pin as the jack is moved.

RE: Forces in scissar lift

I myself am wondering, is the direction of the vector Ra correctly indicated?

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