# Hydraulic force formulas to linkages

• snakeeyes
The figure number is given in post 1. The label for point B is given in post 1 and post 2. The label for point C is given in post 1 and post 2. The elevator is pushing upward. The diagram makes it clear that the effort is downward. Fourth, please state what you mean by "90 deg" and "45 deg." These are undefined terms.thanks for your input.

#### snakeeyes

Hello my name is Scott ,

I was trying to post a question , but after joining the physics forum - I think this site is for posting questions not finding answers until after posting it , so here we go.

Maybe someone can answer this , at work I have a bet ( lunch ) on principles of leverage when applied from a hydraulic cylinder .

I am going to try and attach a sketch and all I am looking for is the formulas to calculate the forces produced at point A and C .

We both agree on how to calculate cylinder force in extend or retract,but all the formulas that we have looked at do not show how to calculate the force at C on figure 3 using the length between B and C or at point A figs 1,2,4,5

Our disagreement is which one produces more force and which one reduces the force.

I say the fulcrum (moving fulcrum)point is C he says no it is B ((fixed fulcrum)point

the lunch team

#### Attachments

• force level drawing 2.jpg
22.2 KB · Views: 839
snakeeyes: You correctly calculated the hydraulic cylinder output force, P = 5154.2 (or 8295.8). The fulcrum appears to be at point B. Point B is an almost stationary pivot point. The force applied to point C on each bar attached at point C is P divided by the number of bars attached at point C. Therefore, in figures 1, 2, 4, and 5, the force applied at point C on each bar attached at point C is 0.5*P; and the gripping force, F, at point A is F = 0.5*P*b/h, where b = horizontal distance from point B to C, and h = vertical distance from point A to line BC. Therefore, in figures 1 and 4, F = 0.5*P*b/h = 0.5*5154.2*6/6 = 2577 lbf. In figure 5, F = 0.5*P*b/h = 0.5*5154.2*6/2 = 7731 lbf. In figure 2, F = 0.5*8295.8*6/6 = 4148 lbf. The force applied at point C in figure 3 is P, and the resulting output force at point C in figure 3 is 5154 lbf. Figures 1 and 2 are first class levers. Figures 3, 4, and 5 are mostly similar to second class levers.

I am sorry I should have been clearer on the drawings , he almost called me on the bet , but I convinced him to let me retry.

thanks for the input

Snake eyes

#### Attachments

• force second class lever sample jpg.JPG
31.9 KB · Views: 731
snakeeyes: Why does your second attached file show the elevator pulling downward? Elevators push upward. They do not pull downward. Please correct the diagram, so we could see if your question is understandable. Also, state the direction of motion of the elevator. Third, state whether your second attached file is intended to correlate to a particular figure number in your first attachment; and if so, try to use the same, corresponding point labels (B and C). Or is your second attached file a completely independent example? Fourth, perhaps state what you mean by "90 deg" and "45 deg," which are undefined.

thank you for responding ,

First did or did not the attached file come out as readable or all rough , I opened it and if I had not been the one who drew it I could not tell what it might represent.

as to the other questions: elevator was representing a hydraulic cylinder rod end not really a elevator (just to show movement ), the direction is down , the chamber/cylinder pressure minus the rod diameter equals 1200 LbF downward , no this drawing is different as to the other drawing, I wiil submit the more accurate drawing as related to the real device and how we need to measure the forces with the right formulas , I am getting hungry hope I when the bet.
thanks
The lunch debate team

#### Attachments

• force second class lever bmp.bmp
244.7 KB · Views: 940
snakeeyes: The image quality and legibility for all three attached files came out well. The active force, also called the effort, is applied to a movable point on a lever; it is the input force or action. In your case, the effort is the hydraulic cylinder force. A fulcrum is a stationary or almost stationary pivot point. The force at a fulcrum is a reaction force. In your third attached file, the fulcrum is the stationary, right-hand pivot point, as described in post 2 for figure 5. The formula is given in post 2.

## 1. What is hydraulic force?

Hydraulic force is the force that is exerted by a fluid, such as water or oil, on an object. It is a result of the pressure applied by the fluid on the surface of the object.

## 2. How is hydraulic force calculated?

The formula for calculating hydraulic force is force = pressure x area. This means that the force is directly proportional to the pressure and the surface area of the object.

## 3. What is the relationship between hydraulic force and linkages?

Hydraulic force and linkages are related because linkages are used to transfer the force from the hydraulic system to the object being moved. This allows for a greater force to be exerted on the object without having to directly apply the force.

## 4. How does the length of a linkage affect hydraulic force?

The length of a linkage can affect hydraulic force by changing the angle at which the force is applied. A longer linkage can create a greater angle and thus a greater force, while a shorter linkage will create a smaller angle and a smaller force.

## 5. What are some common applications of hydraulic force formulas to linkages?

Hydraulic force formulas to linkages are commonly used in various types of machinery and equipment, such as cranes, forklifts, and hydraulic lifts. They are also used in hydraulic systems for vehicles, such as brakes and steering. In addition, they are used in industrial processes, such as metal stamping and pressing.