Question about compound levers

[Philip Potts]

I have been asked to build a cart that can be pulled by a tugger. (A tugger is a small stand-up vehicle.) The cart has to have a ramp to push other wheeled carts up onto the tugger cart. The photo below gives an idea of what I have come up with.

The ramp on the left side is 30 inches from the hinge to the tip. The thick black line is a wire rope that is attached to the ramp 16 inches form the hinge. The incline angle of the ramp is 13 degrees. The ramp weighs 84 pounds. The lever attached to the middle of the cart shows that the lever is to the left when the ramp is in the lowered position and pulling the leaver to the right shows the ramp in the upright position. The length of the lever from the pivot to the wire rope connection point is 30 inches. The distance to the handle on the pulling lever is 42 inches from the pivot point to the handle center.

IF (and that's a big if) I have understood my research on this subject, I have to Class 2 levers. According to the math formulas I found the end result will require around 20 pounds of pulling force to lift the ramp.

Can someone please tell me if I came to the correct conclusions?

Thanks in advance and I apologize if I have asked this question in the wrong section.

 

[BvU]

Hello Philip.

(I haven't spent a second calculating yet, just consulted my gut feelings:)

In the down position, the rope is pulling at an angle wrt the desired motion of the point where it is attached, close to the center of mass of the ramp. So it has to pull (quite a bit) more than 84 Lbf. At the lever handle there is also a (smaller) angle difference. That means I intuitively don't believe your result:

According to the math formulas I found the end result will require around 20 pounds of pulling force to lift the ramp.

so show us what you did and perhaps we can help you work it out

 

[Philip Potts]

Hi BvU,

Thanks for the greeting and your response. I have added some labels to the diagram to make communication easier.

I used the formula for a Class 3 lever of Mechanical Advantage = d2/d1. I got d1 by measuring from point E to the tip of A. Then d2 by measuring from point E to point D. This gave me MA = 16/30 = .53. I then multiplied the total weight of 84 pounds by .53 to get 44 pounds. Then I measured the distance on lever B (class 2 lever I hope) from point F to the tip. Then measured the distance of lever C from point F to the handle. Then I used the formula Mechanical Advantage = d1/d2 This gave me MA = 42/30 = 1.4 Then I divided the 44 pounds by 1.4 to get 31 pounds.

Am I close?

 

[BvU]

Class 3 lever of Mechanical Advantage

Had to look it up (just a physicist...). Mechanical advantage, unfortunately, is the ratio what you have to lift divided by what you have to do for it (so in your case 84 Lbf/0.53 = 158 Lbf). But the weight of the ramp isn't acting (grabbing) at point A but much closer to point D.

Let's look at the ramp and for the sake of simplicity let's look at it in a horizontal position:

mg acts at 15" and a vertical force F_y has to compensate the torque wrt E to prevent the thing from rotating. So F_y * 16"= mg * 15" from which F_y = ##{15\over 16}\times ## 84 Lbf, or 79 Lbf. (that means that in equilibrium, the hinge itself is pulling up as well, to the tune of 5 Lbf)And that is the vertical component of the tension in the cable, so the tension in the cable is considerably greater. If the distance from E to the lower roller is 16" like DE, then the tension in the cable is this 79 divided by 0.71 , some 111 Lbf:

(when the ramp is all the way down, the math is still a little more unfavourable).

Then , at the rollers the levers B are close to pulling at the right angle and from C to B you have a mechanical advantage of C/B.

Note that the torque that the axis of rotation of B and C has to endure, is considerable !

 

[Philip Potts]

And that is the vertical component of the tension in the cable, so the tension in the cable is considerably greater. If the distance from E to the lower roller is 16" like DE, then the tension in the cable is this 79 divided by 0.71 , some 111 Lbf:

Thank you BvU,

I was really off with my design. So, as the saying goes, Back To The Drawing Board. I'm going to move the cable closer to the end of the ramp and see what I can come up with.

Again, thank you for your help.

 

[BvU]

Suggestion: think about a counterbalance weight to lower the force required

 

[Philip Potts]

Suggestion: think about a counterbalance weight to lower the force required

Yep. I was thinking about adding a spring assist. Thanks.

 

[Dr.D]

Some dimensions on your figure would help a lot. This is a fairly simply problem if approached by virtual work, but the kinematics cannot be developed without some dimensions for the various components.

 

[Philip Potts]

Some dimensions on your figure would help a lot. This is a fairly simply problem if approached by virtual work, but the kinematics cannot be developed without some dimensions for the various components.

Hi Dr D, Thanks for responding. I have added dimensions to the drawing. I'm still in the planning stage so none of this is set in stone. My main goal is to provide a way to lift the ramp by the average person. As designed the ramp weighs 84 pounds. I can get that below 55 if I switch some of the material to aluminum. Let me know if I need to add any other information.

 

[Dr.D]

Could you also provide the pulley radii and the distance between the two pulley pivots? That would help.

 

[Philip Potts]

Could you also provide the pulley radii and the distance between the two pulley pivots? That would help.

Yes indeed. The pulley diameter is 2.68 and the distance between pulleys (center to center) is 4.75.

 

[Dr.D]

The picture suggests symmetry in the lever motion, i.e., in raising the platform, the lever goes from 4.l433" from the left side to the same distance from the right side. Is this what you intended (it looks reasonable, but I just wanted to ask before I invest some time in setting this up).

Also, where is the center of mass of the tilting platform? Is it half of 33.347" from the pivot, or somewhere else?

When the tilting platform is all the way down, what is its angle to the ground?

How high is the lever pivot with respect to the ground?

Do you have a total length for the cable?

Since you have the system evidently in CAD, can you produce a figure with the lever about 10 deg either side of vertical (the platform will be partially up, but not fully so). This will help in setting up the kinematics.

 

[Philip Potts]

The picture suggests symmetry in the lever motion, i.e., in raising the platform, the lever goes from 4.l433" from the left side to the same distance from the right side. Is this what you intended (it looks reasonable, but I just wanted to ask before I invest some time in setting this up).

Also, where is the center of mass of the tilting platform? Is it half of 33.347" from the pivot, or somewhere else?

When the tilting platform is all the way down, what is its angle to the ground?

The symmetry was planned because I don't want the levers to extend beyond the sides of the cart when in motion. When the ramp is in the down position it is OK for the lever to extend some but when the ramp is in the up position the cart will be towed across the plant and I don't what the arm to catch on something.

The center of mass is about where I have place the attachment point for the cable. About 16 inches from the hinge.

The ramp has an upward slope of 13.5 degrees. A wheelchair ramp is typically around 10 degrees so I felt this would be acceptable to push a wheeled cart up.

Also, this same pulley system will be attached to the ramp at the opposite end of the cart. This is so the ramp will be supported at both ends to help prevent twisting.

Feel free to adjust these dimensions as needed. I can adapt.

Again, thank you.

 

[Dr.D]

I'll try to set this up tomorrow and get back to you.