# Solving Multiple Leverage Math Problems

• Barstowrat
In summary: Next is the hand lever. This is an unknown, but I am going to go with 5:1. So with 5# of input would be 50# output. The final calculation is then (10:1+5:1=15:1) which would give the braking system a total force of 150#.
Barstowrat
So I am working on a cart and the last thing I can not figure out is the math to the braking system. It is relatively simple, though I have over complicated it. I am lost with how to add the multiple levers in play.

So let's begin.
Problem 1) What is the ratio of force between input and output?

Problem 2) The braking apparatus I am designing has two sets of levers on either side identical to a scissor clamp. I can't even fathom how to figure that out,

Problem 3) From the braking system, it will then be ran by cable to a simple hand lever at about 5:1, or 6:1... How does that add up with the scissor clamp?

Totally lost, I know I have probably not given all the required information, so please tell me what you need to help me figure this out. Also, I am not looking for someone to give me the exact answer, I like to figure things out as well, just have no idea how to actually figure it out.

Thank you for any help.

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• Scissor Clamp.JPG
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In that scissor mechanism, the ratio of input to output is not the same throughout the range of action. The two levers rotate about the centre pivot and what counts is the 'perpendicular distance' of the line of the force and the pivot (you could say it's the radius around which the force is acting). As you pull on the cable and the jaws come together, the angle changes that the two links are pulling and that changes the mechanical advantage (the effective ratio that the lever gives you). Also, the two top links change their angle relative to the cable and that also affects the force on the levers.
So you would need to know the spacing of the jaws then the brake blocks are actually in contact with the wheel rim. Once you know that, you can start on a rough calculation. (I say "rough" because the force from the block is distributed and you can't be sure where it is effectively acting. )
First the levers: The output length would be measured to (say) the centre of the brake block from the pivot centre and the input length would be measured as the shortest line you can draw from the pivot to the centre line of the link (that will be a line at right angles to the link and through the pivot - when fully extended, this perpendicular line may not even intersect with the link itself but an extension of its mid line). The mechanical advantage will then be
(input length) / (output length) and this may be a fraction less than one.
Next the links: A force F pulling vertically upwards will produce a force F1 on the links of
F/2Cos(θ/2) where θ is the angle between the two links (about 120 degrees, in the picture)
That force on the cable is shared between the two links - hence the division by 2. The Cosine in the formula shows that, if the links lie in a straight line (θ is 180) you will get 'infinite' force and that drops to F/2 when the angle between the links is zero.
The overall force F2 on your wheel will be F1 times the lever ratio. There is also the advantage of the hand lever to be considered - which you can also calculate using the length of the lever and the radius of the channel carrying the cable in the lever. Again, you have the problem of deciding exactly how effectively long the handle is and which bit of your hand is doing most. It will give you some idea, though. You could check the credibility of your final answer by finding what sort of weight you can lift with a strap, using your fingers.

I probably should have mentioned this before, but I am a very simple person, never really could get the hang of multiplying the alphabet or symbols. I know and learn what I need too, but usually requires some crayons lol. That said, I am sure what you have said was very informative but not quite down to my layman level. So I have been studying this all day and this is what I have come up with.

Starting with simple numbers for ease of conversation, and please correct where I am wrong. Starting at the front with the first hand lever, saying that it is a 10:1. So with 10# of input would be 100# output. From there we run down the cable to the brake levers. From my rough drawings I am figuring the two levers to lay at about 160 degrees, with about .125 of movement, figured this would allow for max leverage. Again, keeping this very simple I have figured those to be a 2:1 and I am guessing that with two levers that it will divide and not multiply (like I originally thought), the input force between the two levers. I am now concluding that with 100# input to the brake levers that it would equate to 100#s each lever, squeezing toward one another, totaling in 200# of combined braking/clamping/squeezing force.

Sound about right? Where did I go wrong, where did I go right?

Now I am not doing homework or building a spaceship, so exact numbers are not necessary. Just trying to figure how to stop the cart without ripping my shoulder out of socket, or flying off the rails. I suppose if anyone actually cared, I could give exact numbers and they could figure it out for grins and giggles lol.

Again, thank you very much for the information and taking time to deal with my elementary questions.

Joshua

I'm not sure why you asked your original question if you didn't want a way of working out the answer. If you want a more straightforward andswer to this problem you would be better on a cycling forum where people may have the results of measurements and also they will have much more practical experience.
You can rely on the fact that the brake linkage you have was designed to take care of an average load for a wheel of that particular width. The actual force you will have to supply will depend as much on the pedal / hand lever as on anything else.
You are right about multiplying the ratios together - but be prepared for some of the ratios to be less than, or near one.
One point - the two halves of the calliper do not double the force by acting against each other. It's done that way just to avoid sideways forces on the wheel. Theuseful force you produce with the cable is halved, in fact. (as with all cycle brakes)
Weight for weight and, bearing in mind that you will have four wheels (or at least a pair of them) braked then the effort you will need to stop you plus the engine etc on double the number of wheels wouls be about the same as for a bicycle. How do you balance the brake force to each side? That was always a problem for old motor cars before they used hydraulics,

As a scientist, my first suggestion would be to break down the problem into smaller, more manageable parts. Let's start with the first problem: determining the ratio of force between input and output. This can be calculated using the formula for mechanical advantage, which is output force divided by input force. In this case, the output force would be the force needed to stop the cart and the input force would be the force applied by the hand lever. To determine the output force, you will need to consider factors such as the weight of the cart and the speed at which it is moving. Once you have the output force, you can then calculate the mechanical advantage and determine the input force needed.

Moving on to the second problem, where you have two sets of levers on either side of the braking apparatus. This can be thought of as a compound lever system, where the output of one lever becomes the input of the next lever. The mechanical advantage of this system can be calculated by multiplying the mechanical advantages of each individual lever. So if each lever has a mechanical advantage of 2, the total mechanical advantage of the compound system would be 2 x 2 = 4.

Finally, for the third problem, where the braking system is connected to a hand lever with a mechanical advantage of 5:1 or 6:1. This means that for every 5 or 6 units of input force, you will have 1 unit of output force. So if you have a total mechanical advantage of 4 from the compound lever system, you will need to adjust the hand lever to provide an input force that is 4 times smaller than the output force needed to stop the cart.

In summary, the key to solving these multiple leverage math problems is to break them down into smaller parts and use the principles of mechanical advantage to calculate the necessary forces. It may also be helpful to draw diagrams or use physical models to visualize the system and better understand how the forces are acting. I would also recommend seeking assistance from a math or physics tutor if you are still struggling to solve the problems. Good luck!

## 1. What are multiple leverage math problems?

Multiple leverage math problems involve solving equations or inequalities that involve multiple variables and multiple levels of leverage or influence. This means that each variable has a different impact on the overall solution, and the solution may be affected by changes in any or all of the variables.

## 2. How do you approach solving multiple leverage math problems?

The best approach to solving multiple leverage math problems is to first identify all of the variables and their respective levels of influence. Then, use a combination of algebraic manipulation and substitution to solve for each variable one at a time, starting with the variable that has the highest level of influence.

## 3. What are some common strategies for solving multiple leverage math problems?

Some common strategies for solving multiple leverage math problems include isolating each variable on one side of the equation, using the distributive property to simplify equations, and using substitution to solve for each variable one at a time. It is also helpful to check your solution by plugging it back into the original equation or inequalities.

## 4. Can technology be used to solve multiple leverage math problems?

Yes, technology such as graphing calculators or online equation solvers can be used to solve multiple leverage math problems. These tools can help with graphing equations and inequalities, as well as solving systems of equations with multiple variables.

## 5. How can I improve my skills in solving multiple leverage math problems?

The best way to improve your skills in solving multiple leverage math problems is to practice regularly and seek help from a tutor or teacher if needed. It is also helpful to review algebraic concepts and techniques, such as factoring and simplifying expressions, as these skills are often used in solving multiple leverage math problems.

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