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Thread: Homework Question. Please help!

  1. #1
    Associate Engineer
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    Homework Question. Please help!

    I'm trying to learn about engineering. I realized early on that my math skills were lacking so I got Basic Engineering Mathematics by J. Bird. So far I haven't found a problem I couldn't figure out, but the following problem stumped me. The answer is in the back of the book but I cannot figure out how the answer was obtained. A step-by-step answer, showing how it was obtained, would be great, but I'll take any help I can get. Here it goes:

    A simple machine has an effort : load ratio of
    3:37. Determine the effort, in grams, to lift a
    load of 5.55kN.

  2. #2
    Technical Fellow
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    Hi MacB,

    We have rules about not helping with homework, but you seem to be making an effort to work through things, so I can give you a little hint.

    This is all about Moments, as in levers etc.

    I would also have you note that a Newton is a unit of Force, so think about unit conversions.

    That should get you started, but by all means come back and ask questions with examples of what you are doing and maybe, then we can guide you a little more. I can assure you though, there will be no outlined solution to answer your problem as it is against or TOS.

  3. #3
    Associate Engineer
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    MacB, thanks for the reply. I'm actually not a student at any institution so this isn't an attempt to cheat on my homework. I'm sort of trying to teach myself and am very new to the subject (as my question probably indicates) In fact, the answer is in the back of the book I am using, Basic Engineering Mathematics by J. Bird. It's 450g. Before I got into actual engineering, though, I wanted to brush up on my math. I was breezing through the problems until this question, and just do not understand it. I don't understand what is meant by "effort : load" and also haven't used the unit newtons before. I thought that if I could see how the answer was obtained I would understand these concepts. That's why I requested someone work the problem for me so I could see it. This is the only problem I haven't been able to figure out so maybe I'll just skip it and move on until I am able to study the more advanced moments or levers. But I would really appreciate if you or someone else could take the time to show me how the problem is solved (as I mentioned, the answer is provided) so I can investigate exactly what I am missing here. Thanks either way.

  4. #4
    Associate Engineer
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    I misread your name PinkertonD. Sorry.

    newton = the amount of net force to accelerate a mass of one kilogram at a rate of one meter per second squared.

    A simple machine has an effort : load ratio of
    3:37. Determine the effort, in grams, to lift a
    load of 5.55kN.

    Are you saying that both the effort and the load data can be found in 5.55kN? I don't even see how this relates to the other problems in the section. For example,

    The total mass of 120 household bricks is
    57.6kg. Determine the mass of 550 such
    bricks.

    Hooke’s law states that stress is directly pro-portional to strain within the elastic limit of
    a material. When, for copper, the stress is
    60MPa, the strain is 0.000625. Determine
    (a) the strain when the stress is 24MPa and
    (b) the stress when the strain is 0.0005

    It doesn't even seem like the same type of question. I understand these. The "effort, in grams"? Effort is measured in grams? Is there a source you can recommend that may get me up to speed on what I need to know to answer this problem?

  5. #5
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    I figured it out and now I feel totally stupid!
    It's fairly simple problem once you understand it. It was the Newton thing throwing me off.

    3 5.55kN
    -------=------------
    37 x, x = (3*5.55) / 37 = 0.45kg or 450g

    Thanks for the tips!

    Can anyone help out with this next problem. I'm pretty sure the answer the author gives in incorrect. But if someone else could verify that, I would be stoked.

    Boyle’s law states that, for a gas at constant temperature, the volume of a fixed mass of gas is inversely proportional to its absolute pressure. If a gas occupies a volume of 1.5 m3
    at a pressure of 200×10^3 pascals, determine (a) the constant of proportionality, (b) the volume when the pressure is 800×10^3 pascals and (c) the pressure when the volume is 1.25m3.

    My issues is with the "(c)" part of the problem. According to the authors (in the back of the book) the answer is 24 x 10^3 but (300 x 10^3) / 1.25 = 240 x 10^3. Please tell me I'm right and the author is wrong. Or else tell me how stupid I am and why?

    Again, I'm just trying to make sure I understand this stuff. Any help would be appreciated.

  6. #6
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    MacB,

    Don't beat yourself up on these things. You are delving into a new world of terms and relationships and it is not surprising you are getting a little confused.

    I applaud you efforts with the initial problem and resolving it to your understanding.

    All of Engineering is logical and the results can be easily followed through. You just need to slow down and take things in smaller steps. Also consider using extremes cases to get a better understanding of problems presented to you. By that I mean, think about a tiny box with a huge pressure or vice versa. Often using an extreme case will help you kinda-visualize any confusing issues involved.

    I would be surprised if the answers in the book were incorrect, but it is not beyond possibility.

    You need to slow down and think these things through. You seem to be expecting the answers without going through the steps to get there.

    Layout your workings in a neat an orderly fashion and include all steps no matter how tiny, or trivial they seem. That will show you each step during the transpositions. Once you become familiar with the process, then you can skip (combine) the tiny steps. But at this point you need to understand (feel) what is happening at each step.

    Finally, discard the "stupid" word. Just because something is now obvious once you have wandered down each little learning path, does not make you stupid. It is called "Education" and/or "learning." You are doing well.

  7. #7
    Principle Engineer
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    Quote Originally Posted by macbluff619 View Post
    My issues is with the "(c)" part of the problem. According to the authors (in the back of the book) the answer is 24 x 10^3 but (300 x 10^3) / 1.25 = 240 x 10^3. Please tell me I'm right and the author is wrong. Or else tell me how stupid I am and why?
    You are right.
    Hope you are "stoked".

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