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Homework Help: Radians for Inclined PLane?

  1. Aug 29, 2006 #1
    The formula for calculating the force required to move an object (ignoring friction, etc.) up an inclined plane is:


    F = Mgsinθ

    Where:

    F = force (N)
    Mg = gravity (9.81m/s²)
    θ = angle of inclination (degrees out of 360, or radians?)


    The question I have is: what is used to describe the angle of the inclined plane? The SI unit for angles is the radian, but I get the impression that degrees out of 360 are used in the equation above?

    I tried to find an example calculation, but I failed to do so. If anyone knows whether radians are used or degrees, please let me know!
     
  2. jcsd
  3. Aug 29, 2006 #2
    You can use both radians and degrees to describe an angle. The equation will work work with both degrees and radians.
     
  4. Aug 29, 2006 #3
    But the results would vary immensely?

    For example, the sine of 40 degrees is 0.642787609

    And the sine of 40 degrees expressed as radians (0.6981317 rad) is: 0.012184395.

    So the result in Newtons would be very different if radians were used in place of degrees. All other units are SI, as mentioned in the nomenclature.
     
  5. Aug 29, 2006 #4
    If you are using a calculator, then you must put it in radian or degree mode depending on which one your are using.
     
  6. Aug 29, 2006 #5

    I am not following you?
     
  7. Aug 29, 2006 #6

    berkeman

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    Scientific calculators will have a configuration option to allow you to use either radians or degrees. Look for a configuration menu of some kind. On my old HP 42S, it's in the yellow-Modes menu (overloaded on the +/- button). You pick either degrees or radians, which tells the calculator which you will be using for the following trig calculations.

    So, sin(45) = sin(PI/4), as long as you use the appropriate mode for each calculation. As a practical matter, you will leave your calculator in one mode or the other most of the time.
     
  8. Aug 29, 2006 #7

    I really want to understand this without having to rely on a calculator. I am getting a little confused, I admit. All i know is that if radians are used in the formula (along with all the other SI units) then the actual result, the actual numbers generated, would be very different to those that would result from using degrees out of 360!

    Are there any example calculations anywhere for me to see how engineers go about using the above formula?
     
  9. Aug 29, 2006 #8
    The sin of 45 is 0.707106781, and the sin of pi/4 is 0.785398163.

    Two very different results!
     
  10. Aug 29, 2006 #9

    berkeman

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    Please stop doing that -- you're making me dizzy. You must use the correct argument to the trig functions, or of course you will get wrong answers.

    Degree arguments (2*PI radians = 360 degrees) are used for most applications, because they are simpler to visualize. The times you use radians are when you are doing math where the angle is not just in the argument to a trig function, like when you are using complex exponentials. Then, you end up with angles outside trig arguments, multiplying other numbers. It would be a mistake to use degrees in those instances. Have you studied complex exponentials yet? Like [tex]e^{j\theta}[/tex] ?
     
  11. Aug 29, 2006 #10

    berkeman

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    BTW, if you check the help function in Excel, you'll see that by default, the function SIN() expects the argument to be in radians. So if you have some other cell like B2 that has an angle in degrees, you need to be sure to do the conversion in the argument to the SIN() function to get the correct answer, like this:

    =SIN(B2*2*PI/360)

    or use the built-in conversion macro:

    =SIN(RADIANS(B2))
     
  12. Aug 29, 2006 #11
    As you have been told already, if you are using a calculator to get those values, you need to change the mode you're in. For example, if you open the Calculator program that comes with Windows and switch to scientific mode, you'll notice a group of three radio buttons labeled degrees, radians and grads. You can switch between degrees and radians to calculate the sine of an angle.
     
  13. Aug 29, 2006 #12
    I arrived at the dizzying number above by multiplying pi by 2, to give radians in a circle, then I divided that number by 360 and multiplied the result by 45 (being the angle in degrees) to give me the angle in radians.

    Was that wrong?

    So is the answer to my original question '45' for a 45-degree slope, for example? Or, if I were to seek the force in Newtons for a 45 degree slope would I first need to convert that into radians using the method I just described? Or is there another way?

    This is rather confusing!
     
  14. Aug 29, 2006 #13
    OK, I did that and the answer given by entering rads in radian mode was the same as the answer given by entering degrees in degree mode.

    Just what the calculator did is an enigma to me(edit: unless it simply converted the radians back into degrees?). I would like to understand that, before I can possibly have any confidence in my workings.
     
  15. Aug 29, 2006 #14

    berkeman

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    The answer to your original question is still that you can use either one you want. The question does not constrain you. But you have to be sure that you are using your calculator or computer correctly. For example, on my HP calculator, if I accidentally put it in complex mode, I get weird answers to some calculations that are not complex.

    So are you saying that you have your calculator in RAD mode, and take

    sin(2*PI*45/360) and you get 0.785...?

    Then you forgot to push the sin() button. 2*PI*45/360 = 0.785...

    and the sin(0.785) = 0.707....

    Do you see your error now? If not, try the Windows scientific calculator as e(hoOn3 suggests.
     
  16. Aug 29, 2006 #15

    What I did was to calculate 45 degrees in radians by: pi*2/360*45. Then I input that number into windows calculator, whilst in radian mode. The answer it gave me (0.707) was the same answer it gave me when I input 45 into the calculator (Windows) whilst it was in degrees mode.

    If anyone could please workout an example calculation, say for the force in Newtons to move a mass up a 45-degree inclined plane, in order to show me exactly how to do it (pretending that calculators and computers do not exist for a moment), I would be very grateful?
     
  17. Aug 29, 2006 #16

    berkeman

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    You have the correct equation in your original post (OP). If you don't have a calculator handy for figuring out trig functions for you, you can use a math table in something like the CRC math reference book. Or worst case, if you were on a desert island with only pen and paper, you could make a nice big accurate sketch of the sin() and cos() functions, and interpolate numbers with that. You could probably get answers to within say 10%, which isn't bad for being stuck on an island.
     
  18. Aug 29, 2006 #17
    So for a 45-degree inclined plane, do I literally enter the sin of 45 when using that equation?

    Or, as radians are the SI unit for angles, do I use the sin of 0.785398163rad (45 degrees expressed in radians)?
     
  19. Aug 29, 2006 #18

    berkeman

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    Depends on which radio button you have active in the Windows calculator, right?

    Sketch a sine wave on some graph paper, and label the horizontal axis with both degrees and radian tick marks. That should help you keep the picture straight.
     
  20. Aug 29, 2006 #19

    berkeman

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    And from that sketch of the sine wave, you can see how the value of sin(x) varies smoothly from 0 to 1 as you increase the argument from 0 degrees to 90 degrees (or 0 radians to PI/2 radians). And you see that the value of sin(x) = 0.707 at the halfway point (45 degrees or PI/4 radians). In your OP equation, that corresponds to zero force required for frictionless horizontal motion, and full Mg force required to lift the mass vertically. Make sense?
     
  21. Aug 29, 2006 #20

    berkeman

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    Oh, and back to the desert island thing. You can also get the sin(45 degrees) just from remembering that an equalateral right triangle with sides = 1 and hypotenuse = SQRT(2) has 45 degree angles, right? And so the sin(45) = 1/SQRT(2) = 0.707.

    It's good to memorize that 1-1-SQRT(2) triangle, and also the 3-4-5 triangle (quiz question -- what are the angles in a 3-4-5 right triangle?).
     
  22. Aug 29, 2006 #21
    Thank you all for your help. But I am still left without knowing how to use the original formula.

    Perhaps it is the stress I am under to get this project finished, that I am currently working on, that prevents me from understanding your posts.

    I am not a mathematics student; I am not a physics student. I do plan on studying these subjects through home-learning books starting very soon. The project I am working on now is for an engineering course I am starting in just over a month.

    If anyone would be so kind as to show me an example of the formula with working out and the answer in Newtons for any angle you choose, I really would be very grateful.
     
  23. Aug 29, 2006 #22

    berkeman

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    I have to go, but I'll try to give you a couple examples that may help. Sorry that we haven't been able to help you more with our explanations, but I'm sure you'll get it all soon. Just hang in there, and remember to stay positive while you study, and relax and focus.

    Staying in mks units:

    I push a 1kg mass up a 45 degree inclined plane:

    F = Mg sin(45 degrees) = 1kg * 9.8m/s^2 * sin(45) = 9.8 kg*m/s^2 * 0.707 = 6.9 kg*m/s^2 = 6.9N

    remember that the acceleration of gravity g = 9.8 m/s^2, and a Newton N = kg*m/s^2

    Now change the angle to 90 degrees so that I'm pushing the weight straight up:

    F = 1kg * 9.8 m/s^2 * sin(90 degrees) = 9.8N * sin(90) = 9.8N * 1 = 9.8N

    Hope that helps. Gotta run.
     
  24. Aug 29, 2006 #23

    berkeman

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    BTW, you can probably see that I like to carry units along in my calculations. That's an important trick for helping you to catch any mistakes along the way -- the units will often mismatch in a way that you notice (easier than a numerical mismatch usually).
     
  25. Aug 29, 2006 #24

    Thank you very much for that.

    I know now how to use the formula. I am still confused about the radian issue and how using radians or degrees in the same formula renders the same result in Newtons, as radians and degrees are completely different numbers, even though they represent the same thing.

    At least for now I can progress, with the intention of learning more later on. Oddly your 90-degree example is one I worked out myself a couple of months ago when I first found this equation (in the meantime I have been working on other aspects of the project) in order to test it. But I completely forgot that I did that and failed to think about doing the same thing again! Stress and anxiety are real brain-killers. Beware.

    I am sure the stress will not last all that much longer, only until the conclusion of this project (I hope).
     
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