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Easiest way to pull an object

  1. Jul 1, 2009 #1
    I found interesting to me this topic and tryed to analyze.
    http://img505.imageshack.us/img505/6655/49209931.jpg [Broken]

    At which angle it is eseaiest to pull an object?
    I got equation
    [tex]F=\frac{\mu mg}{cos(\alpha)(\mu tan(\alpha)-1)}[/tex] or [tex]F=\frac{\mu mg}{\mu sin(\alpha)+cos(\alpha)}[/tex]


    I dunno this math very well unfortunately, but I would be interested to see a plot:
    how mostly optimum angle is dependent from mu, and for example then mu is 1 alpha is 45, and so on...

    heh sorry for the paint and spelling...
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 1, 2009 #2
    Re: Easyest way to pull an object

    It depends on mu actually. Think of 2 extreme cases. First, mu is 1 (friction is extremely high). In which case X = Y, the answer is 45 degree as you said. But if mu is 0 (frictionless), the answer is clearly 0 degree (the force required to move the object is close to zero).

    I'm not sure though; ha haaa

    Mr Peetiya
     
  4. Jul 2, 2009 #3
    Re: Easyest way to pull an object

    You're looking for a maximum. Use your calculus brain. How does one find the maximum or minimum of a curve?
     
  5. Jul 2, 2009 #4
    Re: Easyest way to pull an object

    I would use calculus if i know how to use it, in school i do not have lessons with it, unfortunately, since I am 11grader. But infact i know quite a lot about it. But still I dunno how to found derirative of the bottom. [tex]f(\alpha)=\mu sin(\alpha)+cos(\alpha) f'(\alpha)=\mu cos(\alpha)-sin(\alpha)[/tex] ?? When to make it to zero and solve it? How to solve what equation when? It's homogenic .. divide by cos alpha and whola? :)
    And does best angle depends from [tex]\mu[/tex] lineraly? In fact this topic is quite clear, just math's is not very clear.
     
  6. Jul 2, 2009 #5
    Re: Easyest way to pull an object

    Well, you seemed to have gleaned the important part of the derivation anyways.

    [tex] f(\alpha) = \frac{\mu mg}{\mu sin(\alpha) + cos (\alpha)}[/tex]

    [tex] f'(\alpha) = \frac{\mu mg (\mu cos(\alpha) - sin(\alpha))} {(\mu sin(\alpha) + cos(\alpha))^2} = 0 [/tex]

    because [tex]\mu mg[/tex] is constant, and the denominator can't be zero, we can simplify this problem a bit by focusing on the only part that can be zero.

    So, you see why I thought you did well on calculating the f' you did:
    [tex]\mu cos(\alpha) - sin(\alpha) = 0 [/tex]

    Then

    [tex]
    \mu cos(\alpha) = sin (\alpha)
    [/tex]

    So,
    [tex]
    tan(\alpha) = \mu
    [/tex]

    Then arc tan both sides to get an explicit value for alpha. By analyzing this function, e.g. you can see that if there is no friction mu = 0, the best pull is the directly horizontal pull (i.e. alpha = 0). If you have mu = 1.0 the best pull is at alpha = 45 degrees.

    If you don't know calculus, this problem probably seems a little out of your league. But I am fairly confident there is a geometrical approach as well, that is within your limits if you're an intelligent high school student, or a bored undergrad. If you have the time, and are still curious, I'd recommend seeking out that way.
     
  7. Jul 3, 2009 #6
    Great, thanks!
     
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