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D'alemberts principle help

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Just asking for a little help on this question

    50kg object is pulled across a rough horizontal surface with a uniform force of 250N for 15m from rest, the surface has a frictional coefficient of 0.4, calculate the acceleration using dealemberts principle



    But this results in a minus answer, so I'm lost, any help would be appreciated!

    2. Relevant equations

    Conservation of energy, d'alemberts principle

    3. The attempt at a solution

    I have worked out the acceleration using conservation of energy using

    PE+KE+Win = PE+KE+Wout+losses

    To find the final velocity, and used

    a=v^2-u^2/2s

    To find acceleration which comes to 1.08m/s^2

    Now the other part of the question is use d'alemberts principle to solve the same problem but I cannot for the life of me work this out, it's probably really simple but I just can't see it

    I thought I would go down the route of force in minus the forces against = 0 to find the inertia force and use that to find acceleration, with F = -ma

    But this results in a minus answer, so I'm lost, any help would be appreciated!
     
  2. jcsd
  3. Apr 19, 2016 #2
    I think you have already used the de Alembert's Principle in a way, after all how do you evaluate losses, Perhaps You think you have used Energy conservation Principle. But the principle that you have used, Newton's Laws of motion and de Alembert's principle are equivalent to each other and derivable from each other. Just think over. Solving a problem mechanically gives you pleasure may be but not insight. try to develop insight.

    Also so called inertial force is not a force at all. That is also creating a conceptual problem, it is the name given to the product ma and because you think it is inertial you give it a minus sign but then you keep it on left hand side only try to equate the sum to zero. Sum of all forces including the so called inertial force = 0 is the principle. Personally I do not like the idea of inertial force at all. It creates more conceptual problems than solve them.
     
  4. Apr 20, 2016 #3
    Completely agree on insight, but I'm struggling atm, just a bit of back ground, I'm not a maths or physics student, I'm doing a career change atm and have to study mechanical principle at this level, we have to go through so many topics that I have to admit I very rarely understand the topics we do but rather answer most of the questions parrot fashion, which so far has been surprisingly successful!

    But more into the working out



    Car Mass: 50kg
    Co-efficient of friction is 0.4
    N = (mg) = 50 x 9.81 = 490
    Frictional Resistance (Fr) = μN = 0.4x490 = 196N
    Distance = 15 meters.

    PE + KE + Win = PE + KE + Wout + Losses
    0 + 0 + (F x Distance) = 0 + (1/2 mv^2)+ 0 + (Fr x distance)

    (F x Distance) - (Fr x distance) = 1/2 x mv^2

    (250N x 15 meters) – (196 x 15 meters) = 1/2 x 50v^2

    (250 x 15 meters) – (196 x 15 meters) / (1/2 x 50) = v^2

    32.4 = v^2

    5.69 = V

    Velocity is 5.69 m/s
     
  5. Apr 20, 2016 #4
    Then using

    a=v^2-u^2/2s

    I come to 1.08m/s^2
     
  6. Apr 20, 2016 #5
    Would it be simply

    Force - frictional force

    250-196= 54

    F=ma

    54 = 50a

    a = 54/50

    This equals 1.08 m/s^2 in my first answer
     
  7. Apr 20, 2016 #6
    However I thought I must use. F = -ma??
     
  8. Apr 20, 2016 #7
    That is the expression for inertial force. Newton's law do not hold good for accelerated frames. So it does not hold good in the frame of the particle where it is at rest. But if we introduce the concept of pseudo force which defined as -ma, where a is the acceleration of the particle in inertial frame, then the law can be made hold true. So in its own frame acceleration = 0 so net force including inertial force = 0 so
    Fnet = Fnet in inertial frame - ma = 0 which is nothing but the expression of Newton's law F net = ma in inertial frame. But note that ma is not the result of any interaction. But all those forces thatyou talked about in this problem relate to some interaction except -ma!
     
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