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Mass being pulled on flat ground

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A case is being pulled along horizontal ground by means of a rope. Show that the force
    required to keep the case moving at constant velocity along the ground is a minimum when
    the rope is at an angle θ to the horizontal, where:

    tanθ=[itex]\mu[/itex]k


    2. Relevant equations

    F=ma
    FF=[itex]\mu[/itex]kN

    3. The attempt at a solution

    So I drew out the free body diagram and got my forces in the x and y directions.
    For the x-direction I got:
    FF-Fx = ma

    For the y-direction I got:
    N-W+FY=0

    I then got the N values on their own and equated the equations but couldn't really get anywhere with it. Any help that could put me in the right direction would be greatly appreciated.
     
  2. jcsd
  3. Apr 29, 2014 #2
    I think showing the steps of getting N and your two resulting equations would help, since I could maybe help you from there. What you essentially want to do is solve for F (force of rope) as a function of theta, and then see where it's a minimum. So replacing F_x and F_y with sine and cosine expressions, and then eliminating the normal force between the two equations.
     
  4. Apr 30, 2014 #3
    So FX = Fcosθ and FY = Fsinθ.

    I have N = FX/[itex]\mu[/itex]

    Where do I go from there?

    Thanks for the quick response by the way.
     
  5. Apr 30, 2014 #4
    No problem -- so now I would solve for N in the second equation (the y component) and then you can combine it with the x equation to eliminate N.
     
  6. Apr 30, 2014 #5

    Notice in the question it asks the force to keep a CONSTANT velocity along the ground ... what would that mean about the sum of your forces?
     
  7. Apr 30, 2014 #6
    Ok so in the y-direction I got N = W - FY.

    In the x-direction equation, FF = [itex]\mu[/itex]N and ma = 0 (a=0 because of constant velocity). I simplified this so N = FX/[itex]\mu[/itex].

    I then equated the two N values and got W - FY = FX/[itex]\mu[/itex].

    FY = Fsinθ and FX = Fcosθ

    I subbed them into the combined equation and got "W - Fsinθ = Fcosθ/[itex]\mu[/itex]"

    From here I didn't know where to go to answer the question.

    Hey Rellek does it mean they are equal to zero?
     
  8. Apr 30, 2014 #7
    If you draw the block and you draw the forces in each direction, it might be easier for you to envision all the forces that are acting on this block.

    Starting with the forces in the Y direction, you already know that the sum of these forces has to equal zero. So, obviously you have gravity acting downward and the Normal force of the ground pushing upward. You also have the Y component of the tension acting against gravity, in the direction of your normal force.

    For the X-direction, you are correct that the sum of these forces will also be equal to zero. The x component of your tension is pulling against the force of friction, and these are you only two forces in the x direction.

    I strongly suggest you draw your block and draw the forces that I have listed so it makes complete sense to you.

    After this, you already know the equation for the friction force. So, try solving for the normal force and use this to solve one of your other equations.

    For the last part, notice that this question is asking for the minimum, or vertex, of this equation, with respect to ##\theta##. This is an important addition in the problem that will allow you to use a calculus concept to solve this equation for its minimum point.
     
  9. Apr 30, 2014 #8
    Hey Rellek, I have done all that, that is where I got the equation W - Fsinθ = Fcosθ/μ

    Its the part at the end where you said to use a calculus concept that I'm stuck on.
     
  10. Apr 30, 2014 #9
    Well, try taking the derivative with respect to ##\theta##.
     
  11. Apr 30, 2014 #10
    And just to reiterate, you want to find the angle that gives you the minimum value for the force. So if you can solve for F in that equation, then you get an function in the form of: ##F(\theta)##, which just says that you give me an angle, and I'll give you a force. You can use differentiation to find the angle that gives the minimum force.
     
  12. Apr 30, 2014 #11
    Ok so differentiating W - Fsinθ = Fcosθ/μ (simplified to Fcosθ = [itex]\mu[/itex]W - [itex]\mu[/itex]Fsinθ) with respect to θ I got:

    -Fsinθ = 0 - [itex]\mu[/itex]Fcosθ

    I then got [itex]\mu[/itex] on its own for [itex]\mu[/itex] = sinθ/cosθ which is equal to tanθ.

    This is what the question shows but is this the final answer for the problem?
     
  13. Apr 30, 2014 #12
    Yup, that's the result they were asking for I think.
     
  14. Apr 30, 2014 #13
    Ok thanks for the all the help, much appreciated.
     
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