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Moving a box with greatest acceleration

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The object is a box with a given mass m. Our person has the choice between pushing the box with a horizontal force, or pulling the box with a wire with an angle of θ=30o. The magnitude of Fo is the force vector he affects the box with in both cases.

    Should the box be pushed or pulled in order to achieve the greatest acceleration?

    2. Relevant equations

    I've made a drawing of each situation adding all the forces acting on the box.

    Situation 1
    http://myupload.dk/handleupload/c5fd9W9J18EBn

    Situation 2
    http://myupload.dk/handleupload/4a037-RJ3a5Ha

    3. The attempt at a solution
    I've been trying to write an equation describing the net-force in both situations. I've here made the assumption that in both cases n and Fg cancels each other out.

    For situation 1 I get:
    Fnet=Fo-ff

    I know that there's some sort of friction acting in the opposite direction and that it's given by ff=[itex]\mu[/itex]n

    For the force vector Fo I can apply newton's 2nd law for the x composant, since the box is not changing direction in the y-axis.
    Therefor I get:
    Fnet=max0-μn

    For situation 2 I get:
    Fnet=Fo*sin(θ)-μn

    The force vector must be dependent on the magnitude from the forces acting in the direction of x and y. By applying newton's 2nd law again I get:
    Fnet=mao*sin(θ)-μn

    Judging situation 1, by pushing with a horizontal force, the acceleration will be constant. However in situation 2, different angles of pulling will result in different accelerations. I believe that he will then achieve the greatest acceleration by pushing it.

    This to me is a difficult one to figure out so I would like some help on what I did wrong, it's important to me to learn.
     
  2. jcsd
  3. Oct 1, 2012 #2
    That assumption is true for the first situation, because there is no other force acting in the y direction. But think about situation 2, you're pulling the box at an angle. That is now applying a force in the y direction as well as the x direction correct? So your Fnety equation is no longer Fg - n
     
    Last edited: Oct 1, 2012
  4. Oct 1, 2012 #3
    I think I understand it. It's in a vertical equilibrium, which means [itex]\Sigma[/itex]Fy=0
    Which then yields:
    n=mg-F*sin(θ)

    Now I know that the object accelerates to the right thereby it's horizontal acceleration is given as:
    Fnet(x)=ma
    Which is the same as:
    ma=F*cos(θ)-μn

    Now I think I get another vision here.
    The magnitude of the acceleration is changed by the value of the friction (μn) since it is impacted by the value of the normal. It is therefor easier to accelerate the box by pulling up, thus decreasing the normal force. Hence situation 2 will yield the greatest acceleration.
     
  5. Oct 1, 2012 #4
    It depends, you have to remember you are also applying less force in the x direction, try plugging in some fake values to verify your prediction.
     
  6. Oct 1, 2012 #5
    Thank you I think I got it now.

    There is one final question that I can't seem to understand.

    Is it possible (if you can freely choose a value for θ) to keep the box moving without applying a force Fo that is greater than the gravety on the box?

    I don't think you can because then it wouldn't be moving? I honestly don't know how to prove it.
     
  7. Oct 1, 2012 #6

    SammyS

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    The normal force, n, depends on F0 and θ .

    You need to include the and solve for acceleration, a.
     
  8. Oct 1, 2012 #7
    Like this?

    n=mg-Fo*sin(θ)

    ma=Fo*cos(θ)-μ(mg-Fo*sin(θ)
    ma=Fo*cos(θ)-μmg+μFo*sin(θ)

    a=(Fo*cos(θ))/m -μg + (μFo*sin(θ))/m
     
  9. Oct 1, 2012 #8

    SammyS

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    Yes. This can be expressed as follows.

    [itex]\displaystyle \frac{F_0}{m}\left(\cos(\theta)+\mu_s\sin(\theta) \right)-\mu_s\,g [/itex]
     
  10. Oct 2, 2012 #9
    Thank you, it really helps :).

    There's only 2 questions left, and I'm having trouble with one of them.

    What is the optimized angle θ he could have pulled the box with to achieve maximum acceleration?

    I decided here to use the equation for a that we just found and defined it as a function of theta. After that I solved a'(θ)=0. Without including all the work I got that θ=arctan(μ). This angle would have given maximum acceleration I believe.

    The question I'm having trouble with:
    Is it possible (if you can freely choose a value for θ) to keep the box moving without applying a force Fo that is greater than the gravity on the box?

    I think both yes and no. Since he's moving it horizontally wouldn't that technically mean he should pull with a force Fo that is greater than the frictional force? However the gravity is pulling the box down, and shouldn't it mean that if he doesn't pull with a force greater than gravity, the box is not moving at all?
     
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