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Work and ramps

  1. Feb 26, 2006 #1
    Recently my AP Physics class had a test on energy. The following is one of the questions verbatim from the test. I just can't work out the problem to any of the choices, let alone the "correct" choice. The problem is:

    Movers want to set the ramp of their truck so that the work they do against gravity and friction is a minimum for crates moving up the ramp with constant velocity. µ is the coefficient of kinetic friction and x is the angle between the ramp and the ground. For the work to be a minimum, they must choose:

    a. tan x = µ
    b. tan x = -µ
    c. tan x = -1/µ
    d. tan x = 1/µ
    e. tan x = 1 - µ


    My reasoning is as follows:
    Since the truck's height is unchanging, the work done against gravity should be constant regardless of the angle of the ramp; work agianst friction is the only variable work. To minimize the work against friction, we should try to minimize the normal force and the actual length of the ramp. The shortest ramp length AND the minimum normal force both occur at x = 90° (this isn't much of a ramp, but oh well). Wouldn't this be the correct answer? The "correct" answer is tan x = 1/µ, but I just don't see how this is possible.

    Questions on these tests are often wrong, which is unfortunate.


    Thanks in advance,
    Andrew
     
  2. jcsd
  3. Feb 27, 2006 #2
    The answer is [itex]tan\theta=\mu[/itex]
     
  4. Feb 27, 2006 #3

    Eus

    User Avatar

    Mmm... I think...

    Hi Ho! :smile:

    The correct answer is [tex]\tan(x)=\frac{1}{\mu}[/tex].

    You're correct.

    Those are synonymous with varying the ramp's angle.

    Well, don't go straight to that point. Using calculus to guide us to a good common sense will prove that your reasoning above is not correct.

    Please look at the attached picture.
    Because the velocity of the crate is constant, there is no acceleration.
    Thus,
    [tex]Fx=-Fpush+Fg(\sin(x))+Ffriction=0[/tex]
    [tex]Fy=N-Fg(\cos(x))=0[/tex]
    Next, doing the [tex]Fy[/tex], [tex]N=Fg(\cos(x))[/tex]
    Therefore, because [tex]Ffriction=\mu N[/tex], substituting it to the [tex]Fx[/tex] equation will result as, [tex]Fg(\sin(x))+\mu N=Fpush[/tex]
    Remember that we have already had the value of [tex]N[/tex] as the result of doing the [tex]Fy[/tex].

    Now we know that [tex]Fpush[/tex] depends on x as shown below.
    [tex]Fpush(x)=Fg(\sin(x))+\mu Fg(\cos(x))[/tex]

    Next, we use calculus to find its minimum value as follows.
    [tex]\frac{d(Fpush)}{dx}=Fg(\frac{d(\sin(x))}{dx}+\mu \frac{d(\cos(x))}{dx})=0[/tex]
    Then, [tex]\frac{d(Fpush)}{dx}=Fg(\cos(x)-\mu \sin(x))=0[/tex]
    [tex]\cos(x)-\mu \sin(x)=0[/tex]
    [tex]\cos(x)=\mu \sin(x)[/tex]
    [tex]\tan(x)=\frac{1}{\mu}[/tex]

    Hopefully, this solves your problem.

    Your welcome :biggrin:
     

    Attached Files:

    Last edited: Feb 27, 2006
  5. Feb 27, 2006 #4
    I'm not sure I get it. :(

    You've minimized the Fpush, which I guess minimizes the work.. but I can't see how the work done against the frictional force is a minimum here. Maybe if you could explain to me why 90° is incorrect, I would understand better.

    Thank you
     
    Last edited: Feb 27, 2006
  6. Mar 1, 2006 #5

    Eus

    User Avatar

    Mentor, we need help!

    Mmm... OK! :smile: Let's assume that [tex]\mu = \frac{1}{2}[/tex]. So,
    [tex]Fpush( \arctan ( \frac{1}{ \frac{1}{2}}))=Fg( \sin ( \arctan ( \frac{1}{ \frac{1}{2}}))+( \frac{1}{2}) \cos ( \arctan ( \frac{1}{ \frac{1}{2}})))[/tex]
    [tex]Fpush( \frac{1}{2})=1.12 Fg[/tex]

    If we use your argument that x should be 90°,
    [tex]Fpush(90°)=Fg( \sin (90°)+( \frac{1}{2}) \cos (90°)[/tex]
    [tex]Fpush(90°)=Fg[/tex]

    It turns out that when using a ramp, at a certain angle that is [tex]\tan (x) = \frac{1}{\mu}[/tex], the [tex]Fpush[/tex] will reach a maximum point.

    Well, I found this out when I try all of the possible value of x using OpenOffice.org Calc.There, the minimum value of [tex]Fpush[/tex] happens when the [tex]x=0[/tex] (using [tex]\mu < 1[/tex] and the maximum value happens when [tex]x= \arctan ( \frac{1}{\mu})[/tex]

    Well, we need a help from a mentor, guys!
    Mentor, we need help, please! :biggrin:
     
    Last edited: Mar 1, 2006
  7. Mar 8, 2006 #6

    Eus

    User Avatar

    Hi Ho!

    I've just discussed this problem with an assistant of school's physics laboratory and it turns out that your answer on choosing x=90° is correct. In fact, when you use a ramp, you'll get an angle that maximizes your Fpush.

    Thank you.

    PS: What's wrong with the LaTex generator? It can't generate the LaTex codes in my previous post correctly. I wonder.
     
  8. Mar 8, 2006 #7
    Thank you :) I would have replied to your other post but I was actually waiting for the latex to be generated this whole time, heh.




    Thank you again :)
     
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