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Homework Help: Work Problem

  1. Apr 17, 2005 #1
    An arm has a mass of 7.0 kg. Treating the arm as if it were a single point mass m attached to a rigid massless rod, determine the work that must be done by the deltoid muscle to move the arm from position 1 to position 2. (Position one and two have an angle of 30 degrees between them and it is 21.0 cm from the shoulder to the elbow of the arm)

    By doing a little trigonometry, I figured out that the distance between position 1 and position 2 is about 0.12m. I will use the equation
    W(for work)=F*delta x*cos(angle)

    How do I find the force applied to move it from position one to two?
  2. jcsd
  3. Apr 17, 2005 #2


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    Your arm is in the earth's gravitational field. You will have to do work against this force. The minimum work you have to do to move your arm 30° is the work done BY GRAVITY on your arm as it moves 30°. So let's find that. If we could find exactly what HEIGHT your arm had moved, it would be easy: W = -mgh. But we can't apparently. So we'll stick with the definition of work done by a force, which is the line integral

    [tex]W = \int_{C}\vec{F}\cdot d\vec{r}[/tex]

    C is a portion of circle. Easy to parametrize. Let's do that..

    [tex]x(\theta) = 0.21 sin(\theta)[/tex]
    [tex]y(\theta) = -0.21 cos(\theta)[/tex]
    [tex]0\leq \theta \leq 30°[/tex]
    [tex]\vec{r}(\theta) = 0.21 sin(\theta)\hat{x} -0.21 cos(\theta)\hat{y}[/tex]
    [tex] \vec{r'}(\theta) = 0.21 cos(\theta)\hat{x} + 0.21 sin(\theta)\hat{y}[/tex]

    [tex]\Rightarrow \int_{C}\vec{F}\cdot d\vec{r} = \int_0^{30°} \vec{F}(\vec{r}(\theta)) \cdot \vec{r'}(\theta)d\theta = \int_0^{30°}-0.21mgsin(\theta)d\theta = 0.21mg(cos(30)-1) [/tex]

    So your muscle will have to do work in the amount -0.21mg(cos(30)-1).
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3

    Andrew Mason

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    The force is mg. What is W in terms of mass, g, and the height?

  5. Apr 17, 2005 #4
    Wow. that makes sense. That was very helpful. Even though I am not in calc based physics, I still understoon your reasoning. Thanks Quasar987 and AM both.
  6. Apr 17, 2005 #5


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    It turns out we can get the height exactly directly from trigonometry. It would require a drawing but the key is

    0.21 - h = 0.21 cos(30°)
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