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Positive vs. negative work in an elevator?

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    You are moving into an apartment and take the elevator to the 6th floor. Does the force exerted on you by the elevator do positive or negative work when the elevator (a) goes up and (b) goes down? Explain your answers.

    Suppose your weight is 685N and that of your belongings is 915N. (a) Determine the work done by the elevator in lifting you and your belongings up to the 6th floor (15.2m) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?


    2. Relevant equations

    W=Fdcos(theta)

    3. The attempt at a solution

    For the concept question, I know that work is negative when it is going in the opposite direction of the force, but I'm thrown off by which force. The question indicates that there is no acceleration, so there is a gravitational pull downwards equal to a normal force upwards. So which force is supposed to be opposed?

    As for the second part, this is what I got:
    685+915=1600N
    1600(15.2)(cos0)=24320J
    685(15.2)(cos180)=-10412J
    BUT here's what confuses me - since displacement is a vector quantity, shouldn't one of the above equations have a negative displacement since it's going in the opposite direction?

    Argh, I've been mulling over this and I'm still so confused. Please help!
     
  2. jcsd
  3. Oct 26, 2011 #2

    cepheid

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    Welcome to PF thepegmeister,

    Here is the answer to that question:


    Say you ascend at a constant speed. You're right that 0 total work is done on you for most of the ride, because the work done depends on the net force on you, and if the normal force on you from the elevator floor just balances your weight, then the net force on you is 0.

    However, the work done by the elevator normal force is non-zero and positive, since it exerts a force on you that is in the same direction as your displacement. It's just that this positive work done by the elevator on you happens to be exactly cancelled by the negative work done on you by gravity.

    Bear in mind also, that in order to even get you moving in the first place, the elevator initially has to exert an upward normal force on you that is higher than your weight, so that there will be a net upward force on you, and hence an upward acceleration. It is only once you have accelerated up to a certain speed that the elevator force is lessened until it just balances your weight and you continue to ascend at a constant speed (which is much more comfortable).

    So, in other words, at the very beginning of the ride, a non-zero and positive amount of work is done on you by the elevator in order to get you in motion in the first place.

    That's the conceptual summary for the ascent. Now can you do it for the descent?
     
  4. Oct 26, 2011 #3

    cepheid

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    The full equation for work is [itex] W = \vec{F} \cdot \vec{d} [/itex]. In other words, the work is the dot product of the force vector with the displacement vector. There are TWO ways to evaluate the dot product. The first way is component-wise. If you resolve each vector into x, y, and z components, you get:

    [tex] \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} [/tex]

    [tex] \vec{d} = d_x \hat{i} + d_y \hat{j} + d_z \hat{k} [/tex]

    where i, j, and k are unit vectors in the x, y, and z directions respectively. The components dx, dy, and dz are the displacements in the x, y, and z directions respectively, and they can each be either positive or negative. Same with the individual components of the force So, anyway, using components, the definition of the dot product is:

    [tex] \vec{F} \cdot \vec{d} = F_x d_x + F_y d_y + F_z d_z [/tex]


    Now, it turns out, that this is a one-dimensional problem. Both vectors are in the "vertical" direction, which we'll call the y-direction. Therefore, the vectors reduce to:

    [tex] \vec{F} = F \hat{j} [/tex]

    [tex] \vec{d} = -d \hat{j} [/tex]

    Where [itex] F = |\vec{F}| [/itex] and [itex] d = |\vec{d}| [/itex]. In other words, F and d are the magnitudes of their respective vectors. In other words, we're saying that the y-component of the force was the entire force, and that it was in the positive y-direction: [itex] F_y = |\vec{F}| [/itex]. We're also saying that the y-component of the displacment was, in fact, the entire displacement, and it happened to be in the negative y-direction: [itex] d_y = -|\vec{d}| [/itex]. With that being done, the dot product becomes:

    [tex] \vec{F} \cdot \vec{d} = F_y d_y = F(-d) = -Fd [/tex]

    With this first method, we explicitly included the negative sign on the displacement vector to indicate that it was in the negative y-direction.

    The second way to evaluate the dot product (which is the way you are doing it) is to use the fact that the dot product is ALSO given by the expression:

    [tex] \vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\theta = Fd\cos\theta [/tex]

    where theta is the angle between the two vectors. I put in the middle part in the equation above to emphasize that, by definition, F and d here are the magnitudes of the vectors. As a result, they are both positive, and explicit signs indicating direction are not put in. Using this method, the fact that the vectors are in opposite directions is already taken care of by the cosine factor, since cos(180°) = -1.

    That was really long-winded, and the actual answer to your question was just the statement in bold and italics above.
     
    Last edited: Oct 26, 2011
  5. Nov 6, 2011 #4
    Thank you guys so much! I wish they would explain these things in the textbook - the small details always get me.
     
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