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How can net force point in the direction of *negative* work?

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    It was just a basic work problem:

    The two ropes seen in the figure below are used to lower a 245 kg piano 6.0 m from a second-story window to the ground. How much work is done by each of the three forces? (T1 = 1910 N andT2 = 1140 N)

    There was a diagram with the angles labeled, and I got the right answers:
    w: 14406 J
    T1: -9925 J
    T2: -4837 J

    2. Relevant equations

    W= Fcos(theta)d

    Fnety= T1y + T2y - w

    3. The attempt at a solution

    Actually I got the problem right- T1 and T2 perform negative work and w performs positive work. But when I tried calculating the net force I found that it pointed upwards. This seems contradictory, since the piano has a displacement downwards. How can you have a net force opposite the displacement?

    I guessed that if the net force pointed downwards too the weight would be exceeding the tension forces and the ropes would snap; in other words, you have to have a net force upwards so that the tension forces are strong enough to keep the piano in place. Is that right? But then I don't know where the displacement is coming from. Would you say that the piano is in dynamic equilibrium and that it's only moving because you're increasing the length of the ropes?

    I'm confused...
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 4, 2009 #2


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    Imagine throwing a ball straight up in the air. Neglect friction. While the ball is going up, the displacement is up but the net force (gravity) is down and opposite to the displacement. Note that the speed of the ball decreases as it moves up.

    Same thing with the piano. As the piano is lowered its downward speed decreases because the net force (and therefore the acceleration) are opposite to the velocity.
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