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Having trouble with applying Newton's 2nd Law

  1. Feb 21, 2006 #1
    Here's the problem:

    An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 66 N when the cab is standing still. What is the reading when the cab is moving upward (a) with a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s while decelerating at a rate of 3.9 m/s2?

    In the sample problem given in my text, they state that for any velocity, you can just plug in the formula: Fn = m(g + a).

    Plugging in the numbers into this formula gets me a pair of wrong answers.

    However, a difference I noticed between the two problems is that their version doesn't give a speed and that in theirs, a man is standing on a scale while in the elevator, as opposed to an object hung from a spring balance.

    Why is my answer wrong, and how do I need to adjust my formula to obtain a correct answer?
  2. jcsd
  3. Feb 21, 2006 #2
    Something that occurs to me is that since the object is hanging as opposed to on the ground, no normal force is involved. That would mean that tension is involved instead since the object is hanging, but again I am unsure as to how to apply this, though to be fair I will try to do it myself pending assistance from the kindly folks here.

    Well, drawing a free-body diagram, I identify two forces at work. The tension on the spring, which is pulling upward, and the force of gravity, which is pulling downward. So only the y-axis is involved, thankfully simplifying the problem. Since the cab is standing still initially, T must equal 66N, but this is where I fall short. I have no idea how to factor in the velocity....
  4. Feb 21, 2006 #3
    Remember that Newton's 2nd law is expressed in terms of vectors, be careful about the signs of acceleration. Define a direction to be positive or negative and make your calculations accordingly.

    [tex]\vec{F} = m(\vec{a} + \vec{g})[/tex]
    You could simply plug that because of the 3rd law of Newton (also take care about what 3rd law states about the direction of forces).
  5. Feb 21, 2006 #4
    I don't understand how that formula is different from mine...
  6. Feb 21, 2006 #5
    The signs. A vector can be negative.
  7. Feb 21, 2006 #6
    But wait. If the scale in question is measuring mass and I'm looking for the reading both ways, couldn't I plug in 66 (the tension force) for F, the accelerations mentioned, and the gravity, then solve for m to get the answer?
  8. Feb 21, 2006 #7
    The scale is measuring force (initially weight), not mass. What's actually going on is this:
    the earth exerts a gravity force, which is [tex]m\vec{g}[/tex]. When elevator accelerates, the elevator also applies a force to that mass (via tension of spring & connection stuff). So you have actually 2 forces.
    [tex]\vec{F_g} = m\vec{g}[/tex]
    [tex]\vec{F_{elevator}} = m\vec{a}[/tex]
    [tex]\vec{F_{total}} = }\vec{F_g} + \vec{F_{elevator}} = m\vec{g} + m\vec{a}[/tex]

    [tex]F_g[/tex] is where you can plug in 66. But again, be careful about the sign of [tex]\vec{a}[/tex], which is due to [tex]\vec{F_{elevator}}[/tex].
    Last edited: Feb 21, 2006
  9. Feb 21, 2006 #8
    But that's the thing. I'm not getting the right answer, signs or otherwise.

    For instance:

    For Fg, I plugin 66 * 9.8, since the elevator is going upwards.

    For Felevator, well, in part a, it mentions that the velocity is constant, which suggests it's 0, and any number times zero is zero, so really part a is just Fg. But when I plug that in I get 646.8. Which is the wrong answer. Changing the sign doesn't make it correct either. I have a sinking suspicion I'm not converting the units properly though..
  10. Feb 21, 2006 #9
    :) you're actually on the right track. In part a, F_g = -F_{elevator}, which is the balance condition. The a in the equations I've written is not [tex]a_{elevator}[/tex]! To clarify, in balance condition, call the a in that equation to be [tex]a_0[/tex], and let [tex]a_{elevator}[/tex] be the acceleartion given if your question. Then, a (i've written) turns out to be [tex]a = a_0 + a_{elevator}[/tex]. I'm sorry for not making my notation clear.

    As for your solution, BTW:

    (choosing down to be positive)
    for the balance condition: [tex]F_{spring} = mg[/tex], which yields m if you consider it without [tex]a_0[/tex] (see below).
    for the second condition: [tex]F_{spring} = mg - ma_{elevator}[/tex].
    Last edited: Feb 21, 2006
  11. Feb 21, 2006 #10
    Well wait, so this problem requires the 3rd law as well, or no?
  12. Feb 21, 2006 #11
    Sort of. Consider the initial balance condition. Suppose you're observing the motion of the mass, on the ground. You see that it's going upwards, with a constant speed. You know 2nd law of Newton and say, "oh yea, since there's no acceleration, there shouldn't be any net force on it". But when you look at the balance, you do see some force. What you just saw is, the force that pulls the spring downward within the balance. But, since the spring is moving with a constant speed, ie has no acceleration, total force should be 0, by the 2nd law. So there should be somewere, with in the system, equal to that force in magnitude, and in the opposite direction (this can be predicted by the 3rd law also), so that the the sums of all forces within the system in the one pulling the elevator. And that is the force pulling m upwards, balancing out the weight.

    More clear?
    Last edited: Feb 21, 2006
  13. Feb 21, 2006 #12
    Well actually, would it be possible to start over? I'm sorry if I don't seem to be getting it, it's just that I have an exam this Thursday and I'm a little nervous and also a little behind in the class. I'm trying to learn about Newton's 2nd law and applying it to problems (without friction), and then motion problems with friction, and then energy and work (3 chapters' worth) within the span of today and tomorrow, so I'm a little confused still.
  14. Feb 21, 2006 #13
    I've edited & tried to clarify the previous post.
    Rather that staring from scratch, tell me where i've lost you, so we can work it out.
  15. Feb 21, 2006 #14
    Well, what I don't understand is how to setup the equation.

    I can identify two forces acting on the object hung from the balance. The first is what's giving the reading, which is 66 N. So while the elevator is still, the object weighs 66N. This makes the 66N the downward force since the object is pushing down on the scale. So this is the weight. Since w = mg. So if I know that w = 66, can't I obtain its mass by taking 66 = m(9.8) and solving for m?

    That said, I'm confused as to what the upward force is, possibly tension force on the spring?
  16. Feb 21, 2006 #15
    Yup, you can get m, which is 66/9.8.

    I guess confusion arose becuase of [tex]a_0[/tex]. I've tried clarifying 9th post by removing [tex]a_0[/tex] it from equation and introduced [tex]F_{spring}[/tex], what you seemed to have liked :)
  17. Feb 21, 2006 #16
    Okay but, now that I have m, going back to the basic equation, I've got:

    Fnet = (66/9.8)a.

    I need one more variable solved but am not sure where to go next and if there's even an upward force at all.
  18. Feb 21, 2006 #17
    Actually nevermind, I suddenly realized what I was doing wrong.

    I was confusing the 66N for the mass when in actuality it is his weight. If I take w = 66, and use the formula w = mg, I can plugin 66 = m(9.8), then get his mass.

    Once I have his mass, I can then plug in the formula you originally told me (F = m(a + g)) and get the answer, and this got the correct answer for both problems.

    So after looking at it, there's only one force acting on the object when the elevator isn't moving, which is the weight. However, when it moves, there's a second downward force, or its weight increases because of gravity. However, while I see how to get it, I don't see why gravity has to be considered again.
  19. Feb 22, 2006 #18
    Whoops! Hmmm, let me try again, from scratch.

    Suppose, we have two pal, Joe and Moe. Joe is watching the event on the ground, and Moe is inside the elevator, can see stuff in detail.

    1st case: constant velocity
    Joe observes that elevator and Moe has a constant speed. By the 2nd law of Newton, he concludes that there should be no net force on the elevator. There might be some internal forces maybe, but overall, they should all cancel and leave no net force. Moe, who's inside can see the mass m closer, seeing the mass standing still, concluding there should be no net force on the mass as well. But he remebers that there should be a force, it's weight, downward! Then why it isn't accelerating. Then he notices the the spring it was attached to, with a rope, suspects whether if this rope is applying a force to it. He concludes that he doesn't know, and calls this force [tex]F_{unknown}[/tex]. Maybe is doesn't exists at, that is it's equal to zero.
    Then Moe writes down the 2nd law:

    [tex]\Sigma F = m.0 = mg + F_{unknown}[/tex]
    (downwards is positive, and upwards is negive)
    Since he knows that [tex]\Sigma F = 0[/tex]

    [tex]0 = mg + F_{unknown}[/tex]
    and [tex]F_{unknown}[/tex] turns out to be [tex]-mg[/tex]. What he's reading on the spring balance is [tex]mg[/tex], so it appears that what we read on the balance is [tex]-F_{unknown}[/tex]. In detail, this is due to the 3rd law, but let's skip that for now.

    2nd case:
    Joe observes that the elevator is accelerating downwards (positive, according to our previous sign agreement), with a. So, he thinks that the whole elevator is under a net force, pointing downward. This time, Joe decides to observe in detail, and notes that everything inside the system, should eventually accelerate at a. So should the mass m. Thinking whether he can write down the 2nd law, and remembering Moe's suspects about the rope that m is connected to, he comes upon this:
    [tex]\Sigma F = ma = mg + F_{unknown}[/tex]
    and this time, the unknown force turns out to be [tex]ma-mg = F_{unknown}[/tex], again non-zero. Remebering what Moe found in the previous experiment, he predicts that Moe should read [tex]-F_{unknown}[/tex] on the balance and says that it should be -(ma-mg), or m(g-a). Moe verifies the result, and that's the happy end!

    So, it turns out that when accelerating downwards, it's weight decreases! But of course, this depends how you depend weight. Weight would decrease if you define it to be what you see on the balance.
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