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Newton's second law problem

  1. Oct 2, 2013 #1
    A clarinetist, setting out for a performance, grabs his 3.41-kg clarinet case (including the clarinet) from the top of the piano and holds it in the air with an upward force of 31.8 N. Find the case\'s vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative.

    _____ m/s^2


    Okay the formula is F(net) = ma
    What I did was 31.8/3.41 but it was wrong and I don't know what is wrong and why it is
     
  2. jcsd
  3. Oct 2, 2013 #2

    BruceW

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    hint: it may be a trick question. When he is holding the clarinet in the air, what is the acceleration of the clarinet? (think intuitively). Also, to explain this, think of what other forces may be acting on the clarinet.
     
  4. Oct 2, 2013 #3
    This is how I thought of the problem first and I had put -9.8 for the gravity but that was obviously wrong too. Then I used the formula and it was wrong as well.
     
  5. Oct 2, 2013 #4
    What formula have you used?
     
  6. Oct 2, 2013 #5
    I used newtons second formula which is the F(net) = am I just simply solved for a
     
  7. Oct 2, 2013 #6
    OK, and what is Fnet? You have two forces acting on the clarinet.
     
  8. Oct 2, 2013 #7
    Wouldn't it just be the 31.8N
     
  9. Oct 2, 2013 #8
    This is just one of the forces. Do you understand what the "net" in the F(net) means?
    It is the vector sum of ALL the forces acting on the body.
    In this case you have TWO forces, as it was pointed out already.
    One is the force of the hand, upward. The other one is the weight of the clarinet, downwards.
    What is the magnitude of this second one?
    And then what is the magnitude of the net force? Don't forget that the two forces are in opposite directions.
     
  10. Oct 2, 2013 #9
    Okay well the hand it using as much force to left the case just as much as the case is putting force in the hand. So would the net force be zero?
     
  11. Oct 2, 2013 #10
    We are talking about the clarinet, not the hand.
    There are two forces acting on the clarinet: weight and force FROM the hand.
    The net force ON THE CLARINET is only due to the force acting ON THE CLARINET. The two mentioned above. Forces on other objects, like hands, pianos, etc do not matter when we calculate net force on the clarinet.

    So let's try one more time.
    We have, on the clarinet, the force of 31.8 N upwards and the weight of the clarinet (calculate it) downwards. What is the resultant or net force?
     
  12. Oct 2, 2013 #11
    28.39 kg?
     
  13. Oct 2, 2013 #12
    No, the force is not measured in kg. The kg is for mass.
    You cannot subtract a mass from a force. Does not make sense.
    You need to find the WEIGHT of the clarinet. You are given the mass, 3.41 kg.
    Do you know how to find the weight, in N?
     
  14. Oct 3, 2013 #13
    W=mg which in my case is (3.41kg * 9.8 m/s^2) which is 33.42N?
     
  15. Oct 3, 2013 #14
    Yeah, so you have 31.8 N up, and 33.42 N down, so what is the net force?
     
  16. Oct 3, 2013 #15

    BruceW

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    yeah. that is right. The way the question is worded, it would suggest the two forces cancel out. But actually, compare the two forces, do they actually cancel out? And so what is the resulting force?

    edit: whoops, beaten to it.
     
  17. Oct 3, 2013 #16
    okay since the net force does not cancel out. The net force is 31.8N - 33.42N = -1.62N
    Would this be correct the net force would be negative?
     
  18. Oct 3, 2013 #17
    Since I found my net force...can I just plug it in to the formula F(net) = ma and solve for my vertical acceleration?
     
  19. Oct 3, 2013 #18

    BruceW

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    yep, go for it! you've got the idea. Although be careful with the number of decimal places in your answer. Remember that the value for g you have used is an approximate value. So when you get to the final answer, remember to use a number of decimal places which is sensible.
     
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