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Newtons 2nd law

  1. Feb 10, 2007 #1
    (1)An elevator is being lifted up and elevator shaft at a constant speed by a steel cable. All frictional effects are negligible.
    In this situation, forces on the elevator are such that?

    1. the upward force by the cable is greater than the downward force of gravity.
    2. the upward force by the cable is smaller than the downward force of gravity.
    3. the upward force by the cable is equal to the downward force of gravity.
    4. None of these. (The elevator goes up because the cable is being shortened, not because an upward force is exerted on the eleva-
    tor by the cable.)
    5. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air.

    Working: my choice is 1, my thought is that since its moving upwards it must mean that the upward force is greater than the downward forces....i neglected 5 since it talks of downward force due to air which is some kind of friction force and the question says to ignore....my choice 1 turned out wrong any idea whats the correct one??

    (2)A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest point and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?
    1. Net constant force up the ramp
    2. Net decreasing force up the ramp
    3. Net increasing force down the ramp
    4. Net constant force down the ramp
    5. Net force of zero
    6. Net increasing force up the ramp
    7. Net decreasing force down the ramp

    Working: i'm confused between them so i did a process of elimination and cancelled 5 thinking that since its moving downwards there must be a force acting on the toy car i also cancelled choices 1, 6 and 7...1 and 6 because it says force acting upwards which i think isnt possible since the car is moving down the ramp and 7 because if its moving down the ramp the force should be increasing right?

    (3)(a)A mass of 1.4 kg lies on a frictionless table, pulled by another mass of 3.1 kg under the influence of Earth's gravity. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration a of the two masses? Answer in units of m/s^2.
    [​IMG]

    Working:
    For the 1.4kg mass: N(normal force) -1.4g=0
    N=1.4g
    T(tension)=1.4g

    For the 3.1kg mass:
    T(tension)-3.1g=-3.1a using T=1.4g and subsituting in
    1.4g-3.1g=-3.1a
    therefore a=5.374193548 m/s^2

    i'm not sure at all if this method is right or if theres something i'm missing

    (b) T, a and g represent positive quantities. Which equation is correct?
    1. T + (1.4 kg) g = (1.4 kg) a
    2. T - (1.4 kg) g = (3.1 kg) a
    3. (3.1 kg) g ¡ T = (3.1 kg) a
    4. T - (3.1 kg) g = (3.1 kg) a
    5. T - (3.1 kg) g = (1.4 kg) a
    6. (3.1 kg) g - T = (1.4 kg + 3.1 kg) a
    7. (3.1 kg) g - T = (1.4 kg) a
    8. T - (3.1 kg) g = (1.4 kg + 3.1 kg) a
    9. T - (1.4 kg) g = (1.4 kg) a
    10. T + (1.4 kg) g = (3.1 kg) a

    Working: i chose eqn4 since thats exactly what i got for part (a) but since theres a possiblity of (a) being wrong i'm not sure
     
  2. jcsd
  3. Feb 10, 2007 #2

    Hootenanny

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    Welcome to the Forums Fang,
    You may wish to reconsider your answer here. Consider Newton's First law and this excerpt from the question;
    I suggest that you draw a free body diagram here and lable all the forces acting.
    What exactly leads you to the conclusion that the tension must be 1.4g? Again, here I suggest you draw a diagram and label all the forces acting.
     
  4. Feb 10, 2007 #3
    newtons first law?? i'm even more confused!! how does newtons 1st law apply here?

    i'm drawing some free body diagrams now
     
  5. Feb 10, 2007 #4

    Hootenanny

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    Newton's First Law;
    "Also called the "law of inertia," Newton's first law states that body at rest remains at rest and a body in motion continues to move at a constant velocity unless acted upon by an external force."
    An alternative version;
    "If no external force acts on a particle, then it is possible to select a set of reference frames observed from which the particle is seen to move without any change in its velocity."
    Note here the words constant and velocity.
    Good :approve:
     
  6. Feb 10, 2007 #5
    for the 2nd question about the toy car, i know that there is a component of the weight acting down the ramp, and that there is no backward force up the ramp since friction is ignored. i'm not sure if we can say that there is a an accelartion force acting down the ramp or not...if there is umm then the answer is 3???

    for the 3rd question i took T=1.4g since that there is no other force acting in the x direction and using F=ma with the force as T then T=1.4a
     
  7. Feb 10, 2007 #6

    Hootenanny

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    But why should the force increase?
    Why should the acceleration be g?
     
  8. Feb 10, 2007 #7
    oooh so for question 3 its supposed to be 1.4a-3.1g=-3.1a then a=6.751111 m/s^2 correct????

    for the 2nd part of question 3 since it says take t a and g as positive quantities i'm not sure but according to my working on paper i'm getting eqn 8 to be correct T-3.1g=(1.4+3.1)a ???? this is too hard :'(
     
  9. Feb 10, 2007 #8
    i'm assuming that the car accelarates down the ramp
     
  10. Feb 10, 2007 #9
    ok so for quetion 1 about the elevator since there is not external force acting on the elevator it will move in constant speed according to newtons law which is the case here because it states in the question that the elevator moves with constant speed therefore the upward force by the cable is equal to the downward force of gravity
     
  11. Feb 10, 2007 #10
    what does it mean by T, a and g are positive quantities? do i take them as positive for this part ???
     
  12. Feb 10, 2007 #11

    Hootenanny

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    Correct :approve:
    Indeed it does, the important part here is to decide whether it accelerates at a constant rate or otherwise.
    Correct :approve:
    That's right. I have just one question, are you sure that option three in the last question is written like this;
     
  13. Feb 10, 2007 #12
    (3.1 kg) g ¡ T = (3.1 kg) a instead of the i it should be minus

    correct one should be like this sorry for typo (3.1 kg) g - T = (3.1 kg) a
     
  14. Feb 10, 2007 #13
    eqn 4 and 8 seem wrong..if i put T a and g postive i dont get any eqn from the 10 choices
     
  15. Feb 10, 2007 #14

    Hootenanny

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    No problem, that makes more sense :smile:
    Okay, lets examine what you written previously;
    So you written;

    [tex]1.4a-3.1g=-3.1a[/tex]

    Now, after a little manipulation (multiply both sides by -1)we have;

    [tex]3.1g - 1.4a = 3.1a[/tex]

    Now, note that [itex]T = 1.4a[/itex]; does this look like any of the options...:wink:
     
  16. Feb 10, 2007 #15
    lol it now makes sense :)
    lol i just reviewed my work i didnt multiply the T by -1 i forgot about it :P silly me
    thanks alot oh and i'm sorry for the confusion there

    gtg study newtons first law :) thanks again
     
  17. Feb 10, 2007 #16
    one more thing..its my first time in any forum so after we're done do i have to stamp something or tell people that we're done in this place? any paper work to fill??
     
  18. Feb 10, 2007 #17

    Hootenanny

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    No problem, twas a pleasure :smile:
     
  19. Feb 10, 2007 #18

    Hootenanny

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    Nope, nothing to fill in or stamp; thanks for asking though, its nice to have a conscientious student for a change...:rolleyes:

    If its your first time, you might want to introduce yourself in General Discussion; there's something that goes on down there with a fish ... :confused: ... never understood it myself...
     
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