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AP physics q

  1. Aug 28, 2005 #1
    First let me say, I have absolutely no idea how to do these. If anyone could show me how I would greatly appreciate it. (Thursday 8/25/05 was my first day in this class)

    Q1) A crane must lift a crate with the mass of 3.5x10^3kg
    a. How much force would be required?
    b. On the moon, do you think it would take more or less force to lift the crate?

    Q2)How much force will it take in order for you to lift a book having a mass of 1350g?

    Q3)How many four liter buckets of water do you need to fill your bathtub, which is 123cm long, 57.2 cm deep, 33.0cm wide?

    Q4) (I have finished the first part) When you buy a 2-liter bottle how many dm^3 have you purchased (2 dm^3)?
    What is the mass in grams if it is mostly water?
    What is its weight in newtons?

  2. jcsd
  3. Aug 28, 2005 #2
    Okay, these questions are, for the most part, testing your knowledge of basic gravity and how forces affect objects.

    The thing to realize is that the gravitational force exerted on an object at sea level is its weight.


    Where weight is W, m is mass, in grams, and g is the gravitational accelleration, 9.8 m/s^2. Weight in Physics is mostly measured in Newtons (N), the measure of force. [itex]N=kg*m/(s^2)[/itex]

    To lift any object, you'd have to apply a force greater than or equal to the weight of that object.

    For the other questions, you'd have to know the following facts about water's density:

    [itex]1 gram H_{2}O = 1 mL H_{2}O = 1 cm^3 H_{2}O[/itex]

    Just remember to convert all units, and you should be fine.
  4. Aug 28, 2005 #3
    1 and 2 need Newton's First Law of Motion (F = ma).
    3 is about volume and convertion.
    4 is, again, about convertion.

    As you have started 4 that is the one I will help with. 2 dm3 is correct. If I said density was equal to mass divided by mililitres and that the density of water is 1 g ml, would that help?

    The Bob (2004 ©)
  5. Aug 28, 2005 #4


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    Q1 would be better phrased to say "What is the minimum force requird to lift a crate..."
  6. Aug 28, 2005 #5
    Could you break down the formula for force [N=kg*m/(s)^2]?
  7. Aug 28, 2005 #6


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    The units you just wrote down are a big hint...Flux made one type-o though...In order for F = m*a to work into Newtons, the mass must be in kilograms, not grams if your units for a are in m/sec^2.

    That being said, Newton's first law again is:

    [tex]F = ma[/tex] compare that to the units you just wrote down.
  8. Aug 28, 2005 #7
    For Q(4) I got 20g and the weight in Newtons is 0.433N. Is this correct?
    For Q(3) I got 5804.37 buckets. (significant figures are not an issue for my teacher)
    Last edited: Aug 28, 2005
  9. Aug 28, 2005 #8
    These are the first force problems I have done and my teacher did not teach us how to do it. I still do not understand it. Do you know of somewhere I can see some example problems?
  10. Aug 28, 2005 #9


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    Q4. How can it be 20 g?
    [tex]1l = 1 dm ^ 3 = 1000 cm ^ 3[/tex].
    Or another way is to remember 1l water has the mass of 1 kg.
    Q3. is also wrong.
    Viet Dao,
  11. Aug 28, 2005 #10
    Q4. was a careless mistake. I tried it again and got 2kg.
    Q3. I did the same thing I got 58.0437 this time.
    I was putting in 1dm^3=10cm^3 instead of 1000cm^3

    PS. I'd hate to do it but I think I will have to give up on Q1 and Q2 b/c I just don't get it. :cry:
  12. Aug 28, 2005 #11


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    When you apply a force that's equal or greater than the object's weight, you can lift it up. Otherwise, you can't.
    * When the object's in the floor, it underacts 2 forces : P, and N. And [tex]\vec{P} + \vec{N} = \vec{0}[/tex]
    When you start exerting force F on it, and F < P, then N decreases, so that:
    [tex]\vec{F} + \vec{P} + \vec{N} = \vec{0}[/tex]
    F, N points upwards, P downward.
    When F > P, the object will go up.
    When F = P, then N = 0, ie: the floor does not exert any force on the object. The object starts to leave the floor.
    So when you exert a force F >= P, you can lift it up.
    And P = mg.
    So [tex]F \geq P = mg[/tex], so what's the minimum force required to lift an object, is it [tex]F = mg[/tex]?
    Where g is the gravitational accelleration, g = 9.8 m/s^2, m is the object's mass (in kg), F is the minimum force required to lift the object (in N).
    In Q1, you have the mass of the crane, can you compute the minimum force required to lift it up?
    In Q2, you have the mass of the book in grams. Remember to change it into kg.
    Viet Dao,
    Last edited: Aug 28, 2005
  13. Aug 28, 2005 #12
    so for Q1 I would say F=(3.5*10^3kg)(9.8m/s^2), and Q2 would be (1.35kg)(9.8m/s^2) correct? Then what do I do w/ the "m/s^2"? (Im sorry to sound dense but I have never seen a force problem)
  14. Aug 28, 2005 #13


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    Yes, that's correct. Force is measured in N (Newton). And:
    F = ma, (m is mass, and a is acceleration), so:
    [tex]1 N = 1 \frac{kg \times m}{s ^ 2}[/tex]
    So in Q1 : F = 3.5 * 10 ^ 3 * 9.8 = 34300 N. Do the same for Q2.
    I think you should re-read the book again to insure that you understand the concepts.
    Viet Dao,
  15. Aug 28, 2005 #14
    Thank you so much! Thats what I got for Q1 and I got 13.23 N for Q2. Thank you again. It has taken me all day to figure out how to do those problems.
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