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3 Dynamic Physics Questions Urgent Help Needed

  1. May 17, 2007 #1

    mix

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    3 Dynamic Physics Questions Urgent Help Needed!!!

    1. The problem statement, all variables and given/known data
    Problem1

    An object of 6kg is moving towards the south. The resulting force that the object has is 12 newtons towards the north. What is the acceleration of the object?


    2. Relevant equations
    f=ma
    fg=mg


    3. The attempt at a solution
    The question is wrong?

    1. The problem statement, all variables and given/known data
    Problem2

    A kid is pulling a wagon of 10kg with 40 newtons of force at an acceleration 3.5m/s^2. There is 3 newtons of friction.

    How do you find the angle at which the child is pulling the wagon at?

    2. Relevant equations
    f=ma
    fg=mg


    3. The attempt at a solution
    don't even know where to start

    1. The problem statement, all variables and given/known data
    Problem3

    There is a helicopter that weighs 15000kg lifting a car of 4500kg with a cable below it. If it where to move upwards at acceleration of 1.4m/s^2 what would be the necessary force required by the helicopters blades?
    What is the force on the cable lifting the car?



    2. Relevant equations
    f=ma
    fg=mg


    3. The attempt at a solution
    add the two weights find the force of gravity and then find the resulting force and add them?



    Any help at all is greatly appreciated thanks for your time and effort!
     
    Last edited: May 17, 2007
  2. jcsd
  3. May 17, 2007 #2

    hage567

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    Homework Helper

    You must have some thoughts on how to solve these. What does Newton's second law state?
     
  4. May 17, 2007 #3

    mix

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    f=ma ( i go to a french school its kinda hard to translate )
     
  5. May 17, 2007 #4

    mix

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    bump update
     
  6. May 17, 2007 #5

    hage567

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    You know the force acting on the object, and you know its mass. So you can find the acceleration by using that equation.
     
  7. May 17, 2007 #6

    mix

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    if its moving towards the south how is it possible that the resulting force is towards the north?
     
  8. May 17, 2007 #7

    mix

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    Problem 2 is my biggest problem i have no clue at all how to get that angle someone please help
     
  9. May 17, 2007 #8

    hage567

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    When the force is acting opposite to the direction of motion, it is decelerating. Deceleration is negative acceleration. So the object is slowing down.
     
  10. May 17, 2007 #9

    mix

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    so is it moving towards the north or the south and what exactly is the acceleration?
     
  11. May 17, 2007 #10

    Office_Shredder

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    mix, imagine slide a box across a floor. It stops after a little bit. Why? Because friction accelerated it in the opposite direction of motion.

    velocity and acceleration don't need to be in the same direction. Another example is a car coming to a stop... the acceleration is in the backwards direction, while the velocity is forward.

    For number two, given the acceleration, you can find the net force. This is going to be a combination of the force the kid pulls in the horizontal direction, and the force of friction. The force of friction is proportional to the normal force, which is equal (in absolute value) to the force of gravity minus (or plus if the kid is pushing down) the vertical force of the kid

    EDIT: So it's moving south, and accelerating north for number 1 (like a car moving south coming to a stop.... though depending on the situation, it might start going backwards afterwards)
     
  12. May 17, 2007 #11

    mix

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    So for problem two, (40N)
    /
    __________ /
    (3N) | |/ ) <--- Angle X
    <-------| 10kg |----------
    L_________|

    Net force = 3.5*10 = 35N Now i am stuck?!?
     
  13. May 17, 2007 #12

    mix

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    That was my attempt to draw the free body diagram of the problem it didnt work out someone please help what he said was close to jibberish to me
     
  14. May 17, 2007 #13

    Office_Shredder

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    The forces are:

    Gravity going down
    Normal force going up
    Kid pulling on wagon going up
    Friction force going backwards
    Kid pulling on wagon going forward

    If the net force of the kid is F, and he's pulling at an angle x wrt to the normal (positive x means he's pulling up),then
    Kid pulling on wagon going up = Fsin(x)
    Kid pulling on wagon going forward = Fcos(x)

    So Fcos(x) - friction force = 35N as you described above.

    So we know F=40N We need x and friction force

    But friction force = 3. So 40cos(x) - 3 = 35

    and cos(x) = 38/40
     
  15. May 17, 2007 #14

    mix

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    So the angle is 18.1 degrees? Correct?
     
  16. May 17, 2007 #15

    Office_Shredder

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    I didn't bother checking it, but at that point it's just using a calculator, so you're probably right.

    Now try number three (same idea.... list out all the forces, and every equation you know. Your answer should just fall out)
     
  17. May 17, 2007 #16

    mix

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    okay so first i add the two weights 15000 + 4500 = 19500kg

    force of gravity fg=mg = (19500)(9.8) = 191100

    force net F=ma = (19500)(1.4) = 27300 Newtons

    So the force that the blades need to put out in order to lift the helicopter and the car at a acceleration of 1.4m/s^2 is 27300+191100 = 218400N

    Look right?
     
  18. May 17, 2007 #17

    mix

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    and for the cable holding the car

    force of gravity

    4500(9.8) = 44100

    F net = 4500*1.4 = 6300

    44100+6300 = 50400N
     
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