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Basic Kinematics questions

  1. Jul 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello there,

    I have answered the first part by calculating the retarden force is equal to 80*cos30. This gave me an answer of 69.3N.

    I was wondering if someone could point me in the right direction to ascertain the force needed to accelerate the mass by 1ms-2 along the 30oto the horizontal line - as shown in the drawing.



    2. Relevant equations
    I used F=mg to convert mass to weight = 196.2 N
    I have looked at rearranging the F=ma equation.
    and also some of the kinematics equations.


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jul 1, 2016 #2

    Doc Al

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    To get such an acceleration, what must be the net force on the lawn mower? (In what direction?)
     
  4. Jul 1, 2016 #3
    I guess the force needs to be equated to the horizontal orientation, but then i think of F=ma and with the mass and acceleration being 20*1 respectively. I end up think about a force of 20N pushing horizontally towards the right.

    It feels like im missing something really obvious to do with the trigonometry side of things, I think.
     
  5. Jul 1, 2016 #4
    I'm a new member too - go easy (!) (I just read i was supposed to announce that first - sorry)
     
  6. Jul 1, 2016 #5

    Doc Al

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    Good. Applying ΣF = ma, you have found that the net force to produce such an acceleration must equal 20 N horizontally.

    Now imagine the force pushing down the handle is X. Write an expression for the horizontal component of that force. Then write an expression for the net force, in terms of X. Set it equal to what you found above and you should be able to solve for X.
     
  7. Jul 1, 2016 #6

    x (30o handle to horizontal ) = horizontal component (20) / cos 30 = 23.1 N ?

    apologies if i come across like an imbecile, but in my head im thinking to move the mass with enough force to accelerate it by 1ms-2, then it needs to receive a horizontal force of 20N, then to apply those 20N from the 30degree handle equates a number that is obviously larger than 20. I kind of reversed the order of the first part of this question to get this result, but it seems not enough, as 20N is like 2kg of force pushing a 20kg mass?
     
  8. Jul 1, 2016 #7

    Doc Al

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    Try doing what I suggested:
    I just want a simple expression for that horizontal component. No numbers (except the angle).
     
  9. Jul 1, 2016 #8

    Doc Al

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    What you want to be doing is applying ΣF = ma in the horizontal direction. We already know that ΣF = 20 N, so we need to identify and add up the horizontal components of all forces to get an expression for ΣF.
     
  10. Jul 1, 2016 #9
    for horizontal component - Horizontal component (H) = x *c os30.

    20N = 20kg x 1ms-2 = horizontal force needed.
    rearrange horizontal component formula to make x the subject - x = H / cos30 = 23N
    23 N + 20N = 43N...?

    Am i going in the right direction roughly?

    Also i greatly appreciate you taking your time to help me. its probably quite a trivial question for you to deal with but I feel once Ive grasped the basic, the rest will become a bit easier
     
  11. Jul 1, 2016 #10

    Doc Al

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    This is good. The horizontal component of the pushing force is Xcos(30). Let's say that force acts to the right.

    Let's do this systematically.

    You found the horizontal component of the pushing force. What other forces act on the mower? (The one you found in part a, for instance!)
     
  12. Jul 1, 2016 #11
    The other force acting would be the vertical component, and that together with the horizontal adds together to make x?

    but can't i take the horizontal component and use the cos function to establish x, like i attempted to do?
     
  13. Jul 1, 2016 #12

    Doc Al

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    We don't care about the vertical component of the pushing force, only the horizontal force. What forces act on the mower? I can identify 4 forces, two of which have horizontal components. One is the pushing force that we called X. What are the other three?
     
  14. Jul 1, 2016 #13
    force and displacement?

    or force and speed?
     
  15. Jul 1, 2016 #14

    Doc Al

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    I want you to name the actual forces that are acting on the mower. (Displacement and speed are not forces.)

    Like this:
    (1) The pushing force down the handle, X
    (2) The weight acting down

    You fill in the other two:
    (3)
    (4)
     
  16. Jul 1, 2016 #15
    the pushing force across the handle
    the weight resisting back

    p.s. I wont blame you if you lose patience with me :(
     
    Last edited: Jul 1, 2016
  17. Jul 1, 2016 #16

    Doc Al

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    We've already covered the force on the handle. That's the first one listed.

    Not sure what you mean by that.

    Hints for the remaining two forces: What's preventing the mower from moving downward? What was the force you calculated in step a?
     
  18. Jul 1, 2016 #17
    ummm gravity, and the force that opposes the gravity, (equal and opposite reaction?)

    gravity and the retard force?
     
    Last edited: Jul 1, 2016
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