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Homework Help: Sprinter resultant force question

  1. Sep 26, 2013 #1

    Just after the gun a sprinter of mass 65kg is pushing against the starting block with a force of 800N. The force acts at an angle of 65 degrees to the horizontal.

    - the resultant horizontal force acting on her
    - the resultant vertical force acting on her
    - the forward acceleration of her centre of gravity
    - the upward acceleration of her centre of gravity

    My attempt:

    I'm guessing the resultant horizontal force is 0, as the starting block would exert an equal force on the runner?

    Then resultant vertical force would be mass x gravity (65 x 9.8 = 637N).

    But I don't know how to attempt the other two questions??
  2. jcsd
  3. Sep 26, 2013 #2
    Why are you guessing ? If the sprinter exerts a force on the block ,the block exerts an equal and opposite force on the sprinter (newtons 3rd law) .The two forces are acting on different bodies,so they do not cancel.You should be concerned only with forces acting on sprinter alone.

    Draw a FBD of the sprinter with all the forces acting on him .What is the net horizontal force on him ?What is the net vertical force on him ?
    Last edited: Sep 26, 2013
  4. Sep 26, 2013 #3
    So if the forces don't cancel out would the horizontal resultant just be 800N then??
  5. Sep 26, 2013 #4
    Is the force acting horizontally ?
  6. Sep 26, 2013 #5
    No, it's at a 65 degree angle...so would you use trig to figure out the components?
  7. Sep 26, 2013 #6
  8. Sep 26, 2013 #7
    Aaah thank you!! Think I've got it now!
  9. Sep 26, 2013 #8
    What are the answers you are getting ?
  10. Sep 26, 2013 #9
    I'm getting horizontal as 338.1N and vertical as 725.04N, however this could be wrong because I never know where to put the angle on the diagram...
  11. Sep 26, 2013 #10
    Look at the figure I have attached.

    If a force F acts at an angle θ with the horizontal .

    1) What is the horizontal component of the force?
    2) What is the vertical component of the force ?

    Attached Files:

    • force.GIF
      File size:
      1,003 bytes
  12. Sep 26, 2013 #11
    Horizontal would be the one touching the 65 degree angle wouldn't it? I calculated the horizontal to be cos65 x 800 and the vertical to be sin65 x 800..
  13. Sep 26, 2013 #12
    Yes...Horizontal component would be Fcosθ and vertical component Fsinθ .

    So,the net horizontal force on the sprinter would be Fcosθ (800cos65).

    The vertical force on the sprinter by the block would be Fsinθ .But what will be the net vertical force on the sprinter ?
  14. Sep 26, 2013 #13
    Would you take away the force of gravity to get the net vertical force? Vertical force downwards would be 9.8 x 65 = 637.
    725-637 = 68N would be the net vertical force?
  15. Sep 26, 2013 #14
    Sorry that was meant to be 88 not 68!
  16. Sep 26, 2013 #15
    The vertical forces acting on the sprinter would be Normal force from the ground(N) acting upwards, force due to gravity (mg) acting downwards , and component of force by block in vertical direction(800 sin65) upwards.

    Is the body accelerating in the vertical direction ?
  17. Sep 26, 2013 #16
    Yes, I would say it is accelerating? As there is a larger force acting upwards than downwards?
  18. Sep 26, 2013 #17
    Are you sure :smile: ?
  19. Sep 26, 2013 #18
    Haha, not now!! I'm so confused?!
  20. Sep 26, 2013 #19
    They must be accelerating though, as the second part of the question asks what her upward acceleration is...
  21. Sep 26, 2013 #20
    Well..The sprinter accelerates only in the horizontal direction .

    The net vertical force is zero .The sum of all the forces in vertical direction will add up to zero.
  22. Sep 26, 2013 #21
    I thought that the normal force from the ground and the force of gravity would be equal, and then you have the upward component of the force from the block?
  23. Sep 26, 2013 #22
    Why do you think like this :rolleyes:?

    If the net acceleration in vertical direction is zero,then the net force in vertical direction has to be zero .

    Consider positive direction upwards .

    N + Fcosθ -mg =0
  24. Sep 26, 2013 #23
    Isn't the Normal force from the ground acting upwards the same thing as the component of force by the block in the vertical direction? In the starting block designs that I am familiar with, the sprinter's feet are pressing entirely on the blocks.

  25. Sep 26, 2013 #24
    Hi Chet

    I think you are right.Thanks for correcting :smile:
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