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Person pushes down the length of the handle of a 12.1kg

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Person pushes down the length of the handle of a 12.1kg lawn spreader. The handle makes an angle 44.1 degrees with the horizontal. Person wishes to accelerate the spreader from rest to 1.33m/s in 1.7s. What forece must person apply to the handle?


    2. Relevant equations

    F=ma

    I imagine a=Δv/Δt

    3. The attempt at a solution

    I started to find the average acceleration, using a =Δv/Δt. a = 0.78m/s^2

    Plug and chug to F=ma and get 9.4N. This seems way too simple of a solution.

    Should I carry on with this:
    Fx=max
    Fy=may

    then just find the resultant of F.

    Fx = -9.47
    Fy = -118.58

    Resultant F = 118.96N...that seems way too much?
     
    Last edited: Jan 29, 2013
  2. jcsd
  3. Jan 29, 2013 #2

    cepheid

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    Re: F=ma

    Yeah, you need to resolve the force into components, because only the horizontal component will contribute to moving the machine horizontally. Obviously the vertical component won't contribute to moving the machine horizontally.

    So the procedure would be to find the force Fx that would produce the required acceleration ax, and from that, and the angle, compute the resulting Fy.

    EDIT: or just use the angle theta plus Fx to compute the resultant directly.
     
  4. Jan 29, 2013 #3

    cepheid

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    Re: F=ma

    Okay, you added this part into your original post later:

    Your Fy is wrong. Plain and simple. Can you post the steps you used to calculate it?
     
  5. Jan 29, 2013 #4
    Re: F=ma

    To find Fy, I used the mass of the object x the negative acceleration due to gravity.

    F= 12.1(-9.8)

    I'm missing the normal force, I think? But now, I am unsure how to calculate it
     
  6. Jan 29, 2013 #5

    cepheid

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    Re: F=ma

    Gravity has nothing to do with anything here. You are computing the y-component of the force that the person is applying to the object. It comes out just as a consequence of the geometry of the situation. If he's pushing with an angle of 44.1 degrees below horizontal, and the x-component is 9.4 N, then the y-component is completely determined.

    Gravity acts on the object too. It tries to pull it towards the centre of the Earth. But it is countered by the normal force from the ground, and so the object stays put. None of this is relevant to the problem at hand. You're considering force on the object from the person, not from gravity, or from the ground.
     
  7. Jan 29, 2013 #6
    Re: F=ma

    Thanks for the clarification about gravity :)

    I went back to find the y component of the vector...to equal 1.33sin41?

    I thought 9.4 was the x component of the Force.

    Does 1.33sin41 equal the y comonent of velocity? If so, then I could go about finiding the y component of acceleration and solve for Fy?
     
  8. Jan 29, 2013 #7

    cepheid

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    Re: F=ma

    No. 1.33 m/s is the speed after 1.7 s, and is entirely horizontal. It has absolutely nothing to do with what you are trying to compute here.

    Yes.

    Again, no. The velocity is entirely horizontal, and it's not the thing that you are trying to compute (so I'm not sure why you keep talking about it). You're trying to compute the force.

    You don't need to find the acceleration first! Just draw a diagram. You have your (overall) force vector. It is angled at 44.1 degrees below the horizontal. Now resolve this vector into component vectors Fx and Fy. Fx is horizontal and Fy is vertical. These three vectors form a right triangle. You know one of the sides (Fx) and you know one of the angles (θ), in addition to the 90 degree angle. Given this information, and the fact that it is a right triangle, how would you solve for Fy? It's entirely trigonometry.
     
  9. Jan 29, 2013 #8
    Re: F=ma

    Did what you said...

    Fx = 9.4
    Fy = 9.1

    F = 13.1N

    Thank you
     
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