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A Lawn Mower and Newton's 2nd Law

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A person pushes a 16.0 kg lawn mower at constant speed with a force of 75.0 N directed along the handle, which is at an angle of θ = 40.0° to the horizontal.

    Calculate the normal force exerted vertically upward on the mower by the ground.
    Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.1 m/s in 2.0 seconds (assuming the same retarding force).


    2. Relevant equations
    Fnet = MA


    3. The attempt at a solution
    To try and solve the first problem I made a free body diagram and then divided the downwards push into horizontal and vertical components so that I could create a triangle. With 75N as the hypotenuse and 40degrees as the angle. I used sine so that I could find the downwards force being exerted by the person. x = sine(40) x 75 which comes out to 55.8834N. I'm not sure what I need to do to figure out the vertical force, wouldn't it just be 55.88? I know its not, but that's what seems logical to me. Do I just need to add the force of gravity pulling down?
    On the second one I am completely lost, so thats where I'd like the most help.
    Thanks
     
    Last edited: Oct 30, 2008
  2. jcsd
  3. Oct 30, 2008 #2
    Think about it this way. If the mower were stationary there would be a normal force pushing back up. If you push down on an object, that normal force must increase to counteract the additional force you have added to the object. So, the answer to the first question cannot simply be 75.0sin(40)
     
  4. Oct 30, 2008 #3
    Alright, I have the answer now. I just had to figure out the normal force of the lawn mower as its pushed at 75N. My final answer (which I checked so I know its correct) is 212.84 N.
    I'm still stuck on the second problem though.
     
  5. Oct 30, 2008 #4
    Ok, well for the second part, the person pushing the mower at that angle is also pushing the mower horizontally at a value of 75.0cos(40). However, since it's moving at a constant horizontal speed, there must be another force pushing back at the same value, otherwise, if there was a net force horizontally, the law mower would accelerate. Now, it's asking you to figure out the force required to accelerate the motor to that final velocity. Using kinematics you ought to be able to find the acceleration and then using F = ma calculate the additional force necessary to cause that acceleration.
     
  6. Oct 30, 2008 #5
    Well to find the acceleration I took the velocity and divided by time so I came up with .55m/s^2 as my acceleration
    This is where I become uncertain
    The frictional force is 55.88
    so I got this fnet = ma equation
    -55.88 + X = (16kg) (.55) Which when I solve it x = 64.68N.
    I checked and thats not the answer, what am I not doing?
     
  7. Oct 30, 2008 #6
    As far as I can tell that is correct, X would be the required force. I must warn you, I am pretty tired right now, so I could be wrong...but it looks good. Does it make logical sense?
     
  8. Oct 30, 2008 #7
    Well I tried plugging that in (we submit these answers to a computer grading system) and I'm told that I'm incorrect.
    Should I add the 64.68 to the 75N force that its originally being pushed at?
     
    Last edited: Oct 30, 2008
  9. Oct 30, 2008 #8
    I think what we forgot to do is incorporate the horizontal force into the force at 40 degrees. So that actual force required to do the acceleration would be a little bigger than the value you calculated. Does that make sense?
     
  10. Jul 1, 2009 #9
    THANK YOU! THIS HELPED ME A LOT!
    Ps: you were right. You need to take your accelerated force, add the anti-retarded force, and then divide by cos(theta). I've spent a good 24 hours stuck on this, god bless you both.
     
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