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Work Problem - Pushing a lawnmower

  1. Feb 11, 2014 #1
    Work Problem -- Pushing a lawnmower

    1. A lawn mower (25kg) is pushed a horizontal distance of 10m by a 50 N downward force directed 60 degrees to the horizontal. The coefficient of kinetic frictional force is .017. What is the work done by each of the external forces on the lawn mower?



    2. Relevant equations



    3. So basically I start off every problem by drawing a free body diagram. I label each force and its direction. I know there is Normal force going up, its Weight going down, and I believe a force going 60 degrees below the horizontal in what would be quadrant 4.

    The part I need help in is friction. If you push on a lawn mower the wheels turn clockwise meaning kinetic friction would be pushing right? (This is assuming you picture the lawn mower is being pushed from left to right as is my diagram) Im confused because my instructor did a similar problem where friction was going left which is not opposite of how the wheel turns.
     
    Last edited by a moderator: Feb 11, 2014
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  3. Feb 11, 2014 #2

    BvU

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    Where exactly is quadrant 4 ?
    If the story says 50 N downward 60 deg wrt horizontal, couldn't it just be someone pushing the handlebar ?
    You mention one external force (the 50 N). Why does it ask for work done by each of the external forces ? Any other forces at work here ?

    What are the relevant equations you can bring in to deal with this problem ?
     

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    Last edited: Feb 11, 2014
  4. Feb 11, 2014 #3
    Quadrant 4 would be the bottom right quadrant in a graph.

    That is true but usually I like to label forces starting the tail from the center of the object. Is that incorrect? Im not sure what you mean by the last question? I believe because each force can do work depending on the angle?

    What are the relevant equations you can bring in to deal with this problem ?

    Work= F*D*cos θ
    Kinetic force= μ*N
     
  5. Feb 11, 2014 #4

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    What about the 25 kg ? Never mind, it's already mentioned.

    Your question is clear now: what about the friction? The exercise is simple: don't bother about the wheels details. There is a weight and an equal and opposite normal force (the mower doesn't move vertically, right?). kinetic friction force you correctly calculate μ*N. Which way is it pointing ? Which way wrt the trajectory ?

    Now the importtant thing about work delivered by a force is that the only component of the force that does work is the component along the trajectory (the path). So the friction force is OK. What about me pushing the thing?

    Can you calculate the component of the 50 N that does work ?
     
    Last edited: Feb 11, 2014
  6. Feb 11, 2014 #5

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    You know I can't see your free body diagram, I hope? Very good practice to draw one. From what you mention it looks OK (starting the tail from the center of the object and such) but might not be totally complete.

    The relevant equations are OK too and now we have to worry about what to fill in where.
     
  7. Feb 11, 2014 #6
    Well I know kinetic friction= μ*N

    Normal force≠ Weight in this case because the 50 N at 60° adds a downward force.

    So to calculate force I would use this equation:

    N-W-Fsin60= 0

    I get Normal force = 293.3 N (I use 10 m/s^2 for gravity)

    So then Work for Kinetic friction = .017 * 10m * cos θ

    Would the angle be 0 or 180?
     
  8. Feb 11, 2014 #7
    I sketched my free body diagram in paint. Does it look correct?
     

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  9. Feb 11, 2014 #8

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    I forgot to welcome you to PF ! Whatever happened to your post #1 ?

    Anyway, bedtime for me (GMT+1 = 01:20 AM here.) I have a free body diagram, but I'd rather we use yours.

    Do you think you can manage ? You know, like finding the component of the 50 N I'm pushing with that actually does work ?

    Some of that work is dissipated (eaten up) by the friction that is pointing against the direction of motion. But I would argue that that is not an external force. Your teacher might disagree, so you be the judge there.

    Let me reveal that gravity isn't doing any work: its direction is perpendicular to the motion, so there is no component along the trajectory.
     
  10. Feb 11, 2014 #9

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    Again, posts cross, for the umpteenth time tonight. More and more annoying.

    You FBD is beautiful.

    Friction work: I miss the 293, otherwise: OK.
    Angle: I would vote for 180 degrees. 0 degrees is then to the right, where the mower is going.

    The other component of the 50 N is more interesting to me: that has to do with the work I am doing.
     
  11. Feb 11, 2014 #10
    Thanks for the welcome!

    Yeah I can manage the rest of the forces, I just needed help with the work done by frictional force.
     
  12. Feb 11, 2014 #11

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    The 180 is relevant: the force and the motion are opposite, so it's not doing any work but it is eating it up. Nearly always the case with fricion (there are exceptions, like making wheels turn. That comes later).
     
  13. Feb 11, 2014 #12

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    I'm almost completely reassured. Just to indulge me: What is the work I have done ?
     
  14. Feb 11, 2014 #13
    Oh I see now. I forgot to include the *293.3 in my Work equation but I got the answer correct. Thanks for the help
     
  15. Feb 11, 2014 #14
    I got 200.2 Joules

    Which would me Work done by force which is= 50*cos 60 * 10m = 250

    Minus 49.8 J from kinetic friction.
     
  16. Feb 11, 2014 #15

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    "Answer correct" ? You mean your book tells you -50 J for the friction work ?
    Does it also say what the work is that I have done ?
     
  17. Feb 11, 2014 #16

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    Ah, good we come to the end of this. No adding up of works here. I have done the full 250 J ! No minus anything!
     
  18. Feb 11, 2014 #17

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    The exercise asks for the work done by each of the external forces. My answer would be 250 J by me and 0 J by gravity (no vertical motion). Also see post #8.

    Did you notice some higher authority has edited post #1 to remove the boldface font?
     
  19. Feb 11, 2014 #18
    No, one of my classmates got the same answer so i kind of assumed. The question says work done by external forces on the lawn mower. Are the forces I have on my diagram external forces? If so, then what are the internal forces?

    Also, is net work the sum of all work no matter if positive or negative?

    edit: I did not notice it was edited. Is that bad?
     
  20. Feb 11, 2014 #19

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    Good questions! Re classmate: if he jumps into the lake, are you going to follow him/her ? (in the latter case i might be mad enough to do so ...;-)

    External forces are at least gravity and my pushing. Somewhat debatable what else is external. If the thing stands still, no friction force is present. I would consider that a kind of reaction, but you could also claim that it's the daisies and the grass resisting being pushed down....

    Normal force is a reaction force: the earth holds back the thing from falling in. External ? Doesn't do work anyway: perp to motion.

    Nice questions to ask teacher! Earns you credit for being inquisitive and curious. Good teachers like that.

    Net sum of work is also sensible to ask for. What does it mean? I push the thing 10 m, costs me 250 J. Friction eats up only 50 J. Where is the 200 ? Did I do -200 J of work to bring it to a halt after 10 m ? Or did I use the brake which dissipated 200 J and changed it into heat to warm up the environment even further ?

    More nice questions to ask teacher! I'm useless: falling asleep.
     
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