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Toaster Tension

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1.3kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.350. To make the toaster start moving, you carelessly pull on its electric cord. (a) For the cord tension to be as small as possible, you should pull at what angle above the horizontal? (b) With this angle, how large must the tension be?

    2. Relevant equations



    3. The attempt at a solution
    Ok I know that there are two components to the tension, TCosθ and TSinθ. I got TSinθ<12.74 and TCosθ>4.459 so θ<70.7. We want TCosθ-4.459 to be as small as possible so

    d(TCosθ-4.459)/dθ=0 and -TSinθ=0 so θ=0.

    Somehow I don't think that is right. And for the second part are you giving T>x or a specified value for T? Thank you.
     
  2. jcsd
  3. Sep 6, 2009 #2

    ideasrule

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    I don't quite understand what you did, so please forgive me if this sounds redundant:

    Try drawing a free-body diagram of the toaster. There are 4 forces: gravity, tension, the normal force, and friction. Don't forget that friction is N*mu, not mg*mu. Write out Newton's second law for the x and y directions, get an expression for tension, and minimize it using calculus. Try it, and post back when you get an answer.
     
  4. Sep 7, 2009 #3
    Ok say there is a force T acting on the toaster at θ above the horizontal. The y component is TSinθ and the x component is TCosθ. We want TCosθ>fstatic so it will move. TCosθ>N(mu) TCosθ>12.74(.350), TCosθ>4.459. Now Cosθ ranges from 1 to 0. The higher Cosθ is, the lower T will be. The highest Cosθ goes is 1 so T>4.459. What am I doing wrong?
     
  5. Sep 7, 2009 #4
    Whats wrong if theta is zero? If you pull horizontally then the tension will be equal to the force of friction. If you make theta larger then your horizontal tension will decrease by a little but you vertical tension will increase by much more. mg is acting downward which is much larger than Nmu.
    [tex]T=\frac{N-mg+F_s}{cos\theta - sin\theta}[/tex]
    If theta is zero N and mg cancel out. If it isn't zero, the two don't cancel out and the denominator is greater than one.
     
  6. Sep 7, 2009 #5

    rl.bhat

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    T*sinθ will act in the opposite direction to mg.. So the normal reaction will be ( mg - T*sinθ )
     
  7. Sep 8, 2009 #6
    Oh dang it, I put N=mg. Well I think I got it now. TCosθ-.35N>0, so T(Cosθ+.35Sinθ)>4.459. We want Cosθ+.35Sinθ to be max so T will be min. -Sinθ+.35Cosθ=0, θ=19.29. Then T>4.2. Right?
     
  8. Sep 9, 2009 #7
    Sorry for being rude, but am I right? Thank you.
     
  9. Sep 9, 2009 #8

    rl.bhat

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    Frictional force fr = μ( mg - T*sinθ ). To start motion T*cosθ > fr.
    So write
    μ( mg - T*sinθ ) = T*cosθ
    Take the derivative and equate it to zεro to get θ
     
  10. Sep 9, 2009 #9
    Isn't that what I did in post 6? Then T=4.2 instead of T>4.2.
     
  11. Sep 9, 2009 #10

    rl.bhat

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    Yes. You are right.
     
  12. Sep 9, 2009 #11
    Thank you come again.
     
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