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Homework Help: A Sliding Block

  1. Sep 24, 2011 #1
    Block 1 as shown in the link below(url) has a mass of 10.3 kg. The coefficient of static friction between the block and the table is μs = 0.36 and the coefficient of kinetic friction is μk = 0.336.

    a) What should the mass of the second block be so that m1 is just on the verge of sliding?
    I ended up getting (10.3 kg)*(0.36) , which was correct

    b) If the block 2 has the mass you just calculated above, what is the tension in the string?
    I got (10.3*0.36)(9.81), and that was also correct

    c) If the block 2 has the mass you just calculated above and then block 1 is given a slight nudge, what is the tension in the string?
    So i did the FBD, to solve for acceleration of the blocks, and i got ((M2 - Uk*M1)/(M1+M2)) *g = a
    i plugged acceleration back into this equation Ft = m1*a + Ffric => where Ffric is (Uk*m1*g)
    and that was not correct please help me

    here's the pic
    https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-06-Forces/tipler_sliding_block/fig_05_10.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 24, 2011 #2

    Doc Al

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    Looks fine to me. Double check your arithmetic.
     
  4. Sep 24, 2011 #3
    yea i looked over everything many times, I'm beginning to think that its the computer program that's wrong...Well if anything my answer for Ft = 35.73 N, if you are getting anything else please let me know
     
  5. Sep 24, 2011 #4

    Doc Al

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    What did you get for the acceleration? (Often the online programs are fussy about significant figures.)
     
  6. Sep 24, 2011 #5
    i got .173 m/s/s
    so i put in the calculator

    (10.3)(.173) + (.336*10.3*9.81) = Ft
     
  7. Sep 24, 2011 #6

    Doc Al

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    Seems a bit off--by a factor of 10. Double check this.
     
  8. Sep 24, 2011 #7
    I can't find my error, thanks anyway Doc, I'll just go find my professor, I'm sure he's made an error in programming this, he's done it before
     
  9. Sep 24, 2011 #8

    Doc Al

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    Before you go to your professor, show the details of your calculation of the acceleration.
     
  10. Sep 24, 2011 #9
    first I made the tensions equal to each other

    m1*a + Uk*m1*g = m2g - m2 a , then i solved for a

    a = ((m2 - uk*m1)/(m1+m2)) * g


    a = (((10.3*.36)-(.336*10.3)/((10.3*.36)+10.3))) * 9.81 = .173
     
  11. Sep 24, 2011 #10

    Doc Al

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    Yep, looks OK. (I misread something earlier.)
     
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