1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    Yep, looks OK. (I misread something earlier.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A Sliding Block
  1. Sliding Block (Replies: 3)

  2. Sliding Block (Replies: 3)

  3. Sliding Block (Replies: 23)

  4. Sliding block (Replies: 4)

  5. Sliding Block (Replies: 3)

Loading...