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!

Newton's Laws; ramps and pulleys

  1. Jul 22, 2007 #1
    1. The problem statement, all variables and given/known data
    The diagram
    -Assuming the system is frictionless, find m2 when it is accelerating down the ramp at 3.4 m/s2.

    2. Relevant equations
    -... F = ma?

    3. The attempt at a solution
    -In this problem, I tried tilting the axis so the sliding-down part aligned with the y-axis and the m1 part on the x-axis. Then, I thought of using the components to the Fw, to figure out the Tension on the rope for the m2 side (which, I guessed, to be m2a.).
    I couldn't figure out what to make the Tension equal to, because m2gcos([tex]\theta[/tex]) wasn't coming out the same as 3.4*m2.

    ...I then realized I was quite retarded at physics (I'm being perfectly honest, now.) and needed help.
     
  2. jcsd
  3. Jul 23, 2007 #2
    did you make the free body diagram???
     
  4. Jul 23, 2007 #3
    oh, sorry! I forgot to attach it... >_<;;;

    here it is.

    I suppose what confuses me is the fact that they're on ramps; I don't know how/where to draw the gravity part.

    (apologies if this is frustrating for you. I'm just really bad at physics.)
     
  5. Jul 23, 2007 #4
    nooooooooooo
    this is what a free body diagram for such a situation looks like.
    now make the other one. write the appropriate equation ie. Fnet = m*a.
    eliminate T from both the equations. you have "a", find "m" and show us what you got. ok??
     

    Attached Files:

    • fbd.PNG
      fbd.PNG
      File size:
      1.3 KB
      Views:
      50
  6. Jul 23, 2007 #5

    andrevdh

    User Avatar
    Homework Helper

    The weight acts at the centre of gravity (mass) of the block.
     
  7. Jul 23, 2007 #6
    i guess OP is asking the direction of gravitational force andrevdh
     
  8. Jul 23, 2007 #7
    (well, my retardation may have reached a new low.)

    I figured out how to do the free-body diagram right this time, and got:

    diagram

    [tex]\Sigma[/tex]Fx = m2gsin[tex]\alpha[/tex] - T = m2a.
    and
    [tex]\Sigma[/tex]Fx = T- m1gsin[tex]\beta[/tex] = m1a.

    so if I solve for T it would be m1a + m1gsin[tex]\beta[/tex], and substitute that into the first equation to figure out m2a.

    with grouping like terms, plugging in numbers, solving, &c, I got 11.0 kg.

    ...would that be right?

    (and a big thank you for the help!)
     
  9. Jul 24, 2007 #8
    i didnt calculate the values, but the method you described, answer should be alright.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newton's Laws; ramps and pulleys
  1. Newton's laws, Pulley (Replies: 5)

Loading...