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 Laws

  1. Jun 23, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\alpha[/tex] = 37, m1 = 10 kg
    Also attachment.

    2. Relevant equations



    3. The attempt at a solution

    ok, the first question is to find m2 when the system is in rest and the second one is to find the acceleration when m1 = m2.
    I found 3 equations:
    n1 = m1 * g * cos([tex]\alpha[/tex] )
    m1 * g * sin([tex]\alpha[/tex] ) = T2
    T1 + T2 = m2 * g
    I need another one I think...

    Could it be that T1 = T2?
    And as for the second one, how can I find the equations between the accelerations? (there is always one...)

    10x in advance.
     

    Attached Files:

    • 1.jpg
      1.jpg
      File size:
      5.5 KB
      Views:
      47
  2. jcsd
  3. Jun 23, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the tensions of lengths of rope connected by frictionless pulleys have the same tension. But in your Newton's law F=ma equations, I only see F parts, I don't see any ma parts. You don't even have an acceleration to solve for...
     
  4. Jun 23, 2008 #3
    Yeah, because the first set of questions is for acceleration = 0 (the system is in rest).
    Here are the equations for the second question:

    T - m1 * g * sin([tex]\alpha[/tex]) = m1 * a1
    m2 * g - 2 * T = m2 * a2

    I need the accelerations equation and I don't know how to come up with that...

    10x.
     
  5. Jun 23, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good point, a=0 for the first part. Sorry, I missed that. But for the second part there is another relation between a1 and a2. Consider the length of rope going through each part of the pulley system. Total length of rope is constant. One acceleration is twice the other. Which is which?
     
  6. Jun 24, 2008 #5
    I have no idea...
    I would like to think m2 is faster cause the g force is stronger with him :smile: But I'm not sure.
    I didn't get you, how can I see it through the length of the rope?

    10x.
     
  7. Jun 24, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If m1 descends 1 meter down the ramp, how much does m2 rise? You can tell just by looking at the length of the rope. The total length of the rope on m2's side of the pulley decreases by 1 meter.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newton Laws
  1. Newtons Laws (Replies: 1)

  2. Newton Laws (Replies: 2)

  3. Newton's Second Law (Replies: 8)

Loading...