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!

Two Blocks with rope and Pulley on Inclined Plane

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    This is a problem that I am trying to help my friend with. I understand how to solve this problem the traditional way by rotating the axes of the block on the inclined plane, but then my friend asked me to show him how the answer is the same with non-rotated axes.
    There are two blocks connected by a rope that goes over a pulley, one block of mass m[tex]_{2}[/tex] hanging by the rope over the edge, and the other block of mass m[tex]_{1}[/tex] on the inclined plane. Given the angle of the plane with the horizontal, and without any type of friction, find the tension in the rope and the acceleration of the two blocks.


    2. Relevant equations
    m[tex]_{1}[/tex]=5kg
    m[tex]_{2}[/tex]=10kg
    [tex]\theta[/tex]=40[tex]^{0}[/tex]
    [tex]\sum[/tex]F[tex]_{x:m[tex]_{1}[/tex]}[/tex]=Tcos[tex]\theta[/tex]+Ncos[tex]\theta[/tex]=-m[tex]_{1}[/tex]acos[tex]\theta[/tex]
    [tex]\sum[/tex]F[tex]_{y:m[tex]_{1}[/tex]}[/tex]=Tsin[tex]\theta[/tex]+Nsin[tex]\theta[/tex]-m[tex]_{1}[/tex]g=m[tex]_{1}[/tex]asin[tex]\theta[/tex]
    [tex]\sum[/tex]F[tex]_{x:m[tex]_{2}[/tex]}[/tex]=0
    [tex]\sum[/tex]F[tex]_{y:m[tex]_{2}[/tex]}[/tex]=T-m[tex]_{2}[/tex]g=-m[tex]_{2}[/tex]a


    3. The attempt at a solution
    I somehow get the Tsin[tex]\theta[/tex]'s to cancel when dividing the force equations to eliminate the m[tex]_{1}[/tex]a's. Maybe it's just the end of the day and I'm tired and suck at this. Can someone please help?[tex]_{}[/tex]
     
  2. jcsd
  3. Oct 13, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Careful with the signs of the components of N and the acceleration. (Make sure your signs for the acceleration are consistent.)
     
  4. Oct 13, 2008 #3
    Well, i made the acceleration of m1 positive y (+sine) and negative x (-cosine) because it is moving "up and to the right." My normal force is acting "up and to the left," so it has a positive y and positive x orientation. The difference is that m2 has an acceleration in the negative y direction. Does this sound like it would be correct? If you need me to, I can explicitly write out what I did, but if you don't need me to I won't because the tex can get tedious.
     
  5. Oct 14, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    OK. For some reason, you take "to the left" as positive. No problem, just be consistent. That means that the x-component of the tension must be negative.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Two Blocks with rope and Pulley on Inclined Plane
Loading...