1. Limited time only! Sign up for a free 30min personal 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!

Derive an expression for the acceleration of each mass and the tensions in the cord

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    1. For the system shown below use Newton’s Laws to clearly derive an expression for the
    acceleration of each mass and the tensions in the cord. The coefficient of friction
    between m1 and the incline is 0.100. The mass m1 is 3.00 kg, m2 = 8.00 kg, and θ = 35.0°.
    The pulley is frictionless solid disk of mass 0.250 kg.


    2. For the system shown below use only Conservation of Energy to clearly derive an
    expression for the speed and acceleration of each mass after m2 falls a distance of
    2.50 m. The coefficient of friction between m1 and the incline is 0.100. The mass m1 is
    3.00 kg, m2 = 8.00 kg, and θ = 35.0°. The pulley is frictionless solid disk of mass
    0.250 k


    2. Relevant equations



    3. The attempt at a solution

    ƩF(x) = m a1

    T1 - F - mgSinθ = m1 * a
    T2 + m2g = m1a

    ƩTorque = IAlpha
    -RT1 + RT2 = IAlpha
    R(T2-T1) = I (alpha/R)

    This is not my homework. This was given to use for practicing for our finals.
    Any help would be really helpful
    Thank You\
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 18, 2011 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Derive an expression for the acceleration of each mass and the tensions in the co

    You have made a decent start to solving the problem. I have a few comments:
    Good, but it looks like you mean to say m1 and not just m on the left side?
    You can also look at ƩF(y), that will help you get the friction force F in terms of other quantities.

    Think about the direction each force acts, and whether a + or - sign should be in front of each force. Also, I think you mean m2a on the right?
    Why did you replace "Alpha" with "Alpha/r"? I think you meant something slightly different... otherwise this looks good.
    That's okay, we pretty much like to see people show an attempt at solving the problem, which you did, even when it's not an assignment.

    Final comment: I think to solve this they need to tell you what R is for the pulley -- or maybe they tell you what I is. Can you check the problem statement again, and see if I or R for the pulley is given?

    Hope that helps.
     
  4. Dec 18, 2011 #3
    Re: Derive an expression for the acceleration of each mass and the tensions in the co

    no there is no R or I is given and only the mass of the pulley is given which is making me confuse. and the part 2 of the quest with the conservation of the energy, I really do not have any idea about that how we do that can you please help me out with that??...:)
     
  5. Dec 18, 2011 #4
    Re: Derive an expression for the acceleration of each mass and the tensions in the co

    If the pulley is frictionless then it won't rotate so it's only function is to redirect the forces, in that case the I and R don't matter and you can essentially forget it is there.
     
  6. Dec 18, 2011 #5
    Re: Derive an expression for the acceleration of each mass and the tensions in the co

    ohhhkk thanks.....and for the conservation energy is

    1/2m1v^2 + 1/2 M2 V^2 = M1g h sin(X) - (mu)M2g hcos(X)- M2gh
    ??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derive an expression for the acceleration of each mass and the tensions in the cord
Loading...