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Linear and Rotational Kinematics System.

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.53. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

    With what acceleration does the mass m3 fall?

    Hint: For each block, apply Newton's 2nd law. For the pulley, apply the rotational analog of Newton's 2nd law. And remember that since the string does not slip on the pulley, there is a relationship between the acceleration of the blocks and the angular acceleration of the pulley. You will end up with three equations and three unknowns (the answers to a,b,c). EDIT: is pulley -> (1/2)m*r*a = T3-T1?

    2. Relevant equations


    3. The attempt at a solution

    I believe I created the correct formulas for each mass.

    Block 1: T1 - Frictional Force = m*a

    Block 3: m3*g - T3 = m*a

    Pulley 2: Torque = I*alpha -> Torque = (1/2)m2*r*a

    I'm however not sure if these are correct, but more importantly, how I would combine them in order to find the acceleration of the falling mass.

    I realize that I need 3 equations and unknowns (T1, T2, and a). I believe my Pulley isn't correct because it also has T. Is my assumption correct? And, How would I change the pulley equation so that it works?

    EDIT: I believe now that my pulley should be, (1/2)m*r*a=T3 - T1. Is this correct?

    EDIT: Still having trouble. Using my above assumptions I came up with this: (m3*g-m1*g*u)/(m3+m1+.5*m2*r)=a
    This unfortunately still doesn't work. Any help would be much appreciated, I'm really trying my hardest to figure this out. :)
    Last edited: Nov 2, 2008
  2. jcsd
  3. Nov 3, 2008 #2


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    Hi CaptainSFS,

    I don't believe this is quite right. The right hand side (T3-T1) is not the torque. What would the torque be? (Think about what the units of torque are.)
  4. Nov 3, 2008 #3
    torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?

    EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/
    Last edited: Nov 3, 2008
  5. Nov 3, 2008 #4


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    Yes, that's right.

    Can you show more of your work? What numbers are you using, and what is your final answer?
  6. Nov 3, 2008 #5

    Block 1: T1 - Frictional Force = m*a

    Block 3: m3*g - T3 = m*a

    Pulley 2: (1/2)m2*r*a = (T3 - T1)*r

    Then, T1 = m1*a + m1*g*u | T3 = m3*g - m3*a <--I plug these two into my 'Pulley 2'

    I then algebraically solve for a and I end up with,

    a = (m3*g - m1*g*u) / (0.5*m2 + m1 + m3)

    I use m1=2kg m2=0.4kg u=.53 m3 = 4kg g=9.81m/ss || The answer I get is 3.51724 m/ss

    EDIT: Okay, So I plugged it in again, and I got 4.6518m/ss, which is the right answer. :P

    Thanks for your help! :)
  7. Nov 3, 2008 #6


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    Glad to help!
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