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Two blocks and a pulley help

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Two blocks are connected by a light string passing over a pulley of radius 0.35 m and moment of inertia I. The blocks move (towards the right) with an acceleration of 1.50 m/s2 along their frictionless inclines (see the figure).

    http://www.webassign.net/gianpse4/10-54alt.gif

    Determine FTA and FTB, the tensions in the two parts of the string.

    Find the net torque (magnitude only) acting on the pulley, and determine its moment of inertia, I.

    2. Relevant equations



    3. The attempt at a solution

    Is there a separate tension for each piece of string and what are the equations if that is the case. If it is not the case, then what is the equation to find the tension?? Please Help!
     
  2. jcsd
  3. Nov 16, 2008 #2

    Doc Al

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    Staff: Mentor

    The tension must be different on each side of the pulley, otherwise there'd be no torque to turn it.

    The equations you need are Newton's 2nd law. Apply it (separately) to each mass and the pulley, then combine those three equations to solve for your unknowns.
     
  4. Nov 16, 2008 #3
    I am not sure how to do that with the tension and all.
     
  5. Nov 16, 2008 #4
    Indeed the tensions are different!
    There are actually three systems here, namely
    1. Block A which accelerates up the incline with an acceleration of 1.5 m/sec2.
    2. The pulley which has a tangential acceleration of 1.5 m/sec2 at r=0,35 and moment of inertia I
    3. Block B which descends with an acceleration of 1.5 m/sec2 down the incline.

    Systems 1 and 3 are related by Newton's law of motion
    F=ma.
    System 2 is related by the analogous equation
    [tex]\tau[/tex]=I[tex]\alpha[/tex]
    where [tex]\tau[/tex]=torque and
    [tex]\alpha[/tex]=angular acceleration=a/r
    r=radius of pulley,
    a=tangential acceleration at the rim of the pulley

    So you have three equations and unknowns, Ta, Tb and I, and you should have no problems solving for them.
     
  6. Nov 16, 2008 #5

    Doc Al

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    Staff: Mentor

    Start by picking one of the systems and identifying the forces acting on it. Drawing a diagram would be wise. Then apply Newton's 2nd law: ∑F = ma (or ∑τ = Iα, for the pulley).
     
  7. Nov 16, 2008 #6
    Block A: F=8.0kg*1.50m/s^2
    Block B: F=10.0kg*1.50m/s^2
    Pulley: T=(2/5Mr^2)(1.50m/s^2)

    Is that correct??
     
  8. Nov 16, 2008 #7
    can we tilt the diagram,61 degree clock wise such that the gravitional force that is acting on block b is 9.8m/s^2. and there will be a downward acceleration for block a?
    can we use that approach to solve this question as well?
     
  9. Nov 16, 2008 #8
    I have no idea.
     
  10. Nov 16, 2008 #9
    What you have got are the net forces (mass +/- string tension) acting on the mass.
    For the pulley, you'd need to convert the linear acceleration into angular acceleration using the radius. I is one of the three unknowns, since you do not know the mass of the pulley.

    The downward force of the 8 and 10 kg masses can be resolved into two components, one parallel to the inclined plane, and the other, normal to it.

    The normal force does not do anything, as the inclines are frictionless. The forces that are parallel to the inclined planes are the ones causing the acceleration. Do not forget to add/subtract the tension of the string from the force parallel to the incline.
     
  11. Nov 17, 2008 #10
    Net force=ma=force of tension-mg

    Block A: (8kg)(1.50m/s^2)=Ft-(8kg)(9.8m/s^2)
    Block B: (10kg)(1.59m/s^2)=Ft-(10kg)(9.8m/s^2)

    Is that what you mean?
     
    Last edited: Nov 17, 2008
  12. Nov 17, 2008 #11
    Correction:

    Block A: Ft-(8kg)(9.8m/s^2)sin32=(8kg)(1.50m/s^2) Ft=53.55N

    Block B: -Ft+(10kg)(9.8m/s^2)sin61=(10kg)(1.5m/s^2) Ft=70.71N
     
    Last edited: Nov 17, 2008
  13. Nov 17, 2008 #12
    So to find the net torque: Block A: (0.35m)(53.55N)=18.74 Block B: -(0.35m)(70.71N)=-24.75
    Net Torque: 18.74-24.75=-6.01N*m
     
    Last edited: Nov 17, 2008
  14. Nov 17, 2008 #13
    Moment of inertia: t=I(alpha) or t=I(a/r)
    -6.01N*m=I(1.50m/s^2)/(0.35m)
    I=-11.45kg*m^2

    What am I doing wrong here?
     
    Last edited: Nov 17, 2008
  15. Nov 17, 2008 #14

    Doc Al

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    Staff: Mentor

    This is fine, except for the minus sign. You just need the magnitude of the net torque.
    Redo your calculation; you made an arithmetic error.
     
  16. Nov 17, 2008 #15
    Ok, I see what I did.

    Thanks for the help ;)
     
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