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Newtons Laws- Double checking

  1. Oct 1, 2006 #1
    A 28.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system begins to move.
    a) Calculate the mass of sand added to the bucket
    b) Calculate the acceleration for the system.


    a)
    28.0 kg block weighs (28.0 kg)(9.80 m/s2) = 274.4 N

    Ffr(max) = µsFN
    Ffr(max) = (.450)(274.4 N) = 123.48 N

    F = ma
    123.48 N = m(9.80 N/kg)=12.6 kg
    12.6 kg – 1.0 kg=11.6 kg of sand

    b)
    Force of Kinetic Friction between Block and Table
    Ffr = µkFN
    Ffr = (.320)(274.4 N) = 87.808 N

    Force of Bucket
    F=ma
    123.48 - T= 12.6a

    Force of Block
    F = ma
    T - 87.808 N= 28a

    Substituting First Expression into Second Expression and Solve for T:
    123.48 - T = 12.6a, T = 123.48 - 12.6a

    T - 87.808 N = 28a
    (123.48 - 12.6a) - 87.808 N = 28a
    35.672 = 40.6a
    a = 0.879 m/s2

    Does this all look correct?
     
  2. jcsd
  3. Oct 1, 2006 #2

    Fermat

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    Homework Helper

    It checks out fine.

    Another way of doing (b) is like this.

    The total force on the system is (Force from bucket - kinetic friction) = 123.48 - 87.808 = 35.672 N
    Total mass (of the system) undergoing acceleration is (mass of block + mass of bucket) = 28 + 12.6 = 40.6 kg

    Then use Newton's 2nd law

    F = Ma
    35.672 = 40.6a
    a = 35.672/40.6 = 0.8786 m/s²
     
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