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    A M2 question I do not understand!!

    A team of 3 husky dogs each of weight W newtons is pulling a sleigh of weight 5W newtons across a rough ice-coverd horizontal plane. The ropes that attach each dog to the sleigh make angles of 30 degrees with the horizontal. The coefficent of friction between the sleigh and the ice is 0.1 and between a dog and the ice is 0.2.

    Assume that there are no other resistive forces, find in terms of W, the rate at which each dog is working when the sleigh is moving at a steady speed of 2 m/s.

    thanks!! :confused:
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    (Original post by John714)
    A M2 question I do not understand!!

    A team of 3 husky dogs each of weight W newtons is pulling a sleigh of weight 5W newtons across a rough ice-coverd horizontal plane. The ropes that attach each dog to the sleigh make angles of 30 degrees with the horizontal. The coefficent of friction between the sleigh and the ice is 0.1 and between a dog and the ice is 0.2.

    Assume that there are no other resistive forces, find in terms of W, the rate at which each dog is working when the sleigh is moving at a steady speed of 2 m/s.

    thanks!! :confused:
    F = ma
    3T - 3Wsin30 - 5Wsin30 - F(s) - 3F(d) = 0
    Resolving pependicular to the plane gives:
    R(s) = 5Wcos30. Hence F(s) = 0.5Wcos30
    R(d) = Wcos30. Hence F(d) = 0.2Wcos30
    3T - 3Wsin30 - 5Wsin30 - 0.5Wcos30 - 0.6Wcos30 = 0
    3T = 1.5W + 2.5W + 0.5Wcos30 + 0.6Wcos30
    P = FV = TV
    P = [1.5W + 2.5W + 1.1Wcos30][2]
    P = 8W + 2.2Wcos30
    P = W[8+2.2cos30]
    It might be wise to briefly check that and then evaluate with your calculator.
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    (Original post by Gaz031)
    F = ma
    3T - 3Wsin30 - 5Wsin30 - F(s) - 3F(d) = 0
    Resolving pependicular to the plane gives:
    R(s) = 5Wcos30. Hence F(s) = 0.5Wcos30
    R(d) = Wcos30. Hence F(d) = 0.2Wcos30
    3T - 3Wsin30 - 5Wsin30 - 0.5Wcos30 - 0.6Wcos30 = 0
    3T = 1.5W + 2.5W + 0.5Wcos30 + 0.6Wcos30
    P = FV = TV
    P = [1.5W + 2.5W + 1.1Wcos30][2]
    P = 8W + 2.2Wcos30
    P = W[8+2.2cos30]
    It might be wise to briefly check that and then evaluate with your calculator.
    WHY perpendicular use cos?
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    (Original post by John714)
    WHY perpendicular use cos?
    The angle made with the horizontal and parallel (of the slope) is equal to that made by the perpendicular and the vertical (as this is simply a rotation of 90degrees).

    Hence the weight makes an angle theta with the line that the normal reaction acts along.

    When you resolve the weight down the slope you use sin theta. When you resolve it perpendicular to the solve you use cos theta as you're resolving it 'along the line it makes the angle with'.

    Please feel free to ask if i can clarify things a bit more.
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    thanks! I will! and good luck to you for the trip to Bristol!!
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    (Original post by John714)
    thanks! I will! and good luck to you for the trip to Bristol!!
    Thanks
 
 
 

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