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Homework Help: Interacting Systems - The Sled Dog!

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    The sled dog in figure (attached) drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10.

    If the tension in rope 1 is 150 N, what is the tension in rope 2?

    2. Relevant equations

    F_a on b = -F_b on a

    3. The attempt at a solution
    The force diagrams I did for sled A and sled B resulted in:

    Sled A: Fnetx = T1 - (mu_k)(m_a)(g) = 52

    Sled B: Fnetx = T2 -T1 - (mu_k)(m_b)(g) = T2 - 228.4

    My solution for T2 = 280 is very close, but not the right solution. Should I be adding another force for sled B, based on the friction of sled A or another component of sled A?
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2007 #2
    can you help me please solving my problem thx
     
  4. Oct 17, 2007 #3
    Apparently the answer is 270N. Still don't know they determined that.
     
  5. Oct 17, 2007 #4
    What are the masses of the sleds? I can't see the picture but I'm imagining that a dog is pulling sled b which is connected to sled a like so.

    Dog--b--a

    I don't know if this is what you meant, but Fnet for sled a and b are NOT the same unless they have the same mass.
     
  6. Nov 4, 2007 #5
    Solution

    T for rope 1:
    100kg * a = 150 N
    a = 1.5 m/s^2

    If the tension in rope 1 is 150 N, what is the tension in rope 2?

    T for rope 2:
    (mass of sled 1 + sled 2)*1.5 m/s^2 = T
    (100kg + 80kg)*1.5m/s^2 = 270N
     
  7. Oct 26, 2009 #6
    kster you ignored the friction force.
    Rope 1:
    T-Mkn=ma
    150-0.10(100*9.8)=ma
    rope 2 just the same expect for the weight is 180kg
     
  8. Oct 30, 2009 #7
    So to Sum it all up...

    Remember when dealing with any type of friction problem, you MUST add it into the eq. usually by finding the F_net along the y-axis.

    F_net_y = m_1 *a_y (a_y = 0, since the object is not moving in the y-direction)

    Find all forces acting on the y-axis to be your F_net_y

    F_net_y = N - m*g
    N - m *g = m_1 *a_y
    N - m *g = 0
    N = m *g

    Also remember what your Friction Force is.

    F_k = U_k * N
    F_k = U_k * (m *g)

    Now you can solve for a

    T_1 - F_k = m_1 *a
    (T_1 - F_k)/(m_1) = a
    (T_1 - (U_k *(m_1 *g)))/(m_1) = a

    Then you solve for T_2

    T_2 - T_1 - F_k = m_2 * a
    T_2 = (m_2 * a) + T_1 + F_k
    T_2 = (m_2 * a) + T_1 + (U_k *( m_2 *g))
     
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