1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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))
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Interacting Systems - The Sled Dog!
  1. Dog Sledding (Replies: 4)

  2. Sled Dogs work? (Replies: 2)

  3. Dog sled (Replies: 2)

  4. Dogs pulling a sled (Replies: 1)

Loading...