Interacting Systems - The Sled Dog

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The discussion revolves around calculating the tension in rope 2 when a sled dog pulls two sleds across snow with a friction coefficient of 0.10. The tension in rope 1 is given as 150 N, and the participants derive equations for the net forces acting on both sleds, factoring in friction. The correct tension in rope 2 is determined to be 270 N, which accounts for the masses of both sleds and the frictional forces. Key equations include the net force calculations along both the x and y axes, emphasizing the importance of incorporating friction into the analysis. The conversation highlights the need to accurately assess all forces to solve for tensions in interconnected systems.
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Homework Statement


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?

Homework Equations



F_a on b = -F_b on a

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?
 

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can you help me please solving my problem thx
 
Apparently the answer is 270N. Still don't know they determined that.
 
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.
 
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
 
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
 
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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