The sled dog in figure 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?
[Sled A, 100kg]---1----[Sled B, 80kg]--2--Dog pulling
This problem was solved here, but I want to know why I come up with different results. I see in the link that they did not account for the friction. Since mu is given, we should use friction, right? Why did they ignore it there and come up with the answer?
Net force and a = F/m
The Attempt at a Solution
First I made axis for A, with normal A pointing up at 980N, F_G_a pointing down at -980, f_k_a (friction) pointing left, away from the dig, at -(.10)(980) = -98N. Tension on the rope to the right was 150 N. Sum, F_net_x_a = (150)-(98)=52 => a = 0.52 m/s^2
But that's not the acceleration given in the link. Like I said, they ignored friction, I did not. Why is ignoring friction resulting in the right answer?
[I won't continue to work the problem here because it will be wrong until I figure out the right acceleration.]