Three blocks of masses 10.0 kg, 5.00 kg, and 3.00 kg are connected by light strings that pass over frictionless pulleys as shown in Figure. The acceleration of the 5.00-kg block is 2.00 m/s^2 to the left, and the surfaces are rough. Find (a) the tension in each string and (b) the coefficient of kinetic friction between blocks and surfaces. (Assume the same mu_ k for both blocks that are in contact with surfaces.)

The problem is complicated right ?. The truth is, no. It is long ... that yes. but uncomplicated, physics has the peculiarity of being separated into simpler parts. Physics problems that look complicated are actually the composition of several small problems that are simpler. in this case we have three blocks. in words, three simple problems joined and looking like hard problem. What unites them? the rope!.
To solve the problem. we will rename the masses as m1, m2 and m3.

we have given direction to the tensions T1 and T2 to the left. because the problem tells us that the m2 block accelerate to the left this acceleration is common to the three blocks in words the whole system is moving to the left.
The first step to get the answers it is to make a free-body diagram (FBD). We begin by the mass m1:

as we said before, there are three more simple problems, we have solved the first one. We got the tension T1. Noting that, in block m1 only acts tension T1 and his weight (acts T1 because is the same rope that is attached to m2 and there is no extra strength eg the friction force, so the tension is the same) only we equate the forces acting on m1 to the product of its mass and system's acceleration.
Now for our second mini problem we work on m2. First we make the free-body diagram. Obviously only the T1 and friction force are the forces acting on m2. As we know who is T1 we will clear the kinetic coefficient of friction:

It is easy to assume that T1 = T2 because the rope remains the same. Truth is not. although the rope is the same over m3 is acting a friction force different from that acting on m2 (remember that although they have the same kinetic friction coefficient the friction force depends on the normal force, which in turn depends on body mass) that makes the tension change.
For our last mini problem, we know who is the kinetic coefficient of friction, we only work on m3 to clear the tension T2:

To facilitate the work we have oriented the XY coordinate system so that the X axis is parallel to the plane in which the block m3 is stand. Thus the weight of the block acts both on the Y axis and the X axis.
We have finished the problem our results are:
part a)
T1 = 78 N and T2 = 55.38 N
part b)
mu_k = 1.387
All right! That's it for this post! see you at the next. if you like our work share it with your friends! remember .. physics is very easy! ;)