**Determine the acceleration of each block:**

In the system of associated blocks shown in figure the coefficient of kinetic friction under blocks A and C is 0.20. Determine the acceleration of each block and the tension in the connecting cable. The pulleys are assumed to be frictionless and of negligible weight.

** Preliminary Discussion**

For simplification, we resolve the weights of A and C into their components acting parallel and perpendicular to the inclined planes, as shown in the free body diagrams. First thing, we must try is to determine the direction of motion of the system. We first suppose that one part of the system does not move and then calculate the tensions necessary to keep it at rest. We then calculate the forces in the remaining system and if any unbalanced force is obtained that part will move in the direction of the force. Also, we take into account the influence of friction in preventing the motion.

In this case, let us assume that block B is at rest. This gives us 2 T = 1200 N or T = 600 N acting at A and C.

If we substitute T = 600 N and sum up the forces parallel to the plane, then frictional force required for equilibrium is 800 - 600 = 200 N, but available friction force on A is 120 N. This is insufficient to keep A at rest and as shown, A shall be moving down the plane. Similarly, block C will move upwards as shown.

Yet we do not know the direction of motion of B because down plane motion of A will tend to raise B upwards whereas up plane motion of C will tend to lower down B. We now assume that let B move downwards as no frictional force is acting on B and incorrect direction of motion will be only importance. The actual value of T will determine its motion.

Now, let us remember the FBDs of A and C. For downward motion of A, T must be less than 800 - 120 = 680 N and for upward motion of C, T should be more than 320 + 48 = 368 N. Thus, we may estimate the value of T as average of (680 + 368) /2 = 524 N.

By using this approximate value of T on FBD of B gives an unbalanced downward force on B and justifies our assumption that B moves down.

We now try to use the concept of method of virtual work to get the kinematic relationships. The entire work done by internal associating forces on a system is zero.

We also keep in mind that work is a product of force and displacement and positive work is done when the displacement is in the direction of force. We have sum of the works done by T on the system of connected blocks.

TS_{C} - TS _{A} - 2 TS _{B } = 0

By cancelling T, we get

S_{C} = S _{A} + 2 S _{B }-------- (a)

∴ v_{C} = v _{A} + 2 v_{B } ------------ (b)

And a_{C} = a _{A} + 2 a_{B }---------- (c)

**Solution**

Now we apply equilibrium equations to each body.

For A 800 - 120 - T - (1000/g) a_{A} = 0 ---------- (1)

For B 1200 - 2 T - (1200/ g) a_{B} = 0 ---------- (2)

For C T - 48 - 320 - (400/ g) a_{C} = 0 ------------- (3)

We replace ac by a A + 2 aB . We obtain

T + 101.94 a _{A} = 680

2 T + 122.324 a_{B} = 1200

T - 40.77 a_{A} - 81.55 a_{B} = 368

∴ a _{A} = 680 /101.94 - T /101.94

And a_{B} = (1200 /122.324) - (2 T /122.324)

T - 40.78 [ (680/101.94) - (T/101.94)] - 81.55 [(1200/122.32) -(2 T/122.324)] = 368

∴ T + 0.4 T - 272.027 + 1.33 T - 800.103 = 368

2.73 T = 1440.13

T = 527.52 N

Substituting these values in Eq. (1) etc., we obtain a A = 1.496 m / sec2 a_{B} = 1.185 m / sec^{2}

a_{C} = 3.912 m/sec^{2}