*Originally posted by sonhouse*

**So there would be a polar co-ordinate vector added to the prime friction number?
**

BTW, wouldn't you have to use 1 minus the co-efficent in your calcs? You went 0.179X (g forces). shouldn't it be 1-.179 or 0.82 times the resultant accel? If I have it right, the 0.179 is the stopping force and if it totally stopped the motion, it would by definition be 1.0 a ...[text shortened]... 821 which would mean the actual acel would be 82% of what it would be without friction, right?

I didn't do the calculation, GregM did. But I would do it like this.

Imagine the blocks are lying flat on a table, connected by a rope, with forces equal to the gravitational forces in the original question being applied to each block. The 350 kg is on the right. Let right be +. Therefore the sum of the forces on the blocks is as follows:

sum(F) = F1 - F2 - Ff

F1 = 350 kg * 9.81 N/kg = 3434 N

F2 = 75 kg * 9.81 N/kg = 736 N

Ff = 0.179 * (350 kg + 75 kg) * 9.81 N/kg = 746 N

sum(F) = 3434 - 736 - 746 = 1952 N

sum(F) = m * a --> a = sum(F)/m

a = 1952 N / (350 kg + 75 kg) = 4.59 m/s^2

Therefore the acceleration will be 4.6 m/s^2 to the right. Bringing this back into the original model, assuming the 75 kg weight is on the left and the 350 kg weight is on the right, the acceleration will be 4.6 m/s^2 in the clockwise direction about the pulley centre.

Without friction, our equation is as follows:

sum(F) = F1 - F2 = 3434 N - 736 N = 2698 N

a = sum(F)/m = 2698 N / (350 kg + 75 kg) = 6.35 m/s^2

Therefore, using the same convention noted above, the acceleration will be 6.4 m/s^2 in the clockwise direction. Taking the ratio of the two answers, we find that:

a(friction)/a(no friction) = (4.59 m/s^2) / (6.35 m/s^2) = 0.723

Therefore the friction reduces the acceleration to 72% of the acceleration in the frictionless system.