A block of mass M = 2m sits on a table and is attached to two other blocks, each having mass m, by strings passing over pulleys as shown (if you need picture let me know, one mass is hanging on the left and the other is hanging on the right). The coefficient of static friction between the large block and the table is 0.55 and the coefficient of kinetic friction between the large block and the table is 0.3. a) If M = 6 kg, what is the tension in the left string? b) What is the friction force on M? c) The string on the right is now cut. Describe in as much detail as you can the subsequent motion of all of the blocks. d) After the string is cut, what is the friction force on the large block?
Fnet = ma
Fsmax = musN
Fk = mukN
The Attempt at a Solution
I believe a and b are connected by principle. Now I'm under the belief that static friction is the friction that must be overcome in order to cause movement, and that kinetic friction is the retained friction opposing the force during movement. Yet for part A since the two masses hanging over the pulley are in direct opposition of each other it seems to me that the friction (in this case static) would be offset in both directions and therefore zero. So the force in T of the left string is = Fg for the hanging mass. Then for part b the force of friction on mass M, right now is 0. Yet he gives us static friction so it seems to me that either it's just there (cause in part c and d we'd be using kinetic friction I believe) or I'm missing something in my thought pattern.
Any help would be greatly appreciated as always! Thank you!