Well, Boxes are identical so their [itex] \mu m [/itex] is the same.
I should allow for the possibility that another assumption is made by your text in approaching this problem. It is really a very unrealistic problem, in my opinion. All the force supporting the cloth is supplied by the table, so the normal force must be constant if it is not sliding, as I stated before. However, for a limp cloth hanging over the edge of the table the normal force is nowhere near uniformly applied to the cloth. It is conceivable that the text will approach the problem by analogy to a mass on the table connected by a string to a mass hanging over a pulley, in which case they may assume a normal force that depends only on the mass still on the table. That is not correct in, my opinion, but the real solution more likely falls somewhere between the two approaches I have mentioned. The most likely thing, it seems to me, is that the edge of the table where most of the normal force is applied will have a reduced coefficient of friction compared to the larger surace area of the cloth on the table. Treating the sitution with constant normal force, and with normal force proportional to the amount oif cloth still on the table will probably give values that surround the realistic solution to the problem. Your text probably does one of the two simple approaches: either constant normal force, or normal force proportional to the mass still on the table.OlderDan said:For number 2, the normal force is constant, so the friction force is constant. The amount of cloth hanging over the edge determines the force opposing the frictional force acting on the part of the cloth on the table. You can treat the cloth like a string with tension in the cloth acting between the two sections of cloth.
In post #21 you hadnperk7288 said:I am still lost.........
You don't know the mass of the block, but in terms of that unknown mass you can writenperk7288 said:So 47.0 N = fs right? but I don't know what N is...