Pyrrhus
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Well, Boxes are identical so their $\mu m$ is the same.

OlderDan
Homework Helper
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.
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.

I am still lost.........

OlderDan
Homework Helper
nperk7288 said:
I am still lost.........
nperk7288 said:
So 47.0 N = fs right? but I don't know what N is...
You don't know the mass of the block, but in terms of that unknown mass you can write

$$47N = f_s = \mu N = \mu Mg$$

where g is the gravitational acceleration constant. You don't need to, but you could solve for

$$\mu = \frac{47N}{Mg}$$

In order to get the bottom block to slide under the top block you must apply enough force to accelerate the bottom block slightly more than the top block. When you achieve this condition, the top block will be accelerated by the frictional force of 47N (slightly less when slipping starts because of the difference between static and kinetic friction; you need not worry about this), and by Newton's third law the top block will exert a force of 47N opposing the motion of the bottom block. The bottom block will have a normal force at its bottom equal to the weight of two blocks. This is found by considering the vertical forces acting on the bottom block, including its weight and the normal force pressing down on its top from the weight of the top block. The applied force must exceed the frictional forces opposing it by enough to accelerate the bottom block by slightly more than the acceleration of the top block. A free body diagram of the two blocks will give you these equations.

Tob block
$$Ma_{top} = 47N = \mu Mg$$

Bottom block vertical
$$N_{bottom} = Mg + N_{top} = 2Mg$$

Bottom block horizontal
$$Ma_{bottom} = F_{applied} - \mu N_{top} - \mu N_{bottom}$$

$$F_{applied} = Ma_{bottom} + \mu N_{top} + \mu N_{bottom}$$

$$F_{applied} = Ma_{bottom} + \mu Mg + 2\mu Mg$$

The condition that $$a_{bottom}$$ must be slightly greater than $$a_{top}$$ combined with the above is all you need to find the minimum applied force.