Moveable pulleys you multiply the mechanical advantage

[Chocolaty]

Question: http://img486.imageshack.us/img486/5593/pulley9dm.jpg

Ok so i understand that for every moveable pulleys you multiply the mechanical advantage by 2. In this scenario there are 3 moveable pulleys, and they all look like they are supported by the ceiling so I need to divide the load by a half (i think...) so does that mean that:

(6000N + 300N)/(2*2*2(MA)*2(ceiling) = 196,9N?

 

[Chocolaty]

Is this right?
cmon, someone has to know this

 

[mukundpa]

whether the uppermost pulley is moveble?

 

[Chocolaty]

No, whether my answer is correct. I'm really not sure if i worked the problem correctly.

 

[mukundpa]

No, whether my answer is correct. I'm really not sure if i worked the problem correctly.

only want to find correct answer or want to learn?

If the uppermost pulley is not movable then how many movable pullies are there and what will be MA?

 

[Chocolaty]

Of course i want to learn, if i have the answer right it means that the calculation is right.

I think that the highest pulley is fixed and the 3 others are moveable. So the mechanical advantage is 2*2*2 = 8. But since the 3 moveable pulleys are attached to the ceiling I think it means that the work force is divised by 2, only once or three times? no clue, i'll say 3 times. So then MA = 8*2*2*2=64
Seems a little high. Is this right?

 

[mukundpa]

no, the mechenical adwantage is 2*2*2 = 8 only
means you have to apply 1/8 of the load.

for a single movable pulley one end of string is attached to the ceiling and the other end is held so that you have to apply a force equal to the tension in the string which is half in case of frictionless system. hence MA is 2.

 

[Chocolaty]

How do you work the math dealing with the ceiling in all this?

Edit: ok i think i understand what you wrote.

 

[mukundpa]

no, the mechenical adwantage is 2*2*2 = 8 only
means you have to apply 1/8 of the load.

for a single movable pulley one end of string is attached to the ceiling and the other end is held so that you have to apply a force equal to the tension in the string which is half in case of frictionless system. hence MA is 2.

it is 2 for a single pulley
in our system we have three pulleies and hence it is 2*2*2 = 8

now what is your answer for the effort, to lift the load?

 

[Chocolaty]

(6000N + 300N)/8MA = 787.5N

So the only thing you have to do is basically ignore the fixed pully and multiply the MA by 2 for each moveable pulleys in any possible system?

This is like childs play... i was complicating things for nothing

 

[mukundpa]

good

As nothing is said about where and how the friction is acting you may add it with load, otherwise the things will be different and a bit complicated.

MP

 

[Gamma]

When I read the problem it sounds like frictional force is a net force. Also note the subscript is 't' as in 'total'. I am wondering if the Ft should be devided by 8 too. If the given 300 N is a net fric. force that the man has to work against, then he really have to apply 6000/8 + 300 = 1050 N.