# Force division of moving pulleys?

I am a second grade at a middle school and I was reading a physics workbook to prepare for a test. And I was solving pulley problems and one problem made my brain stop. The problem asked me what would the minimum force of F would be when the weight of the pulleys were 30N. I checked the answers and the way to solve it. The workbook told me that the force on each string holding the moving pulley equals to 1/n. (n = the numbers of string holding the pulley) Why is it true?

So within a cord/string there is a property called tension which is a measure of the force exerted along the string. If the string stretches homogeneously (the same at all parts of the string) then it turns out that this tension is the same at all parts of the string: you pull with force $m$ Newtons, then everywhere you see the string you need to think of it as a force of $m$ Newtons.

Pulleys, because they "roll" perfectly, allow the tension to come to the same value between the two sides of the string. (In other words, if there is a tension imbalance, then it will pull the rope in one direction, so that the rope will just roll along the wheel of the pulley from low to high tension. This will stretch out the low-tension side and relax the high-tension side, bringing the tension difference closer to 0, until the tension difference is 0 and they're both the same tension.)

Because of this, it's as simple as looking at the big wheel that's suspended in midair and doing a force-balance on that. It is being pulled downward by this weight 450 N and its own gravity 30 N; it is being pulled upward by $3 T$ where $T$ is the force of tension within the string; and $T = F$ according to the diagram, so $3 T = 480\text{ N}$ or $F = T = 160\text{ N}.$

This is a mechanical advantage problem. For fixed pulleys, only the direction of motion is changed, and there is no mechanical advantage. A 1 N force directed downward on one side of the fixed pulleys (the small ones) produces a 1 N force directed upward on the rope on the other side of the fixed pulleys.

For a movable pulley (the large one), there are two supporting strands on it, so the mechanical advantage is 2. Each strand supports 1/2 of the total weight, which is the weight of the load and the weight of the pulley. The mechanical advantage multiplied by the force F must equal the total weight. This means that F equals (450N + weight of the large pulley)/2. If the weight of all the pulleys taken together is 30 N, the question can't be answered. However, if the weight of just the large pulley is 30 N, the answer is F = 240 N.

Note that I don't have to include the weight of the small pulleys because their weight is supported by the beam that they are attached to.

Regarding why this is true: Work = Force x distance. For the movable pulley, F is 1/2 of the total weight, since there are two supporting strands (n=2). When you pull up on one of the ropes supporting the movable pulley, and you pull 2 m of rope through the pulley, the load only rises 1 m, because you remove 1 m of rope from each side of the movable pulley. Work = Force X distance for you when you pull up on the rope, and work = force X distance for the weight that is moving up. This means that work into the device equals work coming out of the device, which (assuming an efficiency of 100%) is a requirement if energy (work) is going to be conserved.

If you have the time to set this pulley arrangement up in a tutorial session, or your teacher gives a demo, the above explanation will make somewhat more sense. Good luck on your test.