# Silly Pully Question

#### Ehwic

5+ Year Member
So Im a little confused as to visualizing how a pulley works. When you have several pulleys connected in series, with a weight at one end of the series, how do you know which pulleys move relative to another. I ask this because one needs to know this in order to find the net force. For some reason, I don't understand how pulleys work.

Also, if youre pulling at an angle, how do you know when to factor that in, and when not to factor it in?

#### ingramw1202

So Im a little confused as to visualizing how a pulley works. When you have several pulleys connected in series, with a weight at one end of the series, how do you know which pulleys move relative to another. I ask this because one needs to know this in order to find the net force. For some reason, I don't understand how pulleys work.

Also, if youre pulling at an angle, how do you know when to factor that in, and when not to factor it in?
Lets look at this example below for a series of pulleys. I assume the pulleys are massless and the string is massless and the distances are all equal.
http://en.wikiversity.org/wiki/File:Pulley3a.png
In the diagram, you have four pulleys connected in a series, and a mass, M, hangs at the right end of the series. Well, all of the pulleys are connected by one rope, and the pulleys, we are assuming, are massless. If this is the case, and the string is massless as well, which is usually the case for the MCAT, then the tension on the rope is the same every where, because there is only one rope. At the right side of the picture, M hangs down, with a force =Mg. When you pull on the end of the rope, depicted on the left hand side of the picture, you are applying a force, F, down. When F is applied down, there is an opposite force, applied up at the first pulley. It counteracts the force you apply. If you look at the picture, the top two pulleys are fixed to the ceiling of the apparatus, so they will not move. Ever. They can't. When you pull down, tension pulls up at the left side of the apparatus. Now, on the same pulley, the one furthest to the left, There is a tension in the right side of the string as well. This tension comes from the mass, M, that is connected to the bottom to pulleys. This tension pulls up, counteracting the mass attached to the bottom two pulleys. So now we know that on the top left pulley, when you pull down, tension pulls up on the left side of the string and on the right side of the sting. If you go to the second pulley, the left bottom one, the string loops underneath it. Since the mass is attached to this pulley, then it feels a force, Mg, downwards. But, since the mass is attached to the bottom right pulley as well, and we are to assume that the mass is hanging between the center of the two bottom pulleys, then the force downwards on the bottom left pulley is actually 1/2Mg Well, since the bottom right pulley feels a force downwards which is equal to 1/2Mg, the tension on the rope will pull upwards on the left side. On the right side of that bottom left pulley, the string pulls up, to counteract the downward force which is also equal to 1/2Mg. So now we have that in the far top left pulley, a force, F, is applied downwards, and the tension pulls upwards on the left side of the pulley. On the right side of the pulley, the tension pulls upwards as well to counteract the downwards force of the mass hanging from the bottom two pulleys. In the bottom left pulley, tension pulls upwards on the left side and the right side to counteract the downward force, 1/2Mg, due to the mass.
The string then goes up and connects to the top right pulley. At the top right pulley, the force due to the hanging mass pulls downward, which means that the tension in the rope will pull upwards to counteract the downward force due to the mass. On the bottom right pulley, the force, 1/2MG is felt. So the tension on the left side of the bottom right pulley pulls up and on the right side it also pulls up, both which counteract the force pulling down on the pulley due to the hanging mass. So now what we have done is essentially broken the entire series into its component parts. Top left pulley feels a force down on the left due to your force and down on the right due to the mass hanging. Tension will counteract both forces. Remember, force is a vector. And since we are only working in the y direction (there are only forces down and up) movement will only be down or up, not left or right. So if Mg is greater than the sum of the forces seeking to pull it up, then Mg will move down. If Mg is less than the sum of the forces seeking to pull it up, then Mg will move up. If Mg = the sum of the forces seeking to pull it up, there will be no net movement because the forces cancel each other.

So, when you pull the rope or string or whatever it is on a single pulley with a single mass attached to it, the sum of the downward forces opposing you is Mg. Even though you are pulling down with a force of F, the pulley converts this downward force into an upward force via tension, hence pulleys make work easier. They essentially cause the force of gravity to work against itself. You pull down against against Mg. If your applied force exceeds the needed applied force to move Mg up, then it will move up. Now, what I am going to do is number the tensions in the rope and right and equation. So starting from left and going right, you have T1, T2,T3,T4, and T5, each is labeled 1/4 on the picture. Now I will explain why they are labeled 1/4.
Notice that the mass is hanging directly underneath the top right pulley and in between the bottom two pulleys. Due to this, you can imagine pulley as a single pulley alone, with a mass M attached to it by a string that loops over the pulley. So that means that the tension in the rope on either side of a single pulley set up in this manner is equal to 1/2Mg. Well, in in this case, we actually have 4 pulleys. And as I previously stated, you only have one rope, so the tension is the same throughout the rope. Well, since there are four pulleys the downward force is divided by four, because the downward force due to the hanging mass does not change. Because the tension in the rope will counter act that constant force, the more tensions you have, the less each tension will be, because their sum will equal the force down due to the mass. Since we have 4 pulleys, the tension on either side each pulley becomes 1/4Mg, and the force felt by each actual pulley downward becomes 1/2 Mg, because the mass itself is suspended between two pulleys. This is how pulleys work. They decrease the amount of force required to lift and object by decreasing the force that must be over come. In this case, instead of instead of applying a force greater than MG down to lift the force, you only have to apply a force greater than 1/4 Mg. So lets write what we have in mathematical terms.

Fnet=[(Fx)^2 + (Fy)^2]^ 1/2.
Since there is no X force, the Fnet simply becomes, Fnet = the summation of all the Y f0orces in the system.
In the situation that the mass and two pulleys it is attached to move upwards we have this:

Fapplied - T1 + T2 - T2 + T3 - T3 + T4 - T4 + T5-T5 = Mtotal x a(Net)

And, since we know we have four pulleys, and Tension(net) = -Mg, we have:

Fapplied -1/4mg + 1/4mg -1/4mg + 1/4mg - 1/4mg + 1/4mg -1/4 mg +1/4mg -1/4mg= Mtotal x Anet.

This simplifies into Fapplied -1/4mg = Mtotal x Anet.
This can be generalized to Fapplied - (1/n)mg=Mtotal x Anet, where n is the number of pulleys.
The left side represents the sum of the forces, the right side represents the total mass of the system and the net acceleration of the system. So we have: Fnet=Mtotal x a(net)
By convention, all of the forces acting upward ( except the force that you apply downward when you pull the rope) are positive those that act downward are negative. As you can see, this will simplify to:
Fup - Fdown= Mtotal x Anet, then
Fnet=Mtotal x Anet, which is always true.

So the key is to break each system into its parts. In the above scenario, if Fnet is positive (up) then the two bottom pulleys as well as the mass will begin to accelerate up. If Fnet is negative (down) then the mass will accelerate downwards. If Fup and Fdown equal each other, there is no net force and the system will exist in static equilibrium (no motion).

Now for the angle question, all you would have to do is find the components of the the force which you apply at that angle. But remember, this angle is only applied to the far left pulley, or T1. So this means that T1 will have X and Y components of the tension. Always add only X's and only Y's together in vector addition. Even though you will now have an added X component to the top left pulleys Force, the other three pulleys still only feel the Y component of that force, because they are aligned in way in which they can only move up or down. IF the top pulleys can slide, then the X component of the F which you will apply will cause the top left pulley to slide in the direction of the X force, which will cause the bottom left pulley to slide in the X direction, which will cause the top right pulley to slide in the X direction, which will cause the bottom right pulley to slide in the X direction. In this case, the problem is EXACTLY THE SAME, accept now each Tension in the rope would have X components as well as why components, and to solve for Fnet, you simply sum all of the X's and Y's and take the square root of the two sums squared added together or says:

Fnet= [[[ Fapplied x cos(theta)] +[(1/n)mg(cos(theta))]]^2 + [[Fapplied x sin(theta)] + [(1/n)mg(sin(theta))]]^2 ]^1/2 = Mtotal x Anet

Hope this helps!

Last edited:

#### ingramw1202

and sorry for all of the typos lol

OP
E

#### Ehwic

5+ Year Member
Yeah that was great. thanks for the help.