This chapter covers the motion of one particle
if it’s dependent on another particle. In other words, just pulleys attached with
cables. The best way to understand is to do some examples,
so let’s start off with some easy ones and then a few hard ones. In this example, we need to find the speed
of block B. So the first thing we have to do is to draw a reference line, also called
a datum. What that means is, for every part that is
moving, or changing in length, we can say its position with respect to the datum. The datum is always placed in a location where
it doesn’t move. In this example, you can see that the top
pulleys are fixed to a wall. That means no matter what cable is pulled,
they don’t move. The next step is to draw position coordinates. These represent a position equation, showing
where each of the moving parts are with respect to the datum at any given time. So we have SA, SB and SC. Notice how we only drew SB to the middle of
the pully. That’s because if you look carefully, you
will see that the length of rope across the pulley itself doesn’t change. It says constant. When we draw position coordinates, we are
only interested in parts that move. When we do other examples, it will make more
sense. Now, we can write an equation to represent
the length of the rope. In this example, it’s a single rope that
goes through all the pulleys, so we only need one equation. If there are 2 ropes, then we would have 2
equations and so forth. So we can say, the total length is equal to
SA plus SB plus SB plus SC. Now, if we take the derivative with respect
to time, we get velocity. The constant lengths become 0. In other words, the positions turn to velocity,
and the total length becomes 0. So we get, VA plus VB plus VB plus VC is equal
to 0. The question says the speed of block a is
2 m/s downwards, and block C is moving up with a speed of 1 m/s. Here, you have to be careful. When we drew our position coordinates, we
drew them downwards, meaning, we assumed any movement downwards to be positive. Block A is moving down, but block C is moving
up, so for block C, it’s -1 m/s. Let’s plug those values into our velocity
equation. Solving gives us VB is equal to -0.5 m/s. So that means it’s going opposite to the
direction we chose to be positive. In other words, block B is moving 0.5 m/s
upwards. Let’s do another example. In this example, we need to find the speed
at which block B rises. So the first step is to draw a datum, which
we can place at the top pully because it’s fixed. Now, we can draw the position coordinates. We have SA, SB and SC. Looking at our pulley system, we see that
it’s made up of 2 cables. That means we will end up with 2 equations. The first cable can be represented with SA
plus SC plus SC which is equal to length 1. The second cable can be represented with SB
plus SB-SC, which is equal to length 2. There are somethings to note in the problem. The first is that even though a chain is holding
on to block B, the chain and the block will move along with the pulley since it’s attached. So if we find the velocity of the pulley that
it’s attached to, the block will have the same velocity. The second is the small metal piece at pulley
C. That piece will move up or down with the pulley. So when we represent the 2nd cable, we are
accounting for that chain piece as well. Let’s tidy up the equations. Now we can take the derivative, giving us
velocity, while the constant lengths become 0. Now we can plug in the velocity at A, and
solve, which gives us our answer. We get -0.5 m/s, or 0.5 m/s upwards. Let’s take a look at this example. In this example, we have to find the time
needed for the load at b to gain a speed to 10 m/s starting from rest. Looking at the pulley system, we see that
the motor pulls the cable to the right, and this in turn pulls the load upwards. So we need 2 datums. One for the horizontal movement and another
for the vertical movement. We can place both of those datums at the pulley
on top, which is fixed to a wall. Now for the position coordinates. We have SA, SB, and SC. Using these 3 position coordinates, we can
express both of the cables in the system. So let’s write our 2 equations. For the first cable, we have SA plus SC is
equal to length 1. For the second cable, we have SB plus 2 times
SB-SC is equal to length 2. Let’s do a bit of tidying up. Now we can take the derivative. That gives us velocity. But notice how the question says the motor
draws the cable with an acceleration of 3 m/s^2. So we actually have to take the derivative
again to get the acceleration, that way, we can plug our value in. So taking the derivative again gives us this. Now we can plug in AA is equal to 3 m/s^2. Solving gives us AB is equal to -2m/s^2. In other words, pulley b has an acceleration
of 2 m/s^2 upwards. To find the time it took for the load to attain
10 m/s, we just need to use an old kinematics equation. Remember that the block has the same velocity
and acceleration as pulley B since it’s attached. So we can use this equation to find the velocity. The initial velocity is 0, final velocity
is 10 m/s, and acceleration is 2 m/s^2. Solving gives us 5 seconds. Let’s look at one last example. This isn’t a hard example, but I wanted
to show what happens when we have fixed lengths that we actually do need to account for. So in this example, we need to find speed
of the block. We can draw the datum at the top pully. Now looking at this diagram, we see that even
if the rope is pulled, the bar that’s attached to the small pulley in the middle does not
move. That’s a fixed length. How can we express the distance from the small
pulley to the big pulley at the bottom when there is a fixed length? For that, we will use a constant. So let’s do what we know already, which
is to draw the position coordinates. We have SA and SB. We also need to label the constant, which
we will label as h. Now, we can write our equation. It’s a single rope, so we only need one. We have SA plus SB plus 2 times SA-a is equal
to the total length. Let’s tidy up the equation. Remember that “a” is a constant, just
like the total length. When we take the derivative of this equation,
it also becomes zero. Now, we can plug in 4m/s in to VB and solve
for VA. We get -1.33 m/s, or 1.33 m/s upwards. When you get a question with pulleys, you
just need to establish a datum, write out position equations for the moving parts, label
any fixed lengths if you need to, and then write equations to represent each of the cables. I hope this helped, and if it did, please
give a like. I’ve included some more links to other solved
problems from this chapter in the description below as well. Best of luck with your studies