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