Let’s look at this image. For these two objects A and B that are connected by a non-elastic rope wrapped around this pulley, we can easily predict that if object A moves upward then B must move downward. And we can also tell just from experience that they will move at the same speed. This is because the rope connecting them has the fixed length therefore the motion of these two objects are dependent on each other. We will use of this very simple example to write a strategy to mathematically solve similar problems involving two particles with depending motion. First step: just like what we learned before for rectilinear motion we need to set up a coordinate system in order to represent the position of the particles. So we choose a fixed point to be the origin, and choose downward to be positive position. But for this type of problem involving pulley systems it’s more convenient to draw a datum, which is a line that any point on it represents a position of zero. Now with the help of the coordinates we can now properly represent the position of the two particles respectively using one dimensional position vectors again: the position of particle A, s_A, and the position of particle B, s_B. Don’t forget s_A and s_B are both variables that change with time. Step three: recognize the constant length in the system. For example for this piece of rope the wraps around the top semicircle of the pulley, although it also moves but the length is always a constant. Let’s call it l_semi for now. Step four: notice that I put a different background color for this a step, since in my opinion this is the most important step and also the easiest for students to make mistakes. Here we need to find out how the positions of the two particles, in this case s_A and s_B, are related. For this example it’s quite simple: the total length of the rope is a constant. And therefore that equals to s_A plus l_semi that we talked about earlier, plus s_B, and that equals to the constant length, l_T. Recognize again in this equation l_semi and l_T are both constants but s_A and s_B are variables with respect to time. And now we can take the previous equation and take the time derivative of the entire equation. And that becomes ds_A/dt plus zero plus ds_B/dt equals to zero. By definition the time derivative of position is the velocity, therefore ds_A/dt is simply the velocity v_A, and ds_B/dt is the velocity v_B. Therefore this becomes v_A plus v_B equals to zero, or v_A equals to negative v_B. And this agrees with our initial guess that the two particles will move at the same speed, but the negative sign indicates that the direction of motion will be opposite: when A moves up. B moves down, and vice versa. And if necessary we can differentiate the velocity equation again to get the relation in the acceleration of the two particles. So now let’s apply our strategy to this example which is a little more complicated than the previous one. Again we have two objects that are connected through this multiple pulley system. We know the velocity of particle A and we need to find the velocity of B. Unlike the previous example, for this one we can see that there are two ropes involved marked with different colors orange and purple. Therefore the depending condition that we’re going to write later will involve that total length of rope one and the total length of rope two are both constants. So let’s follow our strategy to solve this problem. Step one: draw a fixed line, the datum, set the coordinate, s. Step two: represent the positions of the two objects by the position vectors respectively: s_A, and s_B. But since s_A and s_B are not even on the same rope, it is necessary to define another position vector, say, s_C to serve as a middleman to connect the objects A and B together. Step three is to recognize the constant length. Step four, again, the most important step, is to write the depanding relation of the positions. First the total length of rope one, the orange rope, is a constant. That equals to s_A plus s_C, plus another s_C, and then plus some constant length that we discussed earlier, and that equals to the total length l_T1. And then for the second rope, the purple one, its total length equals to, this is s_B minus s_C, and plus s_B, and then plus some constant length again and that equals to the total length l_T2. And now we can combine these two equations by canceling out the middleman s_C, and get: s_A plus 4 s_B equals to a total constant length. And now step five, we differentiate this equation, and then we get v_A plus 4 v_B equals to 0, so v_B equals to negative v_A divided by 4, and that is 0.75 m/s, pointing downwards. Here is a different type of problem that also involves dependent motion. There’s a plate with height of 8 meters moving towards the wall, and a fixed light source casts its shadow on the wall. Obviously when the object moves the top of its shadow moves as well. We know that at this instant the speed of the moving object is 2 m/s, and we need to find the speed of the top of its shadow. For this type of problem, first step, we still need to set up the proper position vectors. Here s_A and s_B are set up along different axes but they are both originated from fixed points, and they represent the position of the moving object and the top of its shadow respectively. And then again the most important step: to find the dependent relation between these two positions. Since these two triangles are similar triangles, therefore we can write the ratio between s_A and 8 equals to the ratio between 40 and s_B. Or these two ratios are the tangent value of the same angle anyway. So from here we get s_B equals to 320 times s_A to the negative first power, or simply 320 over s_A. And now we can differentiate this entire equation with respect to time. The left hand side is simply ds_B/dt, which is the velocity v_B that we are looking for. The right hand side: to take the differentiation with respect to time we can apply the chain rule from calculus. So we find that if the defense Asian of tournament 20 times as a to the Mac and fries power with respect to as a and then multiply So we find that the differentiation of 320 times s_A to the negative first power with respect to s_A, and then multiply that by ds_A/dt. And ds_A/dt is the velocity v_A that’s given. Therefore since we know that s_A equals to 20 m and v_A equals to 2 m/s, we substitute those in. We can calculate v_B to be -1.6 m/s. Negative sign indicates that the top of the shadow moves downwards.