# Resolving vectors along axes

While solving problems involving constrained motion, such as that which the following picture describes, I am always confused about one aspect of the situation in question:

In the above problem, it is given that the ring moves with a velocity of $Vr$ $m/s$ toward the right. This velocity can be resolved into two components, one along the direction of the rope extending from the string(at an angle $θ°$ with the horizontal, as shown)and one along the perpendicular to this direction.

My doubt us this; once we have resolved $Vr$ along the rope, we find that this value is $Vrcosθ$. What is the physical significance of this? I realize that the ring moves in both directions and these vector components give its velocity in those two directions but does this also imply that the rope also moves with a velocity $Vrcosθ$? I am very confused regarding this concept.

Edit: I posted the picture only to illustrate my point better. This is not a homework question because it relates to the basic concept of resolving vectors.

• Okay. But that isn't even the question I had asked; does the fact that the block has a velocity component along the rope mean that the rope moves with that velocity too? What is the physical significance of these components? – user106570 Aug 28 '16 at 6:45
• If the rope is inextensible then each part of the rope must travel at the same speed. – Farcher Aug 28 '16 at 6:54
• Oh no no, I am not supposed to take all that into consideration. I am given θ=53°. Yes, OK, but is it correct to assume that the rope travels with a speed of $Vrcosθ$?! – user106570 Aug 28 '16 at 6:55
• Yes, it might sound like a homework question with the angle given but I only used this problem to illustrate my doubt in the first place. Please understand that I am not required to consider those nuances of the situation that you mentioned in your comment. I'm not even taking friction into consideration! My doubt is simple and unrelated to the question that the above problem is asking; if a body is moving in a given direction with given velocity, will a string/rope attached to it move with the component of the body's velocity along its length? – user106570 Aug 28 '16 at 10:15

Not every point on the rope is moving with the same velocity.

Although the top of the rope, where it is attached to the ring, is moving horizontally with speed $v$, the other end, where it is attached to the block, moves down with speed $v\cos\theta$. You can see this easily by inspection of the cases $\theta=0^{\circ}$ and $\theta=90^{\circ}$, as follows :

When the rope is horizontal ($\theta=0^{\circ}, \cos\theta=1$), if the ring moves to the right by distance $\delta x$ then the block moves down by distance $\delta x$. When the rope is vertical ($\theta=90^{\circ}, \cos\theta=0$), if the ring moves to the right by $\delta x$ then the block does not move down at all as the ring passes vertically above it. Likewise for the velocity $v=\frac{\delta x}{\delta t}$ and acceleration $a=\frac{\delta v}{\delta t}$ of the ring.

To be more precise, in the diagram above, if the ring moves by a small amount from A to B then the length of rope between the ring and pulley has changed from AP to BP. For a small change $AB$ we have that $BP\approx CP$, where ACB is a right angle. So the length of rope between the ring and pulley has shortenend by an amount $AC=AB\cos\theta$. The length of the rope is constant, so the reduction in the length above the pulley increases the length below the pulley, hence the block moves down by the same amount $AC$.

BTW, in your diagram I would draw the section of rope from the ring to the pulley as a straight line, not a curve, to show that it is in tension.

• You are correct about my making an error for which I apologise. I have deleted my erroneous comments. – Farcher Aug 28 '16 at 17:06
• Oh, thank you for explaining it to me so nicely. OK, I will correct my diagram. Instead of using this detailed description of the situation while solving every problem that I encounter, will I be correct in assuming that a rope(/rod) attached to a moving block will move with a velocity equal in magnitude to $Vcosθ$ along the length of the rope(/rod) where $V$ is the speed of the block and $θ$ is the angle b/w the block and the rope? This was my basic doubt. – user106570 Aug 29 '16 at 0:52
• Yes. Resolve the motion along the rope or rod to get $v\cos\theta$. – sammy gerbil Aug 29 '16 at 1:07
• @sammy gerbil Beautiful answer. Helped me a lot. Thanks. I have a quick question. We often say that since the rope is inextensible, each part of the rope moves with the same speed. If they don’t, the rope will either snap or will become slack. You said that the top of the rope where it is attached to the ring is moving horizontally with speed $v$, and the other end where it is attached to the block moves down with speed $vcos\theta$. Then how do I make sense of the statement that because the rope is inextensible, each part of the rope moves with the same speed? – π times e Nov 5 at 2:37
• @sammy gerbil can I say that the top of the rope where it is attached to the ring, is moving horizontally with speed $v$, and its speed has a component along the length of rope that is $vcos\theta$, which is equal to the speed with with the the part of the rope attached to the block is moving down? In other words, speed of approach (or separation) between them is zero which is why length of the rope remains constant? Is my understanding of this situation correct? – π times e Nov 5 at 2:41