0
$\begingroup$

Let's say that a particle moves horizontally with a velocity of $v_x$ and vertically with a velocity of $v_y$. Now, when we represent these two velocities and their resultant velocity using the head to tail method, we draw a horizontal arrow which represents $v_x$ and we draw a vertical arrow whose tip is at the head of the horizontal arrow. The vertical arrow represents $v_y$.

Now, I don't know if this just happens to me but to me, the head to tail method gives the idea that the first velocity acted first and the second velocity acted once the first one was done acting but actually, both of them acted on the particle simultaneously. This just confuses me.

One approach that I have tried was that first I proved that the path that the particle follows due to two velocities acting simultaneously on it is straight. Now, we know what the position of the particle at a certain point in time will be. Let's suppose the $v_1$ represents the horizontal velocity of the particle and $v_2$ represents the vertical velocity of the particle. Then, after a time $t_1$, the horizontal displacement suffered by the particle would be $v_1t_1$ and the vertical displacement suffered by it would be $v_2t_1$. Now, let $A$ be the point that denotes the final position of the particle after time $t_1$. Let $B$ be the point that denotes the initial position of the particle (at time = $0$). Let the angle made by $Ab$ with the horizontal be $\alpha_1$. Now, $\tan\alpha_1 = \dfrac{\text{Vertical displacement in time }t_1}{\text{horizontal displacement in time }t_1} = \dfrac{v_2t_1}{v_1t_1} = \dfrac{v_2}{v_1}$.

If we do the same thing but for time equal to $t_2$, and if we take the angle made this time as $\alpha_2$, then $\tan\alpha_2 = \dfrac{v_2}{v_1} = \tan\alpha_)$

Now, $\alpha_1 \in \Big(0,\dfrac{\pi}{2}\Big)$ and $\alpha_2 \in \Big(0,\dfrac{\pi}{2}\Big)$ and $\tan\alpha_1 = \tan\alpha_2 \implies \alpha_1 - \alpha_2$.

This implies that the path that the particle follows after moving with two velocities simultaneously is straight and now we can evaluate the resultant velocity of the particle which would be the velocity of the particle along that resultant path. Now, after a time $t$, the resultant displacement of the particle $(s)$ becomes : $\sqrt{(v_1t)^2 + (v_2t)^2} = \sqrt{v_1^{~2}t^2 + v_2^{~2}t^2} = \sqrt{t^2\cdot (v_1^{~2} + v_2^{~2})} = t \sqrt{v_1^{~2} + v_2^{~2}}$

Now, the resultant velocity will be $\dfrac{s}{t} = \dfrac{t \sqrt{v_1^{~2} + v_2^{~2}}}{t} = \sqrt{v_1^{~2} + v_2^{~2}}$ which follows the head to tail rule.

Now, first of all, proving that vector quantities follow the head to tail rule for quite a few quantities will get quite tedious and length and still, I wouldn't be able to cover all vector quantities. And even if I somehow do prove this for a lot of vector quantities, I would still not be able to interpret the head to tail rule properly because it just makes it look like one of the vector quantities is acting first and once it is done acting, the second vector quantity comes into play, which is actually not the case.

So, I think that my first step would be to get a clear understanding of the head to tail rule. I want to get rid of the confusion and gain clarity on the statement that I've typed in bold.

Thanks!

$\endgroup$
1
$\begingroup$

There is no order of velocity addition. Vector addition is commutative, ie $\vec{v}_1+\vec{v}_2= \vec{v}_2+ \vec{v}_1 $

You can switch up what the “first” vector is. So this shows that there’s no meaning to order.

However, what the triangle law is saying is actually what you mentioned. Going along one vector and along the other sequentially is equivalent to going simultaneously along both the vectors.

Now, first of all, proving that vector quantities follow the head totail rule for quite a few quantities will get quite tedious and length and still, I wouldn't be able to cover all vector quantities.

This is why we study the abstract quantities called vectors that are not attached to any physical quantities. That subject that’s called linear algebra.

There are a set of criteria for an object to qualify as a vector. These are called the axioms. And if a certain object does indeed qualify as a vector, then all properties obeyed by the abstract vector is automatically obeyed by the physical quantity. Thus we are saved from extra effort every single time!

$\endgroup$
1
  • $\begingroup$ Thanks, first of all. So, does linear algebra explore, in some detail that going along one vector and then the other is equivalent as going along them simultaneously? $\endgroup$ – Rajdeep Sindhu Aug 4 '20 at 4:50
0
$\begingroup$

I understand that your doubt is why the triangle law of vector addition is the way it is.

So, first what do we mean by vector addition. Vector addition is to add two vectors and finding their resultant vector sum. But what does it mean physically?

Basically, it means having one vector that replaces all the vectors we added up, but still have the same effect. But what does it have to do with parallelograms and triangles?

So first thing, when we are adding vectors, the resultant we find is for those specific vectors, not for their future forms that they may acquire due to change in variables with time.

So, as said earlier, the aim of vector sum is to replace existing vectors with the new one, so that the final effect is same. suppose, if we take this case, where your two velocity vectors are perpendicular to each other. now we need a third vector which will have same velocity as two of the vectors acting simultaneously. therefore, it should have same components along x and y directions as the vectors we took initially, the answer would be the third side of triangle. you can find a better figure and derivation in this link

Byju's derivation

Now it looks like you first take a vector, allow it to act, then we allow another vector to act. But this actually does not happen as we basically take x and y component of the vector and add it up to get the new vector, for a better visual representation of direction and magnitude of the vector, we use the triangle law of vector addition. Hope this helps you.

Alternatively, how i remember/understand it is that we actually do that, allow one vector to act after the other but time does not pass while we do it and we directly end up with resultant. It is not right at all, but I find it easier to think about it in this manner.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.