# Vector Addition of velocity [closed]

In the arrangement shown in figure, the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with what speed?

I tried to solve this problem with the following way-

Let the upward speed of the mass M be x. Then, $$x=\sqrt{u^{2}+u^{2}+2u^{2}\cos 2\theta}$$ $$\implies x=u\sqrt{2(1+\cos 2\theta)}$$ $$\implies x=u\sqrt{4\cos^{2}\theta}$$ $$\implies x=2u\cos \theta$$ Hence, the upward speed of mass M is $$2u\cos \theta$$. But the corrcet answer given is $$\dfrac{u}{\cos\theta}$$.

Can anyone please tell me that where i am doing mistake.

• @Steeven See this physics.stackexchange.com/q/244337 Dec 2 '21 at 9:04
• Vector addition rule is- |A vector+B vector|=$\sqrt{a^{2}+b^{2}+2ab\cos \theta}$ Dec 2 '21 at 9:05
• @MathGenius Imagine two horses pulling the same wagon. Both of them are running at the speed $v$. Does the wagon move at speed $2v$? That is the mistake in your approach. Dec 2 '21 at 11:27
• Hi Math Genius. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. Dec 2 '21 at 18:34
• "Homework questions can be on-topic when they are useful to a broader audience." Dec 3 '21 at 23:17

You cannot use vector addition for velocity like that. Imagine two horses running in the same direction at the speed $$v$$ pulling the same wagon. Does the wagon move at speed $$v$$ or $$2v$$? The wagon moves at the speed $$v$$, i.e. you can apply velocity vector addition only to velocity components (vectors) of one body, and two horses are separate bodies.

In your example, vector addition would work for forces but not velocities. Moreover, you have to take into account that $$\theta$$ also changes with time.

Start from the right triangle and steady conditions:

$$H = L \cos\theta , \qquad L^2 = H^2 + D^2$$

where $$H$$ is the vertical component, $$D$$ is the horizontal component and $$L$$ is hypotenuse of the right triangle. However, in your case only the $$D$$ is constant, hence:

$$h(t) = l(t) \cos(\theta(t)) , \qquad l(t)^2 = h(t)^2 + D^2$$

Now take derivative of the right-hand side equation:

$$2 l(t) \frac{d}{dt} l(t) = 2 h(t) \frac{d}{dt} h(t) + \underbrace{\frac{d}{dt} D^2}_{0}$$

where $$v(t) = \frac{d}{dt}h(t)$$ is the speed in the upward direction and $$u(t) = \frac{d}{dt} l(t)$$ is the speed in the downward direction. This finally gives $$l(t) u(t) = h(t) v(t)$$ which is:

$$v(t) = u(t) \frac{l(t)}{h(t)} = \frac{u(t)}{\cos(\theta(t))}$$

• But why cant we do vector addition in the horse case? Dec 3 '21 at 1:00
• @Osmium Because two horses are not the same body. You can add velocities only for the same body. Dec 3 '21 at 8:49

the "work" that done due to the tension force $$~T$$ is

$$T\,\cos(\theta)\,\delta x=T\,\delta u\quad \Rightarrow\\ \delta x=\frac{\delta u}{\cos(\theta)}$$

Note that, deliberately, this is not a completed answer to the question asked.

You are treating $$\cos \theta$$ as a constant. It isn't. What is constant (or almost constant) is the horizontal separation (call it $$2X$$) of the tops of the strings. Then if $$y$$ is the vertical distance of the point where the strings meet below the line joining the tops of the strings, we have $$y^2 + X^2 = r^2$$ in which $$r$$ is the length of each string between pulley and where strings meet.

Using the chain rule, differentiate both sides wrt time, remembering that $$X$$ is a constant.

The given answer follows almost immediately, because you know how to interpret $$\frac{dy}{dt}$$ and $$\frac{dr}{dt}$$.

Addendum For a more pictorial approach, study this diagram showing how the strings change their positions in a small time interval $$dt$$. Note that $$u$$ is the speed at which the side weights are falling, and $$w$$ is the speed at which the centre weight is rising.

A perpendicular, BC, has been dropped from the 'new' point, B, where the strings meet, to the position of a string at the start of the interval $$dt$$. AC is then very nearly the distance by which the sloping strings have shortened in time $$dt$$, and therefore by which the side weights have fallen. [The 'very nearly' approaches exactly as $$dt$$ approaches zero.]

If you look at the tiny right angled triangle, you'll see how the required relationship emerges.

Note that this is essentially a geometric problem; stating it in terms of velocities gives it a Physics flavour!

It's a similar issue to that addressed here Help me in understanding use of vector in this problem

The questioner wondered why the speed that the lower block slid along the table wasn't $$5\cos53$$

Instead it's $$\frac{5}{\cos53}$$

You are working this problem backwards. Starting from the velocity $$u$$ and resolving it on $$M$$ in the vertical direction and adding the two sides up. This is incorrect since velocities do not add up like forces.

While a body might have multiple forces acting on it, there is always only one motion state (velocity + rotation).

So start with the velocity of $$M$$ upwards with $$x$$ and find how fast the length $$MB$$ is changing. This speed is the same as $$u$$ on the other side of the pulley. To match the vertical component of $$u$$ with $$x$$ you need.

$$u \cos \theta = x$$