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A graphical representation of the situation

In the diagram shown, we can find the velocity of $M$ using two methods, one of which is a constraint relation (differentiating the length of the string), or by the method of component of motion, in which we say that if the velocity of $M$ is $v$, then the velocity along the string is $v \cos(\theta)$ which is equal to $u$.

However I tried modifying the question to when the two velocities of point $A$ and $B$ are different. Using the constraint relation, I found that the velocity of the mass $M$ is given by $$v= \frac{u_1+u_2}{2\cos(\theta)}$$ where $u_1$ and $u_2$ are the velocities of the strings. However if I try using the method of components, it does not seem to work here.

We say, $$v\cos(\theta)=u_1=u_2$$

I end up getting that both $u_1$ and $u_2$ are equal.

I concluded that both the velocities have to be same for that method to work. However, I came across another question which looks similar.

The question

I solved this question by splitting $\theta$ into $\theta-\alpha$ and $\alpha$, and using the method of component of velocity, $$v\cos(\theta-\alpha)=u_1,$$ $$v\cos(\alpha)=u_2$$

Eliminating alpha from the two equations, we get that $$v=\frac{\sqrt{(u_1)^2 + (u_2)^2 - 2u_1u_2\cos(\theta)}}{\sin(\theta)}$$

which matches with the answer given. Why does this not match with $v= \frac{u_1+u_2}{2\cos(\theta)}$?

Moreover, the method of components works even though the velocities are different.

I also noticed that if both the velocities are equal, the answer matches the answer in the above case ($v=\frac{u}{\cos\theta}$)

My question is why did the method of components not work in the case where I assumed the velocities to be different? Is it because the angle $\theta$ can not be maintained symmetrically in case of different velocities, or because there will be a component of motion in another direction, or some other reason?

Also can someone clarify why there will be no motion in the direction of the $x$-axis in any case?

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2 Answers 2

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Forget the pulleys for a moment. Imagine that the mass $M$ is tied to two parallel strings and you pull up it with a uniform velocity. First, assume that velocities of each string are equal. So the mass is pulled up gently, without tilting. Then assume that velocities are different. It means you pull one side quick and other side slowly. So one side of the mass is trying to go more height than the other side and so it begins to tilt. In that case it cannot be maintained symmetrically. Thus in your problem you will notice that the angles are different. For your second problem imagine again that the mass is tied with two strings on opposite sides. It will not move unless velocities of the strings are not equal. Hope this helps.

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  • $\begingroup$ OH MY GOD I FINALLY GET IT $\endgroup$
    – sanya
    Sep 4, 2023 at 22:06
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enter image description here

Lets look first at the solution with constraint equations

$$x_2-l_2-y\,\cos(\alpha)=0\\ x_1-l_1-y\,\cos(\beta)=0$$

add this two equations you obtain

$$x_2+x_1-l-y\,(\cos(\alpha)+\cos(\beta))=0$$

this is your constraint equation, from here you obtain that

$$\dot y=\frac{\dot x_1+\dot x_2}{\cos(\alpha)+\cos(\beta)} =\frac{\dot x_1+\dot x_2}{2\cos\left(\frac{\alpha+\beta}{2}\right)\,\cos\left(\frac{\alpha-\beta}{2}\right)}$$

thus if $\alpha=\beta$ and $~\dot x_1=\dot x_2$

$$\dot y=\frac{\dot x_1}{\cos(\theta/2)}$$

which is correct (notice that "my" $\theta/2~$ is "your" $\theta~$ at the first example).

I think that you can't substitute $\beta=\theta-\alpha$ ?


The equation of motions:

from the free body diagram you obtain

$$m\,\ddot x_1=T_1+F_1\\ m\,\ddot y=T_1\cos(\beta)+T_2\,\cos(\alpha)\\ m\,\ddot x=T_1\sin(\beta)-T_2\,\sin(\alpha)\\ m\,\ddot x_2=T_2+F_2$$

and the constraint equations:

$$y\,\cos(\beta)+x_2=l_2\\ y\,\cos(\alpha)+x_1=l_2\\ \Rightarrow\\ \ddot y\,\cos(\beta)+\ddot x_2=0\\ \ddot y\,\cos(\alpha)+\ddot x_1=0$$

you have now 6 equations for 6 unknows

$\ddot x_1~,\ddot x_2~,\ddot y~,\ddot x~$ and the rope tension forces $~T_1~,T_2$

solving those equation and obtain the results

$$\frac{\ddot y}{\ddot x_1+\ddot x_2}=-\frac{1}{\cos(\alpha)+\cos(\beta)} $$

and

\begin{align*} &\ddot x=\frac{no}{de}\\ &no=\left( -\sin \left( \beta+2\,\alpha \right) -3\,\sin \left( \beta \right) \right) F_{{1}}+ \left( \sin \left( \alpha+2\,\beta \right) +3\,\sin \left( \alpha \right) \right) F_{{2}} \\ &de=m \left( \cos \left( 2\,\alpha \right) +4+\cos \left( 2\,\beta \right) \right) \end{align*}

thus if $F_2=F_1$ and $\beta=\alpha$ you obtain $\ddot x=0$ which is correct

notice that i changed the sign of y for the EOM's.

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  • $\begingroup$ Can you please explain how there is no component of motion along the X-direction? $\endgroup$ Jul 5, 2021 at 4:45
  • $\begingroup$ where is your X-direction? $\endgroup$
    – Eli
    Jul 5, 2021 at 6:37
  • $\begingroup$ Along the horizontal $\endgroup$ Jul 6, 2021 at 6:24
  • $\begingroup$ I don’t think that you need it for the constraint equations, otherwise my result for your first example will be wrong $\endgroup$
    – Eli
    Jul 6, 2021 at 8:27
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    $\begingroup$ @Phy_Amatuer I wrote you the EOM's $\endgroup$
    – Eli
    Jul 7, 2021 at 15:55

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