Question

In the arrangement shown \(\mu=0\) everywhere & string & pulley are ideal. For block of mass 'm' to move vertically down we need to find value of m_1 & tension in string

Answer 1

@IrishBoy123

Answer 2

start by drawing the free body diagrams for each of the blocks

Answer 3

In the idealised model, \(T\) is irrelevant. the tension in the string/pulley is just a transmission mechanism.

but \(m_1\) is very relevant.

Imagine \(m_1 = 0\), what might happen?

Then imagine \(m_1 >> m\). In that case \(m_1\) would be in effective free-fall, as would the \(m\) wedge.

Answer 4

wht is ur conclusion @IrishBoy123 ?

Answer 5

do you want to compare solutions?!?!

Answer 6

You don't need T but you do need m_1

FWIW, I **think** the solution in the Lagrangian is this.



we can say the block is moving to left at velocity \(\dot x\), and so the mass \(m_1\) dropping at velocity \(\dot x\)

If we set \(\dot y\) as the **relative velocity** of the block **wrt** to the wedge, ie it is sliding down at \(\dot y \), then it's actual velocity in the main reference frame is:

in horizontal:
\(v_1 = \dot x - \dot y \cos \theta\)

in vertical:
\(v_2 = \dot y \sin \theta\)

and for it to drop vertically, we can say that \(v_1 = 0 \implies \dot x - \dot y \cos \theta = 0\).

So for: \(L = T - V\)
\(T = \frac{1}{2} m_1 \dot x^2 + \frac{1}{2} m \dot x^2 + \frac{1}{2} m (\dot x - \dot y \cos \theta)^2 + \frac{1}{2} m ( \dot y \sin \theta)^2 \)
\( = \frac{1}{2} (m_1 + m ) \dot x^2 + \frac{1}{2} m \dot y^2 \sin^2 \theta \) :)

In terms of the potential:
\(V = - m_1 g x - mg y \sin \theta \)

Meaning that....
\(L = \frac{1}{2} (m_1 + m ) \dot x^2 + \frac{1}{2} m \dot y^2 \sin^2 \theta + m_1 g x + mg y \sin \theta \)

And it's process from here:
\(L_x = m_1 g, \qquad L_y = mg \sin \theta\)
\(L_{\dot x} = (m_1 + m)\dot x , \qquad L_{\dot y} = m \dot y \sin^2 \theta\)
\(\frac{d}{dt} \left( L_{\dot x} \right)= (m_1 + m)\ddot x , \qquad \frac{d}{dt} \left( L_{\dot y} \right) = m \ddot y \sin^2 \theta\)

Pairing them gives the DE's
\( (m_1 + m)\ddot x = m_1 g \qquad m \ddot y \sin^2 \theta = mg \sin \theta\)
\( \ddot x = \dfrac{m_1 g}{(m_1 + m)} \qquad \ddot y = \dfrac{ g}{ \sin \theta}\)

And to see what dropping vertically means, we look at:
\( \dot x - \dot y \cos \theta = 0 \implies \ddot x - \ddot y \cos \theta = 0\)
\( \implies \dfrac{m_1 g}{(m_1 + m)} - \dfrac{ g}{ \sin \theta} \cos \theta= 0 \)
\( \implies \tan \theta = \dfrac{m_1 + m }{ m_1} = 1 + \dfrac{m}{m_1} \)

I don't know how to test this to see if it's right other but this seems to predict that, if \(m_1 >> m\) it happens if the wedge has \(\theta = \pi / 4\); and if \(m_1 << m\) it happens at \(\theta = \pi / 2\). Which sorta makes sense but doesn't make it right

More to the point, you don't need to bother with T. In a FBD analysis it would drop out in the algebra.

We can calculate it, if needed as \(m_1 g - T = m_1 \ddot x\) so \(T = \dfrac{m_1 m g }{(m_1 + m)} \)

Maybe someone else could have a go and see what happens ?!?!