Answer to Question #104637 in Classical Mechanics for alize Magana

Question

A plank of length 2.89 m is to be balanced on a see-saw. On one side of the plank hangs a mass 1 =6.32 kg, and on the other side of the plank there is a massless, frictionless pulley. This pulley has a mass 2​​ =22.2 kg on one side and a mass 3 =6.25 kg on the other side. What is the distance from the see-saw's pivot to the point where mass 1 must hang in order for this plank to be stationary after being released from rest?

Answer 1

As per the given data in the question,

Length of the plank =2.89m

The hanged mass on one side of the plank (m1) =6.32kg

Mass hanged on one side of the pulley (m2) =22.2 Kg

Mass hanged on other side of the pulley(m3) =6.25 kg

Let the acceleration of the blocks, which hanged over the pulley =a

The hanged distance =?

Now, force on the pulley,

$m_2 g-T=m_2 a$

$\Rightarrow 22.2 g-T=22.2 a--------(i)$

Now,

$T-m_3g=m_3 a$

$T-6.25 g=6.25 a--------(ii)$

Adding equation (i) and (ii)

$\Rightarrow 22.2g-6.25g=22.2a+6.25a$

$\Rightarrow 15.95g=28.45a$

$\Rightarrow a=\dfrac{15.95}{28.45}$

$\Rightarrow a=0.56 m/sec^2$

So, tension on the string $(T)=6.25a+6.25g =64.75N$ =

Hence total force on the plank $=2T=64.75\times2=129.5N$

Now, let the block is hanged from the point of rotation,

$m_1g\times (2.89-x)=129.5\times x$

$\Rightarrow 129.5\times x+m_1 g\times x=m_1g\times 2.89$

$\Rightarrow x(129.5+6.32)=6.32\times 9.8\times 2.89$

$\Rightarrow x(135.82)=178.995$

$\Rightarrow x=\dfrac{178.995}{135.82}m$

$x=1.32m$

from the side of pulley.