Answer to Question #125868 in Mechanics | Relativity for Sammy W-S

Question

A pulley system has the following characteristics;
box one has a mass of 2.0×10 to the power of 3 kg, box 2 is at a 90° angle below box one attached by a rope and has a mass of 4.4×10 to the power of 2 kg, the man on top of box 2 has a mass of 6.0×10 to the power of 1 kg. if all surfaces are frictionless, what is the apparent weight of the man?

Answer 1

The problem's solution is based on 2,3 - Newton laws and the connection equation:

$m_1\vec a_1 = m_1\vec g+\vec T_1 + \vec N_2$

$m_2\vec a_2 = m_2\vec g+\vec N_1 + \vec P_h$ h - human

$m_h\vec a_h = m_h\vec g+\vec T_2$

$\vec N_2+\vec N_1 = 0$

$\vec T_2+\vec P_h = 0$

$x_1-x_0 + y_0-y_1 = const$

We can take the second derivative from the last eqution and have had: $a_1 = a_2 = a_h = a$

Then in the projection on XY plane, vector equations become next scalar equations:

$m_1a = -N_2$ (1)

$m_2a = m_2g+P_h-N_1$ (2)

$m_ha_h = m_hg-T_2$ (3)

$N_2-N_1=0$ (4)

$T_2-P_h=0$ (5)

$a_1 = a_2 = a_h = a$ (6)

This system of equations we can solve with respect to a:

$a = \frac{m_1+m_h}{m_2+m_h-m_1}g$

And from (3),(5) get $P_h$

$P_h = \frac{m_1m_h}{m_2+m_h-m_1}g$