Question

A uniform pole 7m long weighs 10kg is supported by a boy 2m from one end and a man 3m from the other end. At what point must a 20kg weight be attached so that the man would support thrice as much weight as the boy

Answer 1

As the pole with the weight attached should be at rest, the resulting force and torque should be zero (we will write down the torque with respect to one end of a pole) :

$F_{man}+F_{boy}=(m_{pole}+M_{weight} )g$

$F_{man}\cdot (l-l_{man})+F_{boy}\cdot l_{boy}=m_{pole}g\cdot\frac{l}{2}+M_{weight}g\cdot x$

Knowing that $F_{man}=3F_{boy}$ we find that

$F_{boy} = \frac{(m_{pole}+M_{weight} )g}{4}$ and thus

$x = \frac{(3l-3l_{man}+l_{boy})\cdot (m_{pole}+M_{weight} )-4m_{pole} \frac{l}{2}}{4M_{weight} }$

$x=\frac{(21-9+2)\cdot 30-4\cdot35}{4\cdot 20} = 3.5m$ i.e. exactly in the middle of the pole.