Answer to Question #161277 in Physics for Valentina

Question

Two .25 kg masses are hung on a mobile above a baby's bed. The mobile is 1.25

meters long and the pivot support is at .5 meters from the left side. The first mass is

hung .25 meters away from the left end and the second mass is hanging at the right

end of the mobile. Where should a third .25 kg. mass be hung on the mobile so that

it hangs motionless (balanced torques on both sides)?

Answer 1

$$\sum \tau=0,$$ $$\tau_{ccw}=\tau_{cw},$$ $$m_3gx+m_1g(0.5\ m-0.25\ m)=m_2g(1.25\ m-0.5\ m),$$ $$x=\dfrac{0.25\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.75\ m-0.25\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.25\ m}{0.25\ kg\cdot9.8\ \dfrac{m}{s^2}}=0.5\ m.$$

Answer:

$x=0.5\ m,$ to the left from the pivot point (at the left end of the mobile).