In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where [tex]m_1 \ and \ m_2[/tex] are hanged ) is attached is zero

Therefore we can say that

[tex]m_1 * 15cm = m_2 * xcm[/tex]

Making x the subject of the formula

[tex]x = \frac{m_1 * 15}{m_2}[/tex]

[tex]= \frac{0.42 * 15}{0.47}[/tex]

[tex]x = 13.4 cm[/tex]

Looking at the diagram we can see that the tension T on the string holding the bar where [tex]m_1 \ and \ m_2[/tex] are hanged is as a result of the masses ([tex]m_1 + m_2[/tex])

Also at equilibrium the moment about the point where the string holding the bar (where ([tex]m_1 +m_2[/tex]) and [tex]m_3[/tex] are hanged ) is attached is zero

So basically

[tex](m_1 + m_2 ) * 20 = m_3 * 30[/tex]

[tex](0.42 + 0.47) * 20 = 30 * m_3[/tex]

Making [tex]m_3[/tex] subject

[tex]m_3 = \frac{(0.42 + 0.47) * 20 }{30 }[/tex]

[tex]m_3 = 0.59 kg[/tex]