Question

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?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   $m_1 = 0.42kg$

      The second mass is  $m_2 = 0.47kg$

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

   Therefore we can say that

               $m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula  

                $x = \frac{m_1 * 15}{m_2}$

                    $= \frac{0.42 * 15}{0.47}$

                     $x = 13.4 cm$

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

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

   So basically

          $(m_1 + m_2 ) * 20 = m_3 * 30$

          $(0.42 + 0.47) * 20 = 30 * m_3$

 Making $m_3$ subject

          $m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

                $m_3 = 0.59 kg$