balance moon stone

A space traveler about to leave for the moon has a spring balance and a 1.0 kg mass A, which when hung on the balance on the Earth gives the reading of 9.8 newtons. Arriving at the moon at a place where the acceleration of gravity is not known exactly but has a value of about 1/6 the acceleration of gravity at the Earth's surface, he picks up a stone B which gives a reading of 9.8 newtons when weighed on the spring balance. He then hangs A and B over a pulley as shown in the figure and observes that B falls with an acceleration of 1.2 m s^–2 . What is the mass of stone B?

Solution by Michael A. Gottlieb

Let 
    gE    = acceleration of gravity on the earth = 9.81  m s–2      
    gM   = acceleration of gravity on the moon ≈ gE / 6,
    mA  = mass of stone A = 1.0 kg,
    mB  = mass of stone B,
      a   = acceleration of stones hung in pulley on moon = 1.2 m s–2.

Stones A and B and the pulley form an Atwood's Machine for which

     a = [ (m_B  - m_A) / (m_B  + m_A) ] g_M .     (1)

Stone B on the moon gives the same scale reading as a 1 Kg stone on Earth, so

     g_M*m_B  = 9.81.                                      (2)

Substituting  m_B  = 9.81/g_M  into (1) yields a quadratic equation in g_M,

     g_M² - 8.61 g_M  + 11.772 = 0,  

which has one solution close to g_E  / 6, namely g_M = 1.7048. Substituting this value into (2) gives m_B  = 5.75434 and rounding to two significant digits (as per the spring balance readings and the acceleration) gives

     m_B  ≈ 5.8.