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 Valentin Eberhardt

Let 
    gE    = acceleration of gravity on the earth,          
    gM   = acceleration of gravity on the moon ≈ gE / 6,
    mA  = mass of stone A = 1.0 kg,
    mB  = mass of stone B = 9.8N / gM,
      a   = acceleration of stones hung in pulley on moon = 1.2 m s–2.

When both masses hang (on the Moon) over the pulley as in the drawing, the system of two masses m_A and m_B connected by the thread, which thread we just think of being not elastic as not mentioned other way, moves with an acceleration of a = 1.2 m s^–2. Newton's second law reads:

    F = (m_A +m_B ) x a.

The resulting force F acting on the system is the gravitational force of the moon which in this case (pulley) is

    F = (m_B - m_A ) x g_M.                     

So we can write

    (m_B - m_A ) x g_M = (m_A + m_B ) x a.  

(where we take m_B > m_A, as B falls and not A). We can rewrite this as

    ( m_B - 1.0 ) x ( 9.8 / m_B ) = ( 1.0 + m_B ) x 1.2.

Multiplying by m_B (m_B ≠ 0), expanding and reordering of the terms gives:

    1.2 m_B² - 8.6 m_B + 9.8 = 0.

This quadratic equation has two solutions,

    m_B = 1.421kg, with  g_M = g_E / 1.421,

and

    m_B = 5.745kg with g_M = g_E / 5.745.

As it is given that g_M ≈ g_E / 6, the second solution must be the correct one. Therefore,

    m_B = 5.745 kg.