**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 g_{E }= acceleration of gravity on the earth, g_{M}= acceleration of gravity on the moon ≈ g_{E }/ 6, m_{A}= mass of stone A = 1.0 kg, m_{B}= mass of stone B = 9.8N / g_{M}, 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_{B}^{2} - 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.