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 = [ (mB - mA) / (mB + mA) ] gM . (1)
Stone B on the moon gives the same scale reading as a 1 Kg stone on Earth, so
gM*mB = 9.81. (2)
Substituting mB =
9.81/g_M into (1) yields a quadratic equation in gM,
gM2 - 8.61 gM + 11.772 = 0,
which has one solution close to gE / 6, namely gM = 1.7048. Substituting this value into (2) gives mB = 5.75434 and rounding to two significant digits (as per the spring balance readings and the acceleration) gives
mB ≈ 5.8.