You\'ve been given the challenge of balancing a uniform, rigid meter-stick with
ID: 1645405 • Letter: Y
Question
You've been given the challenge of balancing a uniform, rigid meter-stick with mass M = 125g on a pivot. Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 2.8g, so that the center of mass of the coins is directly over the end of the meter stick. The pivot point is a distance d from the 0-cm end of the meter stick. Part (a) Determine the distance d = d_1, in centimeters, if there is only one coin on the 0 end of the meter stick and the system is in static equilibrium. d_1 = __________ Part (b) Determine the distance d = d_n, in centimeters, if there are n = 5 coins stacked on the 0 end of the meter stick and the system is in static equilibrium. Part (c) How many coins would you have to stack at the 0 end to achieve equilibrium when the pivot is d = 10.0 cm from that end? Part (d) Is it possible to stack enough coins at the 0 end achieve equilibrium with d = 0?Explanation / Answer
A)
Total mass = 127.8 g and this is supported by the pivot at da from the coins.
The center of gravity of the ruler acts down at 50cm from the coins.
The coins themselves are at zero from the coins.
The moments will be add to zero.
so 127.8 * da - 125 * 50 + 2.8 * 0 = 0
da = 125*50 / 127.8
=48.9 cm
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B)
5 coins mass = 5x2.8+125=139
so 139 * db - 125 * 50 + 14 * 0 = 0
db=44.96 cm
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C)
As we know where the pivot is we might choose to calculate moments from this point.
The coins are 9.99 and the center of mass is (50-9.99) = 40.01 cm from the pivot.
n* 2.8 * 9.99 - 125 * 40.01 + F * 0= 0
n = 125* 40.01 / ( 2.8 * 9.99 ) = 179 coins.
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