47.9| Grade Summary Submissions d 1 = 45.85 d 1 = 47.85 d 1 = 47 d 1 = 47.9 100

Question

47.9|

Grade Summary

Submissions

d1 = 45.85

d1 = 47.85

d1 = 47

d1 = 47.9

100

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You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 105 g on a pivot. Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 4.7 g, 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. 25% Part (a) Determine the distance d = d1, in centimeters, if there is only one coin on the 0 end of the meter stick and the system is in static equilibrium. d1 =

47.9|

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Grade Summary

Deductions 20% Potential 80%

Submissions

Attempts remaining: 6 (5% per attempt) detailed view 1 5%

d1 = 45.85

2 5%

d1 = 47.85

3 5%

d1 = 47

4 5%

d1 = 47.9

Hints: 3% deduction per hint. Hints remaining: 3 Feedback: 3% deduction per feedback. Submission History Answer Hints Feedback Totals 1 d1 = 45.85 5% 0% 0% 5% 2 d1 = 47.85 5% 0% 0% 5% 3 d1 = 47 5% 0% 0% 5% 4 d1 = 47.9 5% 0% 0% 5% Totals 20% 0% 0% 20% 25% Part (b) Determine the distance d = dn, in centimeters, if there are n = 6 coins stacked on the 0 end of the meter stick and the system is in static equilibrium.
25% 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?
25% Part (d) Is it possible to stack enough coins at the 0 end to achieve equilibrium with d = 0?

100

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Explanation / Answer

1) Total mass = 109.7 g and this is supported by the pivot at da from the coins.
The centre of gravity of the ruler acts down at 50 cm from the coins.
The coins themselves are at zero from the coins.
The moments must add to zero.
so 109.7 * da - 105 * 50 + 4.7 * 0 = 0
da = 47.86

2. 6 coins mass = 133.2
so 133.2 * db - 105 * 50 + 28.2* 0 = 0

db = 39.4

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