(25 %) Problem 3: You\'ve been given the challenge of balancing a uniform, rigid

Question

(25 %) Problem 3: You've been given the challenge of balancing a uniform, rigid meter-stick with mass M-85 g on a pivot Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m 3.5 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 e 25 % Part (a) Determine the distance d = dl, in centimeters, if there is only one coin on the 0 end of the meter stick and the system is in static equilibrium Grade Summary 0% 100% Potential cos0 cotan0 asin) acos0 atan)acotan) sinhO Submissions Attempts remaining: 10 (5% per attempt) detailed view sinO cotanhO 0 END Degrees Radians vo Submit Hint I give up! Hints: 3% deduction per hint. Hints remaining: 3 Feedback: 3% deduction per feedback. 25% Part (b) Determine the distance d = dn, in centimeters, if there are n = 5 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?

Explanation / Answer

Part(a)

The length of the meter stick is l = 100 cm

Now,equating the moment of mass in both the sides of the pivot .we get

m*l*d = (l/2 -d)*M

3.5*1*d = (0.5 - d)*85

88.5*d = 42.5

d = 0.480 m

d = 48 cm

Part(b)

3.5*5*d = (l/2 -d)*85

102.5*d = 0.5*85

d = 0.414 m

d = 41.4 cm

part(c)

3.5*10*n = (l/2 -10)*85

35n = 4250 - 850

n = 97 coins

part (d)

no , it is not possible.

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