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January 28, 2015

January 28, 2015

Posted by **Miaow** on Saturday, September 22, 2012 at 1:37pm.

8.) Caluclate the tension in the Atwood's Machine string for the case when m2 is 180 grams and m1 is 220 grams. Assume system is moving freely. The desired tension value is tension in string connected to 220 gram mass.

m1(g) -m2(g) = ma

.220(9.8) -.180(9.8) = .400a

2.156-1.764 = .400a

a=.98m/s^2

T - m1g = ma

T - .220(9.8) = .220(9.8)

T - 2.156 = .2156

T = 2.37

9.) Caluclate the tension in the Atwood's Machine string for the case when m2 is 180 grams and m1 is 220 grams. Assume system is moving freely. The desired tension value is tension in string connected to 180gram mass.

m1(g) -m2(g) = ma

.220(9.8) -.180(9.8) = .400a

2.156-1.764 = .400a

a=.98m/s^2

T - m2g = ma

T - .180(9.8) = .180(9.8)

T - 1.764 = .1764

T = 1.94

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