- #1
Jaimie
- 35
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Hello,
I have a question about a sample problem in the McGraw Hill Physics 12 book. (p. 38-39). "An Atwood machine is made of two objects connected by a rope that runs over a pulley. The objects on the left (m1) has a mass of 8.5 and the object on the right (m2) has a mass of 17kg. a) What is the acceleration of the masses? b) What is the tension of the rope."
Okay. So my question is for b). I understand that to find the tension I can calculate it using either the left of right sides of the diagram (if I were to sketch this out). The acceleration was calculated at 3.27 m/s^2. Therefore w/ left side:
-Fg1 + FT = ma
FT= m1g + m1a
FT= (8.5)(9.81) + (8.5)(3.27) = 111.18N
If I want to check with the right side of the diagram
-FT + Fg2 = m2a
Fg2 - m2a= FT
(17)(9.81) - (17)(3.27) = FT
111.18N
I know that I should be getting a negative value for the latter calculation. Can someone guide me as to what I am doing wrong. Thank you so much for your help!
I have a question about a sample problem in the McGraw Hill Physics 12 book. (p. 38-39). "An Atwood machine is made of two objects connected by a rope that runs over a pulley. The objects on the left (m1) has a mass of 8.5 and the object on the right (m2) has a mass of 17kg. a) What is the acceleration of the masses? b) What is the tension of the rope."
Okay. So my question is for b). I understand that to find the tension I can calculate it using either the left of right sides of the diagram (if I were to sketch this out). The acceleration was calculated at 3.27 m/s^2. Therefore w/ left side:
-Fg1 + FT = ma
FT= m1g + m1a
FT= (8.5)(9.81) + (8.5)(3.27) = 111.18N
If I want to check with the right side of the diagram
-FT + Fg2 = m2a
Fg2 - m2a= FT
(17)(9.81) - (17)(3.27) = FT
111.18N
I know that I should be getting a negative value for the latter calculation. Can someone guide me as to what I am doing wrong. Thank you so much for your help!