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Let me start by stating that this is not a homework question. If I just wanted the answer, it is in the back of my book. What I do need is some understanding of how the answer was reached.
Problem:
Two blocks attached by a string slide down a 10degree incline. Block 1 has a mass m1 = 0.80 kg and block 2 has mass m2 = 0.25 kg. In addition, the kinetic coefficients of friction between the blocks and the incline are 0.30 for block 1 and 0.20 for block 2. Find (a) the magnitude of the acceleration of the blocks, and (b) the tension in the string.
Answers in the back of the book:
(a) 0.96 m/s2
(b) 0.18 N
http://www.mravery.com/images/Physics.jpg
My attempt at a solution:
After drawing free body diagrams for both blocks I have come up with the following:
Block one calculations
Weight force in x direction (Wx)= sin(10) x (.8) x 9.81
Fnety = may
(Normal Force) - cos(10)x (.8) x 9.81 = 0
N = cos(10)x (.8) x 9.81
Friction = .3 x (cos(10)x (.8) x 9.81)
Fnetx = Wx + Tension - FrictionAfter I calculate these, I calculate the same for block 2 ( I will spare you the redundant reading), and obviously the tensional force is opposite. Once i do that I get stuck.
I have been racking my brain for a day and a half. Any help would be greatly appreciated!
Thanks
Problem:
Two blocks attached by a string slide down a 10degree incline. Block 1 has a mass m1 = 0.80 kg and block 2 has mass m2 = 0.25 kg. In addition, the kinetic coefficients of friction between the blocks and the incline are 0.30 for block 1 and 0.20 for block 2. Find (a) the magnitude of the acceleration of the blocks, and (b) the tension in the string.
Answers in the back of the book:
(a) 0.96 m/s2
(b) 0.18 N
http://www.mravery.com/images/Physics.jpg
My attempt at a solution:
After drawing free body diagrams for both blocks I have come up with the following:
Block one calculations
Weight force in x direction (Wx)= sin(10) x (.8) x 9.81
Fnety = may
(Normal Force) - cos(10)x (.8) x 9.81 = 0
N = cos(10)x (.8) x 9.81
Friction = .3 x (cos(10)x (.8) x 9.81)
Fnetx = Wx + Tension - FrictionAfter I calculate these, I calculate the same for block 2 ( I will spare you the redundant reading), and obviously the tensional force is opposite. Once i do that I get stuck.
I have been racking my brain for a day and a half. Any help would be greatly appreciated!
Thanks
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