- #1
Twinflower
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Greetings! This is my first post on this forum :)
("Fasit" means solution in Norwegian)
Inclined plane, 30 degrees from horisontal
Two boxes tied together.
BoxA = 4 kg, friction coefficient = 0,25
BoxB = 8kg, friction coefficient = 0,35
Box A is closest to the bottom of the plane. Box B is slighly above. Not touching.
A: Find each box' acceleration (the solution implies a common acceleration)
B: Find the tension between the boxes
[tex] a = (g * sin(30)) - (\mu * g * cos(30))[/tex]
[tex] \sum F = m * a[/tex]
Regarding Part A
I found both boxes' individual acceleration using the formula above:
[tex]A_A = 9.81 m/s^2 * sin(30) - 0.25 * 9.81 m/s^2 * cos(30) = 2.78 m/s^2[/tex]
[tex]A_B = 9.81 m/s^2 * sin(30) - 0.35 * 9.81 m/s^2 * cos(30) = 1,93 m/s^2[/tex]
However, the solution is supposed to be [tex]2.21m/s^2[/Tex]
What am I doing wrong?
Regarding Part b
I am getting nowhere trying to find a solution for this part.
I have tried multiplying the difference in acceleration with box A's downward force, but I am not getting anything near the solution which is supposed to be 2.27 N
("Fasit" means solution in Norwegian)
Homework Statement
Inclined plane, 30 degrees from horisontal
Two boxes tied together.
BoxA = 4 kg, friction coefficient = 0,25
BoxB = 8kg, friction coefficient = 0,35
Box A is closest to the bottom of the plane. Box B is slighly above. Not touching.
A: Find each box' acceleration (the solution implies a common acceleration)
B: Find the tension between the boxes
Homework Equations
[tex] a = (g * sin(30)) - (\mu * g * cos(30))[/tex]
[tex] \sum F = m * a[/tex]
The Attempt at a Solution
Regarding Part A
I found both boxes' individual acceleration using the formula above:
[tex]A_A = 9.81 m/s^2 * sin(30) - 0.25 * 9.81 m/s^2 * cos(30) = 2.78 m/s^2[/tex]
[tex]A_B = 9.81 m/s^2 * sin(30) - 0.35 * 9.81 m/s^2 * cos(30) = 1,93 m/s^2[/tex]
However, the solution is supposed to be [tex]2.21m/s^2[/Tex]
What am I doing wrong?
Regarding Part b
I am getting nowhere trying to find a solution for this part.
I have tried multiplying the difference in acceleration with box A's downward force, but I am not getting anything near the solution which is supposed to be 2.27 N
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