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1. Homework Statement
A 2.00kg aluminium block and a 6.00kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface, which is fixed, as shown in the figure, with a (let it be alpha lol) = 43.0deg. The coefficient of static friction for aluminum on steel is 0.56, and for copper on steel is 0.83. Determine whether the blocks start to move once any holding mechanism is released. Then, calculate the sum of the magnitudes of the forces of friction acting on the blocks.
2. Homework Equations
F = ma
3. The Attempt at a Solution
The blocks should not move at all according to this equation i've derived (i'm not sure if i should use static or kinetic friction for this case).
a = (m1gsin(a)us1m1gcos(a)us2m2g)/(m1+m2)
Where us1 is the coefficient of static friction of copper, m1 is the mass of copper and us2 is the coefficient of static friction of aluminum.
This gives us a negative acceleration and thus the system won't move.
I've determined that the forces of friction for each of the two blocks are the following:
fr1 = us1m1gcos(a) = 10.9872N
fr2 = us2m2g = 35.7294N
I'm using the coefficient of static friction as the two blocks don't move (again not too sure on this one). I've added these two frictions and i got that the final result (46.7N) was incorrect. Or should i do vector addition and determine the magnitude? However, i also did the same and got a wrong result (44.4N).
What should i do?
By the way: i have the following FBDs:
For Al, i have the normal force going upwards, the weigh, the friction and the tension. For Cu, i have the tension going up the plane, the friction up the plane, the normal force perpendicular to the surface and the weigh.
A 2.00kg aluminium block and a 6.00kg copper block are connected by a light string over a frictionless pulley. They sit on a steel surface, which is fixed, as shown in the figure, with a (let it be alpha lol) = 43.0deg. The coefficient of static friction for aluminum on steel is 0.56, and for copper on steel is 0.83. Determine whether the blocks start to move once any holding mechanism is released. Then, calculate the sum of the magnitudes of the forces of friction acting on the blocks.
2. Homework Equations
F = ma
3. The Attempt at a Solution
The blocks should not move at all according to this equation i've derived (i'm not sure if i should use static or kinetic friction for this case).
a = (m1gsin(a)us1m1gcos(a)us2m2g)/(m1+m2)
Where us1 is the coefficient of static friction of copper, m1 is the mass of copper and us2 is the coefficient of static friction of aluminum.
This gives us a negative acceleration and thus the system won't move.
I've determined that the forces of friction for each of the two blocks are the following:
fr1 = us1m1gcos(a) = 10.9872N
fr2 = us2m2g = 35.7294N
I'm using the coefficient of static friction as the two blocks don't move (again not too sure on this one). I've added these two frictions and i got that the final result (46.7N) was incorrect. Or should i do vector addition and determine the magnitude? However, i also did the same and got a wrong result (44.4N).
What should i do?
By the way: i have the following FBDs:
For Al, i have the normal force going upwards, the weigh, the friction and the tension. For Cu, i have the tension going up the plane, the friction up the plane, the normal force perpendicular to the surface and the weigh.
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