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AnkhUNC
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[SOLVED] Mass of a block via kinetic friction
In Fig, two blocks are connected over a pulley. The mass of block A is 5.9 kg and the coefficient of kinetic friction between A and the incline is 0.36. Angle θ of the incline is 43°. Block A slides down the incline at constant speed. What is the mass of block B?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_34.gif
OK so I'm unsure of how how to compute this question. I'm going to need to find the tension in this problem in order to find the mass of block two correct?
So for m1:
Fy = N-mg+(Tsin43-(Tsin133 [90+43 for the Fk force]*.36))
How do I solve for T here if I don't have a second equation to help me solve? There isn't a normal force on block B seeing as it is in the air so I couldn't impose the two to get rid of N right?
Fx = Tcos(43)-Tcos(133)*.36
Is this correct?
For m2:
Fy = T-mg
Fx = No movement
T>mg or block A would not be in kinetic friction.
Thanks,
Homework Statement
In Fig, two blocks are connected over a pulley. The mass of block A is 5.9 kg and the coefficient of kinetic friction between A and the incline is 0.36. Angle θ of the incline is 43°. Block A slides down the incline at constant speed. What is the mass of block B?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_34.gif
Homework Equations
The Attempt at a Solution
OK so I'm unsure of how how to compute this question. I'm going to need to find the tension in this problem in order to find the mass of block two correct?
So for m1:
Fy = N-mg+(Tsin43-(Tsin133 [90+43 for the Fk force]*.36))
How do I solve for T here if I don't have a second equation to help me solve? There isn't a normal force on block B seeing as it is in the air so I couldn't impose the two to get rid of N right?
Fx = Tcos(43)-Tcos(133)*.36
Is this correct?
For m2:
Fy = T-mg
Fx = No movement
T>mg or block A would not be in kinetic friction.
Thanks,