Frictionless pulleys with unknown masses. ATTEMPTED SOLUTION INSIDE

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SUMMARY

The discussion centers on a physics problem involving two masses connected by a string over a frictionless pulley. The known mass is 9kg, with a coefficient of kinetic friction of 0.25, and the unknown mass accelerates downward at 1m/s². Key equations used include the net force equations for both masses and the frictional force calculation. The confusion arises from the extraneous third equation, which complicates the solution without providing additional value.

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  • Understanding of Newton's laws of motion
  • Knowledge of frictional force calculations, specifically kinetic friction
  • Familiarity with free-body diagrams for analyzing forces
  • Basic algebra for solving simultaneous equations
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  • Review Newton's second law and its application in multi-mass systems
  • Study the principles of friction, focusing on kinetic and static friction coefficients
  • Learn how to construct and analyze free-body diagrams for complex systems
  • Explore methods for solving systems of equations, particularly in physics contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of problem-solving in dynamics involving pulleys and friction.

soccerdude28
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Homework Statement


Two masses are connected by means of a string which passes over a light frictionless pulley. The coefficient of kinetic friction between the 9kg mass and the table is 0.25. The applied force on the 9kg mass is unknown and the acceleration of the unknown mass, m, is 1m.s^2 down. Calculate:
a. The magnitude of the tension in the string
b. The mass
c. The applied force, Fa, on the 9kg mass


Homework Equations


Ff=UkFn
Fnet=ma


The Attempt at a Solution


Alright so i started out with finding frictional force on the 9kg block. Fg=Fn
88.2=Fn
Ff=UkFn
88.2(0.25)
22.05N


Now i found the first equation of just the block with unknown mass
Fnet=ma
Fg-T=ma
9.8m-T=m
T=8.8m

Now second equation of just the 9kg mass
Fnet=ma
T-Fa-Ff=9(1)
T-22.05-Fa=9
T-Fa=31.05

Now the third equation of the full system
Fnet=ma
Fg+Ff+Fa=1(9+m)
9.8m-22.05-Fa=9+m
Fa=-31.05+8.8m

Now from the three equations, i subbed 1 into 2
So: 8.8m-Fa=31.05
Rearrange
8.8m-31.05=Fa

Now i used "Substitution" for the third equation as well as the new one we just found
8.8m-31.05=-31.05+8.8m
and it basically just cancels each other out, which is why i am confused. What did i do wrong?
Thank you:)​
 
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i thought that with an attempted solution, people would be more willing to help, doesn't look like it
please guys, help me out
 
Without a picture, it appears your problem is missing some information, if I am understanding it correctly. It looks like you have a 9kg block on the rough table, connected by a string which wraps around a pulley at the edge of the table and connects to a hanging block of unknown mass m, which accelerates at 1m/s^2 downward. There is also an applied force Fa on the 9kg block, acting to the left. Is that correct? If so, the problem has an infinite number of solutions; I can make m as large as I want, and adjust Fa acordingly to give the proper acceleration of 1m/s^2. That third equation of the system (which I don't encourage using) is extraneous, and doesn't buy you anything. Please clarify the question.
 

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